NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.4) Exercise 6.4

The Central Board of Secondary Education, or CBSE, is a National Board of Secondary Education for public and private schools in New Delhi. The CBSE Board is supervised and controlled by the Government of India. In 1921, it was created as the Uttar Pradesh Board of High School and Intermediate Education, with jurisdiction over Rajputana, Central India, and Gwalior. The government of India formed the Board of High School and Intermediate Education in 1929, which covered the schools of Ajmer, Merwara, Central India, and Gwalior. Later, it was limited to the cities of Ajmer, Bhopal, and Vindhya Pradesh. It was renamed the Central Board of Secondary Education in 1952. The promotion criteria for Class 12 students are 33% overall, along with 33% in both theory and practical examinations. If students do not score well in one subject, they may write the compartment for that subject in July.

The CBSE Board oversees NCERT. The National Council of Educational Research and Training (NCERT), established by the Government of India in 1961, provides support and advice to the sentral and state governments. The organization creates policies and programmes to improve the quality of education in schools. NCERT’s primary mission is to conduct, promote, and coordinate research in the field of school education. Furthermore, NCERT works with international organizations, visits foreign delegations, and trains educational staff from developing nations. NCERT is in charge of creating and releasing model textbooks, supplemental materials, newsletters, journals, educational kits, and multimedia digital products.

The basis of human understanding of the world is mathematics; thus, it is also an essential component of human intellect and logic. In addition to developing mental discipline, mathematics encourages logical reasoning and cognitive clarity. Furthermore, mathematical knowledge is essential for comprehending topics while preparing for the Class 12 board examinations. Class 12 mathematics often requires the use of mathematical concepts, and the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 helps students practice. Mathematical concepts and terminologies are essential to pursuing careers in science, engineering, commerce, astronomy, economics, and other such fields and disciplines.

In addition to mathematics, derivatives are used in a variety of other fields. There are many applications of them in Physics, for example, determining velocity, acceleration, etc. Derivatives are used to determine the rate at which one quantity changes with respect to another. It is widely used in chemistry to determine the rate at which chemical reactions take place. It is also possible to use derivatives to solve the equation of the tangent to a curve. As a result, it is imperative that students thoroughly understand the fundamentals in this chapter to be able to employ them in several other subjects. An introduction to derivatives is provided in this chapter, which focuses on function differentiability. Several derivatives of certain functions as well as their applications are discussed in this chapter.

By solving each sum provided in the six exercises in this chapter, the NCERT Solutions for Mathematics Chapter 6 allows students to study the entire chapter at a steady pace. As a result of the concepts discussed in the NCERT solutions, students will be able to understand the real-world applications of derivatives. Additionally, it can be used to prepare for a variety of competitive examinations.

Extramarks provides a variety of study materials to assist students in understanding the concepts presented in this chapter. The study materials, including the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4, provide a comprehensive review of all the key concepts outlined in the CBSE-recommended syllabus for the board examinations. Many qualified teachers designed the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 to help students prepare for the board examinations that are going to be imposed by the CBSE. The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 can also be used to prepare students for a variety of college entrance exams, such as the JEE and VITEEE. Practising the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 at least twice is recommended for students to fully comprehend the questions.

The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 are prepared to assist students during their preparation for the Class 12 board examinations in the most effective way possible. The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 are discussed in detail, using examples that are relevant and easy to understand. Furthermore, practice questions are supplied at the end of each topic to assist students in performing better in Chapter 6. It is strongly recommended that students keep these NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 ready during exam preparation to utilise them as a quick and effective study tool. The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 are intended to aid students by providing them with solutions to their problems so that they can focus on evaluating their conceptual understanding.

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NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.4) Exercise 6.4

The students who wish to excel in the Class 12 board exams are recommended to download the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 available in PDF format on the Extramarks website. It might not be possible for students to access the internet at all times; hence, the students can use the downloaded PDF of the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 anywhere at their convenience, offline as well.

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Access NCERT Solutions For Class 12 Chapter 6 – Applications of Derivative

Benefits of NCERT Solutions for Class 12 Maths Chapter 6

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The use of Derivatives is discussed in Chapter 6 of the NCERT Mathematics textbook. For students to understand the principles discussed in this chapter, they should complete the preceding chapter on Continuity and Differentiability. Using NCERT Solutions for Mathematics Chapter 5, students can gain a comprehensive understanding of the numerous characteristics of Limits and Continuity of Functions. In reading through each question’s complete solution, students gain a better understanding of Continuity and Differentiability. It is the purpose of this chapter to put the concepts presented in the previous chapter into practice. Exercise 6.4 in this chapter is based on approximation. To answer the questions, students must refer to the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 to verify the solutions. Besides, the students must develop a solid understanding of the principles presented in this exercise. It is usually beneficial to solve exercises from the NCERT book since these problems frequently appear in board exams. It is important to note, however, that some questions in this exercise may be difficult for students to answer. The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 are therefore made available by Extramarks.

