Differentiation (Part-38) | S N De

Differentiation (Part-34) | S N Dey Class 12 Mathematics Solutions

1. State Rolle's theorem. Is this theorem applicable to the function \( f(x) = x \) in the interval \( -1 \le x \le 1 \)?

Rolle's Theorem Statement:

Let \( y = f(x) \) be a single-valued function defined in the closed interval \( [a, b] \) such that:

1. It is continuous in \( [a, b] \),
2. It is differentiable in \( (a, b) \), and
3. \( f(a) = f(b) \).

Then there exists at least one value of \( x \) (say \( x = c \)) between \( a \) and \( b \) (i.e., \( a < c < b \)) such that \( f'(c) = 0 \).

Applicability to \( f(x) = x \) on \( [-1, 1] \):

The function \( f(x) = x \) is continuous on \( [-1, 1] \) and differentiable on \( (-1, 1) \).

However, evaluating at the endpoints gives:

$$ f(-1) = -1 \quad \text{and} \quad f(1) = 1 $$

$$ \implies f(-1) \neq f(1) $$

\( \therefore \) Since the condition \( f(a) = f(b) \) is not satisfied, Rolle's theorem is not applicable.

Verification of Rolle's Theorem

2 (i). Verify the truth of Rolle's theorem for \( f(x) = x^2 + 2x - 8 \) in \( -4 \le x \le 2 \).

Solution:

Since \( f(x) \) is a polynomial function, it is continuous in \( [-4, 2] \) and differentiable in \( (-4, 2) \).

Checking boundary values:

$$ f(-4) = (-4)^2 + 2(-4) - 8 = 16 - 8 - 8 = 0 $$

$$ f(2) = 2^2 + 2(2) - 8 = 4 + 4 - 8 = 0 $$

$$ \implies f(-4) = f(2) = 0 $$

Since all conditions of Rolle's theorem are satisfied, there exists at least one value \( c \in (-4, 2) \) such that \( f'(c) = 0 \).

Differentiating \( f(x) \):

$$ f'(x) = \frac{d}{dx}(x^2 + 2x - 8) = 2x + 2 $$

Setting \( f'(c) = 0 \):

$$ 2c + 2 = 0 \implies 2(c + 1) = 0 \implies c = -1 $$

\( \therefore c = -1 \in (-4, 2) \). Hence, Rolle's theorem is verified.

2 (ii). Verify the truth of Rolle's theorem for \( f(x) = x^3 + 5x^2 - 6x \) in \( 0 \le x \le 1 \).

Solution:

Being a polynomial function, \( f(x) \) is continuous in \( [0, 1] \) and differentiable in \( (0, 1) \).

Checking boundary values:

$$ f(0) = 0^3 + 5(0)^2 - 6(0) = 0 $$

$$ f(1) = 1^3 + 5(1)^2 - 6(1) = 1 + 5 - 6 = 0 $$

$$ \implies f(0) = f(1) = 0 $$

By Rolle's theorem, there exists at least one value \( c \in (0, 1) \) such that \( f'(c) = 0 \).

Differentiating \( f(x) \):

$$ f'(x) = 3x^2 + 10x - 6 $$

Setting \( f'(c) = 0 \):

$$ 3c^2 + 10c - 6 = 0 $$

Solving using the quadratic formula:

$$ c = \frac{-10 \pm \sqrt{10^2 - 4(3)(-6)}}{2(3)} = \frac{-10 \pm \sqrt{172}}{6} = \frac{-5 \pm \sqrt{43}}{3} $$

Evaluating roots:

$$ c_1 = \frac{-5 + \sqrt{43}}{3} \approx 0.52 \in (0, 1) $$

$$ c_2 = \frac{-5 - \sqrt{43}}{3} \approx -3.85 \notin (0, 1) $$

\( \therefore c \approx 0.52 \in (0, 1) \). Hence, Rolle's theorem is verified.

