Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Function and point | Value of \(\displaystyle\frac{dy}{dx}\) at the point | ||
| (A) | \(y = x^{x},\ x = 1\) | (I) | 0 |
| (B) | \(y = xe^{-x},\ x = 1\) | (II) | \(\displaystyle\frac{1}{2}\) |
| (C) | \(y = \sqrt{x^{2} + 9},\ x = 4\) | (III) | \(\displaystyle\frac{4}{5}\) |
| (D) | \(y = \log_e(x^{2} + 3),\ x = 1\) | (IV) | 1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(A) \(y = x^{x},\ x = 1\)
This matches (IV) 1. By logarithmic differentiation, \(\log_e y = x\log_e x\):
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = \log_e x + 1 \)
\( \displaystyle \frac{dy}{dx} = x^{x}(1 + \log_e x) = 1 \times (1 + 0) = 1 \) at x = 1
(B) \(y = xe^{-x},\ x = 1\)
This matches (I) 0. By the product rule:
\( \displaystyle \frac{dy}{dx} = e^{-x} - xe^{-x} = e^{-x}(1 - x) = 0 \) at x = 1
(C) \(y = \sqrt{x^{2} + 9},\ x = 4\)
This matches (III) \(\frac{4}{5}\):
\( \displaystyle \frac{dy}{dx} = \frac{2x}{2\sqrt{x^{2} + 9}} = \frac{x}{\sqrt{x^{2} + 9}} \)
\( \displaystyle = \frac{4}{\sqrt{25}} = \frac{4}{5} \) at x = 4
(D) \(y = \log_e(x^{2} + 3),\ x = 1\)
This matches (II) \(\frac{1}{2}\):
\( \displaystyle \frac{dy}{dx} = \frac{2x}{x^{2} + 3} = \frac{2}{4} = \frac{1}{2} \) at x = 1