NCERT Solutions Class 12 Maths Chapter 6

Mathematics is a subject requiring a strong conceptual understanding of its applications. Hence, students are advised to study the subject from good resources to gain in-depth knowledge about Mathematics chapters with examples, solutions, theorems as well as exercises and to score well in exams.

The Class 12 Mathematics Chapter 6 ‘Application of derivatives’ is a part of the Calculus section of Mathematics. It covers the various applications of derivatives and their role while doing calculations. The vital topics covered in Chapter 6 Class 12 Mathematics include:

Introduction
Rate of Change of Quantities
Increasing and Decreasing Functions
Tangents and Normals
Approximations
Maxima and Minima
Maximum and Minimum Values of a Function in a Closed Interval

The chapter’s important formulas are listed in the NCERT Solutions Class 12 Mathematics Chapter 6. Also, one can find easy ways to carry out calculations after referring to them thoroughly and consistently.

Extramarks is a reliable and trustworthy source for all the NCERT-related study material. One can find NCERT textbooks, NCERT solutions, NCERT exemplars, NCERT revision notes, NCERT formulas, NCERT-based mock tests and the NCERT solutions Class 12 Mathematics Chapter 6 on the Extramarks official website.

Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 6

NCERT Solutions Class 12 Mathematics Chapter 6 is about derivatives and their applications.

This chapter is a part of Calculus and requires the study of differential calculus. One should have a good command of it to excel in this chapter.

If you have a hold on calculus, this chapter will help students to learn to carry out calculations easily. As a result, they will be able to approach problems more logically. The entire chapter is covered thoroughly in the NCERT Solutions Class 12 Mathematics Chapter 6 and is available on the Extramarks website.

After completing Chapter 6 Mathematics Class 12, students will learn to analyse the problems with a better approach and be able to solve them easily.

Introduction

A derivative is the rate of change or the amount on which a particular function changes at one given point.

In the above-given figure, the function is represented in black colour, and a tangent is represented in red colour.

Rate of Change of Quantities

The derivative ds/dt is used to show the rate of change of distance s to the time t.

Assume a particular quantity y varies with another quantity x which satisfies y = f(x), here dy/dx or f’(x) shows the rate of change of y w.r.t. x and [dy/dx]x = x0 or f’(x0) shows the rate of change of y w.r.t. x at x = x0.

Now, assume the two variables x and y are varying w.r.t. another variable t i.e. x = f(t) and y = g(t)

Thus, by chain rule

dy/dx = (dy/dt)/(dx/dt) here dx/dt ≠ 0

So, the rate of change of y to x could be calculated using the rate of change of y at a given time and that of x to t.

Increasing and Decreasing Functions

The function’s derivative might determine if it is increasing or decreasing at any intervals in its domain.

Assume I will be an open interval contained in the domain of the real-valued function f. Then f is said to be

(i) increasing on I when x1 < x2 in I => f(x1) ≤ f(x2) for all x1, x2 Є I.

(ii) strictly increasing on I if x1 < x2 in I => f(x1) < f(x2) for all x1, x2 Є I.

(iii) decreasing on I if x1 < x2 in I => f(x1) ≥ f(x2) for all x1, x2 Є I.

(iv) strictly decreasing on I if x1 < x2 in I => f(x1) > f(x2) for all x1, x2 Є I.

However, some functions are neither increasing nor decreasing.

The graphical representations of all these functions are given below:

function f will be said to be increasing at x0 when there exists an interval I = (x0 – h, x0 + h), h > 0 such that for x1, x2 ∈ I,

x1 < x2 in I => f (x1) ≤ f (x2)

With the help of a theorem listed in the NCERT Solutions Class 12 Mathematics Chapter 6, you can test for increasing and decreasing functions for a given interval.

Tangents and Normals

For the given equation of a straight line that is passing through a given point (x0, y0) having a finite slope, m is represented as y – y0 = m(x – x0)

Assume the given curve y = f(x), and the tangent of the curve at that point (x0, y0) will be

[dy/dx](x0, y0) or f’(x0)

Now, the equation of the tangent of the curve y = f(x) at (x0, y0) will be

y – y0 = f’(x0)(x – x0)

The slope of the normal to curve y = f(x) at (x0, y0) is given by -1/ f’(x0) here f’(x0) ≠ 0

For the equation of normal to the curve, y = f(x) at (x0, y0) will be

Y – y0 = (-1/f’(x0))(x – x0)

(y – y0)* f’(x0) +(x – x0) = 0

Particular cases

(i) If the slope of the tangent line is zero, then tan θ = 0, i.e. θ = 0, which means a tangent line is parallel to the x-axis. For this case, the equation of the given tangent at the point (x0, y0) will be y = y0.

(ii) If θ -> π/2 then tan θ → ∞, that means the tangent line will be perpendicular to the x-axis, that is parallel to the y-axis. For this case, the equation of the tangent at (x0, y0) will be x = x0.

Approximations

An approximation is anything which is similar but not the same as something else.

Assume f : D => R, D Ì R, will be a given function and assume y = f (x). Assume ∆x denotes a small increment in terms of x.

Now, consider the increment in y corresponding to the increment in x will be ∆y = f (x + ∆x) – f (x).

(i) The differential of x is represented as dx = ∆x.

(ii) The differential of y is represented as dy = f’(x) dx or dy = (dy/dx) * ∆x

Maxima and Minima

In the section, you will learn the different methods of calculating a function’s maximum and minimum values in a given domain. You will also learn about the absolute maximum and minimum of a function used to find the solution to many applied problems.

You will clearly understand this topic through the definitions and theorems included in the NCERT Solutions Class 12 Mathematics Chapter 6 available on the Extramarks official website.

Maximum and Minimum Values of a Function in a Closed Interval

You will learn about the two theorems to find out the absolute maximum and minimum values of a function upon a closed interval I.

Theorem: Let f be a continuous function on an interval I = [a, b]. Then f has the absolute maximum value, as well as f attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in I.

Theorem: Let f be a differentiable function on a closed interval I and let c be any interior point of I. Then

(i) f′(c) = 0 when f attains its absolute maximum value at c.

(ii) f′(c) = 0 when f attains its absolute minimum value at c.

NCERT Solutions Class 12 Mathematics Chapter 6 Exercise & Solutions

Find all the NCERT solutions to the exercises covered in the chapter in the NCERT Solutions Class 12 Mathematics Chapter 6. It is prepared by the subject experts while adhering to the NCERT book and following the CBSE guidelines and curriculum.

All the vital topics and key concepts are covered in a point-wise manner. NCERT Solutions Class 12 Mathematics Chapter 6 is written end-to-end, highlighting everything in detail by the faculty experts. . Therefore, it is trusted by students and teachers across the private and government schools.

You can avail of the NCERT Solutions Class 12 Mathematics Chapter 6 from the Extramarks website.

Extramarks provides one-stop solutions to all your problems. To enjoy the maximum benefit of these resources, students just need to register themselves at Extramarks’ official website and stay ahead of the pack.

Click on the below links to view exercise-specific questions and solutions for NCERT Solutions Class 12 Mathematics Chapter 6:

Class 12 Mathematics Chapter 6: Questions and Answers

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NCERT Solutions Class 12 Mathematics Chapter 6 Exercise & Answer Solutions

NCERT Solutions Class 12 Mathematics Chapter 6 Exercise and Answer Solutions are available for students to refer for free on the Extramarks website. The material offers step-by-step solutions that help students understand how to solve problems relating to the chapter. The solution also helps students solve complex questions in a simplified manner. Students may refer to NCERT Solutions Class 12 Mathematics Chapter 6 to strengthen their basics.

Students may refer to the links below to download exercise-specific questions and solutions for NCERT Solutions Class 12 Mathematics Chapter 6 Application of Derivatives:

Chapter 6: Exercise 6.1 Solutions – 18 Questions (10 Long, 6 Short, 2 MCQs)
Chapter 6: Exercise 6.2 Solutions – 19 Questions (10 Long, 7 Short, 2 MCQs)
Chapter 6: Exercise 6.3 Solutions – 27 Questions (14 Long, 11 Short, 2 MCQs)
Chapter 6: Exercise 6.4 Solutions – 9 Questions (7 Short, 2 MCQs)
Chapter 6: Exercise 6.5 Solutions – 29 Questions (15 Long, 11 Short, 3 MCQs)

NCERT Exemplar Class 12 Mathematics

NCERT Exemplar Class 12 Mathematics book is an excellent source for students preparing for JEE Mains, NEET, MHT-CET etc. It has questions according to the topics and concepts covered in the NCERT textbook. This aids students in solving all types of questions confidently.

It has different questions with varying levels of difficulty, which helps students to improve their performance in various tests and exams and definitely builds their confidence level in the process. The questions are selected from different sources for students to prepare and improve their performance. It is also a complete source of information for CBSE students preparing for their 12th standard examinations. The students think logically about a problem after referring to the NCERT Solutions Class 12 Mathematics Chapter 6 and NCERT Exemplar.

The NCERT solutions Class 12 Mathematics Chapter 6 is prepared after analysing CBSE past years’ question papers. It contains extra questions from the NCERT Class 12 Mathematics textbook. Students can refer to NCERT Exemplar for Class 12 Mathematics for more practice and face their examinations with courage. They can assure themselves that nothing remains untouched in the chapter and will score very well in all their examinations. To get good grades in exams students must refer to multiple study resources, practice a lot of questions and stick to a study schedule and follow it religiously to come out with flying colours.

Key Features for NCERT Solutions Class 12 Mathematics Chapter 6

Regular practice is quite necessary for the students to excel. Hence, NCERT Solutions Class 12 Mathematics Chapter 6 helps students develop the habit of regular practice and to clarify their doubts then and there, take regular tests to assess their performance and get proper feedback to step up their learning. . The key features are as follows:

You can find all the topics covered as well as in text to end text exercises from the chapter covered in the NCERT Solutions Class 12 Mathematics Chapter 6.
It helps students analyse the easy and difficult topics in the chapter.
After completing the NCERT Solutions Class 12 Mathematics Chapter 6, students will become experts in solving derivatives problems.

Q.1 Find the rate of change of the area of a circle with respect to its radius r when

(a) r = 3 cm (b) r = 4 cm

Ans.

$\begin{array}{l} The area of a circle (A) with radius (r) is given by, \\ A = {πr}^{2} \\ Now, the rate of change of the area with respect to its radius \\ is given by, \\ \frac{dA}{dr} = \frac{d}{dr} {πr}^{2} = 2 πr \\ (a) When r = 3 cm \\ \frac{dA}{dr} = 2 π (3) = 6 π \\ Hence, the area of the circle is changing at the rate of 6 π {cm}^{2} / cm \\ when its radius is 3 cm . \\ (b) When r = 4 cm \\ \frac{dA}{dr} = 2 π (4) = 8 π \\ Hence, the area of the circle is changing at the rate of \\ 8 π {cm}^{2} / cm when its radius is 8 cm . \end{array}$

Q.2 The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Ans.

$\begin{array}{l} Let x be the length of a side, V be the volume, and S be the \\ surface area of the cube . \\ Then, V = x^{3} and S = 6 x^{2} where x is a function of time t . \\ It is given that, \frac{dV}{dt} = 8 {cm}^{3} / s \\ Then, by using the chain rule, we have : \\ 8 = \frac{dV}{dt} = \frac{d}{dt} (x^{3}) \\ = 3 x^{2} \frac{dx}{dt} \\ \Rightarrow \frac{dx}{dt} = \frac{8}{3 x^{2}} . .. (i) \\ Now, \frac{dS}{dx} = \frac{d}{dx} (6 x^{2}) \\ = 12 x \frac{dx}{dt} \\ = 12 x \times \frac{8}{3 x^{2}} [From equation (i)] \\ = \frac{32}{x} \\ \Rightarrow \frac{dS}{dx} = \frac{32}{12} [Putting x = 12] \end{array}$ $\begin{array}{l} \Rightarrow \frac{d S}{d x} = \frac{8}{3} c m^{2} / s . \\ Hence, if the length of the edge of the cube is 12 cm, then the \\ surface area is increasing at the rate of \frac{8}{3} {cm}^{2} /s . \end{array}$

Q.3 The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Ans.

$\begin{array}{l} The area of a circle (A) with radius (r) is given by, \\ A = {πr}^{2} \\ Now, the rate of change of area (A) with respect to time (t) is \\ given by, \frac{dA}{dt} = \frac{d}{dt} {πr}^{2} \\ = 2 πr \frac{dr}{dt} [By chain rule] \\ = 2 πr \times 3 [∵ \frac{dr}{dt} = 3 cm / s] \\ ∴ \frac{dA}{dt} = 2 π (10) \times 3 [Putting r = 10 cm] \\ = 60 π {cm}^{2} / s \\ Hence, the rate at which the area of the circle is increasing when \\ the radius is 10 cm is 60 π {cm}^{2} / s . \end{array}$

Q.4 An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Ans.

$\begin{array}{l} Let the length of a side be x and V be the volume of the cube . \\ Then, V = x^{3} \\ Differentiating both the sides with respect to t, we get \\ \frac{dV}{dt} = 3 x^{2} \frac{dx}{dt} [By chain rule] \\ = 3 x^{2} (3) [∵ \frac{dx}{dt} = 3 cm / s, Given] \\ = 9 x^{2} \\ Since, edge of cube (x) = 10 cm, so \\ \frac{dV}{dt} = 9 {(10)}^{2} \\ = 900 {cm}^{3} / s \\ Hence, the volume of the cube is increasing at the rate of \\ 900 {cm}^{3} / s when the edge is 10 cm long . \end{array}$

Q.5 A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Ans.

$\begin{array}{l} The area of a circle (A) with radius (r) is given by \\ A = {πr}^{2} \\ Therefore, the rate of change of area (A) with respect to time (t) \\ is given by, \frac{dA}{dt} = \frac{d}{dt} {πr}^{2} \\ = 2 πr \frac{dr}{dt} [By chain rule] \\ = 2 π (8) (5) [\begin{array}{l} ∵ \frac{dr}{dt} = 5 cm / s, \\ r = 8 cm, Given \end{array}] \\ = 80 π \\ Hence, when the radius of the circular wave is 8 cm, the \\ enclosed area is increasing at the rate of 80 π {cm}^{2} / s . \end{array}$

Q.6 The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Ans.

$\begin{array}{l} Let the circumference of a circle be C with radius (r) is \\ given byC = 2 πr \\ Then, the rate of change of circumference (C) with respect \\ to time (t) is given by, \\ \frac{dC}{dt} = \frac{d}{dt} 2 πr \\ = 2 π \frac{dr}{dt} [By chain rule] \\ \Rightarrow \frac{dC}{dt} = 2 π (0 .7) [∵ \frac{dr}{dt} = 0 .7 cm / s] \\ \Rightarrow \frac{dC}{dt} = 1 .4 π \\ Hence, the rate of increase of the circumference is 1 .4 π cm / s . \end{array}$

Q.7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

Ans.

$\begin{array}{l} Since the length (x) is decreasing at the rate of 5 cm / minute \\ and the width (y) is increasing at the rate of 4 cm / minute, \\ we have : \frac{dx}{dt} = - 5 cm / minute and \frac{dy}{dt} = 4 cm / minute \\ (a) Perimeter of rectangle (P) = 2 (x + y) \\ Differentiating w . r . t . t, we get \\ \frac{dP}{dt} = 2 \frac{d}{dt} (x + y) \\ = 2 (\frac{dx}{dt} + \frac{dy}{dt}) \\ = 2 (- 5 + 4) \\ = - 2 cm / minute \\ Thus, the perimeter is decreasing at the rate of 2 cm / \min . \\ (b) Area of rectangle (A) = xy \\ Differentiating w . r . t . t, we get \\ \frac{dA}{dt} = \frac{d}{dt} xy \\ = x \frac{dy}{dt} + y \frac{dx}{dt} [By product rule] \\ = 4 x - 5 y [\begin{array}{l} ∵ \frac{dx}{dt} = - 5 cm / \min and \\ \frac{dy}{dt} = 4 cm / \min \end{array}] \\ = 4 (8) - 5 (6) \\ = 32 - 30 \\ = 2 {cm}^{2} / \min \\ Hence, the area of the rectangle is increasing at the rate \\ of 2 {cm}^{2} / \min . \end{array}$

Q.8 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Ans.

\begin{array}{l} The volume of a sphere (V) with radius (r) is given by, \\ V = \frac{4}{3} {πr}^{3} \\ Differentiating both sides, with respect to t, we get \\ \frac{dv}{dt} = \frac{d}{dt} \frac{4}{3} {πr}^{3} \\ = \frac{4}{3} π \times 3 r^{2} \frac{dr}{dt} \\ 900 = 4 {πr}^{2} \frac{dr}{dt} [∵ \frac{dV}{dt} = 900 {cm}^{3} / \sec] \\ \Rightarrow \frac{dr}{dt} = \frac{900}{4 {πr}^{2}} \\ = \frac{225}{π {(15)}^{2}} [∵ r = 15 ​ cm] \\ = \frac{225}{π (225)} = \frac{1}{π} \\ Hence, the rate at which the radius of the balloon increases \\ when the radius is 15 cm is \frac{1}{π} cm / \sec . \end{array}

Q.9 A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Ans.

\begin{array}{l} Volume of sphere having radius r = \frac{4}{3} {πr}^{3} \\ \Rightarrow V = \frac{4}{3} {πr}^{3} \\ Differentiating with respect to r, we get \\ \Rightarrow \frac{dV}{dr} = \frac{d}{dr} (\frac{4}{3} {πr}^{3}) \\ = \frac{4}{3} π \frac{d}{dr} r^{3} \\ = \frac{4}{3} π \times 3 r^{2} \\ = 4 {πr}^{2} \\ = 4 π {(10)}^{2} \\ = 400 π \\ Hence, the volu e of the balloon is increasing at the rate \\ of 400 π^{} . \end{array}

Q.10 A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Ans.

$\begin{array}{l} Let y m be the height of the wall at which the ladder touches . \\ Also, let the foot of the ladder be x m away from the wall . \\ Then in ΔABC, by Pythagoras theorem, we have: \\ x^{2} {+ y}^{2} =25 [Length of the ladder = 5 m] \\ \Rightarrow y= \sqrt{25 - x^{2}} \\ Differentiating both sides with respect to t, we get \\ \frac{dy}{dt} = \frac{d}{dt} \sqrt{25 - x^{2}} \\ = \frac{- 2x}{2 \sqrt{25 - x^{2}}} \frac{dx}{dt} [By chain rule] \\ = - \frac{x}{\sqrt{{25-x}^{2}}} (2) [\frac{dx}{dt} =2 cm/s] \\ = - \frac{2 (4)}{\sqrt{25 - {(4)}^{2}}} \\ = - \frac{8}{3} \\ Hence, the height of the ladder on the wall is decreasing at the \\ rate of - \frac{8}{3} cm/s . \end{array}$

Q.11 A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y coordinate is changing 8 times as fast as the x-coordinate.

Ans.

\begin{array}{l} The equation of the curve is given as : \\ 6 y = x^{3} + 2 \\ The rate of change of the position of the particle with \\ respect to time (t) is given by, \\ 6 \frac{dy}{dt} = \frac{d}{dt} (x^{3} + 2) \\ 6 \frac{dy}{dt} = 3 x^{2} \frac{dx}{dt} \\ When the y - coordinate of the particle changes 8 times as fast \\ as the x - coordinate i . e ., \frac{dy}{dt} = 8 \frac{dx}{dt}, we have \\ 6 (8 \frac{dx}{dt}) = 3 x^{2} \frac{dx}{dt} \\ \Rightarrow (16 - x^{2}) \frac{dx}{dt} = 0 \\ \Rightarrow 16 - x^{2} = 0 \\ \Rightarrow x = \pm 4 \\ When x = 4, \\ y = \frac{{(4)}^{3} + 2}{6} = 11 \\ When x = - 4, \\ y = \frac{{(- 4)}^{3} + 2}{6} \\ = - \frac{62}{6} = - \frac{31}{3} \\ Hence, the points required on the curve are (4, 11) and (- 4, - \frac{31}{3}) . \end{array}

Q.12 The radius of an air bubble is increasing at the rate of ½ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Ans.

\begin{array}{l} The air bubble is in the shape of a sphere . \\ So, the volume of an air bubble (V) with radius (r) is given by, \\ V = \frac{4}{3} {πr}^{3} \\ The rate of change of volume (V) with respect to time (t) is \\ obtained by differentiating both sides with respect to t, so \\ \frac{dV}{dt} = \frac{d}{dt} (\frac{4}{3} {πr}^{3}) \\ \frac{dV}{dt} = \frac{4}{3} \times 3 {πr}^{2} \frac{dr}{dt} \\ = 4 π {(1)}^{2} (\frac{1}{2}) [\begin{array}{l} ∵ \frac{dr}{dt} = \frac{1}{2} cm / s \\ and r = 1 cm \end{array}] \\ = 2 π {cm}^{3} / s \\ Hence, the rate at which the volume of the bubble increases \\ is 2 π {cm}^{3} / s . \end{array}

Q.13 A balloon, which always remains spherical, has a variable diameter 32 (2x+1). Find the rate of change of its volume with respect to x.

Ans.

\begin{array}{l} The volume (V) of a sphere with radius (r) is given by, \\ V = \frac{4}{3} {πr}^{3} \\ Diameter of balloon = \frac{3}{2} (2 x +1) \\ So, the radiusof balloon = \frac{3}{4} (2 x +1) \\ ∴ V = \frac{4}{3} π {\{\frac{3}{4} (2 x +1)\}}^{3} \\ = \frac{4}{3} π \times \frac{27}{64} (2 x {+1)}^{3} \\ V = \frac{9}{16} π (2 x {+1)}^{3} \\ Hence, the rate of change of volume with respect to x is \\ given by \\ \frac{dV}{dx} = \frac{9}{16} π \frac{d}{dx} (2 x {+1)}^{3} \\ = \frac{9}{16} π \times3 (2 x {+1)}^{2} \frac{d}{dx} (2 x +1) \\ [By chain rule] \\ \frac{dV}{dx} = \frac{27}{16} π (2 x {+1)}^{2} \times 2 \\ \Rightarrow \frac{dV}{dx} = \frac{27}{8} π (2 x {+1)}^{2} \\ Thus, the rate of change of volume with respect to x is \\ \frac{27}{8} π (2 x {+1)}^{2} . \end{array}

Q.14 Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one–sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Ans.

\begin{array}{l} The volume of a cone with radius (r) and height (h) is given by, \\ V = \frac{1}{3} {πr}^{2} h \\ Given, height of cone = \frac{1}{6} r \Rightarrow h = \frac{1}{6} r \Rightarrow r = 6 h \\ ∴ V = \frac{1}{3} π {(6 h)}^{2} h \\ = 12 {πh}^{3} \\ Differentiating both sides with respect to time (t), we get \\ \frac{dV}{dt} = \frac{d}{dt} (12 {πh}^{3}) \\ = 12 π \frac{d}{dt} h^{3} \\ = 36 {πh}^{2} \frac{dh}{dt} [By chain rule] \\ 12 = 36 π {(4)}^{2} \frac{dh}{dt} [∵ \frac{dV}{dt} = 12 and h = 4 cm] \\ \frac{dh}{dt} = \frac{12}{36 π {(4)}^{2}} = \frac{1}{48 π} \\ Hence, when the height of the sand cone is 4 cm, its height is \\ increasing at the rate of \frac{1}{48 π} cm / s . \end{array}

Q.15 The total cost C (x) in Rupees associated with the production of x units of an item is given
by C(x) = 0.007x³ – 0.003 x² + 15x + 4000 Find the marginal cost when 17 units are produced

Ans.

$\begin{array}{l} The total cost C (x) = 0.007 x^{3} - 0.003 x^{2} + 15 x + 4000 \\ Marginal cost is the rate of change of total cost with respect \\ to output . \\ ∴ Marginal cost (MC) = \frac{dC}{dx} \\ = 0.007 (3 x^{2}) - 0.003 (2 x) + 15 \\ When x = 17, MC = 0.021 {(17)}^{2} - 0.006 (17) + 15 \\ = 6 .069 - 0 .102 + 15 \\ = 20 .967 \\ Hence, when 17 units are produced, the marginal cost \\ is Rs . 20.967 . \end{array}$

Q.16 The total revenue in Rupees received from the sale of x units of a product is given by
R(x) = 13x² + 26x + 15 Find the marginal revenue when x = 7.

Ans.

$\begin{array}{l} The total revenue in Rupees received from the sale of x units \\ of a product is \\ R (x) = 13 x^{2} + 26 x + 15 \\ ∴ Marginal Revenue (MR) = \frac{d}{dx} R (x) \\ = \frac{d}{dx} (13 x^{2} + 26 x + 15) \\ = 26 x + 26 \\ = 26 (7) + 26 [When x = 7] \\ = 208 \\ Hence, the required marginal revenue is Rs 208 . \end{array}$

Q.17 The rate of change of the area of a circle with respect to its radius r at r =6 cmis(A)10π (B)12π (C)8π (D)11π

Ans.

$\begin{array}{l} T h e a r e a (A) o f a c i r c l e w i t h r a d i u s (r) i s g i v e n b y, \\ A = π r^{2} \\ D i f f e r e n t i a t i n g b o t h s i d e s w i t h r e s p e c t t o r, w e g e t \\ \frac{d A}{d r} = \frac{d}{d r} π r^{2} \\ = 2 π r \\ {(\frac{d A}{d r})}_{r = 6} = 2 π (6) \\ = 12 π \\ H e n c e, t h e r e q u i r e d r a t e o f c h a n g e o f t h e a r e a o f a c i r c l e \\ i s 12 π . \\ T h e c o r r e c t a n s w e r i s B . \end{array}$

Q.18 The total revenue in Rupees received from the sale of x units of a product is given by Rx = 3x2+36x+5.The marginal revenue, when x = 15 is:(A) 116 (B) 96 (C) 90 (D) 126

Ans.

$\begin{array}{l} ∵ Marginal revenue = The rate of change of total revenue \\ with respect to the number of \\ units sold . \\ ∴ Marginal Revenue (MR) = \frac{d}{dx} R (x) \\ = \frac{d}{dx} (3 x^{2} + 36 x + 5) \\ = 6 x + 36 \\ Putting x = 15, we get \\ Marginal Revenue (MR) = 6 (15) + 36 \\ = 90 + 36 \\ = 126 \\ Hence, the required marginal revenue is Rs 126 . \\ The correct answer is D . \end{array}$

Q.19 Show that the function given by f(x) = 3x + 17 is strictly increasing on R.

Ans.

\begin{array}{l} Function is given by f (x) = 3 x + 17 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 3 > 0 in every interval of R . \\ Thus, the function is strictly increasing on R . \end{array}

Q.20 Show that the function given by f(x) = e^2x is strictly increasing on R

Ans.

\begin{array}{l} Let x_{1} and x_{2} be any two numbers in R . \\ Then, we have : \\ x_{1} < x_{2} \Rightarrow 2 x_{1} < 2 x_{2} \\ \Rightarrow e^{2 x_{1}} < e^{2 x_{2}} \\ \Rightarrow f (x_{1}) < f (x_{2}) \\ Hence, f is strictly increasing on R . \end{array}

Q.21

\begin{array}{l} Show that the function given by f(x) = sin x is \\ (a) strictly increasing in (0, \frac{π}{2}), \\ (b) strictly decreasing in (\frac{π}{2}, π), \\ (c) neither increasing nor decreasing in (0, π) \end{array}

Ans.

\begin{array}{l} The given function is f (x) = \sin x . \\ ∴ f ‘ (x) = cosx \\ (a) Since for each x \in (0, \frac{π}{2}), cosx > 0 \\ we have f ‘ (x) > 0 . \\ Hence, f is strictly increasing in (0, \frac{π}{2}) . \\ (b) Since for each x \in (\frac{π}{2}, π), cosx < 0 \\ we have f ‘ (x) < 0 . \\ Hence, f is strictly decreasing in (\frac{π}{2}, π) . \\ (c) From the results obtained in (a) and (b), it is clear that f is \\ neither increasing nor decreasing in (0, π). \end{array}

Q.22 Find the intervals in which the function f given by f(x) = 2x² − 3x is

(a) strictly increasing (b) strictly decreasing

Ans.

$\begin{array}{l} The given function is f (x) = 2 x^{2} - 3 x . \\ f ‘ (x) = 4 x - 3 \\ Therefore, f ‘ (x) = 0 \Rightarrow 4 x - 3 = 0 \\ \Rightarrow x = \frac{3}{4} \\ Now the point x = \frac{3}{4} divides the real line into two disjoint \\ intervals namely, (- \infty, \frac{3}{4}) and (\frac{3}{4}, \infty) . \end{array}$

$\begin{array}{l} In the interval (- \infty, \frac{3}{4}), \\ f’ (x) = 4 x - 3 < 0 \\ T h e r e f o r e, f is strictly deacreasing in this interval . Also, in \\ the interval (\frac{3}{4}, \infty), f’ (x) > 0 and so the function f is strictly \\ increasing in this interval . \end{array}$

Q.23 Find the intervals in which the function f given by f(x) = 2x³ − 3x² − 36x + 7 is
(a) strictly increasing (b) strictly decreasing

Ans.

