Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Mean Value Theorems Important Questions to help strengthen their preparations for exams.

## TS Inter 1st Year Maths 1B Mean Value Theorems Important Questions

Question 1.

Verify Rolle’s theorem for the function y = f(x) = x^{2} + 4 in [-3, 3]. [Mar. ’19 (TS); Mar. ’17 (AP), ’15 (TS); B.P.; May ’15 (TS)]

Solution:

f is a polynomial function that is continuous and differentiable in [-3, 3]

Given f(x) = x^{2} + 4

∴ f(-3) = (-3)^{2} + 4 = 13

and f(3) = 3^{2} + 4 = 13

∴ f(-3) = f(3).

Hence the function satisfies the conditions of Rolle’s theorem and f'(x) = 2x

∴ f'(c) = 2c = 0

c = 0 ∈ (-3, 3)

Thus Roll’s theorem is verified.

Question 2.

Verify Rolle’s theorem for the following function x^{2} – 1 on [-1, 1]. [Mar. ’14; May ’13]

Solution:

Let f(x) = x^{2} – 1 defined on [-1, 1].

Since the function f(x) is a polynomial, it is continuous on [-1, 1] and differentiable on (-1, 1).

Also f(1) = 1 – 1 = 0,

f(-1) = (-1)^{2} – 1 = 0

f(-1) = f(1).

Hence f satisfies all conditions of Rolle’s theorem.

Now we have to find a point c ∈ (-1, 1) such that f'(c) = 0, f'(x) = 2x

and f'(c) = 0

2c = 0

c = 0 ∈ (-1, 1)

Hence Rolle’s theorem is verified.

Question 3.

Let f(x) = (x – 1) (x – 2) (x – 3). Prove that there is more than one ‘c’ in (1, 3) such that f'(c) = 0. [Mar. ’13]

Solution:

f is a polynomial in x which is continuous over [1, 3] and derivable over (1, 3).

Given f(x) = (x – 1) (x – 2) (x – 3)

We have f'(x) = (x – 1) (x – 2) + (x – 2) (x – 3) + (x – 1) (x – 3)

= x^{2} – 3x + 2 + x^{2} – 5x + 6 + x^{2} – 4x + 3

= 3x^{2} – 12x + 11

f'(c) = 0

3c^{2} – 12c + 11 = 0

c = \(\frac{12 \pm \sqrt{144-132}}{6}\) = 2 ± \(\frac{1}{\sqrt{3}}\)

2 + \(\frac{1}{\sqrt{3}}\) ∈ (1, 3) and 2 – \(\frac{1}{\sqrt{3}}\) ∈ (1, 3)

∴ There exists more than one ‘c’ in (1, 3) such that f'(c) = 0.

Question 4.

Verify Rolle’s theorem for the function f(x) = x(x + 3) \(\mathrm{e}^{-\mathrm{x} / 2}\) in [-3, 0]. [Mar. ’18 (AP)]

Solution:

f is continuous in [-3, 0] and differentiable in (-3, 0).

Also f(-3) = (-3) (-3 + 3) e^{3/2} = 0

f(0) = 0 (0 . 3) e^{3/2} = 0

∴ f(-3) = f (0)

So f satisfies the conditions of Rolle’s theorem.

Given f(x) = x(x + 3) e^{3/2}

Then f'(x) = x(x + 3) \(\left(\frac{-1}{2}\right) \mathrm{e}^{-\mathrm{x} / 2}\) + \(x\left(e^{\frac{-x}{2}}\right)+(x+3) e^{\frac{-x}{2}}\)

f'(e) = 0

-e^{2} + e + 6 = 0

e = -2 or 3

Now -2 ∈ (- 3, 0) and c = -2

∴ Rolle’s theorem was verified.

Question 5.

Verify Lagrange’s mean value theorem for the function f(x) = x^{2} on [2, 4].

Solution:

Being a polynomial function

(i) f(x) = x^{2} is continuous on [2, 4] and

(ii) differentiable in (2, 4)

By Lagrange’s theorem there should exist c ∈ (2, 4) such

Hence Lagrange’s theorem is verified.

Question 6.

On the curve y = x^{2}, find a point at which the tangent is parallel to the chord joining (0, 0) and (1, 1).

Solution:

The slope of the chord joining (0, 0) and (1, 1) is

m_{1} = \(\frac{1-0}{1-0}\) = 1

Given curve is y = x^{2}

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2x

∴ The slope of the tangent at any point on the curve is m_{2} = 2x

If the tangent is parallel to the chord, then m_{1} = m_{2}

2x = 1

x = \(\frac{1}{2}\)

Now, y = x^{2} = \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

∴ The required point on the curve is \(\left(\frac{1}{2}, \frac{1}{4}\right)\)

Question 7.

Verify Rolle’s Theorem for the following function sin x – sin 2x on [0, π].

Solution:

Let f(x) = sin x – sin 2x defined over [0, π].

f is continuous over [0, π] and differentiable over (0, π).

Also f(0) = 0 and f(π) = sin π – sin 2π = 0

f(0) = f(π) = 0

Hence f satisfies the conditions of Rolle’s theorem.

Also f'(x) = cos x – 2 cos 2x and f'(c) = 0

cos c – 2 cos 2c = 0

cos c = 2(2 cos^{2}c – 1)

4 cos^{2}c – cos c – 2 = 0

Hence conditions of Rolle’s theorem are verified.

Question 8.

Show that there is no real number k, for which the equation x^{2} – 3x + k = 0 has two distinct roots in [0, 1].

Solution:

Let f(x) = x^{2} – 3x + k and suppose there exist two different roots α, β (α < β).