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The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 is prepared in accordance with the revised guidelines and syllabus released by the CBSE Board for Class 12 Mathematics. Apart from getting to access the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4 for Class 12 Mathematics Chapter 6 Application of Derivatives, the students can additionally go through the entire Exercise 6.4 and practice the questions for understanding the concepts in the chapter.

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Q.1 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 ) 25 .3 ( i i ) 49 .5 ( i i i ) 0 .6 ( i v ) ( 0 .009 ) 1 3 ( v ) ( 0 .999 ) 1 10 ( v i ) ( 15 ) 1 4 ( v i i ) ( 26 ) 1 3 ( v i i i ) ( 255 ) 1 4 ( i x ) ( 82 ) 1 4 ( x ) ( 401 ) 1 2 ( x i ) ( 0 .0037 ) 1 2 ( x i i ) ( 26 .57 ) 1 3 ( x i i i ) ( 81 .5 ) 1 4 ( x i v ) ( 3 .968 ) 3 2 ( x v ) ( 32 .15 ) 1 5

Ans

(i)25.3Let y=x,  x=25 and Δx=0.3⇒dydx=12xThen,       Δy=x+Δx−x     =25.3−25     =25.3−525.3=Δy+5Now, dy is approximately equal to Δy and is given by,         dy=(dydx)Δx     =12x(0.3)     =1225(0.3)     =12×5×0.3     =0.03Hence, the approximate value of 25.3 is 5+0.03=5.03.(ii)49.5Let y=x,  x=49 and Δx=0.5⇒dydx=12xThen,       Δy=x+Δx−x     =49.5−49     =49.5−749.5=Δy+7Now, dy is approximately equal to Δy and is given by,         dy=(dydx)Δx     =12x(0.5)     =1249(0.5)     =12×7×0.5     =0.035Hence, the approximate value of 49.5 is 7+ 0.035=7.035.(iii)0.6Let y=x,  x=1 and Δx=− 0.4⇒dydx=12xThen,       Δy=x+Δx−x     =0.6−1   0.6=Δy+1Now, dy is approximately equal to Δy and is given by,         dy=(dydx)Δx     =12x(− 0.4)     =121(− 0.4)     =− 0.2Hence, the approximate value of 0.6 is 1− 0.2=0.8(iv) (0.009)13Let y=x13,  x=0.008 and Δx=0.001⇒dydx=13x23Then,            Δy=(x+Δx)13−x13          =(0.009)13−0.2(0.009)13=Δy+0.2Now, dy is approximately equal to Δy and is given by,⇒       dydx=13x23(Δx)       =13(0.008)23(0.001)       =13(0.04)(0.001)=0.0010.12       =0.008Hence, the approximate value of (0.009)23 is 0.2 + 0.008=0.208