$\begin{array}{l} ∴ f ‘ (x) = 0 \Rightarrow 6 (x - 3) (x + 2) = 0 \\ \Rightarrow x = - 2, 3 \\ The points x = - 2 and x = 3 divide the real line into three \\ disjoint intervals i . e ., (- \infty, - 2), (- 2, 3) and (3, \infty) . \end{array}$

$\begin{array}{l} In intervals (- \infty, - 2) and (3, \infty), f ‘ (x) is positive while in interval \\ (- 2, 3), f ‘ (x) is negative . \\ Hence, the given function (f) is strictly increasing in intervals \\ (- \infty, - 2) and (3, \infty), while function (f) is strictly decreasing in \\ interval (- 2, 3) . \end{array}$

Q.24 Find the intervals in which the following functions are strictly increasing or decreasing:

(a) x² + 2x − 5 (b)10 − 6x − 2x²

(c) −2x³ − 9x² − 12x + 1 (d) 6 − 9x − x²

(e) (x + 1)³ (x − 3)³

Ans.

$\begin{array}{l} (a) We have, \\ f (x) = x^{2} + 2 x - 5 \\ Differentiating both sides with respect to x, we get \\ f ‘ (x) = 2 x + 2 \\ ∴ f ‘ (x) = 0 \Rightarrow 2 x + 2 = 0 \\ \Rightarrow x = - 1 \\ Point x = - 1 divides the real line into two disjoint \\ intervals i . e ., (- \infty, - 1) and (- 1, \infty) . \\ In interval (- \infty, - 1), f ‘ (x) = 2 x + 2 < 0 \\ ∴ f is strictly decreasing in interval (- \infty, - 1) . \\ Thus, f is strictly decreasing for x < - 1. \\ In interval (- 1, \infty), f ‘ (x) = 2 x + 2 > 0 \\ ∴ f is strictly increasing in interval (- 1, \infty) . \\ Thus, f is strictly increasing for x > - 1. \\ (b) We have, \\ f (x) = 10 - 6 x - 2 x^{2} \\ Differentiating both sides with respect to x, we get \\ f ‘ (x) = - 6 - 4 x \\ ∴ f ‘ (x) = 0 \Rightarrow - 6 - 4 x = 0 \\ \Rightarrow x = - \frac{3}{2} \\ Point x = - \frac{3}{2} divides the real line into two disjoint intervals \\ i . e ., (- \infty, - \frac{3}{2}) and (- \frac{3}{2}, \infty) . \\ In interval (- \infty, - \frac{3}{2}), f ‘ (x) = - 6 - 4 x > 0 \\ ∴ f is strictly increasing in interval (- \infty, - \frac{3}{2}) . \\ Thus, f is strictly increasing for x < - \frac{3}{2} . \\ In interval (- \frac{3}{2}, \infty), f ‘ (x) = - 6 - 4 x < 0 \\ ∴ f is strictly decreasing in interval (- \frac{3}{2}, \infty) . \\ Thus, f is strictly decreasing for x > - \frac{3}{2} . \\ (c) We have, \\ f (x) = - 2 x^{3} - 9 x^{2} - 12 x + 1 \\ Differentiating both sides with respect to x, we get \\ f ‘ (x) = - 6 x^{2} - 18 x - 12 = - 6 (x^{2} + 3 x + 2) \\ ∴ f ‘ (x) = 0 \Rightarrow - 6 (x^{2} + 3 x + 2) = 0 \\ \Rightarrow (x + 1) (x + 2) = 0 \\ \Rightarrow x = - 1, - 2 \\ The points x = - 1 and x = - 2 divide the real line into three \\ disjoint intervals i . e ., (- \infty, - 2), (- 2, - 1) and (- 1, \infty) . \\ In intervals (- \infty, - 2) and (- 1, \infty), f ‘ (x) is negative . \\ Hence, the given function (f) is strictly decreasing in intervals \\ (- \infty, - 2) and (- 1, \infty), \\ i . e ., function (f) is strictly decreasing when x < - 2 and x > - 1. \end{array}$ $\begin{array}{l} In intervals (- 2, - 1), f ‘ (x) is positive . \\ So, function (f) is strictly increasing in interval (- 2, - 1), \\ i . e ., function (f) is strictly increasing w h e n - 2 < x < - 1. \\ (d) We have, \\ f (x) = 6 - 9x - x^{2} \\ Differentiating both sides with respect to x, we get \\ f’ (x) = - 9 - 2 x \\ ∴ f’ (x) = 0 \Rightarrow - 9 - 2 x = 0 \\ \Rightarrow x = - \frac{9}{2} \\ Point x = - \frac{9}{2} divides the real line into two disjoint intervals \\ i .e ., (- \infty, - \frac{9}{2}) a n d (- \frac{9}{2}, \infty) . \\ In interval (- \infty, - \frac{9}{2}), f ‘ (x) = - 9 - 2 x > 0 \\ ∴ f is strictly increasing in interval (- \infty, - \frac{9}{2}) . \\ Thus, f is strictly increasing for x < - \frac{9}{2} . \\ In interval (- \frac{9}{2}, \infty), f ‘ (x) = - 9 - 2 x < 0 \\ ∴ f is strictly decreasing in interval (- \frac{9}{2}, \infty) . \\ Thus, f is strictly decreasing for x > - \frac{9}{2} . \end{array}$ $\begin{array}{l} (e) We have, \\ f(x) = {(x + 1)}^{3} (x - {3)}^{3} \\ D i f f e r e n t i a t i n g both sides with respect to x, we get \\ f’ (x) = {(x + 1)}^{3} \frac{d}{d x} {(x - 3)}^{3} + {(x - 3)}^{3} \frac{d}{d x} {(x + 1)}^{3} \\ [B y product rule] \\ \Rightarrow f’ (x) = {(x + 1)}^{3} 3 {(x - 3)}^{2} \frac{d}{d x} (x - 3) + {(x - 3)}^{3} 3 {(x + 1)}^{2} \frac{d}{d x} (x + 1) \\ [B y chain rule] \\ \Rightarrow f’ (x) = 3 {(x + 1)}^{3} {(x - 3)}^{2} (1 - 0) + 3 {(x - 3)}^{3} {(x + 1)}^{2} (1 + 0) \\ \Rightarrow f’ (x) = 3 {(x + 1)}^{2} {(x - 3)}^{2} (x + 1 + x - 3) \\ \Rightarrow f’ (x) = 3 {(x + 1)}^{2} {(x - 3)}^{2} (2 x - 2) \\ = 6 {(x + 1)}^{2} {(x - 3)}^{2} (x - 1) \\ N o w, f’ (x) = 0 \Rightarrow x = - 1, 1, 3 \\ The points x = - 1, x = 1, a n d x = 3 divide the real line into \\ four disjoint intervals i .e ., (- \infty, - 1), (- 1, 1), (1, 3) and (3, \infty) . \\ In intervals (- \infty, - 1) and (- 1, 1), \end{array}$ $\begin{array}{l} f’ (x) = 6 {(x + 1)}^{2} {(x - 3)}^{2} (x - 1) < 0 \\ ∴ f is strictly decreasing in intervals (- \infty, - 1) and (- 1, 1) . \\ In intervals (1, 3) and (3, \infty), f’ (x) = 6 {(x + 1)}^{2} {(x - 3)}^{2} (x - 1) > 0 \\ ∴ f is strictly increasing in intervals (1, 3) and (3, \infty) . \end{array}$

Q.25 Showthat y=log(1+x)−2x2+x,x> −1, is an increasing function of x throughout its domain.

Ans.

$\begin{array}{l} We have, y = \log (1 + x) - \frac{2 x}{2 + x} \\ Differentiating both sides with respect to x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \log (1 + x) - \frac{d}{dx} (\frac{2 x}{2 + x}) \\ = \frac{1}{1 + x} \frac{d}{dx} (1 + x) - \frac{(2 + x) \frac{d}{dx} 2 x - 2 x \frac{d}{dx} (2 + x)}{{(2 + x)}^{2}} \\ = \frac{1}{1 + x} \times (0 + 1) - \frac{2 (2 + x) - 2 x (0 + 1)}{{(2 + x)}^{2}} \\ = \frac{1}{1 + x} - \frac{4}{{(2 + x)}^{2}} \\ = \frac{4 + 4 x + x^{2} - 4 - 4 x}{(1 + x) {(2 + x)}^{2}} \\ = \frac{x^{2}}{(1 + x) {(2 + x)}^{2}} \\ Now, \frac{dy}{dx} = 0 \\ \Rightarrow \frac{x^{2}}{(1 + x) {(2 + x)}^{2}} = 0 \\ \Rightarrow x^{2} = 0 [(1 + x) \neq 0 and {(2 + x)}^{2} \neq 0] \\ \Rightarrow x = 0 \\ Since x > - 1, point x = 0 divides the domain (- 1, \infty) in two \\ disjoint intervals i . e ., - 1 < x < 0 and x > 0 . \\ When - 1 < x < 0, we have : \\ x < 0 \Rightarrow x^{2} > 0 \\ x > - 1 \Rightarrow x + 2 > 0 \Rightarrow {(x + 2)}^{2} > 0 \\ ∴ \frac{dy}{dx} = \frac{x^{2}}{{(x + 2)}^{2}} > 0 \\ Also, when x > 0: \\ x > 0 \Rightarrow x^{2} > 0, (2 + x^{2}) > 0 \\ ∴ \frac{dy}{dx} = \frac{x^{2}}{{(x + 2)}^{2}} > 0 \\ Hence, function f is increasing throughout this domain . \end{array}$

Q.26 Find the values of x for whichy=[x(x−2)]2is an increasingfunction.

Ans.

$\begin{array}{l} We have, \\ y = {[x (x - 2)]}^{2} \\ = x^{2} {(x - 2)}^{2} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} x^{2} {(x - 2)}^{2} \\ = x^{2} \frac{d}{dx} {(x - 2)}^{2} + {(x - 2)}^{2} \frac{d}{dx} x^{2} \\ = x^{2} . 2 (x - 2) \frac{d}{dx} (x - 2) + {(x - 2)}^{2} . 2 x \\ = 2 x^{2} (x - 2) (1 - 0) + 2 x {(x - 2)}^{2} \\ = 2 x (x - 2) (x + x - 2) \\ = 2 x (x - 2) (2 x - 2) \\ = 4 x (x - 2) (x - 1) \\ ∴ \frac{dy}{dx} = 0 \Rightarrow x = 0, 1, 2 \\ The points x = 0, x = 1, and x = 2 divide the real line into four \\ disjoint intervals i . e ., (- \infty, 0), (0, 1), (1, 2), (2, \infty) . \\ In intervals (- \infty, 0) and (1, 2), \frac{dy}{dx} < 0 \\ ∴ y is strictly decreasing in intervals (- \infty, 0) and (1, 2) . \\ However, in intervals (0, 1) and (2, \infty), \frac{dy}{dx} > 0 \\ ∴ y is strictly increasing in intervals (0, 1) and (2, \infty) . \\ ∴ y is strictly increasing for 0 < x < 1 and x > 2 . \end{array}$

Q.27 Prove that y=4sinθ(2+cosθ)−θ is an increasing function of θ in [0,π2].

Ans.

$\begin{array}{l} We have, y = \frac{4 sinθ}{(2 + cosθ)} - θ \\ Differentiating w . r . t . θ, we get \\ \frac{dy}{dθ} = \frac{d}{dθ} {\frac{4 sinθ}{(2 + cosθ)} - θ} \\ = \frac{(2 + cosθ) \frac{d}{dθ} (4 sinθ) - 4 sinθ \frac{d}{dθ} (2 + cosθ)}{{(2 + cosθ)}^{2}} - \frac{d}{dθ} θ \\ = \frac{4 (2 + cosθ) cosθ - 4 sinθ (0 - sinθ)}{{(2 + cosθ)}^{2}} - 1 \\ = \frac{4 (2 + cosθ) cosθ + 4 \sin^{2} θ}{{(2 + cosθ)}^{2}} - 1 \\ = \frac{8 cosθ + 4 (\cos^{2} θ + \sin^{2} θ)}{{(2 + cosθ)}^{2}} - 1 \\ = \frac{8 cosθ + 4}{{(2 + cosθ)}^{2}} - 1 \\ Now, \frac{dy}{dθ} = 0 \\ \Rightarrow \frac{8 cosθ + 4}{{(2 + cosθ)}^{2}} - 1 = 0 \Rightarrow \frac{8 cosθ + 4}{{(2 + cosθ)}^{2}} = 1 \\ \Rightarrow 8 cosθ + 4 = {(2 + cosθ)}^{2} \\ \Rightarrow 8 cosθ + 4 = 4 + 4 cosθ + \cos^{2} θ \\ \Rightarrow \cos^{2} θ - 4 cosθ = 0 \\ \Rightarrow cosθ (cosθ - 4) = 0 \\ \Rightarrow cosθ = 0 or cosθ = 4 \\ Since \cos θ \neq 4 \\ ∴ cosθ = 0 \Rightarrow θ = \frac{π}{2} \\ Now, \frac{dy}{dθ} = \frac{8 cosθ + 4}{{(2 + cosθ)}^{2}} - 1 \\ = \frac{8 cosθ + 4 - {(2 + cosθ)}^{2}}{{(2 + cosθ)}^{2}} \\ = \frac{8 cosθ + 4 - 4 - 4 cosθ - \cos^{2} θ}{{(2 + cosθ)}^{2}} \\ = \frac{cosθ (4 - cosθ)}{{(2 + cosθ)}^{2}} \\ In interval (0, \frac{π}{2}), we have cosθ > 0 . Also, 4 > cosθ \\ \Rightarrow 4 - cosθ > 0 . \\ ∴ cosθ (4 - cosθ) > 0 and also {(2 + cosθ)}^{2} > 0 \\ \Rightarrow \frac{cosθ (4 - cosθ)}{{(2 + cosθ)}^{2}} > 0 \Rightarrow \frac{dy}{dθ} > 0 \\ Therefore, y is strictly increasing in interval (0, \frac{π}{2}) . \\ Also, the given function is continuous at x = 0 and x = \frac{π}{2} . \\ Hence, y is increasing in interval (0, \frac{π}{2}) . \end{array}$

Q.28 Provethatthelogarithmicfunctionisstrictly increasingon (0, ∞).

Ans.

$\begin{array}{l} The given function if f (x) = logx \\ ∴ f ‘ (x) = \frac{1}{x} \\ It is clear that for x > 0, f ‘ (x) = \frac{1}{x} > 0 \\ Hence, f (x) = \log x is strictly increasing in interval (0, \infty) . \end{array}$

Q.29 Prove that the function f given by f(x)= x² − x + 1 is neither strictly increasing
nor strictly decreasing on (−1, 1).

Ans.

\begin{array}{l} {The given function is f(x) = x}^{2} - x + 1 . \\ D i f f e r e n t i a t i n g with respect to x, we get \\ f’ (x) = 2 x - 1 \\ N o w, f ‘ (x) = 0 \Rightarrow 2 x - 1 = 0 \Rightarrow x = \frac{1}{2} \\ The point \frac{1}{2} divides the interval (- 1, 1) into two disjoint \\ intervals i .e ., (- 1, \frac{1}{2}) a n d (\frac{1}{2}, 1) . \\ Now, for interval (- 1, \frac{1}{2}), f ‘ (x) = 2 x - 1 < 0 \\ Therefore, f is strictly decreasing in interval (- 1, \frac{1}{2}) . \\ Now, for interval (\frac{1}{2}, 1), f ‘ (x) = 2 x - 1 > 0 \\ So, f is strictly increasing in interval (\frac{1}{2}, 1) . \\ Hence, f is neither strictly increasing nor decreasing in \\ interval (- 1, 1) . \end{array}

Q.30 Whichofthefollowingfunctionsarestrictly decreasingon(0,π2)?(A)cosx (B)cos2x (C)cos3x (D)tanx

Ans.

$\begin{array}{l} (A) L e t f_{1} (x) = \cos x \\ Differentiating w .r .t . x, we get \\ {f’}_{1} (x) = \frac{d}{d x} \cos x \\ = - \sin x \\ I n interval (0, \frac{π}{2}), sinx is positive in first quadrant . \\ So, {f’}_{1} (x) < 0 \\ T h e r e f o r e, f_{1} (x) = \cos x is strictly decreasing in interval (0, \frac{π}{2}) . \\ (B) L e t f_{2} (x) = \cos 2 x \\ Differentiating w .r .t . x, we get \\ {f’}_{2} (x) = \frac{d}{d x} \cos 2 x \\ = - 2 \sin 2 x \\ I n interval (0, \frac{π}{2}), sin2x is positive in first quadrant . \\ So, {f’}_{2} (x) < 0 \\ T h e r e f o r e, f_{2} (x) = \cos 2 x is strictly decreasing in interval (0, \frac{π}{2}) . \\ (C) L e t f_{3} (x) = \cos 3 x \\ Differentiating w .r .t . x, we get \\ {f’}_{3} (x) = \frac{d}{d x} \cos 3 x \\ = - 3 \sin 3 x \\ N o w, {f’}_{3} (x) = 0 \Rightarrow - 3 \sin 3 x = 0 \\ \Rightarrow 3 x = π \Rightarrow x = \frac{π}{3} \\ The point x = \frac{π}{3} divides the interval (0, \frac{π}{2}) into two disjoint \\ intervals i .e ., (0, \frac{π}{3}) and (\frac{π}{3}, \frac{π}{2}) . \\ N o w, in interval (0, \frac{π}{3}), \\ {f’}_{3} (x) = - 3 \sin 3 x < 0 [∵ 0 < x < \frac{π}{3} \Rightarrow 0 < 3 x < π] \\ ∴ f_{3} is strictly decreasing in interval (0, \frac{π}{3}) . \\ N o w, i n interval (\frac{π}{3}, \frac{π}{2}), \\ \frac{π}{3} < x < \frac{π}{2} \Rightarrow π < 3 x < \frac{3 π}{2} \\ S o, sin3x is in third quadrant, where sin3x is negative . \\ ∴ f_{3} (x) = - 3 \sin 3 x > 0 \\ T h e r e f o r e, f_{3} is strictly increasing in interval (\frac{π}{3}, \frac{π}{2}) . \\ {Hence, f}_{3} is neither increasing nor decreasing in interval (0, \frac{π}{2}) . \\ (D) L e t f_{4} (x) = \tan x \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ {f’}_{4} (x) = \frac{d}{d x} \tan x \\ = \sec^{2} x \\ I n int e r v a l (0, \frac{π}{2}), \sec x > 0 \Rightarrow \sec^{2} x > 0 \\ ∴ f_{4} (x) > 0 \\ ∴ f4 is strictly increasing in interval (0, \frac{π}{2}) . \\ Therefore, functions cos x and cos 2x are strictly decreasing \\ in (0, \frac{π}{2}) . \\ Hence, the correct answers are A and B . \end{array}$

Q.31 On which of the following intervals is the function f given byf(x)=x100 +sinx– 1strictly decreasing?(A)(0, 1)(B)(π2, π)(C)(0, π2)(D)None of these

Ans.

$\begin{array}{l} We have, f (x) = x^{100} + sinx - 1 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 100 x^{99} + cosx \\ f ‘ (x) > 0 [∵ 100 x^{99} > 0 and cosx >0 in interval (0, 1)] \\ Thus, function f is strictly increasing in interval (0, 1) . \\ In interval (\frac{π}{2}, π), \\ cosx < 0 and 100 x^{99} > 0 \\ So, f ‘ (x) > 0 \\ Thus, function f is strictly increasing in interval (\frac{π}{2}, π) . \\ In interval (0, \frac{π}{2}), \\ cosx > 0 and 100 x^{99} > 0 \\ \Rightarrow 100 x^{99} + cosx > 0 \\ So, f ‘ (x) > 0 \\ Thus, function f is strictly increasing in interval (0, \frac{π}{2}) . \\ Hence, function f is strictly decreasing in none of the intervals . \\ The correct answer is D . \end{array}$

Q.32 Find the least value of a such that the function f given f(x) =x²+ax+1 is strictly increasing on (1, 2).

Ans.

$\begin{array}{l} We have, \\ f (x) = x^{2} + ax + 1 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 2 x + a \\ Now, function f will be increasing in (1, 2), if \\ f ‘ (x) > 0 \Rightarrow 2 x + a > 0 \\ x > - \frac{a}{2} \\ Therefore, we have to find the least value of a such that \\ x > - \frac{a}{2}, when x \in (1, 2) \\ \Rightarrow x > - \frac{a}{2}, (when 1 < x < 2) \\ Thus, the least value of a for f to be increasing on (1, 2) is \\ given by, \\ - \frac{a}{2} = 1 \Rightarrow a = - 2 \\ Hence, the required value of a is - 2. \end{array}$

Q.33 Let I be any interval disjoint from (−1,1).Prove that the function fgiven byf(x)=x+1x is strictly increasing onI.

Ans.

$\begin{array}{l} we have, \\ f (x) = x + \frac{1}{x} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 1 - \frac{1}{x^{2}} \\ Now, f ‘ (x) = 0 \Rightarrow 1 - \frac{1}{x^{2}} = 0 \\ \Rightarrow 1 = \frac{1}{x^{2}} \Rightarrow x = \pm 1 \\ The points x = 1 and x = - 1 divide the real line in three disjoint \\ intervals i . e ., (- \infty, - 1), (- 1, 1) and (1, \infty) . \\ In interval (- 1, 1), it is noticed that : \\ - 1 < x < 1 \Rightarrow x^{2} < 1 \\ \Rightarrow 1 < \frac{1}{x^{2}}, x \neq 0 \\ \Rightarrow 1 - \frac{1}{x^{2}} < 0, x \neq 0 \\ f ‘ (x) = 1 - \frac{1}{x^{2}} < 0 on (- 1, 1) ~ {0} \\ ∴ f is strictly decreasing on (- 1, 1) ~ {0} \\ In intervals (- \infty, - 1) and (1, \infty), it is seen that \\ x < - 1 or x >1 \\ \Rightarrow x^{2} > 1 \Rightarrow 1 > \frac{1}{x^{2}} \\ \Rightarrow 1 - \frac{1}{x^{2}} > 0 \\ ∴ f ‘ (x) = 1 - \frac{1}{x^{2}} > 0 on (- \infty, - 1) and (1, \infty) . \\ ∴ f is strictly increasing on (- \infty, - 1) and (1, \infty) . \\ Hence, function f is strictly increasing in interval I disjoint from (- 1, 1) . \end{array}$

Q.34 Provethatthefunctionfgivenbyf(x) =logsinxisstrictly increasingon (0,π2)andstrictlydecreasingon(π2,π).

Ans.

$\begin{array}{l} We have, \\ f (x) = logsinx \\ Differentiating both sides w . r . t . x, we get \\ f ‘ (x) = \frac{1}{sinx} \frac{d}{dx} sinx [By chain rule] \\ = \frac{1}{sinx} cosx = cotx \\ f ‘ (x) = cotx > 0 in interval (0, \frac{π}{2}) \\ ∴ f (x) is strictly increasing function in interval (0, \frac{π}{2}) . \\ f ‘ (x) = cotx < 0 in interval (\frac{π}{2}, π) \\ ∴ f (x) is strictly decreasing function in interval (\frac{π}{2}, π) . \end{array}$

Q.35

\begin{array}{l} Prove that the function f given by f(x) = logcos x is strictly \\ decreasing on (0, \frac{π}{2}) and strictly increasing on (\frac{π}{2}, π) . \end{array}

Ans.

$\begin{array}{l} We have, \\ f (x) = logcosx \\ Differentiating both sides w . r . t . x, we get \\ f ‘ (x) = \frac{1}{cosx} \frac{d}{dx} cosx [By chain rule] \\ = \frac{1}{cosx} \times - sinx = - tanx \\ f ‘ (x) = - tanx < 0 in interval (0, \frac{π}{2}) \\ ∴ f (x) is strictly decreasing function in interval (0, \frac{π}{2}) . \\ f ‘ (x) = - tanx > 0 in interval (\frac{π}{2}, π) \\ ∴ f (x) is strictly increasing function in interval (\frac{π}{2}, π) . \end{array}$

Q.36 Prove that the function f given byf(x) = x2−3x2+3x−100is increasinginR.

Ans.

$\begin{array}{l} We have, \\ f (x) = x^{3} - 3 x^{2} + 3 x - 100 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 3 x^{2} - 6 x + 3 \\ = 3 (x^{2} - 2 x + 1) \\ = 3 {(x - 1)}^{2} \\ For any x \in R, {(x - 1)}^{2} > 0 . \\ Thus, f ‘ (x) is always positive in R . \\ Hence, the given function f (x) is increasing in R . \end{array}$

Q.37 The interval in which y=x2e−xis increasing is(A) (−∞, ∞) (B )(−2, 0) (C) (2, ∞) (D) (0, 2)

Ans.

$\begin{array}{l} We have, \\ y = x^{2} e^{- x} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = x^{2} \frac{d}{dx} e^{- x} + e^{- x} \frac{d}{dx} x^{2} \\ = - x^{2} e^{- x} + e^{- x} (2 x) \\ = e^{- x} (- x^{2} + 2 x) \\ Now, \frac{dy}{dx} = 0 \\ \Rightarrow e^{- x} (- x^{2} + 2 x) = 0 \\ \Rightarrow x = 0, 2 [∵ e^{- x} \neq 0] \\ The points x = 0 and x = 2 divide the real line into three disjoint \\ intervals i . e ., (- \infty, 0), (0, 2) and (2, \infty) . \\ f ‘ (x) < 0 in intervals (- \infty, 0) and (2, \infty) . \\ So, f (x) is decreasing in intervals (- \infty, 0) and (2, \infty) . \\ Now, f ‘ (x) > 0 in interval (0, 2) \\ So, f (x) is increasing in intervals (0, 2) \\ Hence, f is strictly increasing in interval (0, 2) . \\ The correct answer is D . \end{array}$

Q.38 Find the slope of the tangent to the curve y = 3x⁴ − 4x at x = 4

Ans.

$\begin{array}{l} The given curve is y = 3 x^{4} - 4 x . \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (3 x^{4} - 4 x) \\ = 12 x^{3} - 4 \\ Then, the slope of the tangent to the given curve at x = 4 is \\ given by, \\ {(\frac{dy}{dx})}_{x = 4} = {(12 x^{3} - 4)}_{x = 4} \\ = 12 {(4)}^{3} - 4 \\ = 768 - 4 \\ = 764 \\ Thus, the slope of tangent to the given curve is 764 . \end{array}$

Q.39

\begin{array}{l} Find the slope of the tangent to the curve y = \frac{x - 1}{x - 2}, x \neq 2 \\ a t x = 10 . \end{array}

Ans.

$\begin{array}{l} The given curve is y = \frac{x - 1}{x - 2} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (\frac{x - 1}{x - 2}) \\ = \frac{(x - 2) \frac{d}{dx} (x - 1) - (x - 1) \frac{d}{dx} (x - 2)}{{(x - 2)}^{2}} \\ = \frac{(x - 2) (1 - 0) - (x - 1) (1 - 0)}{{(x - 2)}^{2}} \\ = \frac{x - 2 - x + 1}{{(x - 2)}^{2}} \\ = \frac{- 1}{{(x - 2)}^{2}} \\ Thus, the slope of the tangent at x = 10 is given by, \\ {(\frac{dy}{dx})}_{x = 10} = {(\frac{- 1}{{(x - 2)}^{2}})}_{x = 10} \\ = \frac{- 1}{{(10 - 2)}^{2}} \\ = \frac{- 1}{64} \\ Thus, the slope of tangent to the given curve is \frac{- 1}{64} . \end{array}$

Q.40 Find the slope of the tangent to curve y = x³−x+1 at the point whose x-coordinate is 2.

Ans.

$\begin{array}{l} The given curve is \\ y = x^{3} - x + 1 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (x^{3} - x + 1) \\ = 3 x^{2} - 1 \\ The slope of tangent to the curve is \\ {(\frac{dy}{dx})}_{x = 2} = 3 {(2)}^{2} - 1 \\ = 12 - 1 = 11 \\ Thus, the slope of the tangent at the point where the \\ x - coordinate is 2 is 11 . \end{array}$

Q.41 Find the slope of the tangent to curve y = x³−3x+2 at the point whose x-coordinate
is 3.

Ans.

$\begin{array}{l} The given curve is \\ y = x^{3} - 3 x + 2 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (x^{3} - 3 x + 2) \\ = 3 x^{2} - 3 \\ The slope of tangent to the curve is \\ {(\frac{dy}{dx})}_{x =3} = 3 {(3)}^{2} - 3 \\ = 27 - 3 = 24 \\ Thus, the slope of the tangent at the point where the \\ x - coordinate is 3 is 24 . \end{array}$

Q.42 Find the slope of the normal to the curve x =acos3θ, y=asin3θat θ=π4.

