Question

Match List-I with List-II:

List-I	List-II
The derivative of \( \log_e x \) with respect to \( \frac{1}{x} \) at \( x = 5 \) is	(I) -5
If \( x^3 + x^2y + xy^2 - 21x = 0 \), then \( \frac{dy}{dx} \) at \( (1, 1) \) is	(II) -6
If \( f(x) = x^3 \log_e \frac{1}{x} \), then \( f'(1) + f''(1) \) is	(III) 5
If \( y = f(x^2) \) and \( f'(x) = e^{\sqrt{x}} \), then \( \frac{dy}{dx} \) at \( x = 0 \) is	(IV) 0

Choose the correct answer from the options given below :

• A. (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
• B. (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
• C. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
• D. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Answer 1

The Correct Option is A

The derivative of $\log x$ with respect to $x$ is computed as:
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$, where $u = \frac{1}{x}$.
$\frac{dy}{dx} = \frac{1}{x} \cdot (-x^{-2}) = -\frac{1}{x^2}$.
At $x=5$, $\frac{dy}{dx} = -\frac{1}{25}$.
Match: (A) $\to$ (I).
Given $x^3 + x^2y + y^2 = 21$, differentiate implicitly:
$3x^2 + 2xy + x^2 \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$.
Rearrange:
$(x^2 + 2y) \frac{dy}{dx} = 21 - 3x^2 - 2xy$.
Substituting $(x, y) = (1, 1)$:
$3 \frac{dy}{dx} = 21 - 3 - 2 = 16 \implies \frac{dy}{dx} = \frac{16}{3}$.
Match: (B) $\to$ (III).
Given $f(x) = x^3 \log_e x$, simplify as $f(x) = x^3 \log x$. Compute $f'(x)$ using the product rule:
$f'(x) = 3x^2 \log x + x^3 \cdot \frac{1}{x} = 3x^2 \log x + x^2$.
Compute $f''(x)$:
$f''(x) = 6x \log x + 3x^2 \cdot \frac{1}{x} + 2x = 6x \log x + 5x$.
At $x = 1$, $\log_e 1 = 0$:
$f'(1) = 1$, $f''(1) = 5 \implies f'(1) + f''(1) = 1 + 5 = 6$.
Match: (C) $\to$ (II).
Given $y = f(x^2)$ and $f'(x) = \sqrt{x}$, use the chain rule:
$\frac{dy}{dx} = f'(x^2) \cdot \frac{d}{dx}(x^2) = f'(x^2) \cdot 2x$.
At $x = 0$:
$\frac{dy}{dx} = f'(0^2) \cdot 2(0) = 0$.
Match: (D) $\to$ (IV).