( v ) ( 0.999 ) 1 10 Let​ y= x 1 10 , x=1 and Δx=−0.001 ⇒ dy dx = 1 10 x 9 10 Then,              Δy= ( x+Δx ) 1 10 − ( x ) 1 10            = ( 0.999 ) 1 10 − ( 1 ) 1 10 ( 0.999 ) 1 10 =Δy+1 Now, dy is approximately equal to Δy and is given by,          dy=( dy dx )Δx     = 1 10 x 9 10 ×−0.001     = 1 10 ( 1 ) 9 10 ×−0.001     =− 0.001 10     =−0.0001 Hence, the approximate value of  ( 0.999 ) 1 10  is 1+( −0.001 )=0.9999 ( vi ) ( 15 ) 1 4 Let y= x 1 4 ,  x=16 and Δx=−1 ⇒ dy dx = 1 4 x 3 4 Then,             Δy= ( x+Δx ) 1 4 − x 1 4 = ( 15 ) 1 4 − ( 16 ) 1 4         ( 15 ) 1 4 =Δy+2 Now, dy is approximately equal to Δy and is given by,          dy=( dy dx )Δx     = 1 4 x 3 4 ×−1     = 1 4 ( 16 ) 3 4 ×−1     = 1 4×8 ×−1=−0.03125 Hence, the approximate value of  ( 15 ) 1 4  is  2+( −0.03125 )=1.96875 ( vii ) ( 26 ) 1 3 Let y= x 1 3 ,  x=27 and Δx=−1 ⇒ dy dx = 1 3 x 2 3 Then,             Δy= ( x+Δx ) 1 3 − x 1 3 = ( 26 ) 1 3 − ( 27 ) 1 3         ( 26 ) 1 3 =Δy+3 Now, dy is approximately equal to Δy and is given by,          dy=( dy dx )Δx      = 1 3 x 2 3 ×−1      = 1 3 ( 27 ) 2 3 ×−1      = 1 3×9 ×−1      =−0.0 370 ¯ Hence, the approximate value of  ( 26 ) 1 3  is  3+( −0.0370 )=2.9629 ( viii ) ( 255 ) 1 4 Let y= x 1 4 ,  x=256 and Δx=−1 ⇒ dy dx = 1 4 x 3 4 Then,             Δy= ( x+Δx ) 1 4 − x 1 4 = ( 255 ) 1 4 − ( 256 ) 1 4       ( 255 ) 1 4 =Δy+4 Now, dy is approximately equal to Δy and is given by,          dy=( dy dx )Δx      = 1 4 x 3 4 ×( −1 )      = 1 4 ( 256 ) 3 4 ×( −1 )      = −1 4× 4 3 =−0.0039 Hence, the approximate value of  ( 255 ) 1 4  is 4+(−0.0039)=3.9961 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( ix )  ( 82 ) 1 4 Let y= x 1 4 ,  x=81 and Δx=1 ⇒ dy dx = 1 4 x 3 4 Then,             Δy= ( x+Δx ) 1 4 − x 1 4 = ( 82 ) 1 4 − ( 81 ) 1 4         ( 82 ) 1 4 =Δy+3 Now, dy is approximately equal to Δy and is given by,          dy=( dy dx )Δx      = 1 4 x 3 4 ×( 1 )      = 1 4 ( 81 ) 3 4 ×( 1 )      = 1 4× 3 3 =0.009 Hence, the approximate value of  ( 82 ) 1 4  is 3+(0.009)=3.009 ( x )   ( 401 ) 1 2 Let y= x 1 2 ,  x=400 and Δx=1 ⇒ dy dx = 1 2 x Then,        Δy= x+Δx − x      = 401 − 400      = 401 −20    401 =Δy+20 Now, dy is approximately equal to Δy and is given by,          dy=( dy dx )Δx      = 1 2 x ( 1 )      = 1 2 400 ( 1 )      = 1 2×20 ×( 1 )      =0.025 Hence, the approximate value of  401  is 20+0.025=20.025. ( xi )   ( 0.0037 ) 1 2 Let y= x 1 2 ,  x=0.0036 and Δx=0.0001 ⇒ dy dx = 1 2 x Then,               Δy= x+Δx − x            = 0.0037 − 0.0036            = 0.0037 −0.06   0.0037 =Δy+0.06 Now, dy is approximately equal to Δy and is given by,          dy=( dy dx )Δx      = 1 2 x ( 0.0001 )      = 1 2 0.0036 ( 0.0001 )      = 1 2×0.06 ×( 0.0001 )      =0.00083 Hence, the approximate value of  0.0037  is 0.06+0.00083=0.06083 ( xii )  ( 26.57 ) 1 3 Let y= x 1 3 ,  x=27 and Δx=−0.43 ⇒ dy dx = 1 3 x 2 3 Then,             Δy= ( x+Δx ) 1 3 − x 1 3 = ( 26.57 ) 1 3 − ( 27 ) 1 3   ( 26.57 ) 1 3 =Δy+3 Now, dy is approximately equal to Δy and is given by,          dy=( dy dx )Δx      = 1 3 x 2 3 ×−0.43      = 1 3 ( 27 ) 2 3 ×−0.43      = 1 3( 9 ) ×−0.43=−0.015 Hence, the approximate value of  ( 26.57 ) 1 3  is 3−0.015=2.984 ( xiii )  ( 81.5 ) 1 4 Let y= x 1 4 ,  x=81 and Δx=0.5 ⇒ dy dx = 1 4 x 3 4 Then,             Δy= ( x+Δx ) 1 4 − x 1 4 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= ( 81.5 ) 1 4 − ( 81 ) 1 4     ( 81.5 ) 1 4 =Δy+3 Now, dy is approximately equal to Δy and is given by,          dy=( dy dx )Δx      = 1 4 x 3 4 ×( 0.5 )      = 1 4 ( 81 ) 3 4 ×( 0.5 )      = 0.5 4× 3 3 =0.0046 Hence, the approximate value of  ( 81.5 ) 1 4  is 3+0.0046=3.0046 ( xiv )  ( 3.968 ) 3 2 Let y= x 3 2 ,  x=4 and Δx=−0.032 ⇒ dy dx = 3 2 x 1 2 Then,             Δy= ( x+Δx ) 3 2 − x 3 2 = ( 3.968 ) 3 2 − ( 4 ) 3 2    ( 3.968 ) 3 2 =Δy+8 Now, dy is approximately equal to Δy and is given by,          dy=( dy dx )Δx      = 3 2 x 1 2 ×( −0.032 )      = 3 2 ( 4 ) 1 2 ×( −0.032 )      = 3 2 ×2×( −0.032 )=−0.096 Hence, the approximate value of  ( 3.968 ) 3 2  is 8−0.096=7.904 ( xv )  ( 32.15 ) 1 5 Let y= x 1 5 ,  x=32 and Δx=0.15 ⇒ dy dx = 1 5 x 4 5 Then,             Δy= ( x+Δx ) 1 5 − x 1 5 = ( 32.15 ) 1 5 − ( 32 ) 1 5     ( 32.15 ) 1 4 =Δy+2 Now, dy is approximately equal to Δy and is given by,          dy=( dy dx )Δx      = 1 5 x 4 5 ×( 0.15 )      = 1 5 ( 32 ) 4 5 ×( 0.15 )      = 0.15 5× 2 4 =0.00187 Hence, the approximate value of  ( 32.15 ) 1 4  is 2+0.00187=2.00187 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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Q.2 Find the approximate value of f 2.01 , where f x =4x 2 +5x+2

