NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.8) Exercise 5.8

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NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.8) Exercise 5.8

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If a function is continuous at a point x0 but not differentiable there, its inverse function need not be continuous at x0. This is because differentiation treats discontinuities as if they were ordinary points on which functions are differentiated. However, inverse functions treat ordinary points like discontinuities by leaving them undifferentiated. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.8 are easy to understand.

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Access NCERT Solutions for Class 12 Maths Chapter 5- Continuity and Differentiability

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NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.8

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What Is The Mean Value Theorem?

The Average Theorem tells us that continuously differentiable functions have an eternal by-product on each purpose in their domain, however, they do not seem to be essentially differentiable all over the domain. In other words, there will be some points wherever the graph does not show any quiet inflection or modification of slope, which means it is continuous while not having a by-product at that time.

The Average Theorem states that for any perform f(x) whose graph passes through 2 given points (a, f(a)), (b, f(b)), there’s a minimum of one purpose (c, f(c)) on the curve wherever the tangent is parallel to the secant passing through the 2 given points. The average theorem is outlined herein in calculus for a perform f(x): [a, b] → R, specified it is continuous And differentiable across an interval.

The perform f(x) is continuous over the interval [a, b].

The perform f(x) is differentiable over the interval (a, b).

There exists a degree c in (a, b) specified f'(c) = [ f(b) – f(a) ] / (b – a)

Here we’ve verified that the tangent at c is parallel to the secant passing through the points (a, f(a)), (b, f(b)). This average theorem is employed to prove an announcement across an interval. Further, the average theorem springs from Rolle’s theorem.

Statement: The average theorem states that if a perform f is continuous over the interval [a, b], and differentiable over the interval (a, b), then there exists a minimum of one purpose c within the interval (a, b) specified f ‘(c) is that the average rate of modification of the perform over [a, b] and it is parallel to the secant line over [a, b].

Proof: Let g(x) be the secant line to f(x) passing through (a, f(a)) and (b, f(b)). Students must know that the equation of the secant line is y – y1 = m (x – x1).

g(x) – f(a) = [ f(b) – f(a) ] / (b – a) (x-a)

g(x) = [ f(b) – f(a) ] / (b – a) (x-a) + f(a) —–>(1)

Let h(x) be f(x) – g(x)

h(x) = f(x) – [[ f(b) – f(a) ] / (b – a) (x-a) + f(a)] (From (1))

h(a) = h(b) = zero and h(x) is continuous on [a, b] and differentiable on (a, b).

Thus applying Rolle’s theorem, there is some x = c in (a, b) specified h'(c) = zero.

h'(x) = f'(x) – [ f(b) – f(a) ] / (b – a)

For some c in (a, b), h'(c) = 0. Thus

h'(c) = f'(c) – [ f(b) – f(a) ] / (b – a) = zero

f'(c) = [ f(b) – f(a) ] / (b – a)

Thus the average theorem is verified.

Note: The result might not hold if the performance is not differentiable, even at one purpose within the interval.

Easy approaches are followed in the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.8.

What Is Rolle’s Theorem?

According to Rolle’s theorem, a real-valued differentiable performance should have a minimum of one mounted purpose wherever the primary by-product is zero if it achieves equal values at 2 different points. A French man of science named Michel Rolle is the subject of Rolle’s theorem. The Average Theorem encompasses a special case called Rolle’s Theorem.

The first average theorem or the average theorem itself is a unit of different names for Lagrange’s Average Theorem. The mean is often thought of because of the average of the values provided, however, once it involves integrals, there is {a totally different|a special|a unique|a distinct} approach to determine the average of 2 different functions. Let us learn Rolle’s theorem and therefore the average of those functions during this article.

Rolle’s theorem states that a perform f is differentiable at a degree x if it is continuous at x, and therefore the differential of f at that time should be zero, i.e., f'(x)=0; conversely, a perform f is continuous at a degree x if it is differentiable at that time, and therefore the differential of f at that time should be non-zero i.e., f'(x)≠0.

Rolle’s theorem states that “If a perform f is outlined within the interval [a, b] in such the simplest way that it satisfies the subsequent condition: i) f is continuous on [a, b], ii) f is differentiable on (a, b), and iii) f (a) = f (b), then there exists a minimum of one price of x, allow us to assume this price to be c, that lies between a and b i.e. (a < c < b ) in such the simplest way that f‘(c) = zero.”

If f(x) is continuous on [a,b],

f(x) is differentiable on (a,b) and

f(a) = f(b)

Then there exists a minimum of on c in (a,b) specified,

f’(c) = 0

Let f: [a, b] R be continuous on [a, b] and differentiable on (a, b), specified f(a) = f(b), wherever a and b area unit some real numbers. this {can be} however Rolle’s theorem can be expressed mathematically. once f′(c) = zero, there’s a c in (a, b) that exists.

