# NCERT Solutions for Class 12 Maths Continuity and Differentiability

Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get the * NCERT Solutions for Class 12 Maths Continuity and Differentiability Exercise 5.8*.

* CBSE Class 12^{th}* is an important school class in your life as you take some serious decision about your career. And out of all subjects, Maths is an important and core subjects. So

*is major role in your exam preparation as it has detailed chapter wise solutions for all exercise. This*

**CBSE NCERT Solutions for Class 12**^{th}Maths**can be downloaded in PDF file. The downloading link is given at last.**

*NCERT Solutions** NCERT Solutions for Class 12 Maths* are not only the solutions of Maths exercise but it builds your foundation of other important subjects. Getting knowledge of depth concept of

*like Algebra, Calculus, Trigonometry, Coordinate Geometry help you to understand the concept of Physics and Physical Chemistry.*

**CBSE Class 12**^{th}MathsAs we know that all the schools affiliated from CBSE follow the NCERT books for all subjects. You can check the **CBSE NCERT Syllabus**. Thus, * NCERT Solutions* helps the students to solve the exercise questions as given in

*.*

**NCERT Books**## NCERT Solutions for Class 12 Maths Continuity and Differentiability Exercise 5.8

**NCERT Solutions for Class 12 Maths Continuity and Differentiability ****Exercise**** 5.8: Ques No 1.**

Verify Rolle’s theorem for the function f (x) = x^{2} + 2x – 8, x ∈ [– 4, 2].

**NCERT Solutions:**

Given that f(x) = x^{2} + 2x – 8 in x ∈ [– 4, 2].

(i) Function is continuous in [-4, 2] as it is a polynomial function and polynomial function is always in continuous.

(ii) f'(x) = 2x + 2 and f'(x) exists in (-4, 2), hence f(x) is derivable.

(iii) f(-4) = 0 and f(2) = 0 ⇒ f(-4) = f(2).

Conditions of Rolle’s theorem are satisfied, hence there exists, at least one c ∈ (– 4, 2) such that f'(c) = 0 ⇒ 2c + 2 = 0 ⇒ c = -1.

**NCERT Solutions for Class 12 Maths Continuity and Differentiability ****Exercise**** 5.8: Ques No 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]

**NCERT Solutions:**

Being greatest integer function the given function is not differentiable and continuous. Hence, Rolle’s theorem is not applicable.

(ii) f (x) = [x] for x ∈ [– 2, 2]

**NCERT Solutions:**

Being greatest integer function the given function is not differentiable and continuous. Hence, Rolle’s theorem is not applicable.

(iii) f (x) = x^{2} – 1 for x ∈ [1, 2]

**NCERT Solutions:**

f(x) is continuous in [1, 2] as it is a polynomial function. f'(x) = 2x, f'(x) exists in (1, 2), hence derivable. f(x) = x^{2} – 1. f(1) = 1 – 1 = 0 and f(2) = 4 – 1 = 3.

Since f(1) ≠ f(2), Rolle’s theorem is not applicable.

**NCERT Solutions for Class 12 Maths Continuity and Differentiability ****Exercise**** 5.8: Ques No 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).

**NCERT Solutions:**

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 ∈ (a, b).

We are given that f is continuous and derivable but f'(c) ≠ 0 ⇒ f(a) ≠ f(b) hence f(-5) ≠ f(5).

**NCERT Solutions for Class 12 Maths Continuity and Differentiability ****Exercise**** 5.8: Ques No 4.**

Verify Mean Value Theorem, if f (x) = x^{2} – 4x – 3 in the interval [a, b], where a = 1 and b = 4.

**NCERT Solutions:**

(i) Function is continuous in [1, 4] as it is a polynomial function and polynomial function is always continuous.

(ii) f'(x) = 2x – 4, f'(x) exist in (1, 4), hence derivable, condition of MVT theorem are satisfied hence, there exists, at least one point c ∈ (1, 4) such that

**NCERT Solutions for Class 12 Maths Continuity and Differentiability ****Exercise**** 5.8: Ques No 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.

**NCERT Solutions:**

(i) Function is continuous in [1, 3] as it is a polynomial function and polynomial function is always continuous.

(ii) f'(x) = 3x^{2} – 10x – 3, f'(x) exist in (1, 3), hence derivable, condition of MVT theorem are satisfied hence, there exists, at least one point c ∈ (1, 3) such that

Hence MVT is applicable and c = 7/3.

**NCERT Solutions for Class 12 Maths Continuity and Differentiability ****Exercise**** 5.8: Ques No 6.**

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

**NCERT Solutions:**

(i) f (x) = [x] for x ∈ [5, 9]

**NCERT Solutions:**

Being greatest integer function the given function is not differentiable and continuous. Hence, MVT is not applicable.

(ii) f (x) = [x] for x ∈ [– 2, 2]

**NCERT Solutions:**

Being greatest integer function the given function is not differentiable and continuous. Hence, MVT is not applicable.

(iii) f (x) = x^{2} – 1 for x ∈ [1, 2]

**NCERT Solutions:**

f(x) is continuous in [1, 2] as it is a polynomial function. f'(x) = 2x, f'(x) exists in (1, 2), hence derivable. f(x) = x^{2} – 1.