Match List I with List II LIST I

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Applications of Derivatives

Question:

Match List I with List II

LIST I		LIST II
A.	$\frac{d}{dx}[x^{-3}(5+3x)]$	I.	$-\frac{2}{(x+1)^2}+\frac{3}{(x-2)^2}$
B.	$\frac{d}{dx}[(x-a)(x-b)]$	II.	$-\frac{15t}{x^4}-\frac{6}{x^3}$
C.	$\frac{d}{dx}\left[\frac{2}{x+1}-\frac{3}{x-2}\right]$	III.	$-\frac{2}{(x-1)^2}$
D.	$\frac{d}{dx}\left[\frac{x+1}{x-1}\right]$	IV.	$2x-a-b$

Choose the correct answer from the options given below :

Options:

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

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

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

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

Correct Answer:

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

Explanation:

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

A. $\frac{d}{dx}\left(x^{-3}(5+3x)\right)=\frac{d}{dx}\left(\frac{5}{x^3}+\frac{3}{x^2}\right)=-\frac{15t}{x^4}-\frac{6}{x^3}$ (II)

B. $\frac{d}{dx}[(x-a)(x-b)]=x-a+x-b=2x-a-b$ (IV)

C. $\frac{d}{dx}\left[\frac{2}{x+1}-\frac{3}{x-2}\right]=\frac{-2}{(x+1)^2}+\frac{3}{(x-2)^2}$ (I)

D. $\frac{d}{dx}\left[\frac{x+1}{x-1}\right]=\frac{x-1-x-1}{(x-1)^2}=\frac{-2}{(x-1)^2}$ (III)