Q.1

Examine if Rolle‘s Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle‘s theorem from these example? i  fx=x  for x ∈5,9ii  fx=x    for x ∈−2,2iii fx=x2−1  for x ∈1,2

 

Ans

(i) Given function is f(x)=[x] for x∈[5, 9]Since the given function f(x) is not continuous at every integral point and in particular it is not continuous at x=5 and x=9.⇒f (x) is not continuous in [5, 9].f(5)=[5]=5f(9)=[9]=9∴f(5)≠f(9)The differentiability of f in (5, 9) is checked as follows.Let n be an integer such that n∈(5, 9).The left hand limit of f at x=n islimh→0−f(n+h)−f(n)h=limh→0−[n+h]−[n]h         =limh→0−n−1−nh[Since, h is very small number.So, limh→0− [n+h]=limh→0−(n−1)Example: [5−0. 3]=[4. 7]=4]         =limh→0−−1h=−∞The right hand limit of f at x=n islimh→0+f(n+h)−f(n)h=limh→0+[n+h]−[n]h         =limh→0+n−nh[Since, h is very small number.So, limh→0+ [n+h]=limh→0+(n)Example: [5+0.3]=[5.3]=5]         =limh→0+0         =0∴Left Hand Derivative≠Right Hand DerivativeSo, f is not differentiable at x=n.∴f is not differentiable in (5, 9).It is observed that f does not satisfy all the conditions of the hypothesis of Roll‘s Theorem.Hence, Roll‘s Theorem is not applicable for f(x)=[x] for x∈[5, 9](ii)Given function is f(x)=[x] for x∈[−2, 2]Since the given function f(x) is not continuous at every integral point and in particular it is not continuous at x=−2 and x=2.⇒f (x) is not continuous in [−2, 2].f(−2)=[−2]=−2   f(2)=[2]=2∴f(−2)≠f(2)The differentiability of f in (−2, 2) is checked as follows.Let n be an integer such that n∈(−2, 2).The left hand limit of f at x=n islimh→0−f(n+h)−f(n)h=limh→0−[n+h]−[n]h         =limh→0−n−1−nh[Since, h is very small number.So, limh→0− [n+h]=limh→0−(n−1)Example: [5−0.3] =[4.7]=4]         =limh→0−−1h=−∞

The right hand limit of f at x=n is lim h→ 0 + f( n+h )−f( n ) h = lim h→ 0 + [ n+h ]−[ n ] h          = lim h→ 0 + n−n h [ Since, h is very small number. So,  lim h→ 0 +  [ n+h ]= lim h→0+ ( n ) Example: [ 5+0.3 ]=[ 5.3 ]=5 ]          = lim h→ 0 + 0          =0 ∴Left Hand Derivative≠Right Hand Derivative So, f is not differentiable at x=n. ∴f is not differentiable in (−2, 2). It is observed that f does not satisfy all the conditions of the hypothesis of Roll’s Theorem. Hence, Roll’s Theorem is not applicable for f( x )=[ x ] for x∈[ −2,2 ] ( iii )The given function, f( x )= x 2 −1, is a polynomial. So, it is continuous in [ 1,  2 ] and is differentiable in ( 1,  2 ). f( 1 )= ( 1 ) 2 −1          =0 f( 2 )= ( 2 ) 2 −1          =4−1          =3  ∴f( 1 )≠f( 2 ) ⇒The values of f( x ) at 1 and 2 do not coincide. It is observed that f does not satisfy a condition of the hypothesis of Rolle’s Theorem. Hence, Rolle’s Theorem is not applicable for f( x )= x 2 −1 for x∈[ 1,2 ] 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Q.2

If f: –5, 5→R is differentiable function and if f‘x does not vanish anywhere, then prove that f –5≠f 5  

 

Ans

Given function f:[−5,5]→R​  is a differentiable function.Since every differentiable function is a continuous function thenwe have(a) f is continuous on [−5, 5].(b) f is differentiable on (−5, 5).Therefore, by the Mean Value Theorem, there exists c∈(−5, 5) such that      f‘(c)=f(5)−f(−5)5−(−5)      =f(5)−f(−5)1010f‘(c)=f(5)−f(−5)It is also given that f‘(x)  does not vanish anywhere.∴            f‘(c)≠0⇒       10f‘(c)≠0f(5)−f(−5)≠0⇒            f(5)≠f(−5)Hence, proved.

Q.3

Verify Mean Value Theorem, if fx=x2−4x−3 in the interval  a, b, where a= 1 and b= 4.

 

Ans

The given function=x2−4x−3Since, function f(x) is polynomial, so it is continuous in [1, 4]and it is differentiable in (1,4).Here, f‘(x)=ddx(x2−4x−3)   =2x−4and,    f‘(c)=2c−4Now,   f(1)=(1)2−4(1)−3  =1−4−3  =−6and     f(4)=(4)2−4(4)−3  =16−16−3  =−3∴f(b)−f(a)b−a=f(4)−f(1)4−1⇒f(b)−f(a)b−a=−3−(−6)3⇒f(b)−f(a)b−a=33⇒f(b)−f(a)b−a=1Mean Value Theorem states that there is a point c∈(1,4)such that f‘(c)=1∵        f‘(c)=1⇒2c−4=1⇒   2c=5⇒          c=52∈(1,4)Hence, Mean Value Theorem is verified for the given function.

