Match List-I with List-II
(Consider \( \log_e x=\log x \))| List-I | List-II | ||
| Integral | Solution: where \( C \) is an arbitrary constant | ||
| (A) | \( \int\displaystyle\frac{f'(x)}{f(x)\log f(x)}dx \) | (I) | \( x\log(\log x)+c \) |
| (B) | \( \int\left(\log(\log x)+\displaystyle\frac{1}{\log x}\right)dx \) | (II) | \( \displaystyle\frac{x}{\log x}+C \) |
| (C) | \( \int\displaystyle\frac{\log x}{(1+\log x)^2}dx \) | (III) | \( \log\left|\log(f(x))\right|+c \) |
| (D) | \( \int\left(\displaystyle\frac{1}{\log x}-\displaystyle\frac{1}{(\log x)^2}\right)dx \) | (IV) | \( \displaystyle\frac{x}{1+\log x}+c \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
(A) \( \int\frac{f'(x)}{f(x)\log f(x)}\,dx \)
This matches (III) \( \log\left|\log(f(x))\right|+c \).
Let \( t=\log f(x) \), so \( dt=\frac{f'(x)}{f(x)}\,dx \):
\( \displaystyle \int\frac{dt}{t}=\log|t|+c \)
\( \displaystyle =\log\left|\log f(x)\right|+c \)
(B) \( \int\left(\log(\log x)+\frac{1}{\log x}\right)dx \)
This matches (I) \( x\log(\log x)+c \).
Differentiating the answer by the product rule:
\( \displaystyle \frac{d}{dx}\Big[x\log(\log x)\Big]=\log(\log x)+x\cdot\frac{1}{\log x}\cdot\frac{1}{x} \)
\( \displaystyle =\log(\log x)+\frac{1}{\log x} \)
(C) \( \int\frac{\log x}{(1+\log x)^2}\,dx \)
This matches (IV) \( \frac{x}{1+\log x}+c \).
By the quotient rule,
\( \displaystyle \frac{d}{dx}\left[\frac{x}{1+\log x}\right]=\frac{(1+\log x)\cdot 1-x\cdot\frac{1}{x}}{(1+\log x)^2} \)
\( \displaystyle =\frac{\log x}{(1+\log x)^2} \)
(D) \( \int\left(\frac{1}{\log x}-\frac{1}{(\log x)^2}\right)dx \)
This matches (II) \( \frac{x}{\log x}+C \).
\( \displaystyle \frac{d}{dx}\left[\frac{x}{\log x}\right]=\frac{\log x\cdot 1-x\cdot\frac{1}{x}}{(\log x)^2} \)
\( \displaystyle =\frac{\log x-1}{(\log x)^2} \)
\( \displaystyle =\frac{1}{\log x}-\frac{1}{(\log x)^2} \)
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).