Since f is a polynomial in x, it is continuous over [0, 1] and differentiable in (0, 1).

f is continuous over [α, β] and differentiable in (α, β).

Also f(α) = α^{2} – 3α + k = 0 and

f(β) = β^{2} – 3β + k = 0

∴ f(α) = f(β) = 0

∴ f satisfies the conditions of Rolle’s theorem.

Also f'(c) = 0

3c^{2} – 3 = 0

c^{2} = 1

c = ±1

This is a contradiction since 0 < α < c < β < 1

Hence there do not exist roots α, β in (0, 1).

So, there is no real number k for which the equation x^{2} – 3x + k = 0 has two distinct roots in [0, 1].

Question 9.

Find a point on the graph of the curve y = (x – 3)^{2} where the tangent is parallel to the chord joining (3, 0) and (4, 1).

Solution:

Let the points be A(3, 0) and B(4, 1)

Slope of chord AB = \(\frac{1-0}{4-3}\) = 1

Given y = (x – 3)^{2}

\(\frac{d y}{d x}\) = 2(x – 3)

Slope of the chord 2(x – 3) = 1

2x = 7

x = \(\frac{7}{2}\)

∴ y = (x – 3)^{2}

= \(\left(\frac{7}{2}-3\right)^2=\frac{1}{4}\)

∴ The point on the curve = \(\left(\frac{7}{2},\frac{1}{4}\right)\)

Question 10.

Find a point on the graph of the curve y = x^{3} where the tangent is parallel to the chord joining (1, 1) and (3, 27).

Solution:

Let the points be A(1, 1) and B(3, 27)

Question 11.

Find ‘c’ so that f'(c) = \(\frac{\mathbf{f}(\mathbf{b})-\mathbf{f}(\mathbf{a})}{b-\mathbf{a}}\) where f(x) = e^{x}; a = 0, b = 1.

Solution:

Question 12.

Verify Rolle’s theorem for the function (x^{2} – 1) (x – 2) on [-1, 2]. Find a point in the interval where the derivative vanishes.

Solution:

Let f(x) = (x^{2} – 1) (x – 2) = x^{3} – 2x^{2} – x + 2 defined over [-1, 2].

f being a polynomial in ‘x’, it is continuous over [-1, 2] and differentiable over (-1, 2).

f(-1) = 0, f(2) = 0

∴ f(-1) = f(2)

∴ f satisfies all the conditions of Rolle’s theorem.

f(x) = 3x^{2} – 4x – 1 and f'(c) = 0

3c^{2} – 4c – 1 = 0

∴ Rolle’s theorem is verified.

Question 13.

Verify the conditions of Lagrange’s mean value theorem for the function x^{2} – 1 on [2, 3]. [Mar. ’18 (TS); Mar. ’16 (AP)]

Solution:

Let f(x) = x^{2} – 1 defined over [2, 3]. This is a polynomial in x, continuous on [2, 3] and differentiable over (2, 3).

So by Lagrange’s mean value theorem, there exists a point ‘c’ ∈ (2, 3) such that

f'(c) = \(\frac{\mathrm{f}(3)-\mathrm{f}(2)}{3-2}\)

f'(x) = 2x

f'(c) = 2c

f(3) = 8, f(2) = 3

∴ 2c = \(\frac{8-3}{1}\) = 5

c = \(\frac{5}{2}\) ∈ (2, 3)

Question 14.

Verify the conditions of Lagrange’s mean value theorem for the function sin x – sin 2x on [0, π].

Solution:

Let f(x) = sin x – sin 2x defined over [0, π].

This is continuous on [0, π] and differentiable over (0, π)

since f'(x) = cos x – 2 cos 2x exists for all x ∈ (0, π).

So by Lagrange’s mean value theorem

f'(c) = \(\frac{\mathrm{f}(\pi)-\mathrm{f}(0)}{\pi-0}\), f(π) = 0, f(0) = 0

∴ cos c – 2 cos 2c = \(\frac{0}{\pi}\) = 0

cos c – 2 cos 2c = 0

cos c – 2(2 cos^{2}c – 1) = 0

4cos^{2}c – cos c – 2 = 0

cos c = \(\frac{1 \pm \sqrt{1+32}}{8}=\frac{1 \pm \sqrt{33}}{8}\)

c = \(\cos ^{-1}\left(\frac{1+\sqrt{33}}{8}\right)\) ∈ (0, π)

Question 15.

Verify the conditions of Lagrange’s mean value theorem for the function log x on [1, 2].

Solution:

Let f(x) = log x defined over [1, 2].

This is continuous over [1, 2] and differentiable over (1, 2) since

f'(x)= \(\frac{1}{x}\) ∀ x ∈ (1, 2)

Hence Lagrange’s mean value theorem is satisfied.

Question 16.

Verify Rolle’s theorem for the function f : [-3, 8] → R be defined by f(x) = x^{2} – 5x + 6. [Mar. ’17 (TS)]

Solution:

Given f(x) = x^{2} – 5x + 6 defined on [-3, 8].

Since the function f(x) is a polynomial,

it is continuous on [-3, 8] and differentiable on [-3, 8)

Also f(-3) = (-3)^{2} – 5(-3) + 6

= 9 + 15 + 6

= 30

f(8) = (8)^{2} – 5(8) + 6

= 64 – 40 + 6

= 30

∴ f(-3) = f(8)

Hence f satisfies all conditions of Rolle’s theorem.

Now we have to find a point C ∈ (-3, 8) such that f'(C) = 0

f'(x) = 2x – 5 and f'(c) = 0

2C – 5 = 0

C = \(\frac{5}{2}\) ∈ (-3, 8)

Hence Rolle’s theorem is verified.