Ans.

\begin{array}{l} Given that x = {acos}^{3} θ and y = {asin}^{3} θ \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = a \frac{d}{dθ} \cos^{3} θ \\ = 3 {acos}^{2} θ \frac{d}{dθ} cosθ = - 3 {acos}^{2} θsinθ \\ \frac{dy}{dθ} = a \frac{d}{dθ} \sin^{3} θ \\ = 3 {asin}^{2} θ \frac{d}{dθ} sinθ \\ = 3 {asin}^{2} θcosθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{3 {asin}^{2} θcosθ}{- 3 {acos}^{2} θsinθ} = - tanθ \\ Therefore, the slope of the tangent at θ = \frac{π}{4}, is \\ {(\frac{dy}{dx})}_{(θ = \frac{π}{4})} = {(- tanθ)}_{(θ = \frac{π}{4})} \\ = - \tan \frac{π}{4} \\ = - 1 \\ Hence, the slope of the normal at θ = \frac{π}{4}, is \\ slope of normal (M) = - \frac{1}{slope of tangent (m)} \\ = - \frac{1}{- 1} \\ = 1 \end{array}

Q.43

\begin{array}{l} Find the slope of the normal to the curve x = 1 - a s i n θ, \\ y = b c o s^{2} θ a t θ = \frac{π}{2} \end{array}

Ans.

$\begin{array}{l} Given that : \\ x = 1 - asinθ and y = {bcos}^{2} θ \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = \frac{d}{dθ} (1 - asinθ) \\ = 0 - acosθ \\ = - acosθ \\ \frac{dy}{dθ} = \frac{d}{dθ} ({bcos}^{2} θ) \\ = 2 bcosθ \frac{d}{dθ} cosθ \\ = - 2 bcosθsinθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{- 2 bcosθsinθ}{- acosθ} \\ = \frac{2 b}{a} sinθ \\ Slope of tangent at θ = \frac{π}{2}, is given by \\ m = {(\frac{dy}{dx})}_{θ = \frac{π}{2}} \\ = \frac{2 b}{a} \sin \frac{π}{2} = \frac{2 b}{a} \\ slope of normal \\ (M) = - \frac{1}{m} \\ = - \frac{1}{(\frac{2 b}{a})} \\ = - \frac{a}{2 b} \\ Thus, slope of normal is - \frac{a}{2 b} . \end{array}$

Q.44 Find points at which the tangent to the curve
y = x³ − 3x² − 9x + 7 is parallel to the x axis.

Ans.

$\begin{array}{l} The equation of the given curve is \\ y = x^{3} - {3x}^{2} - 9x + 7 \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} (x^{3} - {3x}^{2} - 9x + 7) \\ = 3 x^{2} - 6 x - 9 \\ Now, the tangent is parallel to the x-axis if the slope of the \\ tangent is zero . \\ 3 x^{2} - 6 x - 9 = 0 \Rightarrow 3 (x^{2} - 2 x - 3) = 0 \\ \Rightarrow 3 (x - 3) (x + 1) = 0 \\ \Rightarrow x = 3, - 1 \\ W h e n x = 3, then \\ y = {(3)}^{3} - 3 {(3)}^{2} - 9 (3) + 7 \\ = 27 - 27 - 27 + 7 \\ = - 20 \\ W h e n x = - 1, then \\ y = {(- 1)}^{3} - 3 {(- 1)}^{2} - 9 (- 1) + 7 \\ = - 1 - 3 + 9 + 7 \\ = 12 \\ Hence, the points at which the tangent is parallel to the x-axis \\ are (3, - 20) and (- 1, 12) . \end{array}$

Q.45 Find a point on the curve y = (x − 2)² at which the tangent is parallel to the chord joining the
points (2, 0) and (4, 4).

Ans.

$\begin{array}{l} G i v e n curve is \\ y = {(x - 2)}^{2} \\ D i f f e r e n t i a t i n g both sides, w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} {(x - 2)}^{2} \\ = 2 (x - 2) \frac{d}{d x} (x - 2) \\ = 2 (x - 2) \\ The slope of the chord \\ = \frac{4 - 0}{4 - 2} [∵ m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}] \\ = 2 \\ Since, \\ the slope of the tangent = slope of the chord, we have: \\ 2 (x - 2) = 2 \\ \Rightarrow x = 3 \\ W h e n x = 3, y = {(3 - 2)}^{2} = 1 \\ Hence, the required point is (3, 1) . \end{array}$

Q.46 Find the point on the curve y = x³ − 11x + 5
at which the tangent is y = x − 11.

Ans.

$\begin{array}{l} The equation of the given curve is \\ y = x^{3} - 11 x + 5 \\ The equation of the tangent to the given curve is \\ y = x - 11 \\ comparing it with y = mx + c, we get \\ m = 1 \\ ∴ slope of the tangent (m) = 1 \\ Now, the slope of the tangent to the given curve at the \\ point (x, y) is, \\ \frac{d y}{d x} = \frac{d}{d x} (x^{3} - 11 x + 5) \\ = 3 x^{2} - 11 \\ ∵ 3 x^{2} - 11 = 1 [G i v e n] \\ \Rightarrow x^{2} = \frac{12}{3} = 4 \\ \Rightarrow x = \pm 2 \\ When x = 2, y = {(2)}^{3} - 11(2) + 5 \\ = 8 - 22 + 5 \\ = - 9 . \\ When x = - 2, y = (- 2)3 - 11 (- 2) + 5 \\ = - 8 + 22 + 5 \\ = 19 . \\ Hence, the required points are (2, - 9) and (- 2, 19) . \end{array}$

Q.47

\begin{array}{l} Find the equation of all lines having slope - 1 that are \\ tangents to the curve y = \frac{1}{x - 1}, x \neq 1 . \end{array}

Ans.

$\begin{array}{l} Equation of given curve is y = \frac{1}{x - 1}, x \neq 1 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \frac{1}{x - 1} \\ = - \frac{1}{{(x - 1)}^{2}} \frac{d}{dx} (x - 1) \\ = - \frac{1}{{(x - 1)}^{2}} \\ So, m = {(\frac{dy}{dx})}_{(x, y)} \\ = - \frac{1}{{(x - 1)}^{2}} \\ According to question, \\ - \frac{1}{{(x - 1)}^{2}} = - 1 \Rightarrow {(x - 1)}^{2} = 1 \\ \Rightarrow x - 1 = \pm 1 \\ \Rightarrow x - 1 = 1 or x - 1 = - 1 \\ \Rightarrow x = 2, 0 \\ When x = 0, y = - 1 \\ When x = 2, y = 1 \\ Thus, there are two tangents to the given curve having slope \\ - 1 . These are passing through the points (0, -1) and (2, 1) . \end{array}$ $\begin{array}{l} And equation of two tangents are \\ y - 1 = - 1 (x - 2) \\ y - 1 = - x + 2 \\ x + y - 3 = 0 \\ A n d, y + 1 = - 1 (x - 0) \\ y + 1 = - x \\ x + y + 1 = 0 \end{array}$

Q.48

\begin{array}{l} Find the equation of all lines having slope 2 which are \\ tangents to the curve y = \frac{1}{x - 3}, x \neq 3 . \end{array}

Ans.

$\begin{array}{l} Equation of given curve is y = \frac{1}{x - 3}, x \neq 3 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (\frac{1}{x - 3}) \\ = - \frac{1}{{(x - 3)}^{2}} \frac{d}{dx} (x - 3) \\ = - \frac{1}{{(x - 3)}^{2}} \\ So, m = {(\frac{dy}{dx})}_{(x, y)} \\ = - \frac{1}{{(x - 3)}^{2}} \\ According to question, \\ - \frac{1}{{(x - 3)}^{2}} = 2 \\ \Rightarrow {(x - 3)}^{2} = - 2 \\ This is impossible that square of a number is negative . \\ Hence, there is no tangent to the given curve having slope 2 . \end{array}$

Q.49

\begin{array}{l} Find the equations of all lines having slope 0 which are tangent \\ to the curve y = \frac{1}{x^{2} - 2 x + 3} \end{array}

Ans.

$\begin{array}{l} The equation of the given curve is y = \frac{1}{x^{2} - 2 x + 3} \\ The slope of the tangent to the given curve at any point (x, y) \\ {(\frac{d y}{d x})}_{(x, y)} = \frac{d}{d x} \frac{1}{x^{2} - 2 x + 3} \\ = - \frac{1}{{(x^{2} - 2 x + 3)}^{2}} \frac{d}{d x} (x^{2} - 2 x + 3) \\ = - \frac{(2 x - 2)}{{(x^{2} - 2 x + 3)}^{2}} \\ Since, the slope of the tangent is 0, then we have: \\ - \frac{(2 x - 2)}{{(x^{2} - 2 x + 3)}^{2}} = 0 \Rightarrow (2 x - 2) = 0 \\ \Rightarrow x = 1 \\ W h e n x = 1, y = \frac{1}{1 - 2 + 3} = \frac{1}{2} \\ T h e point on the curve is (1, \frac{1}{2}) . \\ ∴ The equation of the tangent through (1, \frac{1}{2}) i s \\ y - \frac{1}{2} = 0 (x - 1) \\ y = \frac{1}{2} \\ Hence, the equation of the required line is y = \frac{1}{2} . \end{array}$

Q.50

\begin{array}{l} Findpointsonthe curve \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 atwhichthe tangents \\ are (i) paralleltox - axis (ii) paralleltoy - axis . \end{array}

Ans.

$\begin{array}{l} The equation of the given curve is \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 \dots (i) \\ On differentiating both sides with respect to x, we have : \\ \frac{2 x}{9} + \frac{2 y}{16} \frac{dy}{dx} = 0 \\ \Rightarrow \frac{dy}{dx} = - \frac{2 x}{9} \times \frac{16}{2 y} = - \frac{16}{9} \frac{x}{y} \\ (i) When the tangent is parallel to x - axis, then \\ slope of tangent (m) = 0 \\ \frac{dy}{dx} = 0 \Rightarrow - \frac{16}{9} \frac{x}{y} = 0 \Rightarrow x = 0 \\ Putting value of x in equation (i), we have \\ \frac{{(0)}^{2}}{9} + \frac{y^{2}}{16} = 1 \Rightarrow y^{2} = 16 \Rightarrow y = \pm 4 \\ Hence, the points at which the tangents are parallel to the \\ x - axis are (0, 4) and (0, - 4) . \\ (b) The tangent is parallel to the y - axis if the slope of the \\ normal (M) is 0, i . e ., M = 0 \\ So, \frac{1}{- \frac{16}{9} \frac{x}{y}} = 0 [∵ M = \frac{1}{m}] \\ \Rightarrow \frac{y}{x} = 0 \Rightarrow y = 0 \\ Putting value of y in equation (i), we have \\ \frac{x^{2}}{9} + \frac{{(0)}^{2}}{16} = 1 \Rightarrow x^{2} = 9 \Rightarrow x = \pm 3 \\ Hence, the points at which the tangents are parallel to the y - axis \\ are (3, 0) and (- 3, 0) . \end{array}$

Q.51 Find the equations of the tangent and normal to the given curves at the indicated points:

(i) y = x⁴− 6x³+13x²−10x + 5 at (0, 5)

(ii) y = x⁴− 6x³+13x²−10x + 5 at (1, 3)

(iii) y = x³ at (1, 1)

(iv) y = x² at (0, 0)

$(v) x = c o s t, y = s i n t a t t = \frac{π}{4} .$

Ans.

$\begin{array}{l} (i) The equation of the curve is y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 . \\ On differentiating with respect to x, we get: \\ \frac{d y}{d x} = \frac{d}{d x} (x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5) \end{array}$ $\begin{array}{l} = 4 x^{3} - 18 x^{2} + 26 x - 10 \\ {(\frac{dy}{dx})}_{(0, 5)} = 4 {(0)}^{3} - 18 {(0)}^{2} + 26 (0) - 10 \\ = - 10 \\ Thus, the slope of the tangent at (0, 5) is - 10 . \\ The equation of the tangent is given as : \\ y - 5 = - 10 (x - 0) \\ \Rightarrow y = - 10 x + 5 \\ \Rightarrow 10 x + y = 5 \\ The slope of the normal (M) at (0, 5) = - \frac{1}{- 10} = \frac{1}{10} \\ There fore, the e quation of the normal at (0, 5) is given as : \\ y - 5 = \frac{1}{10} (x - 0) [∵ y - y_{1} = M (x - x_{1})] \\ \Rightarrow 10 y - 50 = x \\ \Rightarrow x - 10 y + 50 = 0 \\ (ii) The equation of the curve is y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 . \\ On differentiating with respect to x, we get : \\ \frac{dy}{dx} = \frac{d}{dx} (x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5) \\ = 4 x^{3} - 18 x^{2} + 26 x - 10 \\ {(\frac{dy}{dx})}_{(1, 3)} = 4 {(1)}^{3} - 18 {(1)}^{2} + 26 (1) - 10 \\ = 2 \\ Thus, the slope of the tangent at (1, 3) is 2 . \\ The equation of the tangent is given as : \end{array}$

$\begin{array}{l} y - 3 = 2 (x - 1) \\ \Rightarrow y - 3 = 2 x - 2 \\ \Rightarrow y = 2 x + 1 \\ T h e s l o p e o f t h e n o r m a l (M) a t (1, 3) = - \frac{1}{2} \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (1, 3) i s g i v e n a s : \\ y - 3 = - \frac{1}{2} (x - 1) [∵ y - y_{1} = M (x - x_{1})] \\ \Rightarrow x + 2 y - 7 = 0 \\ (i i) T h e e q u a t i o n o f t h e c u r v e i s y = x^{3} \\ O n d i f f e r e n t i a t i n g w i t h r e s p e c t t o x, w e g e t : \\ \frac{d y}{d x} = \frac{d}{d x} (x^{3}) = 3 x^{2} \\ {(\frac{d y}{d x})}_{(1, 1)} = 3 {(1)}^{2} = 3 \\ T h u s, t h e s l o p e o f t h e t a n g e n t a t (1, 1) i s 3 . \\ E q u a t i o n o f t a n g e n t a t (1, 1) \\ o f t h e t a n g e n t i s : y - 1 = 3 (x - 1) \Rightarrow y = 3 x - 2 \\ T h e s l o p e o f t h e n o r m a l a t (1, 1) = - \frac{1}{3} \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (1, 1) i s : \\ y - 1 = - \frac{1}{3} (x - 1) \\ \Rightarrow x + 3 y - 4 = 0 \\ (i v) T h e e q u a t i o n o f t h e c u r v e i s y = x^{2} \\ O n d i f f e r e n t i a t i n g w i t h r e s p e c t t o x, w e g e t : \end{array}$ $\begin{array}{l} \frac{d y}{d x} = \frac{d}{d x} (x^{2}) = 2 x \\ {(\frac{d y}{d x})}_{(0, 0)} = 2 (0) = 0 \\ E q u a t i o n o f t a n g e n t a t (0, 0) : \\ y - 0 = 0 (x - 0) \\ \Rightarrow y = 0 \\ T h e s l o p e o f t h e n o r m a l a t (0, 0) = - \frac{1}{0} = \frac{1}{0} \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (0, 0) i s : \\ y - 0 = \frac{1}{0} (x - 0) \Rightarrow x = 0 \\ (v) T h e e q u a t i o n o f t h e c u r v e i s x = c o s t, y = s i n t . \\ D i f f e r e n t i a t i n g w . r . t . t, w e g e t \\ \frac{d x}{d t} = - \sin t a n d \frac{d y}{d t} = \cos t \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ = \frac{\cos t}{- \sin t} = - \cot t \\ {(\frac{d y}{d x})}_{x = \frac{π}{4}} = - \cot (\frac{π}{4}) \\ = - 1 \\ ∴ E q u a t i o n o f t a n g e n t a t (\cos \frac{π}{4}, \sin \frac{π}{4}) i . e ., (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) : \end{array}$ $\begin{array}{l} y - \frac{1}{\sqrt{2}} = - 1 (x - \frac{1}{\sqrt{2}}) \\ \Rightarrow x + y = \sqrt{2} \\ T h e s l o p e o f t h e n o r m a l a t t = \frac{π}{4} : \\ M = - \frac{1}{{(\frac{d y}{d x})}_{x = \frac{π}{4}}} \\ = - \frac{1}{- 1} = 1 \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) : \\ y - \frac{1}{\sqrt{2}} = 1 (x - \frac{1}{\sqrt{2}}) \Rightarrow y = x \end{array}$

Q.52 Find the equation of the tangent line to the curve y = x² − 2x + 7 which is

(a) parallel to the line 2x – y + 9 = 0

(b) perpendicular to the line 5y −15x = 13.

Ans.

$\begin{array}{l} The equation of the given curve is \\ y = x^{2} - 2 x + 7 . .. (i) \\ On differentiating with respect to x, we get : \\ \frac{dy}{dx} = 2 x - 2 \\ slope of tangent to curve = {(\frac{dy}{dx})}_{(x, y)} = 2 x - 2 \\ (a) The equation of the line is 2 x - y + 9 = 0 \\ \Rightarrow y = 2 x + 9 \\ comparing with y = mx + c, we get \\ m = 2 \\ When tangent is parallel to given line than, \\ {(\frac{dy}{dx})}_{(x, y)} = m \\ 2 x - 2 = 2 \Rightarrow x = 2 \\ Putting x = 2 in equation (i), we get \\ y = {(2)}^{2} - 2 (2) + 7 = 7 \\ Thus, the equation of the tangent passing through (2, 7) is : \\ y - 7 = 2 (x - 2) \\ y - 7 = 2 x - 4 \\ y - 2 x - 3 = 0 \\ Thus, the equation of the tangent line to the given curve (which \\ is parallel to line 2 x - y + 9 = 0) is y - 2 x - 3 = 0 . \\ (b) The equation of the line is 5 y - 15 x = 13. \\ This equation can be written as \\ y = 3 x + \frac{13}{5} which in the form of y = mx + c, \\ So, m = 3 \\ If a tangent is perpendicular to the line 5 y - 15 = 13, \\ then the slope of the tangent is \\ \frac{- 1}{{(\frac{dy}{dx})}_{(x, y)}} = \frac{- 1}{3} \\ \Rightarrow 2 x - 2 = - \frac{1}{3} \\ \Rightarrow x = \frac{5}{6} \\ When x = \frac{5}{6}, y = {(\frac{5}{6})}^{2} - 2 (\frac{5}{6}) + 7 = \frac{217}{36} \\ Thus, the equation of the tangent passing through (\frac{5}{6}, \frac{217}{36}) is \\ y - \frac{217}{36} = \frac{- 1}{3} (x - \frac{5}{6}) \\ \Rightarrow 12 x + 36 y - 227 = 0 \\ Hence, the equation of the required tangent line is \\ 12 x + 36 y - 227 = 0 . \end{array}$

Q.53 Show that the tangents to the curve y = 7x³ + 11 at the points where x = 2 and x = −2 are parallel.

Ans.

$\begin{array}{l} W e have, \\ y = {7x}^{3} + 11 \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = 21 x^{2} \\ S l o p e of tangent at x=2, \\ m_{1} = {(\frac{d y}{d x})}_{x = 2} \\ = 21 {(2)}^{2} \\ = 84 \\ S l o p e of tangent at x= - 2, \\ m_{2} = {(\frac{d y}{d x})}_{x = - 2} \\ = 21 {(- 2)}^{2} \\ = 84 \\ S i n c e, m_{1} = m_{2} \\ Hence, the two tangents are parallel . \end{array}$

Q.54 Find the points on the curve y = x³ at which the slope of the tangent is equal to the y coordinate of the point.

Ans.

$\begin{array}{l} The equation of the given curve is y = x^{3} . \\ Diffferentiating both sides, w . r . t . x, we get \\ \frac{dy}{dx} = 3 x^{2} \\ Slope of tangent at point (x, y) is given by, \\ {(\frac{dy}{dx})}_{(x, y)} = 3 x^{2} \\ ∵ {(\frac{dy}{dx})}_{(x, y)} = y [Given] \\ \Rightarrow 3 x^{2} = y \\ But y = x^{3} [Given] \\ \Rightarrow x^{3} = 3 x^{2} \\ x^{3} - 3 x^{2} = 0 \\ \Rightarrow x^{2} (x - 3) = 0 \\ \Rightarrow x = 0 and 3 \\ When x = 0, then y = 0 and when x = 3, then y = 3 {(3)}^{2} = 27 . \\ Hence, the required points are (0, 0) and (3, 27) . \end{array}$

Q.55 For the curve y = 4x³ − 2x⁵, find all the points at which the tangents

passes through the origin.

Ans.

$\begin{array}{l} The equation of the given curve is \\ y = 4 x^{3} - 2 x^{5} . .. (i) \\ Differentiating with respect to x, we get \\ \frac{dy}{dx} = 12 x^{2} - 10 x^{4} \\ slope of tangent at the point (x_{1}, y_{1}) is \\ {(\frac{dy}{dx})}_{(x_{1}, y_{1})} = 12 {x_{1}}^{2} - 10 {x_{1}}^{4} \\ The equation of the tangent at (x_{1}, y_{1}) is given by \\ y - y_{1} = {(\frac{dy}{dx})}_{(x_{1}, y_{1})} (x - x_{1}) \\ y - y_{1} = (12 {x_{1}}^{2} - 10 {x_{1}}^{4}) (x - x_{1}) \\ When the tangent passes through the origin (0, 0), \\ then x = y = 0 . \\ 0 - y_{1} = (12 {x_{1}}^{2} - 10 {x_{1}}^{4}) (0 - x_{1}) \\ \Rightarrow y_{1} = (12 {x_{1}}^{2} - 10 {x_{1}}^{4}) x_{1} \\ \Rightarrow y_{1} = 12 {x_{1}}^{3} - 10 {x_{1}}^{5} . .. (ii) \\ Point (x_{1}, y_{1}) lies on curve (i), then \\ y_{1} = 4 {x_{1}}^{3} - 2 {x_{1}}^{5} . \dots (iii) \\ From equation (ii) and equation (iii), we have \\ 12 {x_{1}}^{3} - 10 {x_{1}}^{5} = 4 {x_{1}}^{3} - 2 {x_{1}}^{5} \\ \Rightarrow 8 {x_{1}}^{3} - 8 {x_{1}}^{5} = 0 \\ \Rightarrow 8 {x_{1}}^{3} (1 - {x_{1}}^{2}) = 0 \\ \Rightarrow x_{1} = 0, \pm 1 \\ When x = 0, y = 4 {(0)}^{3} - 2 {(0)}^{5} = 0 . \\ When x = 1, y = 4 {(1)}^{3} - 2 {(1)}^{5} = 2 . \\ When x = - 1, y = 4 {(- 1)}^{3} - 2 {(- 1)}^{5} = - 2 . \\ Hence, the required points are (0, 0), (1, 2) and (- 1, - 2) . \end{array}$

Q.56 Find the points on the curve x² + y² − 2x − 3 = 0 at which the tangents are
parallel to the x-axis.

Ans.

$\begin{array}{l} The equation of the given curve is \\ x^{2} {+ y}^{2} - 2x - 3 = 0 … (i) \\ Differentiating both sides, w .r .t . x, we get \\ 2x + 2y \frac{d y}{d x} - 2 = 0 \\ \Rightarrow \frac{d y}{d x} = \frac{2 - 2 x}{2 y} \\ = \frac{1 - x}{y} \\ S l o p e of tangent of the given curve at point (x_{1}, y_{1}) i s \\ \Rightarrow {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1 - x_{1}}{y_{1}} \\ I t is given that \\ {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 0 \\ \frac{1 - x_{1}}{y_{1}} = 0 \Rightarrow 1 - x_{1} = 0 \\ \Rightarrow x_{1} = 1 \\ S i n c e, point (x_{1}, y_{1}) lies on curve (i), we get \\ x_{1}^{2} + y_{1}^{2} - {2x}_{1} - 3 = 0 \\ {(1)}^{2} + y_{1}^{2} - 2 (1) - 3 = 0 \\ y_{1}^{2} = 4 \Rightarrow y_{1} = \pm 2 \\ Hence, the points at which the tangents are parallel to the \\ x-axis are (1, 2) and (1, - 2) . \end{array}$

Q.57 Find the equation of the normal at the point (am², am³) for the curve ay² = x³

Ans.

$\begin{array}{l} The equation of the given curve is {ay}^{2} = x^{3} . \\ On differentiating with respect to x, we have : \\ 2 ay \frac{dy}{dx} = 3 x^{2} \\ \Rightarrow \frac{dy}{dx} = \frac{3 x^{2}}{2 ay} \\ Slope at the point ({am}^{2}, {am}^{3}) \\ = {(\frac{dy}{dx})}_{({am}^{2}, {am}^{3})} \\ = \frac{3 {({am}^{2})}^{2}}{2 a ({am}^{3})} \\ = \frac{3 a^{2} m^{4}}{2 a^{2} m^{3}} \\ = \frac{3}{2} m \\ Slope of normal = - \frac{1}{(\frac{3}{2} m)} \\ = - \frac{2}{3 m} \\ Hence, the equation of the normal at ({am}^{2}, {am}^{3}) is given by, \\ y - {am}^{3} = - \frac{2}{3 m} (x - {am}^{2}) \\ \Rightarrow 3 my - 3 {am}^{4} = - 2 x + 2 {am}^{2} \\ \Rightarrow 2 x + 3 my - {am}^{2} (2 - 3 m^{2}) = 0 \end{array}$

Q.58 Find the equation of the normals to the curve y = x³ + 2x + 6 which are parallel to the line
x + 14y + 4 = 0.

Ans.

$\begin{array}{l} The equation of the given curve is y = x^{3} + 2 x + 6 . \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = 3 x^{2} + 2 \\ The slope of the tangent to the given curve at any point (x_{1}, y_{1}) \\ {(\frac{dy}{dx})}_{(x_{1}, y_{1})} = 3 {x_{1}}^{2} + 2 \\ Slope of normal to the given curve at any point (x_{1}, y_{1}) \\ = - \frac{1}{{(\frac{dy}{dx})}_{(x_{1}, y_{1})}} \\ = - \frac{1}{3 {x_{1}}^{2} + 2} \\ The equation of the given line is \\ x + 14 y + 4 = 0 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = - \frac{1}{14} \\ Slope of line at any point (x_{1}, y_{1}) \\ = {(\frac{dy}{dx})}_{(x_{1}, y_{1})} \\ = - \frac{1}{14} \\ Since normal is parallel to line, so \\ slope of normal = slope of line \\ - \frac{1}{3 x^{2} + 2} = - \frac{1}{14} \\ \Rightarrow 14 = 3 x^{2} + 2 \\ \Rightarrow x^{2} = \frac{12}{3} = 4 \\ \Rightarrow x = \pm 2 \\ When x = 2, y = 8 + 4 + 6 = 18 . \\ When x = - 2, y = - 8 - 4 + 6 = - 6 . \\ Therefore, the equations of normals passing through \\ the points (2, 18) and (- 2, - 6) . \\ y - 18 = - \frac{1}{14} (x - 2) \Rightarrow x + 14 y - 254 = 0 \\ and y + 6 = - \frac{1}{14} (x + 2) \Rightarrow x + 14 y + 86 = 0 \\ Hence, the equations of the normals to the given curve \\ (which are parallel to the given line) are \\ x + 14 y - 254 = 0, x + 14 y + 86 = 0 . \end{array}$

Q.59 Find the equations of the tangent and normal to the parabola y²= 4ax at the point (at², 2at).

Ans.

\begin{array}{l} The equation of parabola is y^{2} = 4 ax \\ Differentiating both sides w . r . t . x, we get \\ 2 y \frac{dy}{dx} = 4 a \\ \Rightarrow \frac{dy}{dx} = \frac{4 a}{2 y} = \frac{2 a}{y} \\ Slope of tangent to parabola at point ({at}^{2}, 2 at) \\ \Rightarrow {(\frac{dy}{dx})}_{({at}^{2}, 2 at)} = \frac{2 a}{2 at} = \frac{1}{t} \\ Slope of normal to parabola at point ({at}^{2}, 2 at) \\ = - \frac{1}{{(\frac{dy}{dx})}_{({at}^{2}, 2 at)}} \\ = - \frac{1}{(\frac{1}{t})} = - t \\ Then, the equation of the tangent at ({at}^{2}, 2 at) \\ y - 2 at = \frac{1}{t} (x - {at}^{2}) \\ \Rightarrow ty = x + {at}^{2} \\ And, the equation of the normal at ({at}^{2}, 2 at) \\ y - 2 at = - t (x - {at}^{2}) \\ \Rightarrow tx + y = 2 at + {at}^{3} \\ Thus, the required equations of tangent and normal are \\ ty = x + {at}^{2} and tx + y = 2 at + {at}^{3} . \end{array}

Q.60 Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1.

