Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Applications of Derivatives Ex 2.3 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3

Question 1.

Check the validity of the Rolle’s theorem for the following functions.

(i) f(x) = x^{2} – 4x + 3, x ∈ [1, 3]

Solution:

The function f given as f(x) = x^{2} – 4x + 3 is polynomial function.

Hence, it is continuous on [1, 3] and differentiable on (1, 3).

Now, f(1) = 1^{2} – 4(1) + 3 = 1 – 4 + 3 = 0

and f(3) = 3^{2} – 4(3) + 3 = 9 – 12 + 3 = 0

∴ f(1) = f(3)

Thus, the function f satisfies all the conditions of Rolle’s theorem.

(ii) f(x) = e^{-x} sin x, x ∈ [0, π].

Solution:

The functions e^{-x} and sin x are continuous and differentiable on their domains.

∴ f(x) = e^{-x} sin x is continuous on [0, π] and differentiable on (0, π).

Now, f(0) = e^{0} sin 0 = 1 × 0 = 0

and f(π) = e^{-π} sin π = e^{-π} × 0 = 0

∴ f(0) = f(π)

Thus, the function f satisfies all the conditions of the Rolle’s theorem.

(iii) f(x) = 2x^{2} – 5x + 3, x ∈ [1, 3].

Solution:

The function f given as f(x) = 2x^{2} – 5x + 3 is a polynomial function.

Hence, it is continuous on [1, 3] and differentiable on (1, 3).

Now, f(1) = 2(1)^{2} – 5(1) + 3 = 2 – 5 + 3 = 0

and f(3) = 2(3)^{2} – 5(3) + 3 = 18 – 15 + 3 = 6

∴ f(1) ≠ f(3)

Hence, the conditions of Rolle’s theorem are not satisfied.

(iv) f(x) = sin x – cos x + 3, x ∈ [0, 2π].

Solution:

The functions sin x, cos x and 3 are continuous and differentiable on their domains.

∴ f(x) = sin x – cos x + 3 is continuous on [0, 2π] and differentiable on (0, 2π).

Now, f(0) = sin 0 – cos 0 + 3 = 0 – 1 + 3 = 2

and f(2π) = sin 2π – cos 2π + 3 = 0 – 1 + 3 = 2

∴ f(0) = f(2π)

Thus, the function f satisfies all the conditions of the Rolle’s theorem.

(v) f(x) = x^{2}, if 0 ≤ x ≤ 2

= 6 – x, if 2 < x ≤ 6.

Solution:

f(x) = x^{2}, if 0 ≤ x ≤ 2

= 6 – x, if 2 < x ≤ 6

∴ f(x) = \(\frac{d}{d x}\left(x^{2}\right)\) = 2x, if 0 ≤ x ≤ 2

= \(\frac{d}{d x}(6-x)\) = -1, if 2 < x ≤ 6

∴ Lf'(2) = 2(2) = 4 and Rf'(2) = -1

∴ Lf'(2) ≠ Rf'(2)

∴ f is not differentiable at x = 2 and 2 ∈ (0, 6).

∴ f is not differentiable at all the points on (0, 6).

Hence, the conditions of Rolle’s theorem are not satisfied.

(vi) f(x) = \(x^{\frac{2}{3}}\), x ∈ [-1, 1].

Solution:

f(x) = \(x^{\frac{2}{3}}\)

∴ \(f^{\prime}(x)=\frac{d}{d x}\left(x^{\frac{2}{3}}\right)=\frac{2}{3} x^{-\frac{1}{3}}\) = \(\frac{2}{3 \sqrt[3]{x}}\)

This does not exist at x = 0 and 0 ∈ (-1, 1)

∴ f is not differentiable on the interval (-1, 1).

Hence, the conditions of Rolle’s theorem are not satisfied.

Question 2.

Given an interval [a, b] that satisfies hypothesis of Rolle’s theorem for the function f(x) = x^{4} + x^{2} – 2. It is known that a = -1. Find the value of b.

Solution:

f(x) = x^{4} + x^{2} – 2

Since the hypothesis of Rolle’s theorem are satisfied by f in the interval [a, b], we have

f(a) = f(b), where a = -1

Now, f(a) = f(-1) = (-1)^{4} + (-1)^{2} – 2 = 1 + 1 – 2 = 0

and f(b) = b^{4} + b^{2} – 2

∴ f(a) = f(b) gives

0 = b^{4} + b^{2} – 2 i.e. b^{4} + b^{2} – 2 = 0.

