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

Get Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Continuity and Differentiability Exercise 5.8 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Continuity and Differentiability Ex 5.8 provided in NCERT Textbook.

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The topics and sub-topics included in the Continuity and Differentiability chapter are the following:

Continuity and Differentiability
Introduction
Algebra of continuous functions
Differentiability
Derivatives of composite functions
Derivatives of implicit functions
Derivatives of inverse trigonometric functions
Exponential and Logarithmic Functions
Logarithmic Differentiation
Derivatives of Functions in Parametric Forms
Second Order Derivative
Mean Value Theorem
Summary

There are total eight exercises and one misc exercise(144 Questions fully solved) in the class 12th maths chapter 5 Continuity and Differentiability.

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

Ex 5.8 Class 12 Maths Question 1.
Verify Rolle’s theorem for the function
f(x) = x² + 2x – 8,x∈ [-4,2]
Solution:
Now f(x) = x² + 2x – 8 is a polynomial
∴ it is continuous and derivable in its domain x∈R.
Hence it is continuous in the interval [-4,2] and derivable in the interval (- 4,2)
f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 = 0,
f(2) = 2² + 4 – 8 = 8 – 8 = 0
Conditions of Rolle’s theorem are satisfied.
f'(x) = 2x + 2
∴ f’ (c) = 2c + 2 = 0
or c = – 1, c = – 1 ∈ [-4,2]
Thus f’ (c) = 0 at c = – 1.

Ex 5.8 Class 12 Maths Question 2.
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) f(x) = [x] for x ∈ [5,9]
(ii) f (x) = [x] for x ∈ [-2,2]
(iii) f (x) = x² – 1 for x ∈ [1,2]
Solution:
(i) In the interval [5, 9], f (x) = [x] is neither continuous nor derivable at x = 6,7,8 Hence Rolle’s theorem is not applicable
(ii) f (x) = [x] is not continuous and derivable at -1, 0, 1. Hence Rolle’s theorem is not applicable.
(iii) f(x) = (x² – 1),f(1) = 1 – 1 = 0,
f(2) = 22 – 1 = 3
f(a)≠f(b)
Though it is continous and derivable in the interval [1,2].
Rolle’s theorem is not applicable.
In case of converse if f (c)=0, c ∈ [a, b] then conditions of rolle’s theorem are not true.
(i) f (x) = [x] is the greatest integer less than or equal to x.
∴f(x) = 0, But fis neither continuous nor differentiable in the interval [5,9].
(ii) Here also, theough f (x) = 0, but f is neither continuous nor differentiable in the interval [-2,2].
(iii) f (x)=x² – 1, f'(x)=2x. Here f'(x) is not zero in the [1,2], So f (2) ≠ f’ (2).

Ex 5.8 Class 12 Maths Question 3.
If f: [-5,5] –>R is a differentiable function and if f (x) does not vanish anywhere then prove that f (- 5) ≠ f (5).
Solution:
For Rolle’s theorem
If (i) f is continuous in [a, b]
(ii) f is derivable in [a, b]
(iii) f (a) = f (b)
then f’ (c)=0, c e (a, b)
∴ f is continuous and derivable
but f (c) ≠ 0 =>f(a) ≠ f(b) i.e., f(-5)≠f(5)

Ex 5.8 Class 12 Maths Question 4.
Verify Mean Value Theorem, if
f (x) = x² – 4x – 3 in the interval [a, b], where a = 1 and b = 4.
Solution:
f (x) = x² – 4x – 3. It being a polynomial it is continuous in the interval [1,4] and derivable in (1,4), So all the condition of mean value theorem hold.
then f’ (x) = 2x – 4,
f’ (c) = 2c – 4
f(4)= 16 – 16 – 3 = – 3,
f(1)= 1 – 4 – 3 = – 6
Then there exist a value c such that
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 4

Ex 5.8 Class 12 Maths Question 5.
Verify Mean Value Theorem, if f (x)=x³ – 5x² – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1,3) for which f’ (c) = 0.
Solution:
f (x)=x³ – 5x² – 3x,
It is a polynomial. Therefore it is continuous in the interval [1,3] and derivable in the interval (1,3)
Also, f'(x)=3x²-10x-3
tiwari academy class 12 maths Chapter 5 Continuity and Differentiability 5

Ex 5.8 Class 12 Maths Question 6.
Examine the applicability of Mean Value theroem for all three functions given in the above Question 2.
Solution:
(i) F (x)= [x] for x ∈ [5,9], f (x) = [x] in the interval [5, 9] is neither continuous, nor differentiable.
(ii) f (x) = [x], for x ∈ [-2,2],
Again f (x) = [x] in the interval [-2,2] is neither continous, nor differentiable.
(iii) f(x) = x²-1 for x ∈ [1,2], It is a polynomial. Therefore it is continuous in the interval [1,2] and differentiable in the interval (1,2)
f (x) = 2x, f(1) = 1 – 1 = 0 ,
f(2) = 4 – 1 = 3, f'(c) = 2c
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 6