Ans.

\begin{array}{l} Given equations of the curves are x = y^{2} and xy = k . \\ Putting x = y^{2} in xy = k, we get \\ y^{2} . y = k \Rightarrow y = k^{\frac{1}{3}} \\ Putting value of y in xy = k, we get \\ k^{\frac{1}{3}} x = k \Rightarrow x = k^{\frac{2}{3}} \\ Thus, the point of intersection of the given curves is (k^{\frac{2}{3}}, k^{\frac{1}{3}}) . \\ Differentiating ​ the curves x = y^{2} and xy = k w . r . t . x, we get \\ \frac{d}{dx} x = \frac{d}{dx} y^{2} \Rightarrow 1 = 2 y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2 y} \\ Slope of tangent to the curves x = y^{2} at point (k^{\frac{2}{3}}, k^{\frac{1}{3}}) \\ m_{1} = {(\frac{dy}{dx})}_{(k^{\frac{2}{3}}, k^{\frac{1}{3}})} \\ = \frac{1}{2 k^{\frac{1}{3}}} \\ and \frac{d}{dx} (xy) = \frac{d}{dx} k \Rightarrow x \frac{dy}{dx} + y \frac{d}{dx} x = 0 \Rightarrow \frac{dy}{dx} = - \frac{y}{x} \\ Slope of tangent to the curve xy = k at point (k^{\frac{2}{3}}, k^{\frac{1}{3}}) \\ m_{2} = {(\frac{dy}{dx})}_{(k^{\frac{2}{3}}, k^{\frac{1}{3}})} \\ = - \frac{k^{\frac{1}{3}}}{k^{\frac{2}{3}}} \\ = - \frac{1}{k^{\frac{1}{3}}} \\ Since, two curves intersect at right angles if \\ m_{1} . m_{2} = - 1 \\ \frac{1}{2 k^{\frac{1}{3}}} \times - \frac{1}{k^{\frac{1}{3}}} = - 1 \\ 1 = 2 k^{\frac{2}{3}} \\ Cubing both sides, we get \\ 1 = 8 k^{2} . \\ Hence, the given two curves cut at right angels if 8 k^{2} = 1. \end{array}

Q.61

\begin{array}{l} Find the equations of the tangent and normal to the \\ hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} =1 at the point (x_{0} {,y}_{0}) . \end{array}

Ans.

\begin{array}{l} Find the equations of the tangent and normal to the \\ hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} =1 at the point (x_{0} {,y}_{0}) . \end{array}

Q.62

\begin{array}{l} Find the equation of the tangent to the curve y= \sqrt{3x - 2} \\ which is parallel to the line 4x - 2y + 5 = 0 . \end{array}

Ans.

\begin{array}{l} The equation of the given curve is y = \sqrt{3 x - 2} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} {(3 x - 2)}^{\frac{1}{2}} \\ = \frac{1}{2} {(3 x - 2)}^{- \frac{1}{2}} \frac{d}{dx} (3 x - 2) [By chain rule] \\ = \frac{1}{2 \sqrt{3 x - 2}} \times 3 \\ = \frac{3}{2 \sqrt{3 x - 2}} \\ The slope of the tangent to the given curve at any point (x_{1}, y_{1}), \\ m_{1} = {(\frac{dy}{dx})}_{(x_{1}, y_{1})} \\ = \frac{3}{2 \sqrt{3 x_{1} - 2}} \\ The equation of the given line is 4 x - 2 y + 5 = 0 . \\ Differentiating w . r . t . x, we get \\ 4 - 2 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = 2 \\ The slope of the tangent to the given line at any point (x_{1}, y_{1}), \\ m_{2} = {(\frac{dy}{dx})}_{(x_{1}, y_{1})} = 2 \\ Since, ​ tangent is parallel to given line . So, \\ m_{1} = m_{2} \\ \Rightarrow \frac{3}{2 \sqrt{3 x_{1} - 2}} = 2 \\ \Rightarrow \sqrt{3 x_{1} - 2} = \frac{3}{4} \\ \Rightarrow 3 x_{1} - 2 = \frac{9}{16} \\ \Rightarrow 3 x_{1} = \frac{9}{16} + 2 \\ \Rightarrow 3 x_{1} = \frac{41}{16} \Rightarrow x_{1} = \frac{41}{48} \\ Since, point (x_{1}, y_{1}) ​ lies on given curve y = \sqrt{3 x - 2} \\ so, y_{1} = \sqrt{3 x_{1} - 2} = \frac{3}{4} \\ ∴ Equation of the tangent passing through the point (\frac{41}{48}, \frac{3}{4}) is \\ y - \frac{3}{4} = 2 (x - \frac{41}{48}) \\ \Rightarrow 48 x - 24 y = 23 \\ Hence, the equation of the required tangent is 48 x - 24 y = 23 . \end{array}

Q.63 The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is
(A) 3
(B) 1/3
(C) – 3
(D) –1/3

Ans.

$\begin{array}{l} T h e equation of given curve is y = {2x}^{2} + 3sinx \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} ({2x}^{2} + 3sinx) \\ = 4 x + 3 \cos x \\ P u t t i n g x= 0, we get \\ m = {(\frac{d y}{d x})}_{x= 0} \\ = 4 (0) + 3 \cos 0 \\ = 3 \\ S l o p e of normal to the given curve is \\ M = - \frac{1}{m} \\ = - \frac{1}{3} \\ The correct answer is D . \end{array}$

Q.64 The line y = x + 1 is a tangent to the curve y² = 4x at the point
(A) (1, 2) (B) (2, 1) (C) (1, −2) (D) (−1, 2)

Ans.

$\begin{array}{l} The equation of the given curve is \\ y^{2} = 4 x \\ Differentiating w . r . t . x, we get \\ 2 y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{4}{2 y} = \frac{2}{y} \\ Therefore, the slope of the tangent to the given curve at any \\ point (x_{1}, y_{1}) = {(\frac{dy}{dx})}_{(x_{1}, y_{1})} \\ = \frac{2}{y_{1}} \\ The given line is y = x + 1 which is in the form of y = mx + c, \\ m = 1 \\ Since, tangent and line are parallel . So \\ {(\frac{dy}{dx})}_{(x_{1}, y_{1})} = m \Rightarrow \frac{2}{y_{1}} = 1 \\ y_{1} = 2 \\ Since, y = x + 1, so \\ x = y - 1 \\ = 2 - 1 = 1 \\ Hence, the line y = x + 1 is a tangent to the given curve at the \\ point (1, 2) . \\ The correct answer is A . \end{array}$

Q.65

\begin{array}{l} Using differentials, find the approximate value o f e a c h o f \\ the following up to 3 places of d e c i m a l : \\ (i) \sqrt{25 .3} (i i) \sqrt{49 .5} (i i i) \sqrt{0 .6} (i v) {(0 .009)}^{\frac{1}{3}} \\ (v) {(0 .999)}^{\frac{1}{10}} (v i) {(15)}^{\frac{1}{4}} (v i i) {(26)}^{\frac{1}{3}} (v i i i) {(255)}^{\frac{1}{4}} \\ (i x) {(82)}^{\frac{1}{4}} (x) {(401)}^{\frac{1}{2}} (x i) {(0 .0037)}^{\frac{1}{2}} (x i i) {(26 .57)}^{\frac{1}{3}} \\ (x i i i) {(81 .5)}^{\frac{1}{4}} (x i v) {(3 .968)}^{\frac{3}{2}} (x v) {(32 .15)}^{\frac{1}{5}} \end{array}

Ans.

$\begin{array}{l} (i) \sqrt{25 .3} \\ Let y = \sqrt{x}, x = 25 and Δx = 0.3 \\ \Rightarrow \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} \\ Then, \\ Δy = \sqrt{x + Δx} - \sqrt{x} \\ = \sqrt{25 .3} - \sqrt{25} \\ = \sqrt{25 .3} - 5 \\ \sqrt{25 .3} = Δy + 5 \\ Now, dy is approximately equal to Δy and is given by, \\ dy = (\frac{dy}{dx}) Δx \\ = \frac{1}{2 \sqrt{x}} (0 .3) \\ = \frac{1}{2 \sqrt{25}} (0 .3) \\ = \frac{1}{2 \times 5} \times 0 .3 \\ = 0 .03 \\ Hence, the approximate value of \sqrt{25 .3} is 5 + 0.03 = 5.03 . \\ (ii) \sqrt{49 .5} \\ Let y = \sqrt{x}, x = 49 and Δx = 0.5 \\ \Rightarrow \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} \\ Then, \\ Δy = \sqrt{x + Δx} - \sqrt{x} \\ = \sqrt{49 .5} - \sqrt{49} \\ = \sqrt{49 .5} - 7 \\ \sqrt{49 .5} = Δy + 7 \\ Now, dy is approximately equal to Δy and is given by, \\ dy = (\frac{dy}{dx}) Δx \\ = \frac{1}{2 \sqrt{x}} (0 .5) \\ = \frac{1}{2 \sqrt{49}} (0 .5) \\ = \frac{1}{2 \times 7} \times 0 .5 \\ = 0 .035 \\ Hence, the approximate value of \sqrt{49 .5} is 7 + 0.035 = 7.035 . \\ (iii) \sqrt{0 .6} \\ Let y = \sqrt{x}, x = 1 and Δx = - 0.4 \\ \Rightarrow \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} \\ Then, \\ Δy = \sqrt{x + Δx} - \sqrt{x} \\ = \sqrt{0 .6} - 1 \\ \sqrt{0 .6} = Δy + 1 \\ Now, dy is approximately equal to Δy and is given by, \\ dy = (\frac{dy}{dx}) Δx \\ = \frac{1}{2 \sqrt{x}} (- 0.4) \\ = \frac{1}{2 \sqrt{1}} (- 0.4) \\ = - 0.2 \\ Hence, the approximate value of \sqrt{0 .6} is 1 - 0.2 = 0.8 \\ (iv) {(0 .009)}^{\frac{1}{3}} \\ Let y = x^{\frac{1}{3}}, x = 0 .008 and Δx = 0 .001 \\ \Rightarrow \frac{dy}{dx} = \frac{1}{3 x^{\frac{2}{3}}} \\ Then, \\ Δy = {(x + Δx)}^{\frac{1}{3}} - x^{\frac{1}{3}} \\ = {(0 .009)}^{\frac{1}{3}} - 0 .2 \\ {(0 .009)}^{\frac{1}{3}} = Δy + 0 .2 \\ Now, dy is approximately equal to Δy and is given by, \\ \Rightarrow \frac{dy}{dx} = \frac{1}{3 x^{\frac{2}{3}}} (Δx) \\ = \frac{1}{3 {(0 .008)}^{\frac{2}{3}}} (0 .001) \\ = \frac{1}{3 (0 .04)} (0 .001) = \frac{0 .001}{0 .12} \\ = 0 .008 \\ Hence, the approximate value of {(0 .009)}^{\frac{2}{3}} is 0.2 + 0 . 008 = 0.208 \end{array}$ $\begin{array}{l} (v) {(0.999)}^{\frac{1}{10}} \\ L e t y = x^{\frac{1}{10}}, x = 1 and Δ x = - 0.001 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{10 x^{\frac{9}{10}}} \\ T h e n, \\ Δ y = {(x + Δ x)}^{\frac{1}{10}} - {(x)}^{\frac{1}{10}} \\ = {(0.999)}^{\frac{1}{10}} - {(1)}^{\frac{1}{10}} \\ {(0.999)}^{\frac{1}{10}} = Δ y + 1 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{10 x^{\frac{9}{10}}} \times - 0.001 \\ = \frac{1}{10 {(1)}^{\frac{9}{10}}} \times - 0.001 \\ = - \frac{0.001}{10} \\ = - 0.0001 \\ Hence, the approximate value of {(0.999)}^{\frac{1}{10}} i s \\ 1 + (- 0.001) = 0.9999 \\ (v i) {(15)}^{\frac{1}{4}} \\ L e t y = x^{\frac{1}{4}}, x = 16 and Δ x = - 1 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{4 x^{\frac{3}{4}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{4}} - x^{\frac{1}{4}} \\ = {(15)}^{\frac{1}{4}} - {(16)}^{\frac{1}{4}} \\ {(15)}^{\frac{1}{4}} = Δ y + 2 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{4 x^{\frac{3}{4}}} \times - 1 \\ = \frac{1}{4 {(16)}^{\frac{3}{4}}} \times - 1 \\ = \frac{1}{4 \times 8} \times - 1 = - 0.03125 \\ Hence, the approximate value of {(15)}^{\frac{1}{4}} i s \\ 2 + (- 0.03125) = 1.96875 \\ (v i i) {(26)}^{\frac{1}{3}} \\ L e t y = x^{\frac{1}{3}}, x = 27 and Δ x = - 1 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{3 x^{\frac{2}{3}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{3}} - x^{\frac{1}{3}} \\ = {(26)}^{\frac{1}{3}} - {(27)}^{\frac{1}{3}} \\ {(26)}^{\frac{1}{3}} = Δ y + 3 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{3 x^{\frac{2}{3}}} \times - 1 \\ = \frac{1}{3 {(27)}^{\frac{2}{3}}} \times - 1 \\ = \frac{1}{3 \times 9} \times - 1 \\ = - 0.0 \bar{370} \\ Hence, the approximate value of {(26)}^{\frac{1}{3}} i s \\ 3 + (- 0.0370) = 2 .9629 \\ (v i i i) {(255)}^{\frac{1}{4}} \\ L e t y = x^{\frac{1}{4}}, x = 256 and Δ x = - 1 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{4 x^{\frac{3}{4}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{4}} - x^{\frac{1}{4}} \\ = {(255)}^{\frac{1}{4}} - {(256)}^{\frac{1}{4}} \\ {(255)}^{\frac{1}{4}} = Δ y + 4 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{4 x^{\frac{3}{4}}} \times (- 1) \\ = \frac{1}{4 {(256)}^{\frac{3}{4}}} \times (- 1) \\ = \frac{- 1}{4 \times 4^{3}} = - 0.0039 \\ Hence, the approximate value of {(255)}^{\frac{1}{4}} i s \\ 4 + (- 0 .0039) = 3 .9961 \end{array}$ $\begin{array}{l} (i x) {(82)}^{\frac{1}{4}} \\ L e t y = x^{\frac{1}{4}}, x = 81 and Δ x = 1 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{4 x^{\frac{3}{4}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{4}} - x^{\frac{1}{4}} \\ = {(82)}^{\frac{1}{4}} - {(81)}^{\frac{1}{4}} \\ {(82)}^{\frac{1}{4}} = Δ y + 3 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{4 x^{\frac{3}{4}}} \times (1) \\ = \frac{1}{4 {(81)}^{\frac{3}{4}}} \times (1) \\ = \frac{1}{4 \times 3^{3}} = 0.009 \\ Hence, the approximate value of {(82)}^{\frac{1}{4}} i s \\ 3 + (0 .009) = 3 .009 \\ (x) {(401)}^{\frac{1}{2}} \\ L e t y = x^{\frac{1}{2}}, x = 400 and Δ x = 1 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \\ Then, \\ Δ y = \sqrt{x + Δ x} - \sqrt{x} \\ = \sqrt{401} - \sqrt{400} \\ = \sqrt{401} - 20 \\ \sqrt{401} = Δ y + 20 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{2 \sqrt{x}} (1) \\ = \frac{1}{2 \sqrt{400}} (1) \\ = \frac{1}{2 \times 20} \times (1) \\ = 0.025 \\ Hence, the approximate value of \sqrt{401} is 20 + 0 .025 = 20 .025 . \\ (x i) {(0.0037)}^{\frac{1}{2}} \\ L e t y = x^{\frac{1}{2}}, x = 0.0036 and Δ x = 0.0001 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \\ Then, \\ Δ y = \sqrt{x + Δ x} - \sqrt{x} \\ = \sqrt{0.0037} - \sqrt{0.0036} \\ = \sqrt{0.0037} - 0.06 \\ \sqrt{0.0037} = Δ y + 0.06 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{2 \sqrt{x}} (0.0001) \\ = \frac{1}{2 \sqrt{0.0036}} (0.0001) \\ = \frac{1}{2 \times 0.06} \times (0.0001) \\ = 0.00083 \\ Hence, the approximate value of \sqrt{0.0037} is \\ 0.06 + 0.00083 = 0 .06083 \\ (x i i) {(26.57)}^{\frac{1}{3}} \\ L e t y = x^{\frac{1}{3}}, x = 27 and Δ x = - 0.43 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{3 x^{\frac{2}{3}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{3}} - x^{\frac{1}{3}} \\ = {(26.57)}^{\frac{1}{3}} - {(27)}^{\frac{1}{3}} \\ {(26.57)}^{\frac{1}{3}} = Δ y + 3 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{3 x^{\frac{2}{3}}} \times - 0.43 \\ = \frac{1}{3 {(27)}^{\frac{2}{3}}} \times - 0.43 \\ = \frac{1}{3 (9)} \times - 0.43 = - 0.015 \\ Hence, the approximate value of {(26.57)}^{\frac{1}{3}} is \\ 3 - 0.015 = 2 .984 \\ (x i i i) {(81.5)}^{\frac{1}{4}} \\ L e t y = x^{\frac{1}{4}}, x = 81 and Δ x = 0.5 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{4 x^{\frac{3}{4}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{4}} - x^{\frac{1}{4}} \end{array}$ $\begin{array}{l} = {(81.5)}^{\frac{1}{4}} - {(81)}^{\frac{1}{4}} \\ {(81.5)}^{\frac{1}{4}} = Δ y + 3 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{4 x^{\frac{3}{4}}} \times (0.5) \\ = \frac{1}{4 {(81)}^{\frac{3}{4}}} \times (0.5) \\ = \frac{0.5}{4 \times 3^{3}} = 0.0046 \\ Hence, the approximate value of {(81.5)}^{\frac{1}{4}} i s \\ 3 + 0.0046 = 3 .0046 \\ (x i v) {(3.968)}^{\frac{3}{2}} \\ L e t y = x^{\frac{3}{2}}, x = 4 and Δ x = - 0.032 \\ \Rightarrow \frac{d y}{d x} = \frac{3}{2} x^{\frac{1}{2}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{3}{2}} - x^{\frac{3}{2}} \\ = {(3.968)}^{\frac{3}{2}} - {(4)}^{\frac{3}{2}} \\ {(3.968)}^{\frac{3}{2}} = Δ y + 8 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{3}{2} x^{\frac{1}{2}} \times (- 0.032) \\ = \frac{3}{2} {(4)}^{\frac{1}{2}} \times (- 0.032) \\ = \frac{3}{2} \times 2 \times (- 0.032) = - 0.096 \\ Hence, the approximate value of {(3.968)}^{\frac{3}{2}} i s \\ 8 - 0.096 = 7 .904 \\ (x v) {(32.15)}^{\frac{1}{5}} \\ L e t y = x^{\frac{1}{5}}, x = 32 and Δ x = 0.15 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{5 x^{\frac{4}{5}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{5}} - x^{\frac{1}{5}} \\ = {(32.15)}^{\frac{1}{5}} - {(32)}^{\frac{1}{5}} \\ {(32.15)}^{\frac{1}{4}} = Δ y + 2 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{5 x^{\frac{4}{5}}} \times (0.15) \\ = \frac{1}{5 {(32)}^{^{\frac{4}{5}}}} \times (0.15) \\ = \frac{0.15}{5 \times 2^{4}} = 0.00187 \\ Hence, the approximate value of {(32.15)}^{\frac{1}{4}} i s \\ 2 + 0.00187 = 2 .00187 \end{array}$

Q.66

\begin{array}{l} Find the approximate value of f (2 .01), where \\ f (x) {=4x}^{2} +5x+2 \end{array}

Ans.

\begin{array}{l} Let x = 2 and Δx = 0.01 . Then, we have : \\ f (2.01) = f (x + Δx) = 4 {(x + Δx)}^{2} + 5 (x + Δx) + 2 \\ Since, Δy = f (x + Δx) - f (x) . ​ Therefore, \\ f (x + Δx) = f (x) + Δy \\ \approx f (x) + f ‘ (x) Δx [∵ dx = Δx] \\ f (2 .01) \approx (4 x^{2} + 5 x + 2) + (8 x + 5) Δx \\ \approx {4 {(2)}^{2} + 5 (2) + 2} + (8 \times 2 + 5) (0 .01) \\ \approx 28 + 21 \times 0 .01 = 28 + 0 .21 \\ \approx 28 .21 \\ Hence, the approximate value of f (2 . 01) is 28.21 . \end{array}

Q.67 Find the approximate value of f (5.001), where f (x) = x³ – 7x²+ 15.

Ans.

\begin{array}{l} Let x = 5 and Δ x = 0 . 001 . Then, we have : \\ f (5 . 001) = f (x + Δ x) = {(x + Δ x)}^{3} - 7 {(x + Δ x)}^{2} + 15 \\ Since, Δy = f (x + Δx) - f (x) . Therefore, \\ f (x + Δx) = f (x) + Δy \\ \approx f (x) + f^{‘} (x) Δx [∵ dx = Δx] \\ f (5 .001) \approx (x^{3} - 7 x^{2} + 15) + (3 x^{2} - 14 x) Δx \\ \approx {{(5)}^{3} - 7 {(5)}^{2} + 15} + {3 {(5)}^{2} - 14 (5)} (0 .01) \\ \approx - 35 + 5 \times (0 .01) = - 35 + 0 .05 \\ \approx - 34 .995 \\ Hence, the approximate value of f (5 . 001) is - 34 .995. \end{array}

Q.68 Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%.

Ans.

\begin{array}{l} Side of cube is x metres . \\ Volume of a cube is given by \\ V = x^{3} \\ dV = (\frac{dV}{dx}) Δx \\ = (\frac{d}{dx} x^{3}) Δx \\ = (3 x^{2}) Δx \\ = (3 x^{2}) (0 .01 x) [1 % of x = 0 .01 x] \\ = 0 .03 x^{3} \\ Hence, the approximate change in the volume of the \\ cube is 0 . 03 x^{3} m^{3} . \end{array}

Q.69 Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.

Ans.

\begin{array}{l} Let side of cube be x metres . Then, \\ Surface area of cube = 6 x^{2} \Rightarrow S = 6 x^{2} \\ \frac{dS}{dx} = \frac{d}{dx} (6 x^{2}) \\ = 12 x \\ dS = (\frac{dS}{dx}) Δx \\ = (12 x) (0 .01 x) [∵ 1 % of x = 0 .01 x] \\ = 0 .12 x^{2} \\ Hence, the approximate change in the surface area of the cube \\ is 0 .12 x^{2} m^{2} . \end{array}

Q.70 If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume.

Ans.

$\begin{array}{l} Let r be the radius of the sphere and Δr be the error in \\ measuring the radius . Then, \\ r = 7 m and Δr = 0 . 02 m \\ Now, the volume V of the sphere is given by, \\ V = \frac{4}{3} {πr}^{3} \\ \frac{dV}{dr} = \frac{d}{dr} (\frac{4}{3} {πr}^{3}) \\ = 4 {πr}^{2} \\ ∴ dV = (\frac{dV}{dr}) Δr \\ = (4 {πr}^{2}) Δr \\ = 4 π {(7)}^{2} \times 0 .02 = 3 .92 π m^{3} \\ Hence, the approximate error in calculating the volume \\ is 3.92 π m^{3} . \end{array}$

Q.71 If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating ‘its surface area’

Ans.

\begin{array}{l} Let r be the radius of the sphere and Δr be the error in \\ measuring the radius . \\ Then, r = 9 m and Δr = 0 . 03 m \\ Now, the surface area of the sphere (S) is given by, \\ S = 4 {πr}^{2} \\ Differentiating ​ w . r . t . r, we get \\ \frac{dS}{dr} = 8 πr \\ ∴ dS = (\frac{dS}{dr}) Δr \\ = (8 πr) \times 0.03 m^{2} \\ = (8 π \times 9) \times 0.03 m^{2} \\ = 2 .16 π m^{2} \\ Hence, the approximate error in calculating the surface area \\ is 2.16 π m^{2} . \end{array}

Q.72 If f (x) = 3x²+ 15x + 5, then the approximate value of f (3.02) is
A. 47.66 B. 57.66 C. 67.66 D. 77.66

Ans.

$\begin{array}{l} Let x = 3 and Δx = 0.02 . Then, we have : \\ f (3.02) = f (x + Δx) = 3 {(x + Δx)}^{2} + 15 (x + Δx) + 5 \\ Since, Δy = f (x + Δx) - f (x) . Therefore, \\ f (x + Δx) = f (x) + Δy \\ \approx f (x) + f ‘ (x) Δx [∵ dx = Δx] \\ f (3 .02) \approx (3 x^{2} +15 x + 5) + (6 x + 15) Δx \\ \approx {3 {(3)}^{2} + 15 (3) + 5} + (6 \times 3 + 15) (0 .01) \\ \approx 77 + 33 \times 0 .02 = 77 + 0 .66 \\ \approx 77 .66 \\ Hence, the approximate value of f (3 . 02) is 77.66 . \\ The correct answer is D . \end{array}$

Q.73 The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is

A. 0.06 x³ m³ B. 0.6 x³ m³

C. 0.09 x³ m³ D. 0.9 x³ m³

Ans.

$\begin{array}{l} Let side of cube is x metres . So \\ The Volume of a cube is given by \\ V = x^{3} \\ dV = (\frac{dV}{dx}) Δx \\ = (\frac{d}{dx} x^{3}) Δx \\ = (3 x^{2}) Δx \\ = (3 x^{2}) (0 .03 x) [3 % of x = 0 .03 x] \\ = 0 .09 x^{3} \\ Hence, the approximate change in the Volume of the \\ cube is 0 .09 x^{3} m^{3} . \\ The correct option is C . \end{array}$

Q.74 Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = (2x − 1)² + 3 (ii) f(x)= 9x²+12x+2

(iii) f(x) = −(x − 1)² + 10 (iv) g(x)= x³ + 1

Ans.

$\begin{array}{l} (i) The given function is \\ f (x) = {(2 x - 1)}^{2} + 3 \\ It is clear that {(2 x - 1)}^{2} \geq 0 for \forall x \in R \\ Therefore, f (x) = {(2 x - 1)}^{2} + 3 \geq 3 for \forall x \in R \\ The minimum value of f is attained \\ when 2 x - 1 = 0 \Rightarrow x = \frac{1}{2} \\ ∴ Minimum value of f = f (\frac{1}{2}) \\ = {(2 \times \frac{1}{2} - 1)}^{2} + 3 \\ = 3 \\ Hence, function f does not have a maximum value . \\ (ii) The given function is \\ f (x) = 9 x^{2} + 12 x + 2 = {(3 x + 2)}^{2} - 2. \\ It is clear that {(3 x + 2)}^{2} \geq 0 for \forall x \in R \\ Therefore, f (x) = {(3 x + 2)}^{2} - 2 \geq - 2 for \forall x \in R \\ The minimum value of f is attained \\ when 3 x + 2 = 0 \Rightarrow x = - \frac{2}{3} \\ ∴ Minimum value of f = f (- \frac{2}{3}) \\ = {(2 \times - \frac{2}{3} + 2)}^{2} - 2 \\ = - 2 \\ Hence, function f does not have a maximum value . \\ (iii) The given function is f (x) = - {(x - 1)}^{2} + 10 \\ It is clear that (x - 1)^{2} \geq 0 for every x \in R \\ Therefore, f (x) = - {(x - 1)}^{2} + 10 \leq 10 for every x \in R \\ The maximum value of f is attained when \\ (x - 1) = 0 \Rightarrow x = 1 \\ ∴ Maximum value of f = f (1) \\ = - {(1 - 1)}^{2} + 10 \\ = 10 \\ Hence, function f does not have a minimum value . \\ (iv) The given function is g (x) = x^{3} + 1 \\ Here, function g has neither a maximum value nor a minimum value . \end{array}$

Q.75

\begin{array}{l} Find the maximum and minimum values, if any, of the \\ following functions given by : \\ (i) |x + 2| - 1 (ii) g (x) = - |x + 1| + 3 \\ (iii) h (x) = \sin 2 x + 5 (iv) f (x) = |\sin 4 x + 3| \\ (v) h (x) = x + 1, x \in (- 1, 1) \end{array}

Ans.