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2 (iii). Verify the truth of Rolle's theorem for \( f(x) = (x-1)(x-2)(x-3) \) in \( [1, 3] \).

Solution:

Expanding the function:

$$ f(x) = (x^2 - 3x + 2)(x - 3) = x^3 - 6x^2 + 11x - 6 $$

Since \( f(x) \) is a polynomial, it is continuous in \( [1, 3] \) and differentiable in \( (1, 3) \).

Checking boundary values:

$$ f(1) = (1-1)(1-2)(1-3) = 0 $$

$$ f(3) = (3-1)(3-2)(3-3) = 0 $$

$$ \implies f(1) = f(3) = 0 $$

By Rolle's theorem, there exists \( c \in (1, 3) \) such that \( f'(c) = 0 \).

Differentiating \( f(x) \):

$$ f'(x) = 3x^2 - 12x + 11 $$

Setting \( f'(c) = 0 \):

$$ 3c^2 - 12c + 11 = 0 \implies c = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{6 \pm \sqrt{3}}{3} $$

Evaluating roots:

$$ c_1 = \frac{6 + \sqrt{3}}{3} \approx 2.58 \in (1, 3) $$

$$ c_2 = \frac{6 - \sqrt{3}}{3} \approx 1.42 \in (1, 3) $$

\( \therefore c \approx 1.42, 2.58 \in (1, 3) \). Hence, Rolle's theorem is verified.

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2 (iv). Verify the truth of Rolle's theorem for \( f(x) = x^3 - x^2 - 4x + 4 \) in \( -2 \le x \le 2 \).

Solution:

Being a polynomial function, \( f(x) \) is continuous in \( [-2, 2] \) and differentiable in \( (-2, 2) \).

Checking boundary values:

$$ f(-2) = (-2)^3 - (-2)^2 - 4(-2) + 4 = -8 - 4 + 8 + 4 = 0 $$

$$ f(2) = 2^3 - 2^2 - 4(2) + 4 = 8 - 4 - 8 + 4 = 0 $$

$$ \implies f(-2) = f(2) = 0 $$

By Rolle's theorem, there exists \( c \in (-2, 2) \) such that \( f'(c) = 0 \).

Differentiating \( f(x) \):

$$ f'(x) = 3x^2 - 2x - 4 $$

Setting \( f'(c) = 0 \):

$$ 3c^2 - 2c - 4 = 0 \implies c = \frac{2 \pm \sqrt{(-2)^2 - 4(3)(-4)}}{2(3)} = \frac{2 \pm \sqrt{52}}{6} $$

Evaluating roots:

$$ c_1 = \frac{2 + \sqrt{52}}{6} \approx 1.54 \in (-2, 2) $$

$$ c_2 = \frac{2 - \sqrt{52}}{6} \approx -0.87 \in (-2, 2) $$

\( \therefore c \approx -0.87, 1.54 \in (-2, 2) \). Hence, Rolle's theorem is verified.

2 (v). Verify the truth of Rolle's theorem for \( f(x) = \sin^2 x \) in \( 0 \le x \le \pi \).

Solution:

The trigonometric function \( f(x) = \sin^2 x \) is continuous in \( [0, \pi] \) and differentiable in \( (0, \pi) \).

Checking boundary values:

$$ f(0) = \sin^2 0 = 0 \quad \text{and} \quad f(\pi) = \sin^2 \pi = 0 $$

$$ \implies f(0) = f(\pi) = 0 $$

By Rolle's theorem, there exists \( c \in (0, \pi) \) such that \( f'(c) = 0 \).

Differentiating \( f(x) \):

$$ f'(x) = 2\sin x \cos x = \sin 2x $$

Setting \( f'(c) = 0 \):

$$ \sin 2c = 0 $$

Since \( c \in (0, \pi) \implies 2c \in (0, 2\pi) \):

$$ 2c = \pi \implies c = \frac{\pi}{2} $$

\( \therefore c = \frac{\pi}{2} \in (0, \pi) \). Hence, Rolle's theorem is verified.