Search Previous Year Questions

Calculation:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2}, & x < 0 \ 9, & x = 0 \ \frac{\sqrt{x}}{\sqrt{(16 + \sqrt{-4})}}, & x > 0 \end{cases}$

To evaluate the limit $\lim_{x \to 0} f(x)$, we consider the left-hand and right-hand limits separately.

For $x \to 0^-$:

⇒ $\lim_{x \to 0^-} \frac{1 - \cos 2x}{x^2} = \lim_{x \to 0^-} \frac{2 \sin^2 x}{x^2} = 2$

Hence, the correct answer is Option 1.

Calculation:

We evaluate the right-hand limit of the function $f(x)$ as $x \to 0^+$:

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{(16 + \sqrt{x})} - 4}$

To simplify, multiply numerator and denominator by the conjugate of the denominator:

$= \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{(16 + \sqrt{x})} + 4)}{(16 + \sqrt{x}) - 16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{(16 + \sqrt{x})} + 4)}{\sqrt{x}}$

Canceling $\sqrt{x}$ from numerator and denominator:

$= \lim_{x \to 0^+} \sqrt{(16 + \sqrt{x})} + 4$

As $x \to 0^+$, $\sqrt{x} \to 0$, so:

$\lim_{x \to 0^+} f(x) = \sqrt{16} + 4 = 4 + 4 = 8$

Hence, the correct answer is option 3.

Calculation:

The function $f(x) = x|x|$ is defined as:

$f(x) = \begin{cases} x^2, & x \geq 0 \ -x^2, & x < 0 \end{cases}$

Evaluating the limit as $x \to -1$:

$\lim_{x \to -1} f(x) = \lim_{x \to -1} -x^2 = -1$

Hence, the correct answer is Option 1.

Calculation:

Required area:

$\left| \int_{-2}^{0} x^2 , dx \right| + \int_{0}^{1} x^2 , dx$

\( = -\left[ \frac{x^3}{3} \right]{-2}^{0} + \left[ \frac{x^3}{3} \right]{0}^{1} \)

$= \left[ 0 + \frac{8}{3} \right] + \left[ \frac{1}{3} - 0 \right] = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3$

Hence, the correct answer is Option 4.

Calculation:

$f(x) = \begin{cases} x^2, & x \ge 0 \ - x^2, & x < 0 \end{cases}$

It is clear from the graph that $f(x)$ is increasing in the interval $(-\infty, \infty)$.

$\therefore \text{L.H.D.} = 0 = \text{R.H.D.}$

$\Rightarrow f(x) \text{ is differentiable at } x = 0$

Hence both I and II are correct.

Calculation:

Given the function:

$f(x) = \frac{x}{1 - x}$

Evaluate $f(x + 1)$:

⇒ $f(x + 1) = \frac{x + 1}{1 - (x + 1)} = \frac{x + 1}{-x}$

Now compute $\frac{f(x)}{f(x + 1)}$:

⇒ $\frac{f(x)}{f(x + 1)} = \frac{\frac{x}{1 - x}}{\frac{x + 1}{-x}} = \frac{1 - x}{x + 1} \cdot \frac{x}{-x}$

$= \frac{x^2}{x^2 - 1} = -f(x^2)$

Hence, the correct answer is option 1.

Calculation:

Given the function:

$f(x) = \frac{x}{1 - x} \Rightarrow f(\sqrt{x}) = \frac{\sqrt{x}}{1 - \sqrt{x}}$

Evaluate $f(\sqrt{x} + 1)$:

⇒ $f(\sqrt{x} + 1) = \frac{\sqrt{x} + 1}{1 - \sqrt{x} - 1} = \frac{\sqrt{x} + 1}{-\sqrt{x}}$

Now compute the expression:

⇒ $(1 - x) \cdot \frac{\sqrt{x}}{1 - \sqrt{x}} - x \cdot \frac{\sqrt{x} + 1}{\sqrt{x}}$

Simplifying the numerator:

$= \frac{(1 - x)\sqrt{x}(1 - \sqrt{x}) - x(1 - \sqrt{x})(\sqrt{x} + 1)}{\sqrt{x}(1 - \sqrt{x})}$

$= \frac{x - x^2 - x + x^2}{\sqrt{x}(1 - \sqrt{x})} = 0$

Hence, the correct answer is Option 4.

Calculation:

Given the function:

$y = f(x) = x \cdot \frac{\sin^{-1}x}{\sqrt{1 - x^2}} + \ln \sqrt{1 - x^2}$

Differentiating $f(x)$ step-by-step:

⇒ $f'(x) = \left( \sin^{-1}x + \frac{x}{\sqrt{1 - x^2}} \right) \cdot \sqrt{1 - x^2} + x \cdot \sin^{-1}x + \frac{2x}{2\sqrt{1 - x^2}}$

$+ \frac{1}{\sqrt{1 - x^2}} \cdot \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x)$

Simplifying:

$= \sqrt{1 - x^2} \cdot \sin^{-1}x + x + \frac{x^2 \cdot \sin^{-1}x}{\sqrt{1 - x^2}} - \frac{x}{1 - x^2}$

Combining terms:

⇒​ $f'(x) = \frac{(1 - x^2)\sin^{-1}x + x\sqrt{1 - x^2} + x^2 \sin^{-1}x - x\sqrt{1 - x^2}}{(1 - x^2)\sqrt{1 - x^2}}$

$= \frac{\sin^{-1}x}{(1 - x^2)\sqrt{1 - x^2}}$

Evaluating the derivative at $x = 0.5$:

⇒​ $f'(0.5) = \frac{\sin^{-1}(0.5)}{(1 - 0.25)\sqrt{1 - 0.25}} = \frac{\frac{\pi}{6}}{\frac{3}{4} \cdot \frac{\sqrt{3}}{2}} = \frac{\pi}{6} \cdot \frac{8}{3\sqrt{3}} = \frac{4\pi\sqrt{3}}{27}$

Thus, the slope of $y = f(x)$ at $x = 0.5$ is $\frac{4\pi\sqrt{3}}{27}$.

Calculation:

$\frac{dy}{dx} = \frac{\sin^{-1}x}{(1 - x^2)^{3/2}}$

⇒ $\frac{dy}{dx^2} = \frac{\frac{1}{\sqrt{1 - x^2}}(1 - x^2)^{3/2} - \sin^{-1}x \cdot \frac{3}{2}(1 - x^2)^{1/2}(-2x)}{(1 - x^2)^3}$

$= \frac{\sqrt{1 - x^2} + 3x\sqrt{1 - x^2} \sin^{-1}x}{(1 - x^2)^3}$

$\left( \frac{d^2 y}{dx^2} \right)_{x=0} = \frac{1 + 0}{(1 - 0)^3} = 1$

Hence, the correct answer is Option 3.

Calculation:

$\frac{dy}{dx} = \frac{\sin^{-1}x}{(1 - x^2)^{3/2}}$

$\left( \frac{dy}{dx} \right) = \sin \theta = \frac{\sin^{-1}(\sin \theta)}{(1 - \sin^2 \theta)^{3/2}}$

$= \frac{\theta}{\cos^3 \theta} = \theta \cdot \sec^3 \theta$

Hence, the correct answer is Option 3.