Differentiation (Part-38) | S N De

Differentiation (Part-36) | S N Dey Class 12 Mathematics Solutions

3. Show that Rolle's theorem is not applicable to the following functions in the specified intervals:

(i) \( f(x) = 1 - \sqrt[3]{x^2} \) in \( -1 \le x \le 1 \)

Solution:

The given function is \( f(x) = 1 - x^{2/3} \).

Clearly, \( f(x) \) has a definite and unique real value at every point in the closed interval \( [-1, 1] \). Therefore, \( f(x) \) is continuous in \( [-1, 1] \).

Differentiating \( f(x) \) with respect to \( x \):

$$ f'(x) = \frac{d}{dx}\left(1 - x^{2/3}\right) = 0 - \frac{2}{3}x^{\frac{2}{3}-1} = -\frac{2}{3x^{1/3}} $$

At \( x = 0 \in (-1, 1) \), the derivative \( f'(x) \) becomes undefined (division by zero).

Since \( f'(x) \) does not exist at \( x = 0 \in (-1, 1) \), the condition of differentiability on the open interval is violated. Hence, Rolle's theorem is not applicable.

(ii) \( f(x) = \tan x \) in \( 0 \le x \le \pi \)

Solution:

Consider \( f(x) = \tan x \) on the interval \( [0, \pi] \).

At \( x = \frac{\pi}{2} \in [0, \pi] \):

$$ \lim_{x \to \frac{\pi}{2}^-} \tan x = +\infty \quad \text{and} \quad \lim_{x \to \frac{\pi}{2}^+} \tan x = -\infty $$

Since \( f\left(\frac{\pi}{2}^+\right) \neq f\left(\frac{\pi}{2}^-\right) \), the function \( f(x) \) is discontinuous at \( x = \frac{\pi}{2} \).

The function is neither continuous on \( [0, \pi] \) nor differentiable on \( (0, \pi) \). Hence, Rolle's theorem is not applicable.

(iii) \( f(x) = |x - 1| \) in \( 0 \le x \le 2 \)

Solution:

By definition, the absolute value function is defined as:

$$ f(x) = \begin{cases} x - 1 & \text{if } x \ge 1 \\ -(x - 1) & \text{if } x < 1 \end{cases} $$

Since polynomial sub-functions are continuous and differentiable everywhere, we check differentiability at the critical boundary point \( x = 1 \in (0, 2) \):

Right-Hand Derivative (RHD):

$$ Rf'(1) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{[(1+h) - 1] - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 $$

Left-Hand Derivative (LHD):

$$ Lf'(1) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{-[(1+h) - 1] - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 $$

$$ \therefore Rf'(1) \neq Lf'(1) $$

Since \( f(x) \) is not differentiable at \( x = 1 \in (0, 2) \), the condition of differentiability on the open interval is not satisfied. Hence, Rolle's theorem is not applicable.

(iv) \( f(x) = 2 + (x - 2)^{2/3} \) in \( 1 \le x \le 3 \)

Solution:

The function \( f(x) = 2 + (x - 2)^{2/3} \) yields a well-defined real value at all points in \( [1, 3] \), making it continuous in \( [1, 3] \).

Differentiating \( f(x) \):

$$ f'(x) = \frac{d}{dx}\left[2 + (x - 2)^{2/3}\right] = 0 + \frac{2}{3}(x - 2)^{\frac{2}{3}-1} = \frac{2}{3(x - 2)^{1/3}} $$

At \( x = 2 \in (1, 3) \), the derivative \( f'(x) \) becomes undefined due to a zero denominator.

Since \( f'(x) \) does not exist at \( x = 2 \in (1, 3) \), the differentiability requirement fails. Hence, Rolle's theorem is not applicable.

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(v) Piecewise function \( f(x) \) in \( 0 \le x \le 2 \):

$$ f(x) = \begin{cases} x^2 + 1 & \text{when } 0 \le x \le 1 \\ 3 - x & \text{when } 1 < x \le 2 \end{cases} $$

Solution:

Each component polynomial is continuous and differentiable within its respective domain. We examine differentiability at the transition point \( x = 1 \in (0, 2) \):

Right-Hand Derivative (RHD):

$$ Rf'(1) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{[3 - (1+h)] - (1^2 + 1)}{h} = \lim_{h \to 0^+} \frac{2 - h - 2}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1 $$

Left-Hand Derivative (LHD):

$$ Lf'(1) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{[(1+h)^2 + 1] - (1^2 + 1)}{h} = \lim_{h \to 0^-} \frac{1 + 2h + h^2 + 1 - 2}{h} = \lim_{h \to 0^-} \frac{h(2 + h)}{h} = 2 $$

$$ \therefore Rf'(1) \neq Lf'(1) $$

Since \( Rf'(1) \neq Lf'(1) \), \( f(x) \) is not differentiable at \( x = 1 \in (0, 2) \). Hence, Rolle's theorem is not applicable.