Since, b = 1 satisfies this equation, b = 1 is one of the roots of this equation.

Hence, b = 1.

Question 3.

Verify Rolle’s theorem for the following functions.

(i) f(x) = sin x + cos x + 7, x ∈ [0, 2π]

Solution:

The functions sin x, cos x and 7 are continuous and differentiable on their domains.

∴ f(x) = sin x + cos x + 7 is continuous on [0, 2π] and differentiable on (0, 2π)

Now, f(0) = sin 0 + cos 0 + 7 = 0 + 1 + 7 = 8

and f(2π) = sin 2π + cos 2π + 7 = 0 + 1 + 7 = 8

∴ f(0) = f(2π)

Thus, the function f satisfies all the conditions of Rolle’s theorem.

∴ there exists c ∈ (0, 2π) such that f'(c) = 0.

Now, f(x) = sin x + cos x + 7

∴ f'(x) = \(\frac{d}{d x}\) (sin x + cos x + 7)

= cos x – sin x + 0

= cos x – sin x

∴ f'(c) = cos c – sin c

∴ f'(c) = 0 gives, cos c – sin c = 0

∴ cos c = sin c

∴ c = \(\frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4}, \ldots\)

But \(\frac{\pi}{4}, \frac{5 \pi}{4}\) ∈ (0, 2π)

∴ c = \(\frac{\pi}{4} \text { or } \frac{5 \pi}{4}\)

Hence, the Rolle’s theorem is verified.

(ii) f(x) = sin(\(\frac{x}{2}\)), x ∈ [0, 2π]

Solution:

The function f(x) = sin(\(\frac{x}{2}\)) is continuous on [0, 2π] and differentiable on (0, 2π).

Now, f(0) = sin 0 = 0

and f(2π) = sin π = 0

∴ f(0) = f(2π)

Thus, the function f satisfies all the conditions of Rolle’s theorem.

∴ there exists c ∈ (0, 2π) such that f'(c) = 0.

Hence, Rolle’s theorem is verified.

(iii) f(x) = x^{2} – 5x + 9, x ∈ [1, 4].

Solution:

The function f given as f(x) = x^{2} – 5x + 9 is a polynomial function.

Hence it is continuous on [1, 4] and differentiable on (1, 4).

Now, f(1) = 1^{2} – 5(1) + 9 = 1 – 5 + 9 = 5

and f(4) = 4^{2} – 5(4) + 9 = 16 – 20+ 9 = 5

∴ f(1) = f(4)

Thus, the function f satisfies all the conditions of the Rolle’s theorem.

∴ there exists c ∈ (1, 4) such that f'(c) = 0.

Now, f(x) = x2 – 5x + 9

∴ f'(x) = \(\frac{d}{d x}\) (x^{2} – 5x + 9)

= 2x – 5 × 1 + 0

= 2x – 5

∴ f'(c) = 2c – 5

∴ f'(c) = 0 gives, 2c – 5 = 0

∴ c = 5/2 ∈ (1, 4)

Hence, the Rolle’s theorem is verified.

Question 4.

If Rolle’s theorem holds for the function f(x) = x^{3} + px^{2} + qx + 5, x ∈ [1, 3] with c = 2 + \(\frac{1}{\sqrt{3}}\), find the values of p and q.

Solution:

The Rolle’s theorem holds for the function f(x) = x^{3} + px^{2} + qx + 5, x ∈ [1, 3]

∴ f(1) = f(3)

∴ 1^{3} + p(1)^{2} + q(1) + 5 = 3^{3} + p (3)^{2} + q(3) + 5

∴ 1 + p + q + 5 = 27 + 9p + 3q + 5

∴ 8p + 2q = -26

∴ 4p + q = -13 ….. (1)

Also, there exists at least one point c ∈ (1, 3) such that f'(c) = 0.

Now, f'(x) = \(\frac{d}{d x}\) (x^{3} + px^{2} + qx + 5)

= 3x^{2} + p × 2x + q × 1 + 0

= 3x^{2} + 2px + q

But f'(c) = 0

∴ \(4 p+\frac{2 p}{\sqrt{3}}+q+13+\frac{12}{\sqrt{3}}=0\)

∴ (4√3 + 2)p + √3q + (13√3 + 12) = 0

∴ (4√3 + 2)p + √3q = -13√3 – 12 ……. (2)

Multiplying equation (1) by √3, we get

4√3p + √3q= -13√3

Subtracting this equation from (2), we get

2p = -12 ⇒ p= -6

∴ from (1), 4(-6) + q = -13 ⇒ q = 11

Hence, p = -6 and q = 11.

