Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

Match List-I with List-II

List-I List-II
(A) $y=log(sin x)$ (I) $\frac{d^2y}{dx^2}=-\frac{1}{x^2}$
(B) $y=e^{(1+logx)}$ (II) $\frac{d^2y}{dx^2}=2$
(C) $y=log|x|$ (III) $\frac{d^2y}{dx^2}=0$
(D) $y=x^2+4x-1$ (IV) $\frac{d^2y}{dx^2}=-cosec^2x$

Choose the correct answer from the options given below :

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is option (4) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

(A) $y=\log(\sin x)$

$\frac{dy}{dx}=\cot x⇒\frac{d^2y}{dx^2}=-cosec^2x$ (IV)

(B) $y=e^{(1+\log x)}$

$⇒\log y=1+\log x$

$\log y=\log ex⇒y=ex$

so $\frac{d^2y}{dx^2}=0$ (III)

(C) $y=\log x⇒\frac{d^2y}{dx^2}=\frac{-1}{x^2}$ (I)

(D) $y=x^2+4x-1⇒\frac{d^2y}{dx^2}=2$ (II)