Match the LIST-I with LIST-II, LIST-I Function f(x), LIST-II Second order derivative of f(x) w.r.t 'x', A., x^2 _e x, I., e^x(2 + 4x + x^2), B., e^x _e x, II., e^x(_e x + 2/x - 1/x^2), C., x^2 e^x, III., 2_e x + 3, D., x^(-2) e^x, IV., e^x/x^2(6/x^2 - 4/x + 1), Choose the correct answer from the options given below: Option 1: A-III, B-IV, C-I, D-II Option 2: A-III, B-II, C-I, D-IV Option 3: A-I, B-III, C-II, D-IV Option 4: A-III, B-I, C-II, D-IV

Answer: A-III, B-II, C-I, D-IV. Solution: Differentiating twice by the product rule: x^2 x gives 2 x+3; e^x x gives e^x( x+2/x-1/x^2); x^2e^x gives e^x(x^2+4x+2); and x^(-2)e^x gives e^x/x^2(6/x^2-4/x+1).

Calculus

Continuity & Differentiability

Medium

applied

Match the LIST-I with LIST-II

LIST-I
Function $f(x)$ LIST-II
Second order derivative of $f(x)$ w.r.t $'x'$
A. $x^2 \log_e x$ I. $e^x(2 + 4x + x^2)$
B. $e^x \log_e x$ II. $e^x\left(\log_e x + \dfrac{2}{x} - \dfrac{1}{x^2}\right)$
C. $x^2 e^x$ III. $2\log_e x + 3$
D. $x^{-2} e^x$ IV. $\dfrac{e^x}{x^2}\left(\dfrac{6}{x^2} - \dfrac{4}{x} + 1\right)$

Choose the correct answer from the options given below:

	LIST-I Function $f(x)$		LIST-II Second order derivative of $f(x)$ w.r.t $'x'$
A.	$x^2 \log_e x$	I.	$e^x(2 + 4x + x^2)$
B.	$e^x \log_e x$	II.	$e^x\left(\log_e x + \dfrac{2}{x} - \dfrac{1}{x^2}\right)$
C.	$x^2 e^x$	III.	$2\log_e x + 3$
D.	$x^{-2} e^x$	IV.	$\dfrac{e^x}{x^2}\left(\dfrac{6}{x^2} - \dfrac{4}{x} + 1\right)$

A-III, B-IV, C-I, D-II

A-III, B-II, C-I, D-IV

A-I, B-III, C-II, D-IV

A-III, B-I, C-II, D-IV

✅ Correct Option: 2

Differentiating twice by the product rule: $x^2\log x$ gives $2\log x+3$ ; $e^x\log x$ gives $e^x\left(\log x+\frac{2}{x}-\frac{1}{x^2}\right)$ ; $x^2e^x$ gives $e^x(x^2+4x+2)$ ; and $x^{-2}e^x$ gives $\frac{e^x}{x^2}\left(\frac{6}{x^2}-\frac{4}{x}+1\right)$ .

2026: 22 May Shift 2

If $y=\tan^{-1}(x+y)$ , then $\frac{d^2y}{dx^2}$ is equal to:

2025: 30 May Shift 2

If $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , then $\frac{d^2y}{dx^2}$ is equal to

2022: 25 May Shift 1

If $y = \frac{1}{x+1}$ , then $\frac{d^2y}{dx^2}$ at $x = 2$ is:

2025: 15 May Shift 1

If $x = 4t$ and $y = \frac{4}{t}$ , then $\frac{d^2y}{dx^2}$ is

2025: 16 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $f(x) = x \sin x$	(I) is not continuous at $x = -3$
(B) $f(x) = \frac{\vert x\vert }{x}, x \neq 0$ and $f(x) = 1 \text{ at } x = 0$	(II) is continuous everywhere
(C) $f(x) = x - [x]$ , $[x]$ denotes greatest integer function	(III) is not differentiable at $x = 1$
(D) $f(x) = e^{\vert x - 1\vert }$	(IV) is not continuous at $x = 0$

Choose the correct answer from the options given below:

2026: 30 May Shift 2

If $x = at^2$ and $y = 4at$ , then the value of $\frac{d^2y}{dx^2}$ at $t = 1/a$ , is

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