Question 5.

If Rolle’s theorem holds for the function f(x) = (x – 2) log x, x ∈ [1, 2], show that the equation x log x = 2 – x is satisfied by at least one value of x in (1, 2).

Solution:

The Rolle’s theorem holds for the function f(x) = (x – 2) log x, x ∈ [1, 2].

∴ there exists at least one real number c ∈ (1, 2) such that f'(c) = 0.

Now, f(x) = (x – 2) log x

∴ f'(c) = 0 gives 1 – \(\frac{2}{c}\) + log c = 0

∴ c – 2 + c log c = 0

∴ c log c = 2 – c, where c ∈ (1, 2)

∴ c satisfies the equation x log x = 2 – x, c ∈ (1, 2).

Hence, the equation x log x = 2 – x is satisfied by at least one value of x in (1, 2).

Question 6.

The function f(x) = \(x(x+3) e^{-\frac{x}{2}}\) satisfies all the conditions of Rolle’s theorem on [-3, 0]. Find the value of c such that f'(c) = 0.

Solution:

The function f(x) satisfies all the conditions of Rolle’s theorem, therefore there exist c ∈ (-3, 0) such that f'(c) = 0.

Question 7.

Verify Lagrange’s mean value theorem for the following functions:

(i) f(x) = log x on [1, e].

Solution:

The function f given as f(x) = log x is a logarithmic function that is continuous for all positive real numbers.

Hence, it is continuous on [1, e] and differentiable on (1, e).

Thus, the function f satisfies the conditions of Lagrange’s mean value theorem.

∴ there exists c ∈ (1, e) such that

Hence, Lagrange’s mean value theorem is verified.

(ii) f(x) = (x – 1)(x – 2)(x – 3) on [0, 4].

Solution:

The function f given as

f(x) = (x – 1)(x – 2)(x – 3)

= (x – 1)(x^{2} – 5x + 6)

= x^{3} – 5x^{2} + 6x – x^{2} + 5x – 6

= x^{3} – 6x^{2} + 11x – 6 is a polynomial function.

Hence, it is continuous on [0, 4] and differentiable on (0, 4).

Thus, the function f satisfies the conditions of Lagrange’s, mean value theorem.

∴ there exists c ∈ (0, 4) such that

Hence, Lagrange’s mean value theorem is verified.

(iii) f(x) = x^{2} – 3x – 1, x ∈ \(\left[\frac{-11}{7}, \frac{13}{7}\right]\)

Solution:

The function f given as f(x) = x^{2} – 3x – 1 is a polynomial function.

Hence, it is continuous on \(\left[\frac{-11}{7}, \frac{13}{7}\right]\) and differentiable on \(\left(\frac{-11}{7}, \frac{13}{7}\right)\).

Thus, the function f satisfies the conditions of LMVT.

∴ there exists c ∈ \(\left(\frac{-11}{7}, \frac{13}{7}\right)\) such that

Hence, Lagrange’s mean value theorem is verified.

(iv) f(x) = 2x – x^{2}, x ∈ [0, 1].

Solution:

The function f given as f(x) = 2x – x^{2} is a polynomial function.

Hence, it is continuous on [0, 1] and differentiable on (0, 1).

Thus, the function f satisfies the conditions of Lagrange’s mean value theorem.

∴ there exists c ∈ (0, 1) such that

Hence, Lagrange’s mean value theorem is verified.

(v) f(x) = \(\frac{x-1}{x-3}\) on [4, 5].

Solution:

The function f given as

f(x) = \(\frac{x-1}{x-3}\) is a rational function which is continuous except at x = 3.

But 3 ∉ [4, 5]

Hence, it is continuous on [4, 5] and differentiable on (4, 5).

Thus, the function f satisfies the conditions of Lagrange’s mean value theorem.

∴ there exists c ∈ (4, 5) such that

f'(c) = \(\frac{f(5)-f(4)}{5-4}\) ……..(1)

Hence, Lagrange’s mean value theorem is verified.