# NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.8

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-5 (Continuity And Differentiability)Exercise 5.8 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

### Exercise 5.8

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

So, f(x) is continuous in the interval [-4,2] and differentiable in the interval (- 4,2)

f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 = 0

f(2) = 2² + 4 – 8 = 8 – 8 = 0

f(-4) = f(2)

As Conditions of Rolle’s theorem are satisfied.

Then there exists some c in (-4, 2) such that f′(c) = 0

f'(x) = 2x + 2

f’ (c) = 2c + 2 = 0

c = – 1,

and -1  [-4,2]

Hence, f’ (c) = 0 at c = – 1.

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]

Solution:

In the interval [5, 9],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = 6,7,8

Hence, Rolle’s theorem is NOT applicable

(ii) f(x) = [x] for x  [– 2, 2]

Solution:

In the interval [– 2, 2],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = -1,0,1

Hence, Rolle’s theorem is NOT applicable

(iii) f(x) = x2– 1 for x  [1, 2]

Solution:

Now f(x) = x² – 1 is a polynomial

So, f(x) is continuous in the interval [1, 2] and differentiable in the interval (1,2)

f(1) = (1)²  – 1 = 0

f(2) = 2² – 1 = 3

f(-4) ≠ f(2)

As Conditions of Rolle’s theorem are NOT satisfied.

Hence, Rolle’s theorem is NOT applicable

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

f is continuous in [a, b] ………(1)

f is derivable in [a, b] ………(2)

f (a) = f (b)  ………(3)

then f’ (c)=0, c  (a, b)

So as, f is continuous and derivable

but f ‘(c) ≠ 0

It concludes, f(a) ≠ f(b)

f(-5) ≠ f(5)

Question 4. Verify Mean Value Theorem, if f(x) = x2– 4x – 3 in the interval [a, b], where a = 1 and b = 4.

Solution:

Now f(x) = x² – 4x -3 is a polynomial

So, f(x) is continuous in the interval [1,4] and differentiable in the interval (1,4)

f(1) = (1)² – 4(1) – 3 = -6

f(4) = 4² – 4(4) – 3 = -3

f′(c) = 2c – 4

As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,4) such that

f′(c) =  = 1

2c – 4 = 1

c = 5/2

and c = 5/2 ∈ (1,4)

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:

Now f(x) = x3– 5x2– 3x is a polynomial

So, f(x) is continuous in the interval [1,3] and differentiable in the interval (1,3)

f(1) = (1)3– 5(1)2– 3(1) = -7

f(3) = 33– 5(3)2– 3(3) = -27

f′(c) = 3c2 – 5(2c) – 3

f′(c) = 3c2 – 10c – 3

As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,3) such that

f′(c) =    3c2 – 10c – 3 = -10

3c2 – 10c + 7 = 0

3c2 – 3c – 7c + 7 = 0

3c (c-1) – 7(c -1) = 0

(3c -7) (c-1) = 0

c = 7/3 or c = 1

As, 1 ∉ (1,3)

So, c = 7/3 ∈ (1,3)

According to the Rolle’s Theorem

As, f(3) ≠ f(1), Then there does not exist some c ∈ (1,3) such that f′(c) = 0

Question 6. Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

(i) f(x) = [x] for x  [5, 9]

Solution:

In the interval [5, 9],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = 6,7,8

Hence, Mean value theorem is NOT applicable

(ii) f(x) = [x] for x  [– 2, 2]

Solution:

In the interval [– 2, 2],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = -1,0,1

Hence, Mean value theorem is NOT applicable

(iii) f(x) = x2– 1 for x  [1, 2]

Solution:

Now f(x) = x² – 1 is a polynomial

So, f(x) is continuous in the interval [1,2] and differentiable in the interval (1,2)

f(1) = (1)² – 1 = 0

f(2) = 2² -1 = 3

f′(c) = 2c

As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,2) such that

f′(c) =   = 3

2c = 3

c = 3/2

and c = 3/2 ∈ (1,4)