** NCERT Solutions for Maths Class 12 Exercise 5.8**

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**NCERT Solutions for Maths Class 12 Exercise 5.8** helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

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**NCERT Solutions for Maths Class 12 Exercise 5.1**

**NCERT Solutions for Maths Class 12 Exercise 5.2**

**NCERT Solutions for Maths Class 12 Exercise 5.3**

**NCERT Solutions for Maths Class 12 Exercise 5.4**

**NCERT Solutions for Maths Class 12 Exercise 5.5**

**NCERT Solutions for Maths Class 12 Exercise 5.6**

**NCERT Solutions for Maths Class 12 Exercise 5.7**

**NCERT Solutions for Maths Class 12 Exercise 5.8**

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**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] = 5 and f(9) = [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] = –2 and f(2) = [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) = x^{3}– 5x

^{2}– 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) = x^{3} –
5x^{2} – 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'(c) = −10

⇒ 3c^{}2 − 10c – 3 = −10

⇒ 3c^{}2 − 10c + 7 = 0

⇒ 3c^{}2 − 3c − 7c + 7 = 0

⇒ 3c(c − 1) − 7(c − 1) = 0

⇒ (c − 1) (3c − 7) = 0

⇒ c = 1, 7/3 [where c = 7/3 ∈ (1, 3)]

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)

Then there exists some c ∈ (a, b) such that

**(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) = x^{2} − 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) = x^{2} − 1 for x ∈ [1, 2].

It can be proved as follows.

We have, f(x) = x^{2} − 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]]

**NCERT Solutions for Maths Class 12 Exercise 5.1**

**NCERT Solutions for Maths Class 12 Exercise 5.2**

**NCERT Solutions for Maths Class 12 Exercise 5.3**

**NCERT Solutions for Maths Class 12 Exercise 5.4**

**NCERT Solutions for Maths Class 12 Exercise 5.5**