Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(m) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(m)

Question 1.

Verify Rolle’s theorem for the function

f(x) = x (x – 2)^{2}, 0 ≤ x ≤ 2.

Solution:

f(x) = x (x – 2)^{2}, 0 ≤ x ≤ 2

Here a = 0, b = 2

f(x) is a polynomial function hence it is continuous and also differentiable.

∴ f is continuous on [0. 2]

f is differentiable on (0, 2)

f(0) = 0 = f(2)

Thus conditions of Rolle’s theorem are satisfied.

f'(x) = (x – 2)^{2} + 2x (x – 2)

= (x – 2) (x – 2 + 2x)

= (x – 2) (3x – 2)

f'(x)= 0 ⇒ x = 2, x = \(\frac{2}{3}\)

But x = 2 ∉ (0, 2). Thus c = \(\frac{2}{3}\) such that f'(c) = 0

Thus Rolle’s theorem is verified.

Question 2.

Examine if Rolle’s theorem is applicable to the following functions:

(i) f(x) = |x| on [-1, 1]

Solution:

f(x) = |x| on [-1, 1]

As f(x) = |x| is not differentiable at x = 0 ∈ (-1, 1)

We have Rolle’s theorem is not applicable.

(ii) f(x) = [x] on [-1, 1]

Solution:

f(x) = [x] on [-1, 1]

f(x) = [x] is not continuous at 0 ∈ [-1, 1]

Rolle’s theorem is not applicable.

(iii) f(x) = sin x on [0, π]

Solution:

f(x) = sin x on [0, π]

f is a trigonometric function hence continuous and differentiable on its domain.

∴ f is continuous on [0, π]

f is differentiable on (0, π]

f(0) = f(π)

Thus Rolle’s theorem is applicable for f(x) = sin x on [0, π]

(iv) f(x) = cot x on [0, π]

Solution:

f(x) = cot x on [0, π]

Clearly cot (0) and cot (π] are not defined hence Rolle’s theorem is not applicable.

Question 3.

Verify Lagrange’s Mean-Value theorem for

F(x) = x^{3} – 2x^{2} – x + 3 on [1, 2]

Solution:

f(x) = x^{3} – 2x^{2} – x + 3 on [1, 2]

f is a polynomial function hence continuous as well as differentiable.

∴ f is continuous on [1, 2]

f is differentiable on (1, 2)

Thus Largange’s mean value theorem is applicable.

Now f(1) = 1 – 2 – 1 + 3 = 1

f(2) = 8 – 8 – 2 + 3 = 1

∴ f(x) = 2x^{2} – 4x – 1

Thus Lagrange’s mean value theorem is verified.

Question 4.

Test if Lagrange’s mean value theorem holds for the functions given in question no. 2.

Solution:

(i) f(x) = |x| is not differentiable at x = 0 ∈ (-1, 1)

Thus Lagrange’s mean value theorem, does not hold.

(ii) f(x) = [x] is discontinuous at 0 ∈ [-1, 1]

Thus Lagrange’s mean value theorem is not applicable.

(iii) f(x) = sin x is a trigonometric function, which is continuous as well as differentiable in its domain.

∴ f is continuous on [0, π]

f is differentiable on (0, π)

Thus conditions of Lagrange’s mean value theorem are satisfied.

Hence mean value theorem is applicable.

(iv) f(x) = cot x

Which is undefined x = 0 and x = π

Thus Lagrange’s mean value theorem is not applicable.

Question 5.

(Not for examination) Verify Cauchy’s mean value theorem for the functions x^{2} and x^{3}

in [1, 2].

Solution:

Let f(x) = x^{2}, and g(x) = x^{3} on [1, 2]

Both f and g are polynomial functions, hence continuous and differentiable.

∴ f and g are continuous on [1, 2]

f and g are differentiable on (1, 2)

g'(x) = 3x^{2} ≠ 0 ∀ x ∈ (1, 2)

Thus conditions of Cauchy’s mean value theorem are satisfied.

Now f(1) = 1, f(2) = 4, g(1) = 1, g(2) = 8

f'(x) = 2, and g'(x) = 3x^{2}