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 12th 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 CBSE NCERT Solutions for Class 12th Maths is major role in your exam preparation as it has detailed chapter wise solutions for all exercise. This NCERT Solutions can be downloaded in PDF file. The downloading link is given at last.
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 CBSE Class 12th Maths like Algebra, Calculus, Trigonometry, Coordinate Geometry help you to understand the concept of Physics and Physical Chemistry.
As 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) = x2 + 2x – 8, x ∈ [– 4, 2].
NCERT Solutions:
Given that f(x) = x2 + 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) = x2 – 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) = x2 – 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) = x2 – 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) = x3 – 5x2 – 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) = 3x2 – 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) = x2 – 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) = x2 – 1.
