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 :
Answer & explanation
Correct answer: option 3
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)