Match List I with List II
| LIST I | LIST II | ||
| A. | \(\displaystyle\frac{d}{dx}\left[x^{-3}(5+3x)\right]\) | I. | \(-\displaystyle\frac{2}{(x+1)^2}+\frac{3}{(x-2)^2}\) |
| B. | \(\displaystyle\frac{d}{dx}\left[(x-a)(x-b)\right]\) | II. | \(-\displaystyle\frac{15}{x^4}-\frac{6}{x^3}\) |
| C. | \(\displaystyle\frac{d}{dx}\left[\frac{2}{x+1}-\frac{3}{x-2}\right]\) | III. | \(-\displaystyle\frac{2}{(x-1)^2}\) |
| D. | \(\displaystyle\frac{d}{dx}\left[\frac{x+1}{x-1}\right]\) | 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) \(\frac{d}{dx}\left[x^{-3}(5+3x)\right]\)
This matches (II) \(-\frac{15}{x^4}-\frac{6}{x^3}\).
Multiplying out first and using the power rule:
\(x^{-3}(5+3x) = 5x^{-3} + 3x^{-2}\)
\( \displaystyle \frac{d}{dx}\left(5x^{-3} + 3x^{-2}\right) = -15x^{-4} - 6x^{-3} = -\frac{15}{x^4}-\frac{6}{x^3} \)
(B) \(\frac{d}{dx}\left[(x-a)(x-b)\right]\)
This matches (IV) \(2x-a-b\).
By the product rule:
\(1\cdot(x-b) + (x-a)\cdot 1 = 2x - a - b\)
(C) \(\frac{d}{dx}\left[\frac{2}{x+1}-\frac{3}{x-2}\right]\)
This matches (I) \(-\frac{2}{(x+1)^2}+\frac{3}{(x-2)^2}\).
Since \(\frac{d}{dx}\left(\frac{1}{x+k}\right) = -\frac{1}{(x+k)^2}\):
\( \displaystyle \frac{d}{dx}\left[\frac{2}{x+1}-\frac{3}{x-2}\right] = -\frac{2}{(x+1)^2}+\frac{3}{(x-2)^2} \)
(D) \(\frac{d}{dx}\left[\frac{x+1}{x-1}\right]\)
This matches (III) \(-\frac{2}{(x-1)^2}\).
By the quotient rule:
\( \displaystyle \frac{(x-1)\cdot 1 - (x+1)\cdot 1}{(x-1)^2} = \frac{-2}{(x-1)^2} \)