$\begin{array}{l} (i) f (x) = | x + 2 | - 1 \\ We know that | x + 2 | \geq 0 \forall x \in R \\ Therefore, f (x) = | x + 2 | - 1 \geq - 1 \forall x \in R \\ The minimum value of f is attained when \\ | x + 2 | = 0 \Rightarrow x = - 2 \\ ∴ Minimum value of f = f (- 2) \\ = | - 2 + 2 | - 1 \\ = - 1 \\ Thus, function f does not have a maximum value . \\ (ii) g (x) = - | x + 1 | + 3 \\ Since, - | x + 1 | \leq 0 \forall x \in R \\ Therefore, g (x) = - | x + 1 | + 3 \leq 3 \forall x \in R \\ The maximum value of g is attained when \\ - | x + 1 | = 0 \Rightarrow x = - 1 \\ Maximum value of g = g (- 1) \\ = - | - 1 + 1 | + 3 \\ = 3 \\ Hence, function g does not have a minimum value . \\ (iii) f (x) = | \sin 4 x + 3 | \\ Since, - 1 \leq \sin 4 x \leq 1 \\ \Rightarrow 2 \leq \sin 4 x + 3 \leq 4 \\ Thus, the maximum and minimum values of f are 4 \\ and 2 respectively . \\ (v) h (x) = x + 1, x \in (- 1, 1) \\ Here, if a point x_{0} is closest to - 1, then we find \\ \frac{x_{0} + 1}{2} < x_{0} + 1 for all x_{0} \in (- 1, 1) \\ Also, if x_{1} is closest to 1, then x_{1} + 1 < \frac{x_{1} + 1}{2} + 1 \forall x \in (- 1, 1) \\ Hence, function h (x) has neither maximum nor \\ minimum value in (- 1, 1) . \end{array}$

Q.76 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

$\begin{array}{l} (i) f (x) = x^{2} (i i) g (x) = x^{3} - 3 x \\ (i i i) h (x) = s i n x + c o s x, 0 < x < \frac{π}{2} \\ (i v) f (x) = s i n x - c o s x, 0 < x < \frac{π}{2} \\ (v) f (x) = x^{3} - 6 x^{2} + 9 x + 15 \\ (v i) g (x) = \frac{x}{2} + \frac{2}{x}, x > 0 \\ (v i i) g (x) = \frac{1}{x^{2} + 2} (v i i i) f (x) = x \sqrt{1 - x}, x > 0 \end{array}$

Ans.

$\begin{array}{l} (i) f (x) = x^{2} . .. (i) \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 2 x . .. (ii) \\ For maxima or minima, \\ f ‘ (x) = 0 \\ \Rightarrow 2 x = 0 \Rightarrow x = 0 \\ Thus, x = 0 is the only critical point which could possibly \\ be the point of local maxima or local minima of f . \\ differentiating equation (ii) w . r . t . x, we get \\ f ” (x) = 2, which is positive . \\ Therefore, by second derivative test, x = 0 is a point of local \\ minima and local minimum value of f at x = 0 is f (0) = 0. \\ (ii) g (x) = x^{3} - 3 x . .. (i) \\ Differentiating w . r . t . x, we get \\ g ‘ (x) = 3 x^{2} - 3 . .. (ii) \\ For maxima or minima, we have \\ g ‘ (x) = 0 \Rightarrow 3 x^{2} - 3 = 0 \\ \Rightarrow x = \pm 1 \\ Differentiating equation (ii) w . r . t . x, we get \\ g ‘ ‘ (x) = 6 x \\ Putting x = 1, we get \\ g ‘ ‘ (1) = 6 (1) = 6 > 0 \end{array}$

$\begin{array}{l} S o, g (x) is minimum at x = 1 and minimum value is \\ g (1) = {(1)}^{3} - 3 (1) = - 2 \\ P u t t i n g x = - 1, we get \\ g ‘ ‘ (- 1) = 6 (- 1) = - 6 < 0 \\ S o, g (x) is maximum at x = - 1 and maximum value is \\ g (- 1) = {(- 1)}^{3} - 3 (- 1) = 2. \\ (i i i) h (x) = sinx + cosx, 0 < x < \frac{π}{2} … (i) \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ h’ (x) = cosx - sinx … (i i) \\ F o r maxima or minima, we have \\ h’ (x) = 0 \Rightarrow cosx - sinx = 0 \\ \Rightarrow \cos x = \sin x \\ \Rightarrow \frac{\cos x}{\sin x} = 1 \\ \Rightarrow \cot x = \cot 45 ° \Rightarrow x = 45 ° \\ D i f f e r e n t i a t i n g equation (i i) w . r . t . x, we get \\ h” (x) = - sinx - cosx \\ putting x = 45 °, w e get \\ h” (45 °) = - sin 45 ° - cos 45 ° \\ = - (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) < 0 \\ T h u s, h (x) is maximum at x = 45°, then maximum value is \\ h (45 °) = sin 45 ° + cos 45 ° \\ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \\ = \frac{2}{\sqrt{2}} = \sqrt{2} . \\ (i v) f (x) = \sin x - \cos x, 0 < x < 2 π \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ f’ (x) = c o x + \sin x \\ F o r maxima or minima, \\ f’ (x) = 0 \Rightarrow c o x + \sin x = 0 \\ \Rightarrow \sin x = - \cos x \\ \Rightarrow \tan x = - 1 \Rightarrow x = \frac{3 π}{4}, \frac{7 π}{4} \in (0, 2 π) \\ a n d f ‘ ‘ (x) = - \sin x + \cos x \\ \Rightarrow f ‘ ‘ (\frac{3 π}{4}) = - \sin \frac{3 π}{4} + \cos \frac{3 π}{4} \\ = - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \\ = - \sqrt{2} < 0 \\ Therefore, by second derivative test,x = \frac{3 π}{4} is a point of local \\ maxima and the local maximum value of f at x = \frac{3 π}{4} i s \\ f (\frac{3 π}{4}) = \sin \frac{3 π}{4} - \cos \frac{3 π}{4} \end{array}$

$\begin{array}{l} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \\ a n d \\ f ‘ ‘ (\frac{7 π}{4}) = - \sin \frac{7 π}{4} + \cos \frac{7 π}{4} \\ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \\ = \sqrt{2} > 0 \\ Therefore, by second derivative test,x = \frac{7 π}{4} is a point of local \\ minima and the local minimum value of f at x = \frac{7 π}{4} i s \\ f (\frac{7 π}{4}) = \sin \frac{7 π}{4} - \cos \frac{7 π}{4} \\ = - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = - \sqrt{2} \\ (v) f (x) = x^{3} - {6x}^{2} + 9x + 15 \\ Differentiating w .r .t . x, we get \\ f’ (x) = 3 x^{2} - 12x + 9 \\ For maxima or minima, we have \\ f’ (x) = 0 \Rightarrow 3 (x^{2} - 4x + 3) = 0 \\ \Rightarrow (x - 3) (x - 1) = 0 \\ \Rightarrow x = 3, 1 \\ N o w, \\ f ‘ ‘ (x) = 6 x - 12 \\ Putting x=3 in f” (x), we get \\ f ‘ ‘ (3) = 6 (3) - 12 = 6>0 \\ So, f (x) i s l o c a l minimum at x = 3, then l o c a l \\ minimum value = f (3) \\ = {(3)}^{3} - 6 {(3)}^{2} + 9 (3) +15 \\ = 15 \\ Putting x = 1 in f” (x), we get \\ f ‘ ‘ (1) = 6 (1) - 12 = - 6<0 \\ So, f (x) i s l o c a l maximum at x = 1, then l o c a l \\ maximum value = f (1) \\ = {(1)}^{3} - 6 {(1)}^{2} + 9 (1) +15 \\ = 19 \\ (v i) g (x) = \frac{x}{2} + \frac{2}{x} \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ g’ (x) = \frac{1}{2} - \frac{2}{x^{2}} a n d g ‘ ‘ (x) = \frac{4}{x^{3}} \\ F o r \max i m a or minima, we have \\ g’ (x) = 0 \Rightarrow \frac{1}{2} - \frac{2}{x^{2}} = 0 \\ \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2 \\ S i n c e, x>0 so, x = 2 \\ then, g” (2) = \frac{4}{2^{3}} = \frac{1}{2} > 0 \\ Therefore, by second derivative test, x = 2 is a point of local \\ minima and the local minimum value of g at x = 2 is \end{array}$

$\begin{array}{l} g(2) = \frac{2}{2} + \frac{2}{2} = 2 \\ (v i i) g (x) = \frac{1}{x^{2} + 2} \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ g’ (x) = - \frac{1}{{(x^{2} + 2)}^{2}} \frac{d}{d x} (x^{2} + 2) \\ = - \frac{2 x}{{(x^{2} + 2)}^{2}} \\ F o r maxima or minima, we have \\ g’ (x) = 0 \Rightarrow - \frac{2 x}{{(x^{2} + 2)}^{2}} = 0 \\ \Rightarrow x = 0 \\ Now, for values close to x = 0 and to the left of 0, \\ g ‘ (x) > 0. Also, for values close to x = 0 and to the \\ right of 0, g ‘ (x) < 0. \\ Therefore, by first derivative test, x = 0 is a point of local \\ maxima and the local maximum value of g (0) = \frac{1}{0 + 2} = \frac{1}{2} \\ (v i i i) f (x) = x \sqrt{1 - x}, x > 0 \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ f ‘ (x) = \frac{d}{d x} (x \sqrt{1 - x}) \\ = x \frac{d}{d x} \sqrt{1 - x} + \sqrt{1 - x} \frac{d}{d x} x \\ = x (- \frac{1}{2 \sqrt{1 - x}}) + \sqrt{1 - x} \\ = \frac{- x + 2 - 2 x}{2 \sqrt{1 - x}} \\ = \frac{2 - 3 x}{2 \sqrt{1 - x}} \\ F o r maximum or minimum, we have \\ f ‘ (x) = 0 \Rightarrow \frac{2 - 3 x}{2 \sqrt{1 - x}} = 0 \\ \Rightarrow 2 - 3 x = 0 \Rightarrow x = \frac{2}{3} \\ f ‘ ‘ (x) = \frac{d}{d x} (\frac{2 - 3 x}{2 \sqrt{1 - x}}) \\ = \frac{2 \sqrt{1 - x} \frac{d}{d x} (2 - 3 x) - (2 - 3 x) \frac{d}{d x} 2 \sqrt{1 - x}}{{(2 \sqrt{1 - x})}^{2}} \\ = \frac{2 \sqrt{1 - x} (0 - 3) - (2 - 3 x) (- \frac{2}{2 \sqrt{1 - x}})}{{(2 \sqrt{1 - x})}^{2}} \\ = \frac{- 6 \sqrt{1 - x} + \frac{(2 - 3 x)}{\sqrt{1 - x}}}{4 (1 - x)} \\ = \frac{\frac{- 6 (1 - x) + 2 - 3 x}{\sqrt{1 - x}}}{4 (1 - x)} \\ = \frac{3 x - 4}{4 {(1 - x)}^{\frac{3}{2}}} \\ P u t t i n g x = \frac{2}{3} in f” (x), we get \\ f” (\frac{2}{3}) = \frac{3 (\frac{2}{3}) - 4}{4 {(1 - \frac{2}{3})}^{\frac{3}{2}}} \\ = \frac{2 - 4}{4 {(\frac{1}{3})}^{\frac{3}{2}}} \\ = \frac{- 2}{(\frac{4}{3 \sqrt{3}})} \\ = \frac{- 3 \sqrt{3}}{2} < 0 \\ S o, f (x) is maximum at x = \frac{2}{3} and maximum value is \\ f (\frac{2}{3}) = \frac{2}{3} \sqrt{1 - \frac{2}{3}} \\ = \frac{2}{3} \times \frac{1}{\sqrt{3}} = \frac{2 \sqrt{3}}{9} \end{array}$

Q.77 Prove that the following functions do not have maxima or minima:
(i) f(x) = e^x (ii) g(x) = logx

(iii) h(x) = x³ + x² + x + 1

Ans.

$\begin{array}{l} (i) We have, \\ f (x) = e^{x} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = e^{x} \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow e^{x} = 0 \\ Which is impossible, because the exponential function can never \\ assume 0 for any value of x . \\ Therefore, there is no number c for that f ‘ (c) = 0 . \\ Thus, function f is neither maxima nor minima . \\ (ii) We have, \\ g (x) = \log x \\ Differentiating w . r . t . x, we get \\ g ‘ (x) = \frac{1}{x} \\ For maxima or minima, \\ g ‘ (x) = 0 \Rightarrow \frac{1}{x} = 0 \\ \Rightarrow x = \frac{1}{0} \\ Thus, there is no point for maxima or minima . \\ Hence, function g (x) does not have maxima or minima . \\ (iii) We have, \\ h (x) = x^{3} + x^{2} + x + 1 \\ Differentiating w . r . t . x, we get \\ h ‘ (x) = 3 x^{2} + 2 x + 1 \\ For maxima or minima, we have \\ h ‘ (x) = 0 \Rightarrow 3 x^{2} + 2 x + 1 = 0 \\ \Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ = \frac{- 2 \pm \sqrt{2^{2} - 4 (3) (1)}}{2 (3)} \\ = \frac{- 2 \pm \sqrt{- 8}}{6} \\ = \frac{- 1 \pm i \sqrt{2}}{3} = \notin R \\ Hence, function h does not have maxima or minima . \end{array}$

Q.78

\begin{array}{l} Find the absolute maximum value and the absolute minimum \\ value of the following functions in the given intervals : \\ (i) f (x) = x^{3}, x \in [- 2, 2] (ii) f (x) = sinx + cosx, x \in [, π] \\ (iii) f (x) = 4 x - \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}] (iv) f (x) = {(x - 1)}^{2} + 3, x \in [- 3, 1] \end{array}

Ans.

$\begin{array}{l} (i) The given function is f (x) = x^{3} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 3 x^{2} \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow 3 x^{2} = 0 \Rightarrow x = 0 \\ Then, we evaluate the value of f at critical point x = 0 and at \\ end points of the interval [- 2, 2] . \\ f (0) = 0 \\ f (- 2) = {(- 2)}^{3} = - 8 \\ f (2) = {(2)}^{3} = 8 \\ Hence, we can conclude that the absolute maximum value of f \\ on [- 2, 2] is 8 occurring at x = 2 . Also, the absolute minimum \\ value of f on [- 2, 2] is - 8 occurring at x = - 2. \\ (ii) The given function is f (x) = \sin x + \cos x . \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = \cos x - sinx \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow \cos x - sinx = 0 \\ \Rightarrow tanx = 1 \Rightarrow x = \frac{π}{4} \\ Then, we evaluate the value of f at critical point x = \frac{π}{4} \\ and at the end points of the interval [0, π] . \\ f (\frac{π}{4}) = \sin \frac{π}{4} + \cos \frac{π}{4} \\ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \\ = \frac{2}{\sqrt{2}} \\ = \sqrt{2} \\ f (0) = \sin 0 + \cos 0 \\ = 0 + 1 \\ = 1 \\ f (π) = sinπ + cosπ \\ = 0 - 1 \\ = - 1 \\ Hence, we can conclude that the absolute maximum value of \\ f on [0, π] is \sqrt{2} occurring at x = \frac{π}{4} and the absolute minimum \\ value of f on [0, π] is - 1 occurring at x = π . \\ (iii) The given function is f (x) = 4 x - \frac{1}{2} x^{2} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 4 - x \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow 4 - x = 0 \\ \Rightarrow x = 4 \\ Then, we evaluate the value of f at critical point x = 4 and at \\ the end points of the interval [- 2, \frac{9}{2}] . \\ f (4) = 4 (4) - \frac{1}{2} {(4)}^{2} = 16 - 8 = 8 \\ f (- 2) = 4 (- 2) - \frac{1}{2} {(- 2)}^{2} = - 8 - 2 = - 8 \\ f (\frac{9}{2}) = 4 (\frac{9}{2}) - \frac{1}{2} {(\frac{9}{2})}^{2} = 18 - \frac{81}{8} = \frac{63}{4} \\ Hence, we can conclude that the absolute maximum value \\ of f on [- 2, \frac{9}{2}] is 8 occurring at x = 4 and the absolute minimum \\ value of f on [- 2, \frac{9}{2}] is - 10 occurring at x = - 2. \\ (iv) The given function is f (x) = {(x - 1)}^{2} + 3 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 2 (x - 1) \\ For maximum and minimum, we have \\ f ‘ (x) = 0 \Rightarrow 2 (x - 1) = 0 \\ \Rightarrow x = 1 \\ Then, we evaluate the value of f at critical point x = 1 and at \\ the end points of the interval [- 3, 1] . \\ f (1) = {(1 - 1)}^{2} + 3 = 3 \\ f (- 3) = {(- 3 - 1)}^{2} + 3 = 19 \\ Hence, we can conclude that the absolute maximum value of f \\ on [- 3, 1] is 19 occurring at x = - 3 and the minimum value of \\ f on [- 3, 1] is 3 occurring at x = 1. \end{array}$

Q.79 Find the maximum profit that a company can make, if the profit function is given by

p(x) = 41 − 24x − 18x²

Ans.

\begin{array}{l} We have, \\ p (x) = 41 - 24 x - 18 x^{2} \\ Differentiating w . r . t . x, we get \\ p ‘ (x) = - 24 - 36 x \\ and \\ p ” (x) = - 36 \\ For ​​ maxima or minima, we have \\ p ‘ (x) = 0 \Rightarrow - 24 - 36 x = 0 \\ \Rightarrow x = \frac{24}{- 36} = - \frac{2}{3} \\ So, p ” (- \frac{2}{3}) = - 36 < 0 \\ By second derivative test, x = - \frac{2}{3}, is the point of local \\ maxima of p . \\ ∴ maximum profit = 41 - 24 (- \frac{2}{3}) - 18 {(- \frac{2}{3})}^{2} \\ = 41 + 16 - 8 \\ = 49 \\ Hence, the maximum profit that the company can make is \\ 49 units . \end{array}

Q.80

\begin{array}{l} Find both the maximum value and the minimum value of \\ {3x}^{4} - {8x}^{3} {+12x}^{2} - 48x+25 on the interval [0,3] . \end{array}

Ans.

$\begin{array}{l} Let f (x) = 3 x^{4} - 8 x^{3} + 12 x^{2} - 48 x + 25 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 12 x^{3} - 24 x^{2} + 24 x - 48 \\ f ‘ ‘ (x) = 36 x^{2} - 48 x + 24 \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow 12 x^{3} - 24 x^{2} + 24 x - 48 = 0 \\ \Rightarrow (x - 2) (x^{2} + 2) = 0 \\ \Rightarrow x = 2, \sqrt{- 2} \\ \sqrt{- 2} is not real, so x = 2. \\ Then, we evaluate the value of f at critical point x = 2 and at \\ the end points of the interval [0, 3] . \\ f (0) = 3 {(0)}^{4} - 8 {(0)}^{3} + 12 {(0)}^{2} - 48 (0) + 25 = 25 \\ f (2) = 3 {(2)}^{4} - 8 {(2)}^{3} + 12 {(2)}^{2} - 48 (2) + 25 = - 39 \\ f (3) = 3 {(3)}^{4} - 8 {(3)}^{3} + 12 {(3)}^{2} - 48 (3) + 25 = 16 \\ So, the functin f (x) is minimum at x =2 and minimum value \\ is - 39 . The function is maximum at x =0 and maximum value \\ is 25 . \end{array}$

Q.81 At what point in the interval [0, 2π], does the function sin 2x attain its maximum value?

Ans.

$\begin{array}{l} Let f (x) = \sin 2 x \\ Differentiating w . r . t . x, we get \\ f^{‘} (x) = 2 \cos 2 x \\ For maxima or minima, we have \\ f^{‘} (x) = 0 \Rightarrow 2 \cos 2 x = 0 \\ \Rightarrow 2 x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2} \\ \Rightarrow x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4} \\ Then, we evaluate the values of f at critical points \\ x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4} and at the end point s o f the interval [0, 2 π] . \\ f (\frac{π}{4}) = \sin 2 (\frac{π}{4}) = \sin (\frac{π}{2}) = 1, \\ f (\frac{3 π}{4}) = \sin 2 (\frac{3 π}{4}) = \sin (\frac{3 π}{2}) = - 1, \\ f (\frac{5 π}{4}) = \sin 2 (\frac{5 π}{4}) = \sin (\frac{5 π}{2}) = 1, \\ f (\frac{7 π}{4}) = \sin 2 (\frac{7 π}{4}) = \sin (\frac{7 π}{2}) = - 1, \\ f (0) = \sin 2 (0) = \sin (0) = 0, \\ f (2 π) = \sin 2 (2 π) = \sin 4 π = 0, \\ Hence, we can conclude that the ab solute maximum value of f \\ on [0, 2 π] is o ccurring at x = \frac{π}{4} and at x = \frac{5 π}{4} . \end{array}$

Q.82 What is the maximum value of the function sin x + cos x?

Ans.

\begin{array}{l} Let f (x) = sinx + cosx \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = cosx - sinx \\ f ‘ ‘ (x) = - sinx - cosx \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow cosx - sinx = 0 \\ \Rightarrow tanx = 1 \\ \Rightarrow x = \frac{π}{4}, \frac{5 π}{4}, . .. \\ So, \\ f ‘ ‘ (\frac{π}{4}) = - \sin \frac{π}{4} - \cos \frac{π}{4} \\ = - (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) \\ = - \sqrt{2} < 0 \\ By second derivative test, f will be the maximum at x = \frac{π}{4}, \\ So maximum value = f (\frac{π}{4}) \\ = \sin \frac{π}{4} + \cos \frac{π}{4} \\ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \end{array}

Q.83 Find the maximum value of 2x³ – 24x + 107 in the interval [1, 3]. Find the maximum
value of the same function in [- 3, -1].

Ans.

$\begin{array}{l} Let f (x) = 2 x^{3} - 24 x + 107 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 6 x^{2} - 24 \\ f ‘ ‘ (x) = 12 x \\ For maxima or minima, we get \\ f ‘ (x) = 0 \Rightarrow 6 x^{2} - 24 = 0 \\ \Rightarrow 6 (x^{2} - 4) = 0 \\ \Rightarrow x = \pm 2 \\ We first consider the interval [1, 3] . \\ Then, we evaluate the value of f at the critical point \\ x = 2 \in [1, 3] and at the end points of the interval [1, 3] . \\ f (2) = 2 (8) - 24 (2) + 107 = 16 - 48 + 107 = 75 \\ f (1) = 2 (1) - 24 (1) + 107 = 2 - 24 + 107 = 85 \\ f (3) = 2 (27) - 24 (3) + 107 = 54 - 72 + 107 = 89 \\ Hence, the absolute maximum value of f (x) in the interval \\ [1, 3] is 89 occurring at x = 3. \\ Next, we consider the interval [- 3, - 1] . \\ Now, we find the value of f (x) at the critical point \\ x = - 2 \in [- 3, - 1] and at the end points of \\ the interval [- 3, - 1] . \\ f (- 3) = 2 (- 27) - 24 (- 3) + 107 \\ = - 54 + 72 + 107 \\ = 125 \\ f (- 1) = 2 (- 1) - 24 (- 1) + 107 \\ = - 2 + 24 + 107 \\ = 129 \\ f (- 2) = 2 (- 8) - 24 (- 2) + 107 \\ = - 16 + 48 + 107 \\ = 139 \\ Hence, the absolute maximum value of f (x) in the interval \\ [- 3, - 1] is 139 occurring at x = - 2. \end{array}$

Q.84 It is given that x = 1, the function x⁴ – 62 x² + ax + 9 attains its maximum value, on the interval [0, 2].

Find the value of a.

Ans.

$\begin{array}{l} Let f (x) = x^{4} - 62 x^{2} + ax + 9 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 4 x^{3} - 124 x + a \\ Since, the given that function f attains its maximum value on the \\ interval [0, 2] at x = 1. \\ ∴ f ‘ (1) = 0 \\ \Rightarrow 4 {(1)}^{3} - 124 (1) + a = 0 \\ 4 - 124 + a = 0 \\ - 120 + a = 0 \Rightarrow a = 120 \\ Hence, the value of a is 120 . \end{array}$

Q.85 Find the maximum and minimum values of x + sin2x on [0, 2π].

Ans.

$\begin{array}{l} Let f (x) = x + \sin 2 x . \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 1 + 2 \cos 2 x \\ f ‘ ‘ (x) = - 4 \sin 2 x \\ Now, for maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow 1 + 2 \cos 2 x = 0 \\ \Rightarrow \cos 2 x = - \frac{1}{2} = - \cos \frac{π}{3} \\ = \cos (π - \frac{π}{3}) \\ \Rightarrow \cos 2 x = \cos \frac{2 π}{3} \\ \Rightarrow 2 x = 2 nπ \pm \frac{2 π}{3}, n \in Z \\ \Rightarrow x = nπ \pm \frac{π}{3}, n \in Z \\ \Rightarrow x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3} \in (0, 2 π) \\ Then, we evaluate the value of f at critical points \\ x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3} and at the end points of the interval [0, 2 π]. \\ f (\frac{π}{3}) = \frac{π}{3} + \sin (\frac{2 π}{3}) \\ = \frac{π}{3} + \sin (π - \frac{π}{3}) \\ = \frac{π}{3} + \sin (\frac{π}{3}) \\ = \frac{π}{3} + \frac{\sqrt{3}}{2} \\ f (\frac{π}{3}) = \frac{π}{3} + \sin (\frac{2 π}{3}) = \frac{2 π}{3} + \frac{\sqrt{3}}{2} \\ f (\frac{2 π}{3}) = \frac{2 π}{3} + \sin (\frac{4 π}{3}) = \frac{2 π}{3} - \frac{\sqrt{3}}{2} \\ f (\frac{4 π}{3}) = \frac{4 π}{3} + \sin (\frac{8 π}{3}) = \frac{4 π}{3} + \frac{\sqrt{3}}{2} \\ f (\frac{5 π}{3}) = \frac{5 π}{3} + \sin (\frac{10 π}{3}) = \frac{5 π}{3} - \frac{\sqrt{3}}{2} \\ f (0) = 0 + \sin 0 = 0 \\ f (2 π) = 2 π + \sin 2 π = 0 \\ Hence, we can conclude that the absolute maximum value of \\ f (x) in the interval [0, 2 π] is 2 π occurring at x = 2 π and the \\ absolute minimum value of f (x) in the interval [0, 2 π] \\ is 0 occurring at x = 0 . \end{array}$

Q.86 Find two numbers whose sum is 24 and whose product is as large as possible.

Ans.

$\begin{array}{l} Let two numbers be x and (24 - x) andP (x) denotes the product \\ of two numbers, then \\ P (x) = x (24 - x) = 24 x - x^{2} \\ Differentiating w . r . t . x, we get \\ \frac{dP (x)}{dx} = 24 - 2 x \\ P ‘ ‘ (x) = - 2 \\ For maxima or minima, we have \\ P ‘ (x) = 0 \Rightarrow 24 - 2 x = 0 \\ \Rightarrow x = 12 \\ So, P ‘ ‘ (12) = - 2 < 0 \\ ∴ By second derivative test, x = 12 is the point of local maxima \\ of P . Hence, the product of the numbers is the maximum when \\ the numbers are 12 and 24 - 12 = 12. \end{array}$

Q.87 Find two positive numbers x and y such that x + 6 = 60 and xy³ is maximum.

Ans.

$\begin{array}{l} The two numbers are x and y such that x + y = 60 . \\ ∴ y = 60 - x \\ L e t f (x) = x y^{3} \Rightarrow f (x) = x {(60 - x)}^{3} \\ D i f f e r e n t i a t i n g f (x) w .r .t . x, we get \\ f’ (x) = \frac{d}{d x} {x {(60 - x)}^{3}} \\ = x \frac{d}{d x} {(60 - x)}^{3} + {(60 - x)}^{3} \frac{d}{d x} x \\ = x .3 {(60 - x)}^{2} (0 - 1) + {(60 - x)}^{3} \times 1 \\ = {(60 - x)}^{3} - 3 x {(60 - x)}^{2} \\ = {(60 - x)}^{2} (60 - x - 3 x) \\ = {(60 - x)}^{2} (60 - 4 x) \\ f ‘ ‘ (x) = {(60 - x)}^{2} \frac{d}{d x} (60 - 4 x) + (60 - 4 x) \frac{d}{d x} {(60 - x)}^{2} \\ = {(60 - x)}^{2} (0 - 4) + (60 - 4 x) .2 (60 - x) (0 - 1) \\ = - 4 {(60 - x)}^{2} - 2 (60 - 4 x) (60 - x) \\ = - 2 (60 - x) (120 - 2 x + 60 - 4 x) \\ = - 2 (60 - x) (180 - 6 x) \\ = - 12 (60 - x) (30 - x) \\ F o r maxima or minima, we have \\ f’ (x) = 0 \Rightarrow {(60 - x)}^{2} (60 - 4 x) = 0 \\ \Rightarrow x = 60, \frac{60}{4} = 60, 16 \\ W h e n, x = 60 \\ f” (60) = - 12 (60 - 60) (30 - 60) = 0 \\ T h u s, at x = 60, f (x) is neither maximum nor minimum . \\ W h e n, x = 15 \\ f” (60) = - 12 (60 - 15) (30 - 15) < 0 \\ S o, f (x) is maximum when x = 15 and y = 60 - 15 = 45 \\ Hence, the required numbers are 15 and 45 . \end{array}$

Q.88 Find two positive numbers x and y such that their sum is 35 and the product x²y⁵ is a maximum.

Ans.

