Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

Match List-I with List-II

List-I List-II
(A) $\frac{d}{dx}(sin\, x^2)$ (I) $\frac{1}{5}$
(B) $\frac{d}{dx}(e^{sin\, x})$ (II) 0
(C) $f(x)=tan^{-1}x$ then $f'(2)$ (III) $2xcosx^2$
(D) If $y=3 \, cos x-2 \, sin x , $ then $\frac{d^2y}{dx^2}+y$ (IV) $e^{sinx}.cosx$

Choose the correct answer from the options given below :

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A) $\frac{d}{dx}(\sin\, x^2)=2x\cos x^2$ (III)

(B) $\frac{d}{dx}(e^{\sin x})=\cos e^{\sin x}$ (IV)

(C) $f(x)=\tan^{-1}x$

so $f'(x)=\frac{1}{1+x^2}⇒f'(2)=\frac{1}{5}$ (I)

(D) $y=3\cos x-2\sin x$

$\frac{dy}{dx}=-3\sin x-2\cos x$

$\frac{d^2y}{dx^2}=-3\cos x+2\sin x⇒\frac{d^2y}{dx^2}+y=0$ (II)