Match List-I with List-II. List-I

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Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Calculus

Question:

Match List-I with List-II.

List-I		List-II
(A)	If $y=log_e\left(\frac{x^2}{e^3}\right),$ then $\frac{d^2y}{dx^2}$ is,	(I)	$x^2(7+12log_ex)$
(B)	If $y=x^4log_ex,$ then $\frac{d^2y}{dx^2}$ is,	(II)	$\frac{2log_ex-3}{x^3}$
(C)	If $y=x^3e^x,$ then $\frac{d^2y}{dx^2}$ is,	(III)	$\frac{-2}{x^2}$
(D)	If $y=\frac{log_ex}{x},$ then $\frac{d^2y}{dx^2}$ is,	(IV)	$xe^x(x^2+6x+6)$

Choose the correct answer from the options given below :

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (2) → (A)-(III),(B)-(I),(C)-(IV),(D)-(II)

(A) $y=\log_e\left(\frac{x^2}{e^3}\right)⇒\frac{dy}{dx}=\frac{e^3}{x^2}×\frac{2x}{e^3}=\frac{2}{x}⇒\frac{d^2y}{dx^2}=\frac{-2}{x^2}$ (III)

(B) $y=x^4\log_ex⇒\frac{dy}{dx}4x^3\log_ex+x^3⇒\frac{d^2y}{dx^2}=12x^2\log x+4x^2+3x^2=12x^2\log x+7x^2$ (I)

(D) $y=\frac{\log_ex}{x}⇒\frac{dy}{dx}=\frac{\frac{1}{x}×x+\frac{\log x}{x^2}}{x^2}⇒\frac{d^2y}{dx^2}=\frac{2\log x-3}{x^3}$ (II)