$\begin{array}{l} Let one number be x . Then, the other number is y = (35 - x) . \\ Let P(x) = x^{2} y^{5} . Then, we have: \\ P (x) = x^{2} {(35 - x)}^{5} \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ P’ (x) = x^{2} \frac{d}{d x} {(35 - x)}^{5} + {(35 - x)}^{5} \frac{d}{d x} x^{2} \\ = x^{2} \times 5 {(35 - x)}^{4} \frac{d}{d x} (35 - x) + {(35 - x)}^{5} \times 2 x \\ = 5 x^{2} {(35 - x)}^{4} (0 - 1) + 2 x {(35 - x)}^{5} \\ = x {(35 - x)}^{4} (- 5 x + 70 - 2 x) \\ = {(35 - x)}^{4} (70 x - 7 x^{2}) \\ P ‘ ‘ (x) = {(35 - x)}^{4} \frac{d}{d x} (70 x - 7 x^{2}) + \{\frac{d}{d x} {(35 - x)}^{4}\} (70 x - 7 x^{2}) \\ P ‘ ‘ (x) = {(35 - x)}^{4} (70 - 14 x) + \{4 {(35 - x)}^{3} \frac{d}{d x} (35 - x)\} (70 x - 7 x^{2}) \\ P ‘ ‘ (x) = {(35 - x)}^{4} (70 - 14 x) + 4 {(35 - x)}^{3} (0 - 1) (70 x - 7 x^{2}) \\ = 7 {(35 - x)}^{3} \{(35 - x) (10 - 2 x) - 4 (10 x - x^{2})\} \\ = 7 {(35 - x)}^{3} (6 x^{2} - 120 x + 350) \\ F o r maximum or minimum, we have \end{array}$

$\begin{array}{l} P’ (x) = 0 \Rightarrow {(35 - x)}^{4} (70 x - 7 x^{2}) = 0 \\ \Rightarrow 7 x {(35 - x)}^{4} (10 - x) = 0 \Rightarrow x = 0, 10, 35 \\ A t x = 0, \\ P ‘ ‘ (0) = 7 {(35 - 0)}^{3} (6 \times 0^{2} - 120 \times 0 + 350) \\ = 7 {(35)}^{3} (350) > 0 \\ S o, function is minimum at x = 0 . \\ A t x = 35, \\ y = 35 - 35 = 0 \\ S o, P (x) = 35 \times 0 = 0 \\ T h u s, x=35 will not be possible value of x . \\ A t x = 10, \\ P ‘ ‘ (10) = 7 {(35 - 10)}^{3} (6 \times 10^{2} - 120 \times 10 + 350) \\ = 7 {(25)}^{3} (600 - 1200 + 350) < 0 \\ S o, function is maximum at x = 10 . \\ Therefore, the two positive numbers are \\ x = 10 a n d y = 35 - 10 = 25 \\ H e n c e, the required numbers are 10 and 25 . \end{array}$

Q.89 Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Ans.

$\begin{array}{l} Let one number be x . Then, the other number is y = (16 - x) . \\ Let S (x) = x^{3} + y^{3} . Then, we have : \\ S (x) = x^{3} + {(16 - x)}^{3} \\ Differentiating w . r . t . x, we get \\ S ‘ (x) = \frac{d}{dx} x^{3} + \frac{d}{dx} {(16 - x)}^{3} \\ = 3 x^{2} + 3 {(16 - x)}^{2} \frac{d}{dx} (16 - x) \\ = 3 x^{2} + 3 {(16 - x)}^{2} (0 - 1) \\ = 3 x^{2} - 3 {(16 - x)}^{2} \\ S ” (x) = \frac{d}{dx} 3 x^{2} - \frac{d}{dx} 3 {(16 - x)}^{2} \\ = 6 x - 6 (16 - x) \frac{d}{dx} (16 - x) \\ = 6 x - 6 (16 - x) (0 - 1) \\ = 6 x + 6 (16 - x) = 96 \\ For maxima or minima, \\ f ‘ (x) = 0 \Rightarrow 3 x^{2} - 3 {(16 - x)}^{2} = 0 \\ \Rightarrow 3 x^{2} - 3 (256 - 32 x + x^{2}) = 0 \\ \Rightarrow 3 (x^{2} - 256 + 32 x - x^{2}) = 0 \\ \Rightarrow 3 (32 x - 256) = 0 \Rightarrow x = \frac{256}{32} = 8 \\ Then, \\ S ‘ ‘ (8) = 96 > 0 \\ By second derivative test, x = 8 is the point of local minima of S . \\ Hence, the sum of the cubes of the numbers is the minimum when \\ the numbers are 8 and 16 - 8 = 8. \end{array}$

Q.90 A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Ans.

$\begin{array}{l} Let the side of the square to be cut off be x cm . Then, the \\ length and the breadth of the box will be (18 - 2x) cm each \\ and the height of the box is x cm . \end{array}$

$\begin{array}{l} \end{array}$

$\begin{array}{l} Therefore, the volume V(x) of the box is given by, \\ V (x) = x(18 - {2x)}^{2} \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ V’ (x) = x \frac{d}{d x} {(18 - 2x)}^{2} + {(18 - 2x)}^{2} \frac{d}{d x} x \end{array}$

$\begin{array}{l} = x . 2 (18 - 2 x) \frac{d}{dx} (18 - 2 x) + {(18 - 2 x)}^{2} \times 1 \\ = 2 x (18 - 2 x) (0 - 2) + {(18 - 2 x)}^{2} \\ = (18 - 2 x) (- 4 x + 18 - 2 x) \\ = 2 (9 - x) \times 6 (3 - x) \\ = 12 (9 - x) (3 - x) \\ V ” (x) = 12 {(9 - x) \frac{d}{dx} (3 - x) + (3 - x) \frac{d}{dx} (9 - x)} \\ = 12 {(9 - x) (0 - 1) + (3 - x) (0 - 1)} \\ = 12 (- 9 + x - 3 + x) \\ = 24 (x - 6) \\ For maxima or minima, we have \\ v ‘ (x) = 0 \Rightarrow 12 (9 - x) (3 - x) = 0 \\ \Rightarrow x = 3, 9 \\ If x = 9, then the length and the breadth will become 0 . \\ \Rightarrow x \neq 9 . \\ When x = 3, \\ V ” (3) = 24 (3 - 6) = - 72 < 0 \\ ∴ By second derivative test, x = 3 is the point of maxima of V . \\ Hence, if we remove a square of side 3 cm from each corner \\ of the square tin and make a box from the remaining sheet, \\ then the volume of the box obtained is the largest possible . \end{array}$

Q.91 A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the ‘volume of the box is maximum’ ?

Ans.

$\begin{array}{l} Let the side of the square to be cut off be x cm . Then, the \\ height of the box is x, the length is (45 - 2x), and the breadth \\ is (24 - 2x) . \end{array}$

$\begin{array}{l} Therefore, the volume V (x) of the box is given by, \\ V (x) = x (45 - 2 x) (24 - 2 x) \\ = x (1080 - 90 x - 48 x + 4 x^{2}) \\ = 1080 x - 138 x^{2} + 4 x^{3} \\ Differentiating w . r . t . x, we get \\ V ‘ (x) = \frac{d}{dx} (1080 x - 138 x^{2} + 4 x^{3}) \end{array}$

$\begin{array}{l} = 1080 - 276 x + 12 x^{2} \\ = 12 (x^{2} - 23 x + 90) \\ V ‘ ‘ (x) = 12 (2 x - 23) \\ F o r maxima or minima, we have \\ V’ (x) = 0 \Rightarrow 12 (x^{2} - 23 x + 90) = 0 \\ \Rightarrow (x - 18) (x - 5) = 0 \Rightarrow x = 5, 18 \\ It is not possible to cut off a square of side 18 cm from each \\ corner of the rectangular sheet . Thus, x cannot be equal to 18 . \\ When x = 5, \\ V ‘ ‘ (5) = 12 (2 \times 5 - 23) \\ = 12 (10 - 23) = - 156 < 0 \\ ∴ By second derivative test, x = 5 is the point of maxima . \\ Hence, the side of the square to be cut off to make the volume \\ of the box maximum is 5 cm . \end{array}$

Q.92 Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Ans.

$\begin{array}{l} Let a rectangle of length l and breadth b be inscribed in the \\ given circle of radius a .Then, the diagonal passes through the \\ centre and is of length 2a cm . \end{array}$

$\begin{array}{l} Now, by applying the Pythagoras theorem, we have: \\ {(2 a)}^{2} = l^{2} + b^{2} \\ \Rightarrow b^{2} = 4 a^{2} - l^{2} \\ \Rightarrow b = \sqrt{4 a^{2} - l^{2}} \\ ∴ Area of the rectangle, \\ A = l b \\ = l \sqrt{4 a^{2} - l^{2}} \\ D i f f e r e n t i a t i n g w .r .t . l, we get \\ \frac{d A}{d l} = \frac{d}{d l} (l \sqrt{4 a^{2} - l^{2}}) \\ = \sqrt{4 a^{2} - l^{2}} \frac{d}{d l} l + l \frac{d}{d l} \sqrt{4 a^{2} - l^{2}} \\ = \sqrt{4 a^{2} - l^{2}} + l \frac{1}{2 \sqrt{4 a^{2} - l^{2}}} (- 2 l) \\ = \sqrt{4 a^{2} - l^{2}} - \frac{l^{2}}{\sqrt{4 a^{2} - l^{2}}} \\ = \frac{4 a^{2} - l^{2} - l^{2}}{\sqrt{4 a^{2} - l^{2}}} \\ = \frac{4 a^{2} - 2 l^{2}}{\sqrt{4 a^{2} - l^{2}}} \\ \frac{d^{2} A}{d l^{2}} = \frac{d}{d l} (\frac{4 a^{2} - 2 l^{2}}{\sqrt{4 a^{2} - l^{2}}}) \\ = \frac{\sqrt{4 a^{2} - l^{2}} \frac{d}{d l} (4 a^{2} - 2 l^{2}) - (4 a^{2} - 2 l^{2}) \frac{d}{d l} \sqrt{4 a^{2} - l^{2}}}{{(\sqrt{4 a^{2} - l^{2}})}^{2}} \\ = \frac{\sqrt{4 a^{2} - l^{2}} (0 - 4 l) - (4 a^{2} - 2 l^{2}) \times \frac{1}{2 \sqrt{4 a^{2} - l^{2}}} (- 2 l)}{4 a^{2} - l^{2}} \\ = \frac{- 4 l \sqrt{4 a^{2} - l^{2}} + \frac{4 a^{2} l - 2 l^{3}}{\sqrt{4 a^{2} - l^{2}}}}{(4 a^{2} - l^{2})} \\ = \frac{- 4 l (4 a^{2} - l^{2}) + 4 a^{2} l - 2 l^{3}}{{(4 a^{2} - l^{2})}^{\frac{3}{2}}} \\ = \frac{- 16 a^{2} l + 4 l^{3} + 4 a^{2} l - 2 l^{3}}{{(4 a^{2} - l^{2})}^{\frac{3}{2}}} \\ = \frac{2 l^{3} - 12 a^{2} l}{{(4 a^{2} - l^{2})}^{\frac{3}{2}}} \\ F o r maxima or minima, we have \\ \frac{d A}{d l} = 0 \Rightarrow \frac{4 a^{2} - 2 l^{2}}{\sqrt{4 a^{2} - l^{2}}} = 0 \\ \Rightarrow 4 a^{2} - 2 l^{2} = 0 \Rightarrow l^{2} = \frac{4 a^{2}}{2} = 2 a^{2} \\ \Rightarrow l = \sqrt{2} a a n d b = \sqrt{4 a^{2} - 2 a^{2}} = \sqrt{2} a \\ W h e n l= \sqrt{2} a, \\ \frac{d^{2} A}{d l^{2}} = \frac{2 \sqrt{2} a (2 a^{2} - 6 a^{2})}{{(4 a^{2} - 2 a^{2})}^{\frac{3}{2}}} = \frac{2 \sqrt{2} a (- 4 a^{2})}{{(2 a^{2})}^{\frac{3}{2}}} < 0 \\ ∴ By the second derivative test, when l= \sqrt{2} a, then the area of \\ the rectangle is the maximum . \\ Since l = b = \sqrt{2} a, the rectangle is a square . \\ Hence, it has been proved that of all the rectangles inscribed \\ in the given fixed circle, the square has the maximum area . \end{array}$

Q.93 Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.

Ans.

\begin{array}{l} Let r and h be the radius and height of the cylinder respectively . \\ Then, the surface area (S) of the cylinder is given by, \\ S = 2 {πr}^{2} + 2 πrh \Rightarrow h = \frac{S - 2 {πr}^{2}}{2 πr} = \frac{S}{2 π} (\frac{1}{r}) - r \\ Let V be the volume of the cylinder . Then, \\ V = 2 {πr}^{2} h \\ = 2 {πr}^{2} {\frac{S}{2 π} (\frac{1}{r}) - r} \\ = \frac{Sr}{2} - {πr}^{3} \\ Differentiating w . r . t . r, we get \\ \frac{dV}{dr} = \frac{S}{2} - 3 {πr}^{2} and \frac{d^{2} V}{{dr}^{2}} = - 6 πr \\ For maxima or minima, we have \\ \frac{dV}{dr} = 0 \Rightarrow \frac{S}{2} - 3 {πr}^{2} = 0 \Rightarrow r^{2} = \frac{S}{6 π} \\ \Rightarrow r = \pm \sqrt{\frac{S}{6 π}} = \sqrt{\frac{S}{6 π}}, - \sqrt{\frac{S}{6 π}} (Neglect) \\ When r = \sqrt{\frac{S}{6 π}} \\ \frac{d^{2} V}{{dr}^{2}} = - 6 π (\sqrt{\frac{S}{6 π}}) < 0 \\ ∴ By second derivative test, the volume is the maximum when \\ r = \sqrt{\frac{S}{6 π} \Rightarrow} S = 6 {πr}^{2} . \\ ∴ h = \frac{6 {πr}^{2} - 2 {πr}^{2}}{2 πr} \\ = \frac{4 {πr}^{2}}{2 πr} = 2 r \\ Hence, the volume is the maximum when the height is twice \\ the radius i . e ., when the height is equal to the diameter . \end{array}

Q.94 Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions
of the can which has the minimum surface area?

Ans.

$\begin{array}{l} Let r and h be the radius and height of the cylinder respectively . \\ Then, volume (V) of the cylinder is given by, \\ V = {πr}^{2} h = 100 \\ \Rightarrow h = \frac{100}{{πr}^{2}} \\ Surface area (S) of the cylinder is given by, \\ S = 2 {πr}^{2} + 2 πrh = 2 {πr}^{2} + 2 πr (\frac{100}{{πr}^{2}}) \\ = 2 {πr}^{2} + (\frac{200}{r}) \\ Differentiating w . r . t . x, we get \\ \frac{dS}{dr} = \frac{d}{dr} (2 {πr}^{2}) + \frac{d}{dr} (\frac{200}{r}) \\ = 4 πr - \frac{200}{r^{2}} \\ \frac{d^{2} S}{{dr}^{2}} = \frac{d}{dr} 4 πr - \frac{d}{dr} \frac{200}{r^{2}} \\ = 4 π + \frac{400}{r^{3}} \\ For maxima or minima, we have \\ \frac{dS}{dr} = 0 \Rightarrow 4 πr - \frac{200}{r^{2}} = 0 \\ 4 πr = \frac{200}{r^{2}} \Rightarrow r^{3} = \frac{200}{4 π} = \frac{50}{π} \\ When r = {(\frac{50}{π})}^{\frac{1}{3}} \\ So, \frac{d^{2} S}{{dr}^{2}} > 0 \\ ∴ By second derivative test, the surface area is the minimum \\ when the radius of the cylinder is {(\frac{50}{π})}^{\frac{1}{3}} cm . \\ When r = {(\frac{50}{π})}^{\frac{1}{3}} cm, h = \frac{100}{π {(\frac{50}{π})}^{\frac{2}{3}}} = 2 {(\frac{50}{π})}^{\frac{1}{3}} cm \\ Hence, the required dimensions of the can which has the \\ minimum surface area is given by radius = {(\frac{50}{π})}^{\frac{1}{3}} and \\ height = 2 {(\frac{50}{π})}^{\frac{1}{3}} cm . \end{array}$

Q.95 A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Ans.

$\begin{array}{l} Letcirclepiece length = x m \\ Thensquarepiecelength = (28 - x) m \\ For circle, \\ Perimeterofcircle = 2 πr = x \Rightarrow r = \frac{x}{2 π} \\ Andfor square, \\ Perimeterof square = 28 - x = 4 a \Rightarrow a = \frac{28 - x}{4} \\ Nowthesumoftheareas A ofcircleandsquare \\ A = π r^{2} + a^{2} \\ = π {(\frac{x}{2 π})}^{2} + {(\frac{28 - x}{4})}^{2} \\ = \frac{x^{2}}{4 π} + {(\frac{28 - x}{4})}^{2} \\ \frac{dA}{dx} = \frac{x}{2 π} - \frac{28 - x}{8} \\ \frac{d^{2} A}{d x^{2}} = \frac{1}{2 π} + \frac{1}{8} \\ For max ima or minima, we have \\ \frac{dA}{dx} = 0 \Rightarrow \frac{x}{2 π} - \frac{28 - x}{8} = 0 or x = \frac{28 π}{4 + π} \\ \frac{d^{2} A}{d x^{2}} > 0 at x = \frac{28 π}{4 + π} \\ ∴ Aisminimum at circle piece = x \\ = \frac{28 π}{4 + π} \\ Squarepiece = 28 - x = \frac{112}{4 + π} \end{array}$

Q.96

\begin{array}{l} Prove that the volume of the largest cone that can be \\ inscribed in a sphere of radius R is \frac{8}{27} of the volume of \\ the sphere . \end{array}

Ans.

$\begin{array}{l} DAB is the cone inscribed in a sphere of radius R . \\ Volume of cone V = \frac{1}{3} {πr}^{2} h \\ Where r is the base radius of cone and h its height \\ Height of cone h = OC + OD = R + OC = R + x \\ Radius of Cone = r where r^{2} = R^{2} - {OD}^{2} = R^{2} - x^{2} \\ ∴ V = \frac{1}{3} π (R^{2} - x^{2}) (R + x) = \frac{1}{3} π [R^{3} + R^{2} x - x^{2} R - x^{3}] \\ \frac{dV}{dx} = \frac{1}{3} π [R^{2} - 2 xR - 3 x^{2}] \\ = \frac{1}{3} π [R^{2} - 3 xR - xR - 3 x^{2}] \\ = \frac{1}{3} π (R - 3 x) (R + x) \\ \frac{d^{2} V}{{dx}^{2}} = \frac{1}{3} π {(R - 3 x) \frac{d}{dx} (R + x) + (R + x) \frac{d}{dx} (R - 3 x)} \\ = \frac{1}{3} π (R - 3 x - 3 R - 3 x) \\ = \frac{1}{3} π (- 2 R - 6 x) \\ = - \frac{2}{3} π (R + 3 x) \\ \frac{dV}{dx} = 0 \Rightarrow x = \frac{R}{3} or - R \\ \frac{d^{2} V}{{dx}^{2}} < 0 at x = \frac{R}{3} \\ Hence x = \frac{R}{3} gives maximum volume . \\ V = \frac{1}{3} π (R^{2} - \frac{R^{2}}{9}) (R + \frac{R}{3}) \\ = \frac{1}{3} π \times \frac{8 R^{2}}{9} \times \frac{4 R}{3} \\ = \frac{8}{27} [\frac{4}{3} {πR}^{3}] \\ Hence the volume of cone = \frac{8}{27} (Volume of sphere) is the \\ largest cone that could be inscribed in a given sphere . \end{array}$

Q.97

\begin{array}{l} Show that the right circular cone of least curved surface \\ and given volume has an altitude equal to \sqrt{2} time the \\ radius of the base . \end{array}

Ans.

$\begin{array}{l} Here, Volume of the cone i s ‘ V ’ . \\ ∴ V = \frac{1}{3} π r^{2} h \Rightarrow r^{2} = \frac{3 V}{π h} \\ and surface area be ‘ S ’ of cone . \\ ∴ S = π r l = π r \sqrt{h^{2} + r^{2}} \\ Where h = height of the cone \\ r = radius of the cone \\ l = slant height of the cone \\ S^{2} = π^{2} r^{2} (h^{2} + r^{2}) \\ Let S_{1} = S^{2} then \\ S_{1} = π^{2} r^{2} (h^{2} + r^{2}) \\ S_{1} = \frac{3 π V}{h} (h^{2} + \frac{3 V}{π h}) = 3 π V h + \frac{9 V^{2}}{h^{2}} [∵ r^{2} = \frac{3 V}{π h}] \\ \frac{d S_{1}}{d h} = 3 π V + 9 V^{2} (\frac{- 2}{h^{3}}) \\ \frac{d^{2} S_{1}}{d h^{2}} = \frac{54 V^{2}}{h^{4}} \\ \frac{d S_{1}}{d h} = 0 for maxima/minima \\ 3 π V + 9 V^{2} (\frac{- 2}{h^{3}}) = 0 \\ \Rightarrow 3 π V = 9 V^{2} (\frac{2}{h^{3}}) \\ \Rightarrow h^{3} = \frac{6 V}{π} \\ \frac{d^{2} S_{1}}{d h^{2}} > 0 at h^{3} = \frac{6 V}{π^{}} \\ Therefore curved surface area is minimum at \frac{π h^{3}}{6} = V . \\ T h u s, \frac{π h^{3}}{6} = \frac{1}{3} π r^{2} h \Rightarrow h^{2} = 2 r^{2} \\ \Rightarrow h = \sqrt{2} r \\ Hence for least curved surface the altitude is \sqrt{2} times radius . \end{array}$

Q.98

\begin{array}{l} Show that the semi-vertical angle of the cone of the \\ maximum volume and of given slant height is {tan}^{- 1} \sqrt{2} . \end{array}

Ans.

$\begin{array}{l} Let θ be the semi-vertical angle of the cone . \\ It is clear that θ \in [0, \frac{π}{2}] . \\ Let r, h, and l be the radius, height, and the slant height of \\ the cone respectively . The slant height of the cone is given \\ as constant . \end{array}$

$\begin{array}{l} Now, r = l \sin θ and h = l \cos θ \\ The volume (V) of the cone is given by, \\ V = \frac{1}{3} {πr}^{2} h \\ = \frac{1}{3} π {(l \sin θ)}^{2} (l \cos θ) \\ = \frac{1}{3} {πl}^{3} \sin^{2} θcosθ \\ Differentiating w . r . t . θ, we get \\ \frac{dV}{dθ} = \frac{1}{3} {πl}^{3} (\sin^{2} θ \frac{d}{dθ} cosθ + cosθ \frac{d}{dθ} \sin^{2} θ) \\ = \frac{1}{3} {πl}^{3} (- \sin^{3} θ + 2 {sinθcos}^{2} θ) \\ & \frac{d^{2} V}{{dθ}^{2}} = \frac{1}{3} {πl}^{3} (- 3 \sin^{2} θcosθ + 3 \cos^{3} θ - 2 \sin^{2} θcosθ) \\ For maxima or minima, we have \\ \frac{dV}{dθ} = 0 \Rightarrow \frac{1}{3} {πl}^{3} (- \sin^{3} θ + 2 {sinθcos}^{2} θ) = 0 \\ \Rightarrow \sin^{3} θ = 2 {sinθcos}^{2} θ \\ \Rightarrow \tan^{2} θ = 2 \Rightarrow tanθ = \sqrt{2} \\ \Rightarrow θ = \tan^{- 1} \sqrt{2} \\ When θ = \tan^{- 1} \sqrt{2} \Rightarrow \sin^{2} θ = 2 \cos^{2} θ \\ \frac{d^{2} V}{{dθ}^{2}} = \frac{1}{3} {πl}^{3} (- 3 \times 2 \cos^{2} θ . cosθ + 3 \cos^{3} θ - 2 \times 2 \cos^{2} θ . cosθ) \\ = \frac{1}{3} {πl}^{3} (- 6 \cos^{3} θ + 3 \cos^{3} θ - 4 \cos^{3} θ) \\ = \frac{1}{3} {πl}^{3} (- 4 \cos^{3} θ) < 0 \\ ∴ By second derivative test, the volume (V) is the maximum \\ when θ = \tan^{- 1} \sqrt{2} . \end{array}$

Q.99

\begin{array}{l} Show that semi-vertical angle of right circular cone of \\ {given surface area and maximum volume is sin}^{-1} (\frac{1}{3}) . \end{array}

Ans.

$\begin{array}{l} Here in ΔAOC, l^{2} = h^{2} + r^{2} \\ The total surface area of the cone is \\ S = {πr}^{2} + πrl \Rightarrow \frac{S - {πr}^{2}}{πr} = l \\ or l = \frac{S}{πr} - r \\ Volume of the cone is given by \\ V = \frac{1}{3} {πr}^{2} h = \frac{1}{3} {πr}^{2} \sqrt{l^{2} - r^{2}} \\ Let \\ V_{1} = V^{2} = \frac{1}{9} π^{2} r^{4} (l^{2} - r^{2}) \\ \Rightarrow V_{1} = \frac{1}{9} π^{2} r^{4} ({[\frac{S}{πr} - r]}^{2} - r^{2}) \\ \Rightarrow V_{1} = \frac{1}{9} π^{2} r^{4} (\frac{S^{2}}{π^{2} r^{2}} - \frac{2 S}{π}) = \frac{1}{9} (S^{2} r^{2} - 2 {Sπr}^{4}) \\ Differentiating w . r . t . r, we get \\ V_{1} ‘ = \frac{1}{9} (2 S^{2} r - 8 {Sπr}^{3}) \\ V_{1} ‘ ‘ = \frac{1}{9} (2 S^{2} - 24 {Sπr}^{2}) \\ For maxima or minima, \\ V_{1} ‘ = 0 \\ \Rightarrow \frac{1}{9} (2 S^{2} r - 8 {Sπr}^{3}) = 0 \\ \Rightarrow 2 S^{2} r = 8 {Sπr}^{3} \\ \Rightarrow \frac{S}{4 π} = r^{2} \\ V_{1} ‘ ‘ = \frac{1}{9} (2 S^{2} - 24 {Sπr}^{2}) < 0 at r^{2} = \frac{S}{4 π} \\ \Rightarrow V is maximum at r^{2} = \frac{S}{4 π} \\ ∵ 4 {πr}^{2} = S \\ \Rightarrow 4 {πr}^{2} = {πr}^{2} + πrl \\ \Rightarrow 3 {πr}^{2} = πrl \\ or \frac{r}{l} = \frac{1}{3} \\ i . e \sin ∠ AOC = \frac{1}{3} [∵ InΔAOC, \sin (∠ OAC) = \frac{r}{l}] \\ \Rightarrow semi - vertical angle is \sin^{- 1} (\frac{1}{3}) . \end{array}$

Q.100

\begin{array}{l} The point on the curve x^{2} = 2 y which is nearest to the \\ point (0, 5) is \\ (A) (2 \sqrt{2}, 4) (B) (2 \sqrt{2}, 0) (C) (0,0) (D) (2,2) \end{array}

Ans.

\begin{array}{l} The given curve is x^{2} = 2 y . \\ Let P (x, y) be a point on x^{2} = 2 y and A (0, 5) be the given point . \\ Then, {AP}^{2} = {(x - 0)}^{2} + {(y - 5)}^{2} \\ = x^{2} + {(\frac{x^{2}}{2} - 5)}^{2} [∵ x^{2} = 2 y \Rightarrow y = \frac{x^{2}}{2}] \\ Let Z = {AP}^{2} . Then, Z is maximum or minimum according as AP \\ is maximum or minimum . \\ Now, Z = x^{2} + {(\frac{x^{2}}{2} - 5)}^{2} \\ Differentiating w . r . t . x, we get \\ \frac{dZ}{dx} = \frac{d}{dx} {x^{2} + {(\frac{x^{2}}{2} - 5)}^{2}} \\ = 2 x + 2 (\frac{x^{2}}{2} - 5) \frac{d}{dx} (\frac{x^{2}}{2} - 5) \\ = 2 x + 2 (\frac{x^{2}}{2} - 5) (x - 0) \\ = 2 x + x^{3} - 10 x \\ = x^{3} - 8 x \\ \frac{d^{2} Z}{{dx}^{2}} = 3 x^{2} - 8 \\ For maxima or minima, we have \\ \frac{dZ}{dx} = 0 \Rightarrow x^{3} - 8 x = 0 \\ \Rightarrow x (x^{2} - 8) = 0 \\ \Rightarrow x = 0, \pm 2 \sqrt{2} \\ When x = 0, \\ \frac{d^{2} Z}{{dx}^{2}} = 3 {(0)}^{2} - 8 = - 8 < 0 \\ So, Z is maximum at x = 0 i . e ., point P is not nearest to curve . \\ When x = \pm 2 \sqrt{2} \\ \frac{d^{2} Z}{{dx}^{2}} = 3 {(\pm 2 \sqrt{2})}^{2} - 8 \\ = 24 - 8 = 16 > 0 \\ So, Z is minimum at x = \pm 2 \sqrt{2} i . e ., point P is nearest to curve . \\ Putting x = \pm 2 \sqrt{2} in x^{2} = 2 y, we get \\ y = \frac{x^{2}}{2} = \frac{{(\pm 2 \sqrt{2})}^{2}}{2} = \frac{8}{2} = 4 \\ Hence, the point (\pm 2 \sqrt{2}, 4) on x^{2} = 2 y is the nearest to the \\ point (0, 5) . \\ Thus, the option A is correct . \end{array}

Q.101

\begin{array}{l} For all real values of x, the minimum value of \frac{1 - x + x^{2}}{1 + x + x^{2}} is \\ (A) 0 (B) 1 (C) 3 (D) \frac{1}{3} \end{array}

Ans.

