Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| Function \( f(x) \) | Second order derivative of \( f(x) \) w.r.t \( 'x' \) | ||
| (A) | \( x^2\log_e x \) | (I) | \( e^x(2+4x+x^2) \) |
| (B) | \( e^x\log_e x \) | (II) | \( e^x\left(\log_e x+\displaystyle\frac{2}{x}-\frac{1}{x^2}\right) \) |
| (C) | \( x^2e^x \) | (III) | \( 2\log_e x+3 \) |
| (D) | \( x^{-2}e^x \) | (IV) | \( \displaystyle\frac{e^x}{x^2}\left(\frac{6}{x^2}-\frac{4}{x}+1\right) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
Each function is a product, so the product rule is used twice.
(A) \( x^2\log_e x \)
This matches (III) \( 2\log_e x+3 \).
\( \displaystyle f'(x)=2x\log_e x+x^2\cdot\frac{1}{x} \)
\( \displaystyle =2x\log_e x+x \)
\( \displaystyle f''(x)=2\log_e x+2x\cdot\frac{1}{x}+1 \)
\( \displaystyle =2\log_e x+3 \)
(B) \( e^x\log_e x \)
This matches (II) \( e^x\left(\log_e x+\frac{2}{x}-\frac{1}{x^2}\right) \).
\( \displaystyle f'(x)=e^x\left(\log_e x+\frac{1}{x}\right) \)
\( \displaystyle f''(x)=e^x\left(\log_e x+\frac{1}{x}\right)+e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) \)
\( \displaystyle =e^x\left(\log_e x+\frac{2}{x}-\frac{1}{x^2}\right) \)
(C) \( x^2e^x \)
This matches (I) \( e^x(2+4x+x^2) \).
\( \displaystyle f'(x)=e^x(x^2+2x) \)
\( \displaystyle f''(x)=e^x(x^2+2x)+e^x(2x+2) \)
\( \displaystyle =e^x(2+4x+x^2) \)
(D) \( x^{-2}e^x \)
This matches (IV) \( \frac{e^x}{x^2}\left(\frac{6}{x^2}-\frac{4}{x}+1\right) \).
\( \displaystyle f'(x)=e^x(x^{-2}-2x^{-3}) \)
\( \displaystyle f''(x)=e^x(x^{-2}-2x^{-3})+e^x(-2x^{-3}+6x^{-4}) \)
\( \displaystyle =e^x(x^{-2}-4x^{-3}+6x^{-4}) \)
\( \displaystyle =\frac{e^x}{x^2}\left(1-\frac{4}{x}+\frac{6}{x^2}\right) \)
So the matching is (A)-(III), (B)-(II), (C)-(I), (D)-(IV).
Shortcut: (A) has no \( e^x \), so it can only match (III); (B) has \( \log_e x \) together with \( e^x \), so it can only match (II). Only Option (2) has (A)-(III) and (B)-(II).