Q.4

Verify Mean Value Theorem,if f(x)=x3−5x2 – 3x in the interval [a,b],wherea= 1and b= 3. Find all c∈(1, 3)for which f‘(c) = 0.

 

Ans

The given function=x3−5x2−3xSince, function f(x) is polynomial, so it is continuous in [1, 3]and it is differentiable in (1,3).Here, f‘(x)=ddx(x3−5x2−3x)   =3x2−10x−3and,    f‘(c)=3c2−10c−3Now,   f(1)=(1)3−5(1)2−3(1)  =1−5−3  =−7and     f(3)=(3)3−5(3)2−3(3)  =27−45−9  =−27∴f(b)−f(a)b−a=f(3)−f(1)3−1⇒f(b)−f(a)b−a=−27−(−7)2⇒f(b)−f(a)b−a=−202⇒f(b)−f(a)b−a=−10Mean Value Theorem states that there is a point c∈(1,3)such that f‘(c)=−10∵        f‘(c)=−10⇒3c2−10c+7=0⇒3c2−3c−7c+7=0⇒3c(c−1)c−7(c−1)=0⇒(c−1)(3c−7)=0⇒          c=1,  73,  where c=73∈(1,3)Hence, Mean Value Theorem is verified for the given functionc=73∈(1,3) is the only point for which f‘(c)=0.

Q.5

Examine the applicability of Mean Value Theorem for all threefunctions given below.(i)f(x)=[x]forx∈[5,9](ii)f(x)=[x]forx∈[−2,2](iii)f(x)=x2−1forx∈[1,2]

 

Ans

(i) Given function is f(x)=[x] for x∈[5,9]Since the given function f(x) is not continuous at every integral point and in particular it is not continuous at x=5 and x=9.⇒f (x) is not continuous in [5, 9].f(5)=[5]=5f(9)=[9]=9∴f(5)≠f(9)The differentiability of f in (5, 9) is checked as follows.Let n be an integer such that n∈(5, 9).The left hand limit of f at x=n islimh→0−f(n+h)−f(n)h=limh→0−[n+h]−[n]h         =limh→0−n−1−nh[Since, h is very small number.So, limh→0− [n+h]=limh→0−(n−1)Example: [5−0.3]=[4.7]=4]         =limh→0−−1h=−∞The right hand limit of f at x=n islimh→0+f(n+h)−f(n)h=limh→0+[n+h]−[n]h         =limh→0+n−nh[Since, h is very small number.So, limh→0+ [n+h]=limh→0+(n)Example: [5+0.3]=[5.3]=5]         =limh→0+0         =0∴Left Hand Derivative≠Right Hand DerivativeSo, f is not differentiable at x=n.∴f is not differentiable in (5, 9).It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.Hence, Mean Value Theorem is not applicable for f(x)=[x] for x∈[5,9](ii)Given function is f(x)=[x] for x∈[−2,2]Since the given function f(x) is not continuous at every integral point and in particular it is not continuous at x=−2 and x=2.⇒f (x) is not continuous in [−2, 2].The differentiability of f in (−2, 2) is checked as follows.Let n be an integer such that n∈(−2, 2).The left hand limit of f at x=n islimh→0−f(n+h)−f(n)h=limh→0−[n+h]−[n]h         =limh→0−n−1−nh[Since, h is very small number.So, limh→0− [n+h]=limh→0−(n−1)Example: [5−0.3]=[4.7]=4]         =limh→0−−1h=−∞The right hand limit of f at x=n islimh→0+f(n+h)−f(n)h=limh→0+[n+h]−[n]h         =limh→0+n−nh[Since, h is very small number.So, limh→0+ [n+h]=limh→0+(n)Example: [5+0.3]=[5.3]=5]         =limh→0+0         =0∴Left Hand Derivative≠Right Hand DerivativeSo, f is not differentiable at x=n.∴f is not differentiable in (−2, 2).It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.Hence, Mean Value Theorem is not applicable for f(x)=[x] for x∈[−2,2]

( iii )The given function, f( x )= x 2 −1, is a polynomial. So, it is continuous in [ 1,  2 ] and is differentiable in ( 1,  2 ). It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem. Hence, Mean Value Theorem is applicable for f( x )= x 2 −1 for x∈[ 1,2 ] It can be proved as:     f( 1 )= ( 1 ) 2 −1=0    f( 2 )= ( 2 ) 2 −1=3 ∴ f( b )−f( a ) b−a = f( 2 )−f( 1 ) 2−1        = 3−0 1        =3 ∵       f‘( x )=2x ∴       f‘( c )=3 ⇒ 2c=3 ⇒ c= 3 2 =1.5∈[ 1,2 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@D192@

Q.6 Differentiate w.r.t. x the function:
(3x2 – 9x + 5)9.

Ans

Let y=(3x2−9x+5)9Differentiating w.r.t. x, we get   dydx=ddx(3x2−9x+5)9=9(3x2−9x+5)8ddx(3x2−9x+5)[By chain rule]=9(3x2−9x+5)8(6x−9)=9×3(2x−3)(3x2−9x+5)8=27(2x−3)(3x2−9x+5)8

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