NCERT Solutions for Maths Class 12 Exercise 5.8

# NCERT Solutions for Maths Class 12 Exercise 5.8

## NCERT Solutions for Maths Class 12 Exercise 5.8

### Maths Class 12 Ex 5.8 Question 1.

Verify Rolle’s theorem for the function f(x) = x² + 2x – 8, x [-4, 2].

Solution:
Here, f(x) = x² + 2x – 8 is a polynomial.
Therefore, it is continuous and differentiable in its domain x
R.
Hence, it 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 and f(2) = 2² + 4 – 8 = 8 – 8 = 0
Conditions of Rolle’s theorem are satisfied because f(–4) = f(2) = 0.
f'(x) = 2x + 2
f’(c) = 2c + 2 = 0 or c = –1

Hence, Rolle’s theorem is verified at c = –1 [–4, 2].

### Maths Class 12 Ex 5.8 Question 2.

Examine if Rolle’s theorem is applicable to any of the following functions. Can you say something 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:

By Rolle’s theorem, f : [a, b] → R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

(c) f(a) = f(b)

Then, there exists some c (a, b) such that f′(c) = 0.

Thus, Rolle’s theorem is not applicable to those functions that do not satisfy any of the three conditions given above.

(i) We have, f(x) = [x] for x [5, 9]

Since, the given function f(x) is not continuous at every integral point.

In general, f(x) is not continuous at x = 5 and x = 9.

Therefore, f(x) is not continuous in [5, 9]

Also, f(5) =  = 5 and f(9) =  = 9

Thus, f(5) ≠ f(9)

Since the LHD and RHD of the function f(x) at n (5, 9) are not equal, the function f(x) is not differentiable in (5, 9).

Hence, Rolle’s theorem is not applicable to the function f(x) = [x] for x [5, 9].

(ii) We have, f(x) = [x] for x
[2, 2]

Since, the given function f(x) is not continuous at every integral point.

In general, f(x) is not continuous at x = 2 and x = 2.

Therefore, f(x) is not continuous in [2, 2]

Also, f(2) =  = 2 and f(2) =  = 2

Thus, f(2) ≠ f(2)

Since the LHD and RHD of the function f(x) at n (–2, 2) are not equal, the function f(x) is not differentiable in (–2, 2).

Hence, Rolle’s theorem is not applicable to the function f(x) = [x] for x [–2, 2].

(iii) We have, f(x) = x² – 1 for x [1, 2]

Since f is a polynomial, it is continuous in [1, 2] and differentiable in (1, 2).

Now, f(1) = 1 – 1 = 0 and f(2) = 4 – 1 = 3
Here, f(a) ≠ f(b)
Since the function does not satisfy the conditions of Rolle’s theorem

Hence, Rolle’s theorem is not applicable to the given function f(x) = x² – 1 for x [1, 2].

In case of converse of Rolle’s theorem, if f(c) = 0, c [a, b], then the 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 f is neither continuous nor differentiable in the interval [5, 9].
(ii) Here also, 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 interval [1, 2]. So, f(2) ≠ f’(2).

### Maths Class 12 Ex 5.8 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:
We have, f : [−5, 5→ R

Since every differentiable function is a continuous function, we obtain

f is continuous on [−5, 5]

f is continuous on (−5, 5)

Thus, by the Mean Value Theorem, there exists c (5, 5) such that

f’(c) = [f(5) – f(–5)]/[5 – (–5)]

10f’(c) = [f(5) – f(–5)]

It is also given that f′(x) does not vanish anywhere.

Therefore, f′(c) ≠ 0

Thus, 10f′(c) ≠ 0

f(5) − f(−5) ≠ 0

f(5) ≠ f(−5)

10f(c) 0

f(5) f(5) 0

f(5) f(5)

Hence proved.

### Maths Class 12 Ex 5.8 Question 4.

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

Solution:
We have, f(x) = x² – 4x – 3. f being a polynomial, it is continuous in the interval [1, 4] and differentiable in (1, 4). So, all the conditions of mean value theorem hold.
Now, f’(x) = 2x – 4,
Then, f’(c) = 2c – 4
f(4) = 16 – 16 – 3 = –3,
f(1) = 1 – 4 – 3 = –6
Then there exist a value c such that

Hence, the mean value theorem is verified for the given function.

### Maths Class 12 Ex 5.8 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:
We have, f(x) = x3 – 5x2 – 3x,
It is a polynomial. Therefore, it is continuous in the interval [1, 3] and differentiable in the interval (1, 3).
Also, f'(x) = 3x² – 10x – 3, then  f'(c) = 3c² – 10c – 3

f(1) = 1 – 5 – 3 = –7 and f(3) = 27 – 45 – 9 = –27

f'(c10



⇒ 3c2 − 10– 3 10



⇒ 3c− 100



⇒ 3c− 3− 70



⇒ 3c(− 1− 7(− 1



⇒ (− 1(3− 70



⇒ 17/3   [where 7/3 ∈ (13)]



Thus, Mean Value Theorem is verified for the given function and c = 7/3 ∈ (1, 3)

### Maths Class 12 Ex 5.8 Question 6.

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

Solution:
Mean Value Theorem states that for a function f : [a, b] → R, if

a.      f is continuous on [a, b]

b.     f is continuous on (a, b)

Thus, Mean Value Theorem is not applicable to those functions that do not satisfy any of three conditions given above.

(i) We have f(x) = [x] for x [5, 9]

Since, the given function f(x) is not continuous at every integral point.

In general, f(x) is not continuous at x = 5 and x = 9

Therefore, f(x) is not continuous in [5, 9]

The differentiability of f in (5, 9) is checked as follows.

Let n be an integer such that n (5, 9)

Since LHD and RHD of f at x = n are not equal, f is not differentiable at x = n.

Therefore, f is not differentiable in (5, 9).

It is observed that f does not satisfy all the conditions of the Mean Value Theorem.

Thus, Mean Value Theorem is not applicable for f(x) = [x] for x [5, 9]

(ii) We have, f(x) = [x] for x [−2, 2]

Since, the given function f(x) is not continuous at every integral point.

In general, f(x) is not continuous at x = −2 and x = 2

Therefore, 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)

Since LHD and RHD of f at x = n are not equal, f is not differentiable at x = n

Therefore, 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.

Thus, Mean Value Theorem is not applicable for f(x) = [x] for x [−2, 2].

(iii) We have, f(x) = x2 − 1 for x [1, 2]

Since, f being a polynomial function 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) = x2 − 1 for x [1, 2].

It can be proved as follows.

We have, f(x) = x2 − 1

Then, f(1) = (1)2 – 1 = 0 and f(2) = (2)2 – 1 = 3

Hence, f′(x) = 2x

Thus, f′(c) = 3

2c = 3

c = 3/2

c = 1.5 [where 1.5 [1, 2]]