Ans

Let x=2 and Δx=0.01. Then, we have:f(2.01)=f(x + Δx)=4(x + Δx)2+ 5(x + Δx)+ 2Since,   Δy=f(x+Δx)−f(x).​ Therefore,f(x+Δx)=f(x)+Δy≈f(x)+f‘(x)Δx[∵dx=Δx]    f(2.01)≈(4x2+5x+2)+(8x+5)Δx≈{4(2)2+5(2)+2}+(8×2+5)(0.01)≈28+21×0.01=28+0.21≈28.21Hence, the approximate value of f (2.01) is 28.21.

Q.3 Find the approximate value of f (5.001), where f (x) = x3 – 7x2 + 15.

Ans

Letx=5 and Δx=0.001.Then, wehave:f(5.001)=f(x+Δx)=(x+Δx)3−7(x+Δx)2+15Since,Δy =f(x+Δx)−f(x).Therefore,f(x+Δx)=f(x)+Δy≈f(x)+f‘(x)Δx[∵dx=Δx]f(5.001)≈(x3−7x2+15)+(3x2−14x)Δx≈{(5)3−7(5)2+15}+{3(5)2−14(5)}(0.01)≈−35+5×(0.01)=−35+0.05≈−34.995Hence, the approximate value of f(5.001) is −34.995.

Q.4 Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%.

Ans

Side of cube is x metres.Volume of a cube is given byV=x3dV=(dVdx)Δx=(ddxx3)Δx=(3x2)Δx=(3x2)(0.01x)[1%ofx=0.01x]=0.03x3Hence,the approximate change in the volume of thecube is 0.03x3m3.

Q.5 Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.

Ans

Let side of cube be x metres.Then,Surface area of cube=6x2⇒S=6x2   dSdx=ddx(6x2)=12x   dS=(dSdx)Δx=(12x)(0.01x)[∵1% of x=0.01x]=0.12x2Hence, the approximate change in the surface area of the cube is 0.12x2 m2.

Q.6 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

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 mNow, the volume V of the sphere is given by,   V=43πr3dVdr=ddr(43πr3)        =4πr2∴        dV=(dVdr)Δr        =(4πr2)Δr        =4π(7)2×0.02=3.92π m3Hence, the approximate error in calculating the volume is 3.92π m3.

Q.7 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

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 mNow, the surface area of the sphere (S) is given by,S=4πr2Differentiating​ w.r.t. r, we get     dSdr=8πr∴dS=(dSdr)Δr=(8πr)×0.03  m2=(8π×9)×0.03  m2=2.16π m2Hence, the approximate error in calculating the surface area is 2.16π m2.

Q.8 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

Let x=3 and Δx=0.02. Then, we have:f(3.02)=f(x + Δx)=3(x + Δx)2+ 15(x + Δx)+ 5Since,   Δy=f(x+Δx)−f(x).​ Therefore,f(x+Δx)=f(x)+Δy≈f(x)+f‘(x)Δx[∵dx=Δx]    f(3.02)≈(3x2 +15x + 5)+(6x+15)Δx≈{3(3)2+15(3)+5}+(6×3+15)(0.01)≈77+33×0.02=77+0.66≈77.66Hence, the approximate value of f (3.02) is 77.66.The correct answer is D.

Q.9 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

Let side of cube is x metres. SoThe Volume of a cube is given by V=x3       dV=(dVdx)Δx    =(ddxx3)Δx   =(3x2)Δx   =(3x2)(0.03x)[3% of x=0.03x]   =0.09x3Hence, the approximate change in the Volume of the cube is 0.09x3 m3.The correct option is C.

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