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Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.8
  • Last Updated : 18 Mar, 2021

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) = \frac{f(b)-f(a)}{b-a}

\frac{-3-(-6)}{4-1}

= 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) = \frac{f(b)-f(a)}{b-a}

\frac{f(3)-f(1)}{b-a}

\frac{-27-(-7)}{3-1}

\frac{-20}{2}

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) = \frac{f(b)-f(a)}{b-a}

\frac{f(2)-f(1)}{2-1}

\frac{3-0}{2-1}

= 3

2c = 3

c = 3/2

and c = 3/2 ∈ (1,4)

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