$\begin{array}{l} Let f (x) = \frac{1 - x + x^{2}}{1 + x + x^{2}} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = \frac{d}{dx} (\frac{1 - x + x^{2}}{1 + x + x^{2}}) \\ = \frac{(1 + x + x^{2}) \frac{d}{dx} (1 - x + x^{2}) - (1 - x + x^{2}) \frac{d}{dx} (1 + x + x^{2})}{{(1 + x + x^{2})}^{2}} \\ = \frac{(1 + x + x^{2}) (0 - 1 + 2 x) - (1 - x + x^{2}) (0 + 1 + 2 x)}{{(1 + x + x^{2})}^{2}} \\ = \frac{- 1 - x - x^{2} + 2 x + 2 x^{2} + 2 x^{3} - (1 - x + x^{2} + 2 x - 2 x^{2} + 2 x^{3})}{{(1 + x + x^{2})}^{2}} \\ = \frac{- 1 - x - x^{2} + 2 x + 2 x^{2} + 2 x^{3} - 1 + x - x^{2} - 2 x + 2 x^{2} - 2 x^{3}}{{(1 + x + x^{2})}^{2}} \\ = \frac{- 2 + 2 x^{2}}{{(1 + x + x^{2})}^{2}} \\ = \frac{2 (x^{2} - 1)}{{(1 + x + x^{2})}^{2}} \\ f ‘ ‘ (x) = \frac{{(1 + x + x^{2})}^{2} \frac{d}{dx} 2 (x^{2} - 1) - 2 (x^{2} - 1) \frac{d}{dx} {(1 + x + x^{2})}^{2}}{{(1 + x + x^{2})}^{4}} \\ = \frac{{(1 + x + x^{2})}^{2} 2 (2 x - 0) - 2 (x^{2} - 1) 2 (1 + x + x^{2}) \frac{d}{dx} (1 + x + x^{2})}{{(1 + x + x^{2})}^{4}} \\ = \frac{4 x {(1 + x + x^{2})}^{2} - 4 (x^{2} - 1) (1 + x + x^{2}) (0 + 1 + 2 x)}{{(1 + x + x^{2})}^{4}} \\ = \frac{4 (1 + x + x^{2}) {(1 + x + x^{2}) x - (x^{2} - 1) (1 + 2 x)}}{{(1 + x + x^{2})}^{4}} \\ = \frac{4 {x + x^{2} + x^{3} - x^{2} + 1 - 2 x^{3} + 2 x}}{{(1 + x + x^{2})}^{3}} \\ = \frac{4 (- x^{3} + 3 x + 1)}{{(1 + x + x^{2})}^{3}} \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow \frac{2 (x^{2} - 1)}{{(1 + x + x^{2})}^{2}} = 0 \\ \Rightarrow x^{2} - 1 = 0 \Rightarrow x = \pm 1 \\ When x = 1, \\ f ‘ ‘ (1) = \frac{4 (- 1^{3} + 3 \times 1 + 1)}{{(1 + 1 + 1^{2})}^{3}} \\ = \frac{4 (3)}{27} = \frac{4}{9} > 0 \\ f ‘ ‘ (- 1) = \frac{4 {- {(- 1)}^{3} + 3 (- 1) + 1}}{{1 + (- 1) + {(- 1)}^{2}}^{3}} \\ = \frac{4 (- 1)}{1} = - 4 < 0 \\ ∴ By second derivative test, f is the minimum at x = 1 and \\ the minimum value is given by \\ f (1) = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3} \\ Thus, the correct answer is D . \end{array}$

Q.102

\begin{array}{l} The maximum value of {[x (x-1) +1]}^{\frac{1}{3}}, 0£x£1 is \\ (A) {(\frac{1}{3})}^{\frac{1}{3}} (B) \frac{1}{2} (C) 1 (D) 0 \end{array}

Ans.

\begin{array}{l} Let f (x) = {[x (x - 1) + 1]}^{\frac{1}{3}}, 0 \leq x \leq 1 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = \frac{d}{dx} {[x (x - 1) + 1]}^{\frac{1}{3}} \\ = \frac{1}{3} {[x (x - 1) + 1]}^{- \frac{2}{3}} \frac{d}{dx} [x (x - 1) + 1] \\ = \frac{1}{3} {[x (x - 1) + 1]}^{- \frac{2}{3}} (\frac{d}{dx} x^{2} - \frac{d}{dx} x + \frac{d}{dx} 1) \\ = \frac{1}{3} {[x (x - 1) + 1]}^{- \frac{2}{3}} (2 x - 1) \\ = \frac{(2 x - 1)}{3 {[x (x - 1) + 1]}^{\frac{2}{3}}} \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow \frac{(2 x - 1)}{3 {[x (x - 1) + 1]}^{\frac{2}{3}}} = 0 \\ \Rightarrow 2 x - 1 = 0 \Rightarrow x = \frac{1}{2} \\ Then, we evaluate the value of f at critical point x = \frac{1}{2} \\ and at the end points of the interval [0, 1] . \\ f (0) = {[(0) (0 - 1) + 1]}^{\frac{1}{3}} = 0 \\ f (\frac{1}{2}) = {[(\frac{1}{2}) (\frac{1}{2} - 1) + 1]}^{\frac{1}{3}} = {(\frac{3}{4})}^{\frac{1}{3}} \\ f (1) = {[(1) (1 - 1) + 1]}^{\frac{1}{3}} = 1 \\ Hence, we can conclude that the maximum value of f in the \\ interval [0, 1] is 1 . \\ The correct answer is C . \end{array}

Q.103 Using differentials, find the approximate value of each of the following:

$(a) {(\frac{17}{81})}^{\frac{1}{4}} (b) {(33)}^{- \frac{1}{5}}$

Ans.

$\begin{array}{l} (a) Let y = x^{\frac{1}{4}} where x = \frac{16}{81} and Δx = \frac{1}{81} \\ \frac{dy}{dx} = \frac{1}{4 x^{\frac{3}{4}}} \\ Then, Δy = {(x + Δx)}^{\frac{1}{4}} - x^{\frac{1}{4}} \\ = {(\frac{17}{81})}^{\frac{1}{4}} - {(\frac{16}{81})}^{\frac{1}{4}} \\ = {(\frac{17}{81})}^{\frac{1}{4}} - \frac{2}{3} \\ {(\frac{17}{81})}^{\frac{1}{4}} = Δy + \frac{2}{3} \\ Now, dy is approximately equal to Ny and is given by, \\ dy = (\frac{dy}{dx}) Δx \\ = (\frac{1}{4 x^{\frac{3}{4}}}) Δx \\ = {\frac{1}{4 {(\frac{16}{81})}^{\frac{3}{4}}}} \frac{1}{81} \\ = \frac{27}{4 \times 8} \times \frac{1}{81} \\ = \frac{1}{96} = 0 .010 \\ Hence, the approximate value of {(\frac{17}{81})}^{\frac{1}{4}} \\ = 0 .010 + \frac{2}{3} \\ = 0 .010 + 0 .667 = 0 .677. \\ (b) Let y = x^{- \frac{1}{5}} and x = 32 and Δx = 1 \\ ∴ {(33)}^{- \frac{1}{5}} = \frac{1}{2} + Δy \\ Now, dy is approximately equal to Δy and is given by, \\ dy = (\frac{dy}{dx}) Δx = - \frac{1}{5 x^{\frac{6}{5}}} Δx \\ = - \frac{1}{5 {(32)}^{\frac{6}{5}}} \times 1 \\ = - \frac{1}{5 {(2)}^{6}} = - 0 .003 \\ Hence, the approximate value of {(33)}^{- \frac{1}{5}} \\ = \frac{1}{2} - 0 .003 \\ = 0 .5 - 0 .003 = 0 .497 \end{array}$

Q.104

\begin{array}{l} Show that the function given by f (x) = \frac{logx}{x} has maximum \\ at x = e . \end{array}

Ans.

\begin{array}{l} Given : f (x) = \frac{logx}{x} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = \frac{d}{dx} (\frac{logx}{x}) \\ = \frac{x \frac{d}{dx} logx - logx \frac{d}{dx} x}{x^{2}} \\ = \frac{x . \frac{1}{x} - logx . 1}{x^{2}} \\ = \frac{1 - logx}{x^{2}} \\ f ” (x) = \frac{d}{dx} (\frac{1 - logx}{x^{2}}) \\ = \frac{x^{2} \frac{d}{dx} (1 - logx) - (1 - logx) \frac{d}{dx} x^{2}}{{(x^{2})}^{2}} \\ = \frac{x^{2} (0 - \frac{1}{x}) - (1 - logx) \times 2 x}{x^{4}} \\ = \frac{- x - 2 x + 2 xlogx}{x^{4}} \\ f ” (x) = \frac{- 3 + 2 logx}{x^{3}} \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow \frac{1 - logx}{x^{2}} = 0 \\ \Rightarrow 1 - logx = 0 \Rightarrow logx = 1 \\ \Rightarrow logx = loge \\ \Rightarrow x = e \\ Now, f ” (x) = \frac{- 3 + 2 loge}{e^{3}} \\ = \frac{- 3 + 2}{e^{3}} = \frac{- 1}{e^{3}} < 0 \\ Therefore, f (x) is maximum at x = e . \end{array}

Q.105 The two equal sides of an isosceles triangle with fixed base b and decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Ans.

$\begin{array}{l} Let Δ ABC be isosceles where BC is the base of fixed length b . \\ Let the length of the two equal sides of Δ ABC be a . \\ Draw AD ⊥ BC \end{array}$

$\begin{array}{l} Now, in Δ ADC, by applying the Pythagoras theorem, we have: \\ AD = \sqrt{a^{2} - \frac{b^{2}}{4}} \\ = \frac{1}{2} \sqrt{4 a^{2} - b^{2}} \\ A r e a of triangle (A) = \frac{1}{2} b \times \frac{1}{2} \sqrt{4 a^{2} - b^{2}} \\ The rate of change of the area with respect to time (t) is \\ given by, \\ \frac{d A}{d t} = \frac{1}{4} b \times \frac{8 a}{2 \sqrt{4 a^{2} - b^{2}}} \frac{d a}{d t} \\ It is given that the two equal sides of the triangle are \\ decreasing at the rate of 3 cm per second, i .e ., \frac{d a}{d t} = - 3 c m / s \\ \frac{d A}{d t} = b \times \frac{a}{\sqrt{4 a^{2} - b^{2}}} \times - 3 = \frac{- 3 a b}{\sqrt{4 a^{2} - b^{2}}} \\ W h e n a = b, we have \\ \frac{d A}{d t} = \frac{- 3 b^{2}}{\sqrt{4 b^{2} - b^{2}}} \\ = \frac{- 3 b^{2}}{\sqrt{3} b} = - \sqrt{3} b \\ Hence, if the two equal sides are equal to the base, then \\ the area of the triangle is decreasing at the rate of \sqrt{3} b c m^{2} / s . \end{array}$

Q.106 Find the equation of the normal to curve y² = 4x at the point (1, 2).

Ans.

\begin{array}{l} The given curve ​ ​​ is \\ y^{2} = 4 x \\ Differenetiating w . r . t . x, we get \\ 2 y \frac{dy}{dx} = 4 \\ \frac{dy}{dx} = \frac{4}{2 y} = \frac{2}{y} \\ m = {(\frac{dy}{dx})}_{(1, 2)} \\ = \frac{2}{2} = 1 \\ Slope of Normal is \\ M = - \frac{1}{m} \\ = - \frac{1}{1} = - 1 \\ Equation of normal at the point (1, 2) is \\ y - 2 = (- 1) (x - 1) \\ \Rightarrow y - 2 = - x + 1 \\ \Rightarrow x + y = 3 \end{array}

Q.107

\begin{array}{l} Show that the normal at any point θ to the curve \\ x = acos θ + a θsinθ, y = asinθ - a θcosθ \\ is at a constant distance from the origin . \end{array}

Ans.

\begin{array}{l} We have, \\ x = acosθ + a θsinθ and y = asinθ - a θcosθ \\ Then differentiating w . r . t . x, we get \\ \frac{dx}{dθ} = a \frac{d}{dθ} cosθ + a \frac{d}{dθ} (θsinθ) \\ = - asinθ + a (θcosθ + sinθ) \\ = aθcosθ \\ \frac{dy}{dθ} = a \frac{d}{dθ} sinθ - a \frac{d}{dθ} (θcosθ) \\ = acosθ - a (- θsinθ + cosθ) \\ = aθsinθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{aθsinθ}{aθcosθ} \\ = tanθ \\ Slope of normal = - \frac{1}{tanθ} \\ The equation of the normal at a given point (x, y) is given by, \\ y - asinθ + a θcosθ = - \frac{1}{tanθ} (x - acosθ - a θsinθ) \\ \Rightarrow y - asinθ + a θcosθ = - \frac{cosθ}{sinθ} (x - acosθ - a θsinθ) \\ \Rightarrow ysinθ - {asin}^{2} θ + aθsinθcosθ = - xcosθ + {acos}^{2} θ + aθsinθcosθ \\ \Rightarrow ysinθ - {asin}^{2} θ = - xcosθ + {acos}^{2} θ \\ \Rightarrow xcosθ + ysinθ - a (\sin^{2} θ + \cos^{2} θ) = 0 \\ \Rightarrow xcosθ + ysinθ - a = 0 [∵ \sin^{2} θ + \cos^{2} θ = 1] \\ Now, the perpendicular distance of the normal from the origin : \\ p = \frac{| 0 . cosθ + 0 . sinθ - a |}{\sqrt{\cos^{2} θ + \sin^{2} θ}} = a, \\ which is independent of θ . \\ Hence, the perpendicular distance of the normal from the \\ origin is constant . \end{array}

Q.108

\begin{array}{l} Find the intervals in which the function f given by \\ f (x) = \frac{4 sinx - 2 x - xcosx}{2 + cosx} \\ is (i) increasing (ii) decreasing . \end{array}

Ans.

\begin{array}{l} The function is \\ f (x) = \frac{4 sinx - 2 x - xcosx}{2 + cosx} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = {\frac{\begin{array}{l} (2 + cosx) \frac{d}{dx} (4 sinx - 2 x - xcosx) \\ - (4 sinx - 2 x - xcosx) \frac{d}{dx} (2 + cosx) \end{array}}{{(2 + cosx)}^{2}}} \\ \Rightarrow f ‘ (x) = {\frac{\begin{array}{l} (2 + cosx) (4 cosx - 2 + xsinx - cosx) \\ - (4 sinx - 2 x - xcosx) (0 - sinx) \end{array}}{{(2 + cosx)}^{2}}} \\ \Rightarrow f ‘ (x) = {\frac{(2 + cosx) (3 cosx - 2 + xsinx) + sinx (4 sinx - 2 x - xcosx)}{{(2 + cosx)}^{2}}} \\ \Rightarrow f ‘ (x) = \frac{\begin{array}{l} 6 cosx + 3 \cos^{2} x - 4 - 2 cosx + 2 xsinx + xsinxcosx \\ + 4 \sin^{2} x - 2 xsinx - xsinxcosx \end{array}}{{(2 + cosx)}^{2}} \\ \Rightarrow f ‘ (x) = \frac{4 cosx + 3 \cos^{2} x - 4 + 4 \sin^{2} x}{{(2 + cosx)}^{2}} \\ \Rightarrow f ‘ (x) = \frac{4 cosx + 3 \cos^{2} x - 4 + 4 (1 - \cos^{2} x)}{{(2 + cosx)}^{2}} \\ = \frac{4 cosx + 3 \cos^{2} x - 4 + 4 - 4 \cos^{2} x}{{(2 + cosx)}^{2}} \\ = \frac{4 cosx - \cos^{2} x}{{(2 + cosx)}^{2}} \\ = \frac{cosx (4 - cosx)}{{(2 + cosx)}^{2}} \\ Now, f ‘ (x) = 0 \\ \Rightarrow cosx (4 - cosx) = 0 \Rightarrow cosx = 0 or 4 - cosx = 0 \\ \Rightarrow cosx = 0 but cosx \neq 4 \\ \Rightarrow x = \frac{π}{2}, \frac{3 π}{2}, which divides interval (0, 2 π) into three disjoint \\ intervals, \\ i . e ., (0, \frac{π}{2}), (\frac{π}{2}, \frac{3 π}{2}) and (\frac{3 π}{2}, 2 π) . \\ In intervals, (0, \frac{π}{2}), (\frac{3 π}{2}, 2 π), f ‘ (x) > 0 \\ Thus, f (x) is increasing for 0 < x < \frac{π}{2} ​​ and \frac{3 π}{2} < x < 2 π . \\ In the interval (\frac{π}{2}, \frac{3 π}{2}), f ‘ (x) < 0 . \\ Thus, f (x) is decreasing for \frac{π}{2} < x < \frac{3 π}{2} . \end{array}

Q.109

\begin{array}{l} Find the intervals in which the function f given by \\ f (x) {=x}^{3} + \frac{1}{x^{3}}, x¹0 is \\ (i) increasing (ii) decreasing . \end{array}

Ans.

\begin{array}{l} The function is \\ f (x) = x^{3} + \frac{1}{x^{3}} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = \frac{d}{dx} (x^{3} + \frac{1}{x^{3}}) \\ = 3 x^{2} - \frac{3}{x^{4}} \\ Then, f ‘ (x) = 0 \Rightarrow 3 x^{2} - \frac{3}{x^{4}} = 0 \\ \Rightarrow \frac{3 (x^{6} - 1)}{x^{4}} = 0 \Rightarrow x^{6} - 1 = 0 \\ \Rightarrow x = \pm 1 \\ Now, the points x = 1 and x = - 1 divide the real line into \\ three disjoint intervals \\ i . e ., (- \infty, - 1), (- 1, 1) ​ and (1, \infty) . \\ In intervals (- \infty, - 1) and (1, \infty) i . e ., x < - 1 and x >1, f ‘ (x) > 0 \\ Thus, when x < - 1 and x > 1, f is increasing . \\ In interval (- 1, 1) i . e ., when - 1 < x < 1, F ‘ (x) < 0 \\ Thus, when - 1 < x < 1, f is decreasing . \end{array}

Q.110

\begin{array}{l} Find the maximum area of an isosceles triangle inscribed in the \\ ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 with \\ its vertex at one end of the major axis . \end{array}

Ans.

$\begin{array}{l} The given ellipse is \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . \\ Let the major axis be along the x - axis . \\ Let ABC be the triangle inscribed in the ellipse where vertex C \\ is at (a, 0) . Since the ellipse is symmetrical with respect to the \\ x - axis and y - axis, we can assume the coordinates of A to be \\ (- x_{1}, y_{1}) and the coordinates of B to be (- x_{1}, - y_{1}). \\ Now, we have y_{1} = \pm \frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}} \\ ∴ Coordinates of A are (- x_{1}, \frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}}), the coordinates of \\ B are (- x_{1}, - \frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}}) and the coordinates of C are (a, 0) . \\ As the point (x_{1}, y_{1}) lies on the ellipse, the area (A) of ΔABC : \\ A = \frac{1}{2} | \begin{array}{l} a (- \frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}} - \frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}}) + (- x_{1}) (\frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}} - 0) \\ + (- x_{1}) (0 - \frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}}) \end{array} | \\ A = \frac{1}{2} | a (- \frac{2 b}{a} \sqrt{a^{2} - {x_{1}}^{2}}) - x_{1} (\frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}}) - x_{1} (- \frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}}) | \\ A = b \sqrt{a^{2} - {x_{1}}^{2}} + x_{1} \frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}} . .. (i) \\ Differentiating w . r . t . x_{1}, we get \\ \frac{dA}{{dx}_{1}} = \frac{d}{{dx}_{1}} (b \sqrt{a^{2} - {x_{1}}^{2}} + x_{1} \frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}}) \\ = \frac{- {bx}_{1}}{\sqrt{a^{2} - {x_{1}}^{2}}} + \frac{b}{a} \sqrt{a^{2} - {x_{1}}^{2}} - \frac{{bx}_{1}^{2}}{\sqrt{a^{2} - {x_{1}}^{2}}} \\ = \frac{b}{a \sqrt{a^{2} - {x_{1}}^{2}}} (- {ax}_{1} + a^{2} - {x_{1}}^{2} - {x_{1}}^{2}) \\ = \frac{b (- 2 {x_{1}}^{2} - {ax}_{1} + a^{2})}{a \sqrt{a^{2} - {x_{1}}^{2}}} \\ Now, \frac{dA}{{dx}_{1}} = 0 \\ \Rightarrow - 2 {x_{1}}^{2} - {ax}_{1} + a^{2} = 0 \\ \Rightarrow x_{1} = \frac{a \pm \sqrt{a^{2} - 4 (- 2) a^{2}}}{2 (- 2)} = \frac{a \pm \sqrt{9 a^{2}}}{- 4} \\ \Rightarrow x_{1} = \frac{a \pm 3 a}{- 4} = - a, \frac{a}{2} \\ But, x_{1} cannot be equal to a . \\ ∴ x_{1} = \frac{a}{2} \Rightarrow y_{1} = \frac{b}{a} \sqrt{a^{2} - \frac{a^{2}}{4}} = \frac{ba}{2 a} \sqrt{3} = \frac{\sqrt{3} b}{2} \\ Now, \frac{d^{2} A}{{dx}_{1}^{2}} = \frac{b}{a} {\frac{\sqrt{a^{2} - {x_{1}}^{2}} (- 4 x_{1} - a) - (- 2 {x_{1}}^{2} - x_{1} a + a^{2}) \frac{(- 2 x_{1})}{2 \sqrt{a^{2} - {x_{1}}^{2}}}}{(a^{2} - {x_{1}}^{2})}} \\ = \frac{b}{a} {\frac{(a^{2} - {x_{1}}^{2}) (- 4 x_{1} - a) + x_{1} (- 2 {x_{1}}^{2} - x_{1} a + a^{2})}{{(a^{2} - {x_{1}}^{2})}^{\frac{3}{2}}}} \\ = \frac{b}{a} {\frac{2 {x_{1}}^{3} - 3 a^{2} x - a^{3}}{{(a^{2} - {x_{1}}^{2})}^{\frac{3}{2}}}} \\ Also, when x_{1} = \frac{a}{2}, then \\ \frac{d^{2} A}{{dx}_{1}^{2}} = \frac{b}{a} {\frac{2 {(\frac{a}{2})}^{3} - 3 a^{2} (\frac{a}{2}) - a^{3}}{{(a^{2} - {\frac{a}{2^{2}}}^{2})}^{\frac{3}{2}}}} \\ = \frac{b}{a} \times \frac{\frac{a^{3}}{4} - \frac{3 a^{3}}{2} - a^{3}}{{({\frac{3 a}{2^{2}}}^{2})}^{\frac{3}{2}}} = - \frac{b}{a} {\frac{\frac{9}{4} a^{3}}{{({\frac{3 a}{2^{2}}}^{2})}^{\frac{3}{2}}}} < 0 \\ Thus, the area is the maximum when x_{1} = \frac{a}{2} . \\ ∴ Maximum area of the triangle is : \\ A = b \sqrt{a^{2} - \frac{a^{2}}{4}} + (\frac{a}{2}) \frac{b}{a} \sqrt{a^{2} - \frac{a^{2}}{4}} \\ = ab \frac{\sqrt{3}}{2} + \frac{ab}{4} \sqrt{3} = \frac{3 \sqrt{3}}{4} ab \end{array}$

Q.111 A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs ₹ 70 per sq meters for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?

Ans.

\begin{array}{l} Let l, b, and h represent the length, breadth, and height of the \\ tank respectively . \\ Then, we have height (h) = 2 m \\ Volume of the tank = 8 m^{3} \\ Volume of the tank = l \times b \times h \\ ∴ 8 = l \times b \times 2 \\ \Rightarrow lb =4 \Rightarrow b = \frac{4}{l} \\ Now, area of the base = lb =4 \\ Area of the 4 walls (A)=2 h (l + b) \\ ∴ A =2\times2 (l + \frac{4}{l}) \\ Differentiating w . r . t . x, we get \\ \frac{dA}{dl} = 4 (1 - \frac{4}{l^{2}}) \\ For maxima or minima, \\ \frac{dA}{dl} = 0 \Rightarrow 4 (1 - \frac{4}{l^{2}}) = 0 \\ \Rightarrow l^{2} = 4 \Rightarrow l =\pm2 \\ However, the length cannot be negative . \\ Therefore, we have l = 4 . \\ ∴ b = \frac{4}{l} = \frac{4}{2} = 2 \\ Now, \frac{d^{2} A}{{dl}^{2}} = \frac{32}{l^{3}} \\ When l =2, \frac{d^{2} A}{{dl}^{2}} = \frac{32}{2^{3}} = 4 >0 \\ Thus, by second derivative test, the area is the minimum \\ when l = 2 . \\ We have l = b = h = 2 . \\ ∴ Cost of building the base = ₹ 70\times(lb) \\ = ₹ 70 (4) \\ = ₹ 280 . \\ Cost of building the walls = ₹ 2 h (l + b)\times45 \\ = ₹ 90 (2) (2 +2) \\ = ₹ 720 \\ Required total cost = ₹ (280+720) \\ = ₹ 1000 \\ Hence, the total cost of the tank will be ₹ 1000 . \end{array}

Q.112 The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the
side of square is double the radius of the circle.

Ans.

$\begin{array}{l} Let r be the radius of the circle and a be the side of the \\ square .Then, we have: \\ 2 π r + 4 a = k \Rightarrow a = \frac{k - 2 π r}{4} \\ The sum of the areas of the circle and the square = A \\ A = π r^{2} + a^{2} \\ = π r^{2} + {(\frac{k - 2 π r}{4})}^{2} \\ = π r^{2} + \frac{{(k - 2 π r)}^{2}}{16} \\ ∴ \frac{d A}{d r} = \frac{d}{d r} π r^{2} + \frac{d}{d r} \frac{{(k - 2 π r)}^{2}}{16} \\ = 2 π r + \frac{2 (k - 2 π r)}{16} (0 - 2 π) \\ \frac{d A}{d r} = 2 π r - \frac{π (k - 2 π r)}{4} \\ F o r maximum or minimum, we have \\ \frac{d A}{d r} = 0 \Rightarrow 2 π r - \frac{π (k - 2 π r)}{4} = 0 \\ \Rightarrow 2 π r = \frac{π (k - 2 π r)}{4} \Rightarrow 8 π r = π (k - 2 π r) \\ \Rightarrow r = \frac{k}{2 (4 + π)} \\ N o w, \frac{d^{2} A}{d r^{2}} = 2 π + \frac{π^{2}}{2} > 0 \\ W h e n r = \frac{k}{2 (4 + π)}, \frac{d^{2} A}{d r^{2}} = 2 π + \frac{π^{2}}{2} > 0 \\ ∴ The sum of the areas is least when r = \frac{k}{2 (4 + π)} . \\ P u t t i n g value of r in a = \frac{k - 2 π r}{4}, w e get \\ a = \frac{k - 2 π (\frac{k}{2 (4 + π)})}{4} \\ = \frac{4 k + π k - π k}{4 (4 + π)} \\ = \frac{k}{(4 + π)} = 2 r \\ Hence, it has been proved that the sum of their areas is least \\ when the side of the square is double the radius of the circle . \end{array}$

Q.113 A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find
the dimensions of the window to admit maximum light through the whole opening.

Ans.

$\begin{array}{l} Let x and y be the length and breadth of the rectangular \\ window .Radius of the semicircular opening \frac{x}{2} . \end{array}$

$\begin{array}{l} It is given that the perimeter of the window is 10 m . \\ ∴ P e r i m e t e r of window = x + 2 y + \frac{π x}{2} \\ \Rightarrow x + 2 y + \frac{π x}{2} = 10 \\ \Rightarrow x (1 + \frac{π}{2}) + 2 y = 10 \\ \Rightarrow y = 5 - \frac{1}{2} x (1 + \frac{π}{2}) \\ N o w, \\ area (A) of window = x y + \frac{1}{2} π {(\frac{x}{2})}^{2} \\ \Rightarrow A = x y + \frac{1}{2} π {(\frac{x}{2})}^{2} \\ \Rightarrow A = x {5 - \frac{1}{2} x (1 + \frac{π}{2})} + \frac{1}{8} π x^{2} \\ = 5 x - \frac{1}{2} x^{2} (1 + \frac{π}{2}) + \frac{1}{8} π x^{2} \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ \Rightarrow \frac{d A}{d x} = 5 - x (1 + \frac{π}{2}) + \frac{1}{4} π x \\ a n d \frac{d^{2} A}{d x^{2}} = - (1 + \frac{π}{2}) + \frac{1}{4} π \\ = - 1 - \frac{1}{2} π \\ F o r maxima or minima, we have \\ \frac{d A}{d x} = 0 \\ \Rightarrow 5 - x (1 + \frac{π}{2}) + \frac{1}{4} π x = 0 \\ \Rightarrow x = \frac{20}{π + 4} \\ Thus, when x = \frac{20}{π + 4}, \frac{d^{2} A}{d x^{2}} < 0 \\ Therefore, by second derivative test, the area is the maximum \\ when length x = \frac{20}{π + 4} m . \\ S o, y = 5 - \frac{20}{π + 4} (\frac{2 + π}{4}) \\ = 5 - \frac{5 (2 + π)}{π + 4} \\ = \frac{5 π + 20 - 10 - 5 π}{π + 4} = \frac{10}{π + 4} m \\ Hence, the required dimensions of the window to admit maximum \\ light is given length = \frac{20}{π + 4} m and breadth = \frac{10}{π + 4} m . \end{array}$

Q.114 A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle.Show that the minimum length of the
hypotenuse is (a^2/3+ b^2/3) .

Ans.

$\begin{array}{l} Let Δ ABC be right-angled at B . Let AB = x and BC = y .Let P be \\ a point on the hypotenuse of the triangle such that P is at a \\ distance of a and b from the sides AB and BC respectively . \end{array}$

$\begin{array}{l} L e t ∠ C = θ \\ I n Δ A B C, A C = \sqrt{x^{2} + y^{2}} \\ N o w, P C = b c o s e c θ \\ A n d, A P = a s e c θ \\ \Rightarrow A C = A P + P C \\ \Rightarrow A C = b c o s e c θ + a s e c θ . . . (1) \\ D i f f e r e n t i a t i n g w . r . t . θ, w e g e t \\ \frac{d}{d θ} (A C) = - b cosec θ \cot θ + a \sec θ \tan θ \\ a n d \\ \frac{d^{2}}{d θ^{2}} (A C) = - b (- {cosec}^{3} θ - cosec θ \cot^{2} θ) + a (\sec^{3} θ + \sec θ \tan^{2} θ) \\ = b ({cosec}^{3} θ + cosec θ \cot^{2} θ) + a (\sec^{3} θ + \sec θ \tan^{2} θ) \end{array}$

$\begin{array}{l} F o r m a x i m a o r m i n i m a, w e h a v e \\ \frac{d}{d θ} (A C) = 0 \Rightarrow - b cosec θ \cot θ + a \sec θ \tan θ = 0 \\ \Rightarrow b cosec θ \cot θ = a \sec θ \tan θ \\ \Rightarrow \frac{b}{\sin θ} \times \frac{\cos θ}{\sin θ} = \frac{a}{\cos θ} \times \frac{\sin θ}{\cos θ} \\ \Rightarrow a \sin^{3} θ = b \cos^{3} θ \\ \Rightarrow \tan^{3} θ = \frac{b}{a} \Rightarrow \tan θ = {(\frac{b}{a})}^{\frac{1}{3}} \\ ∴ \sin θ = \frac{b^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}} a n d c o s θ = \frac{a^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}} . . . (i i) \\ S o, \frac{d^{2}}{d θ^{2}} (A C) > 0 f o r \tan θ = {(\frac{b}{a})}^{\frac{1}{3}} \\ T h e r e f o r e, b y s e c o n d d e r i v a t i v e t e s t, t h e l e n g t h o f t h e \\ h y p o t e n u s e i s t h e m i n i m u m w h e n \tan θ = {(\frac{b}{a})}^{\frac{1}{3}} . \\ S o, A C = b c o s e c θ + a s e c θ \\ = \frac{b}{\sin θ} + \frac{a}{\cos θ} \\ = \frac{b \sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}}{b^{\frac{1}{3}}} + \frac{a \sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}}{a^{\frac{1}{3}}} \end{array}$

$\begin{array}{l} = b^{\frac{2}{3}} \sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}} + a^{\frac{2}{3}} \sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}} \\ = \sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}} (b^{\frac{2}{3}} + a^{\frac{2}{3}}) \\ = {(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{\frac{3}{2}} \\ H e n c e, t h e m i n i m u m l e n g t h o f t h e h y p o t e n u s e i s {(a^{\frac{2}{3}} + b^{\frac{2}{3}})}^{\frac{3}{2}} . \end{array}$

Q.115

\begin{array}{l} Find the points at which the function f given by \\ f (x) = {(x-2)}^{4} {(x+1)}^{3} has (i) local maxima (ii) local minima \\ (iii) point of inflexion \end{array}

Ans.

$\begin{array}{l} T h e given function is \\ f (x) = {(x - 2)}^{4} {(x+1)}^{3} \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ f’ (x) = {(x - 2)}^{4} .3 {(x+1)}^{2} + 4 {(x - 2)}^{3} {(x+1)}^{3} \\ = {(x - 2)}^{3} {(x+1)}^{2} (3 x - 6 + 4 x + 4) \\ = {(x - 2)}^{3} {(x+1)}^{2} (7 x - 2) \\ N o w, f’ (x) = 0 \Rightarrow {(x - 2)}^{3} {(x+1)}^{2} (7 x - 2) = 0 \\ \Rightarrow x = 2, - 1 a n d \frac{2}{7} \\ Now, for values of x close to \frac{2}{7} and to the left of \frac{2}{7}, f ‘ (x) > 0. \\ Also, for values of x close to \frac{2}{7} and to the right of \frac{2}{7}, f ‘ (x) < 0. \\ Thus, x = \frac{7}{2} is the point of local maxima . \\ Now, for values of x close to 2 and to the left of 2, f ‘ (x) < 0. \\ Also, for values of x close to 2 and to the right of 2, f ‘ (x) > 0. \\ Thus, x = 2 is the point of local minima . \\ Now, as the value of x varies through - 1, f’ (x) does not changes \\ its sign . Thus, x = -1 is the point of inflexion . \end{array}$

Q.116

\begin{array}{l} Find the absolute maximum and minimum values of the \\ function f given by f (x) = \cos^{2} x + sinx, x \in [0, π] \end{array}

Ans.

$\begin{array}{l} T h e given function is \\ f (x) = \cos^{2} x + \sin x \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ f’ (x) = - 2 \cos x \sin x + \cos x \\ N o w, f ‘ (x) = 0 \\ \Rightarrow - 2 \cos x \sin x + \cos x = 0 \\ \Rightarrow \cos x (- 2 \sin x + 1) = 0 \\ \Rightarrow \cos x = 0 and - 2 \sin x + 1 = 0 \\ \Rightarrow x = \frac{π}{2} \in [0, π] \\ and sinx = \frac{1}{2} \Rightarrow x = \frac{π}{6} \in [0, π] \\ Now, evaluating the value of f at critical points x = \frac{π}{2} \\ and x = \frac{π}{6} and at the end points of the interval [0, π], w e have \\ f (\frac{π}{6}) = \cos^{2} \frac{π}{6} + \sin \frac{π}{6} = \frac{5}{4} \\ f (0) = \cos^{2} 0 + \sin 0 = 1 \\ f (π) = \cos^{2} π + \sin π = 1 \\ f (\frac{π}{2}) = \cos^{2} \frac{π}{2} + \sin \frac{π}{2} = 1 \\ Hence, the absolute maximum value of f is \frac{5}{4} occurring at x = \frac{π}{6} \\ a n d absolute minimum value of f is 1 occurring at x = 0, \frac{π}{2} a n d π . \end{array}$

Q.117

\begin{array}{l} Show that the altitude of the right circular cone of \\ maximum volume that can be inscribed in a sphere of \\ radius r is \frac{4 r}{3} . \end{array}

Ans.

$\begin{array}{l} A sphere of fixed radius (r) is given . Let R and h be the radius \end{array}$

and the height of the cone respectively.

$\begin{array}{l} The volume (V) of the cone = \frac{1}{3} π R^{2} h \\ Now, from the right triangle BCD, we have: \\ B C = \sqrt{r^{2} - R^{2}} \Rightarrow h = \sqrt{r^{2} - R^{2}} \\ A C = r + \sqrt{r^{2} - R^{2}} \\ ∴ V = \frac{1}{3} π R^{2} (r + \sqrt{r^{2} - R^{2}}) \\ = \frac{1}{3} π r R^{2} + \frac{1}{3} π R^{2} \sqrt{r^{2} - R^{2}} \\ D i f f e r e n t i a t i n g w .r .t . R, we get \\ \frac{d V}{d R} = \frac{2}{3} π r R + \frac{2}{3} π R \sqrt{r^{2} - R^{2}} + \frac{1}{3} π R^{2} \times \frac{1}{2 \sqrt{r^{2} - R^{2}}} \times (0 - 2 R) \\ \frac{d V}{d R} = \frac{2}{3} π r R + \frac{2}{3} π R \sqrt{r^{2} - R^{2}} - \frac{1}{3 \sqrt{r^{2} - R^{2}}} π R^{3} \\ = \frac{2}{3} π r R + \frac{2 π R (r^{2} - R^{2}) - π R^{3}}{3 \sqrt{r^{2} - R^{2}}} \\ = \frac{2}{3} π r R + \frac{2 π R r^{2} - 3 π R^{3}}{3 \sqrt{r^{2} - R^{2}}} \\ \frac{d^{2} V}{d R^{2}} = \frac{d}{d R} (\frac{2}{3} π r R) + \frac{d}{d R} (\frac{2 π R r^{2} - 3 π R^{3}}{3 \sqrt{r^{2} - R^{2}}}) \\ = \frac{2}{3} π r \\ + \frac{3 \sqrt{r^{2} - R^{2}} \frac{d}{d R} (2 π R r^{2} - 3 π R^{3}) - (2 π R r^{2} - 3 π R^{3}) \frac{d}{d R} 3 \sqrt{r^{2} - R^{2}}}{{(3 \sqrt{r^{2} - R^{2}})}^{2}} \\ = \frac{2}{3} π r + \frac{3 \sqrt{r^{2} - R^{2}} (2 π r^{2} - 9 π R^{2}) - (2 π R r^{2} - 3 π R^{3}) \frac{3 (0 - 2 R)}{2 \sqrt{r^{2} - R^{2}}}}{9 (r^{2} - R^{2})} \\ = \frac{2}{3} π r + \frac{6 (r^{2} - R^{2}) (2 π r^{2} - 9 π R^{2}) + 6 R (2 π R r^{2} - 3 π R^{3})}{18 {(r^{2} - R^{2})}^{\frac{3}{2}}} \\ F o r maxima or minima, we have \\ \frac{d V}{d R} = 0 \Rightarrow \frac{2}{3} π r R + \frac{2 π R r^{2} - 3 π R^{3}}{3 \sqrt{r^{2} - R^{2}}} = 0 \\ \Rightarrow \frac{2}{3} π r R = \frac{3 π R^{3} - 2 π R r^{2}}{3 \sqrt{r^{2} - R^{2}}} \\ \Rightarrow 2 r \sqrt{r^{2} - R^{2}} = 3 R^{2} - 2 r^{2} \\ S q u a r r i n g both sides, we get \\ 4 r^{2} (r^{2} - R^{2}) = {(3 R^{2} - 2 r^{2})}^{2} \\ \Rightarrow R^{2} = \frac{8 r^{2}}{9} \\ W h e n R^{2} = \frac{8 r^{2}}{9}, \frac{d^{2} V}{d R^{2}} < 0 \\ ∴ The volume is the maximum when R^{2} = \frac{8 r^{2}}{9} . \\ W h e n R^{2} = \frac{8 r^{2}}{9}, \\ the height of the cone \\ = r + \sqrt{r^{2} - \frac{8 r^{2}}{9}} \\ = r + \frac{r}{3} = \frac{4 r}{3} \\ Hence, it can be seen that the altitude of the right circular \\ cone of maximum volume that can be inscribed in a sphere \\ of radius r is \frac{4 r}{3} . \end{array}$

Q.118

\begin{array}{l} Let f be a function defined on [a, b] such that \\ f ’ (x) > 0, \forall \in (a, b) . \\ Then prove that f is an increasing function on (a, b) . \end{array}

Ans.

\begin{array}{l} Let (x_{1}, x_{2}) \in (a, b) such that x_{1} < x_{2} . Consider the sub - interval \\ [x_{1}, x_{2}] . Since f (x) is differentiable on (a, b) and [x_{1}, x_{2}] \subset (a, b) . \\ Therefore, f (x) is continuous on [x_{1}, x_{2}] and differentiable at \\ (x_{1}, x_{2}) . By the Lagrange ‘ s mean value theorem, there exists \\ c \in (x_{1}, x_{2}) such that \\ f ‘ (c) = \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}} . .. (i) \\ Since, f ‘ (x) > 0 for all x \in (a, b), so in particular, f ‘ (c) > 0 \\ Now, f ‘ (c) > 0 \\ \Rightarrow \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}} > 0 \\ \Rightarrow f (x_{2}) - f (x_{1}) > 0 \\ \Rightarrow f (x_{2}) > f (x_{1}) \Rightarrow f (x_{1}) < f (x_{2}) \\ Since, x_{1}, x_{2} are arbitrary points in (a, b) . Therefore, \\ x_{1} < x_{2} \Rightarrow f (x_{1}) < f (x_{2}) ​ for all x_{1}, x_{2} \in (a, b) \\ Hence, f (x) is increasing on (a, b) . \end{array}

Q.119

\begin{array}{l} Show that the height of the cylinder of maximum volume \\ that can be inscribed in a sphere of radius R is \frac{2R}{\sqrt{3}} . Also find \\ the maximum volume . \end{array}

Ans.

\begin{array}{l} A sphere of fixed radius (R) is given . Let r and h be the radius \\ and the height of the cylinder respectively . \end{array}

$\begin{array}{l} From the given figure, we have h = 2 \sqrt{R^{2} - r^{2}} \\ The volume (V) of the cylinder = {πr}^{2} h \\ V = 2 {πr}^{2} \sqrt{R^{2} - r^{2}} \\ Differentiating w . r . t . r, we get \\ \frac{dV}{dr} = 2 π \frac{d}{dr} (r^{2} \sqrt{R^{2} - r^{2}}) \\ = 2 π (r^{2} \frac{d}{dr} \sqrt{R^{2} - r^{2}} + \sqrt{R^{2} - r^{2}} \frac{d}{dr} r^{2}) \\ = 2 π (r^{2} \frac{(0 - 2 r)}{2 \sqrt{R^{2} - r^{2}}} + 2 r \sqrt{R^{2} - r^{2}}) \\ = 2 π (\frac{2 r (R^{2} - r^{2}) - r^{3}}{\sqrt{(R^{2} - r^{2})}}) = \frac{4 {πrR}^{2} - 6 {πr}^{3}}{\sqrt{R^{2} - r^{2}}} \\ Now, \\ \frac{d^{2} V}{{dr}^{2}} = \frac{d}{dr} \frac{4 {πrR}^{2} - 6 {πr}^{3}}{\sqrt{R^{2} - r^{2}}} \\ = \frac{\sqrt{R^{2} - r^{2}} \frac{d}{dr} (4 {πrR}^{2} - 6 {πr}^{3}) - (4 {πrR}^{2} - 6 {πr}^{3}) \frac{d}{dr} \sqrt{R^{2} - r^{2}}}{{(\sqrt{R^{2} - r^{2}})}^{2}} \\ = \frac{\sqrt{R^{2} - r^{2}} (4 {πR}^{2} - 18 {πr}^{2}) - (4 {πrR}^{2} - 6 {πr}^{3}) \frac{- 2 r}{2 \sqrt{R^{2} - r^{2}}}}{R^{2} - r^{2}} \\ = \frac{(R^{2} - r^{2}) (4 {πR}^{2} - 18 {πr}^{2}) + r (4 {πrR}^{2} - 6 {πr}^{3})}{{(R^{2} - r^{2})}^{\frac{3}{2}}} \\ For maxima or minima, we have \\ \frac{dV}{dr} = 0 \Rightarrow \frac{4 {πrR}^{2} - 6 {πr}^{3}}{\sqrt{R^{2} - r^{2}}} = 0 \\ \Rightarrow 4 {πrR}^{2} - 6 {πr}^{3} = 0 \Rightarrow r^{2} = \frac{2 R^{2}}{3} \\ When r^{2} = \frac{2 R^{2}}{3}, \frac{d^{2} V}{{dr}^{2}} < 0 \\ ∴ The volume is the maximum when r^{2} = \frac{2 R^{2}}{3} . \end{array}$

$\begin{array}{l} When r^{2} = \frac{2 R^{2}}{3}, \\ the height of the cylinder = 2 \sqrt{R^{2} - \frac{2 R^{2}}{3}} \\ = \frac{2 R}{\sqrt{3}} \\ Hence, the volume of the cylinder is the maximum when \\ the height of the cylinder is \frac{2 R}{\sqrt{3}} . \\ Maximum Volume = π [R^{2} \frac{2 R}{\sqrt{3}} - \frac{1}{4} . \frac{4 R^{2}}{3} . \frac{2 R}{\sqrt{3}}] \\ = {πR}^{2} \frac{2 R}{\sqrt{3}} [1 - \frac{1}{3}] \\ = \frac{4 {πR}^{3}}{3 \sqrt{3}} \end{array}$

Q.120

$\begin{array}{l} Show that height of the cylinder of greatest volume which \\ can be inscribed in a right circular cone of height h and semi \\ vertical angle θ is one - third that of the cone and the \\ greatest volume of cylinder is \frac{4}{27} {πh}^{3} \tan^{2} α . \end{array}$

Ans.

\begin{array}{l} In given right circular cone of fixed height is h and \\ semi - vertical angle be α . \\ Then a cylinder of radius R and height H is inscribed in \\ the cone . Then ∠ GAO = α, OG = r, OA = h, OE = R ​ and CE = H . \\ We have, r = htanα \\ Now, since ΔAOG is similar to ΔCEG, we have : \\ \frac{AO}{OG} = \frac{CE}{EG} \Rightarrow \frac{h}{r} = \frac{H}{r - R} [EG = OG - OE \end{array}

$\begin{array}{l} \Rightarrow H = \frac{h}{r} (r - R) \\ = \frac{h}{htan α} (htan α - R) \\ = \frac{1}{tan α} (htan α - R) \\ Now, the volume (V) of the cylinder \\ V = π R^{2} h \\ = \frac{π R^{2}}{tan α} (htan α - R) \\ = π R^{2} h - \frac{π R^{3}}{tan α} \\ D i f f e r e n t i a t i n g w .r .t . R, we get \\ \frac{d V}{d R} = 2 π R h - \frac{3 π R^{2}}{tan α} \\ a n d \frac{d^{2} V}{d R^{2}} = 2 π h - \frac{6 π R}{tan α} \\ F o r maxima or minima, we have \\ \frac{d V}{d r} = 0 \Rightarrow 2 π R h - \frac{3 π R^{2}}{tan α} = 0 \\ \Rightarrow π R (2 h - \frac{3 R}{tan α}) = 0 \Rightarrow R = \frac{2 h \tan α}{3} \\ S o, \frac{d^{2} V}{d R^{2}} < 0 f o r R = \frac{2 h \tan α}{3} \\ ∴ By second derivative test, the volume of the cylinder is \\ the greatest when R = \frac{2 h \tan α}{3} . \\ T h e n, H = \frac{1}{tan α} (htan α - R) \\ = \frac{1}{tan α} (htan α - \frac{2 h \tan α}{3}) \\ = \frac{h}{3} \\ Thus, the height of the cylinder is one-third the height of the \\ cone when the volume of the cylinder is the greatest . \\ Now, the maximum volume of the cylinder \\ V = π R^{2} H \\ = π {(\frac{2 h \tan α}{3})}^{2} (\frac{h}{3}) \\ ∴ V = \frac{4}{27} π h^{3} \tan^{2} α \end{array}$

Q.121 A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic mere per hour. Then the depth of the wheat is increasing at the rate of
(A) 1 m³/h (B) 0.1 m³/h
(C) 1.1 m³/h (D) 0.5 m³/h

Ans.

\begin{array}{l} Let r be the radius of the cylinder . \\ Then, volume (V) ofthe cylinder = π r^{2} h \\ Putting r = 10 m, we get \\ V = π {(10)}^{2} h \\ = 100 πh \\ Differentiating with respect to time t, we get : \\ \frac{dV}{dt} = \frac{d}{dt} 100 πh \\ = 100 π \frac{dh}{dt} \\ The tank is being filled with wheat at the rate of 314 cubic \\ metres per hour . \\ ∴ \frac{dV}{dt} = 314 m^{3} / h \\ Thus, wehave : \\ 314 = 100 π \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{314}{100 \times 3 .14} = 1 \\ Hence, the depth of wheat is increasing at the rate of 1 m / h . \\ The correct answer is A . \end{array}

Q.122 The slope of the tangent to the curve x = t² + 3t – 8, y = 2t² – 2t – 5 at the point (2, -1) is
(A) 22/7 (B)6/7 (C) 7/6 (D) – 6/7

Ans.

\begin{array}{l} The given curve is x = t^{2} + 3 t - 8 and y = 2 t^{2} - 2 t - 5 \\ Differentiating w . r . t . x, we get \\ \frac{dx}{dt} = 2 t + 3 and \frac{dy}{dt} = 4 t - 2 \\ ∴ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4 t - 2}{2 t + 3} \\ Given is point (2, - 1) . \\ When x = 2, \\ 2 = t^{2} + 3 t - 8 \Rightarrow t^{2} + 3 t - 10 = 0 \\ \Rightarrow (t + 5) (t - 2) = 0 \Rightarrow t = - 5, 2 \\ At y = - 1, \\ - 1 = 2 t^{2} - 2 t - 5 \Rightarrow 2 t^{2} - 2 t - 4 = 0 \\ \Rightarrow t^{2} - t - 2 = 0 \Rightarrow (t - 2) (t + 1) = 0 \\ \Rightarrow t = 2, - 1 \\ The common value of t is 2, so \\ slope = {(\frac{dy}{dx})}_{t = 2} = \frac{4 (2) - 2}{2 (2) + 3} \\ = \frac{8 - 2}{4 + 3} = \frac{6}{7} \\ Thus, the correct answer is B . \end{array}

Q.123 The line y = mx + 1 is a tangent to the curve y² = 4x if the value of m is
(A) 1 (B) 2
(C) 3 (D) 1/2

Ans.

\begin{array}{l} The equation of the tangent to the given curve is y = mx + 1 . \\ Substituting y = mx + 1 in y^{2} = 4 x, we get : \\ {(mx + 1)}^{2} = 4 x \Rightarrow m^{2} x^{2} + 2 mx + 1 = 4 x \\ \Rightarrow m^{2} x^{2} + 2 (m - 2) x + 1 = 0 . .. (i) \\ Since a tangent touches the curve at one point, the roots of \\ equation (i) must be equal . Therefore, we have : \\ D = 0 \Rightarrow b^{2} - 4 ac = 0 \\ \Rightarrow {2 (m - 2)}^{2} - 4 (m^{2}) 1 = 0 \\ \Rightarrow 4 {(m - 2)}^{2} = 4 m^{2} \\ \Rightarrow m^{2} - 4 m + 4 = m^{2} \Rightarrow m = 1 \\ Hence, the required value of m is 1 . \\ The correct answer is A . \end{array}

Q.124 The normal at the point (1, 1) on the curve 2y + x² = 3 is
(A) x + y = 0 (B) x – y = 0
(C) x + y + 1= 0 (D) x – y = 0

Ans.

$\begin{array}{l} {The equation of the given curve is 2y + x}^{2} = 3 . \\ Differentiating with respect to x, we have: \\ 2 \frac{d y}{d x} + 2 x = 0 \\ \frac{d y}{d x} = - x \\ ∴ {(\frac{d y}{d x})}_{(1, 1)} = - 1 \\ The slope of the normal to the given curve at point (1, 1) \\ = - \frac{1}{{(\frac{d y}{d x})}_{(1, 1)}} \\ = - \frac{1}{- 1} = 1 \\ Hence, the equation of the normal to the given curve \\ at (1, 1) is \\ y - 1 = 1 (x - 1) \Rightarrow y - 1 = x - 1 \\ \Rightarrow x - y = 0 \\ T h u s, the correct answer is B . \end{array}$

Q.125 The normal to the curve x² = 4y passing (1, 2) is

(A) x + y = 3 (B) x – y = 3
(C) x + y = 1 (D) x – y = 1

Ans.

\begin{array}{l} The equation of the given curve is x^{2} = 4 y . \\ Differentiating with respect to x, we have : \\ 2 x = 4 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x}{2} \\ The slope of the normal to the given curve at point (h, k) is : - \\ (\frac{dy}{dx}) (h, k) = - \frac{2}{h} \\ ∴ Equation of the normal at point (h, k) is : - \\ y - k = - \frac{2}{h} (x - h) \\ Now, it is given that the normal passes through the point (1, 2) . \\ Therefore, we have : \\ 2 - k = - \frac{2}{h} (1 - h) \Rightarrow k = 2 + \frac{2}{h} (1 - h) . .. (i) \\ Since (h, k) lies on the curve x^{2} = 4 y, we have h^{2} = 4 k . \\ \Rightarrow k = \frac{h^{2}}{4} \\ From equation (i), we have : \\ \frac{h^{2}}{4} = 2 + \frac{2}{h} (1 - h) \Rightarrow h^{3} = 8 h + 8 - 8 h \\ \Rightarrow h = 2 \\ ∴ k = \frac{2^{2}}{4} = 1 \\ Hence, the equation of the normal is given as : \\ y - 1 = - \frac{2}{2} (x - 2) \\ \Rightarrow y - 1 = - x + 2 \\ \Rightarrow x + y = 3 \\ Thus, the correct answer is A . \end{array}

Q.126

\begin{array}{l} The points on the curve 9 y^{2} = x^{3}, where the normal \\ to the curve makes equal intercepts with the axes are \\ (A) (4,\pm \frac{8}{3}) (B) (4, \frac{- 8}{3}) (C) (4,\pm \frac{3}{8}) (D) (4,\pm \frac{3}{8}) \end{array}

Ans.

\begin{array}{l} The equation of the given curve is 9 y^{2} = x^{3} . \\ Differentiating with respect to x, we have : \\ 9 (2 y) \frac{dy}{dx} = 3 x^{2} \Rightarrow \frac{dy}{dx} = \frac{3 x^{2}}{18 y} = \frac{x^{2}}{6 y} \\ The slope of the tangent to the given curve at point (h, k) is \\ {(\frac{dy}{dx})}_{(h, k)} = \frac{h^{2}}{6 k} \\ The slope of the normal to the given curve at point (h, k) is \\ - \frac{1}{{(\frac{dy}{dx})}_{(h, k)}} = - \frac{6 k}{h^{2}} \\ The equation of the normal to the curve at (h, k) is \\ y - k = - \frac{6 k}{h^{2}} (x - h) \\ \Rightarrow h^{2} y - h^{2} k = - 6 kx + ykh \\ \Rightarrow h^{2} y + 6 kx = h^{2} k + ykh \\ \Rightarrow \frac{x}{\frac{h^{2} k + ykh}{6 k}} + \frac{y}{\frac{h^{2} k + ykh}{h^{2}}} = 1 \\ Since, the normal makes equal intercepts with the axes . \\ Therefore, We have : \\ \frac{h^{2} k + ykh}{6 k} = \frac{h^{2} k + ykh}{h^{2}} \\ \Rightarrow \frac{h (h + y)}{6} = \frac{(h + y) k}{h} \\ \Rightarrow \frac{h}{6} = \frac{k}{h} \\ \Rightarrow h^{2} = 6 k . .. (ii) \\ Also, the point (h, k) lies on the curve 9 y^{2} = x^{3}, so we have \\ 9 k^{2} = h^{3} . .. (ii) \\ From equation (i) and (ii), we have \\ 9 {(\frac{h^{2}}{6})}^{2} = h^{3} \Rightarrow \frac{h^{4}}{4} = h^{3} \\ \Rightarrow h = 4 \\ From equation (ii), ​ we have \\ 9 k^{2} = 4^{3} \Rightarrow k^{2} = \frac{64}{9} \Rightarrow k = \pm \frac{8}{3} \\ Hence, the required points are (4, \pm \frac{8}{3}) . \\ Thus, the correct answer is A . \end{array}

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Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

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Chapter 9 - Differential Equations

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Chapter 3 - Matrices

Chapter 4 - Determinants

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Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

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Chapter 1 - Relations and Functions

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Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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