Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Function | Derivative | ||
| (A) | \(y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}\,;\ |x| \lt 1\) | (I) | \(\displaystyle\frac{dy}{dx}=\frac{1}{2y-1}\) |
| (B) | \(y=\sqrt{x+y}\), \(x+y \gt 0\) and \(y \ne \displaystyle\frac{1}{2}\) | (II) | \(\displaystyle\frac{dy}{dx}=10^x\log_e 10\) |
| (C) | \(y = \log_{10} x,\ x \gt 0\) | (III) | \(\displaystyle\frac{dy}{dx} = 0\) |
| (D) | \(y = 10^x\) | (IV) | \(\displaystyle\frac{dy}{dx}=\frac{1}{x\log_e 10}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(A) \(y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}\)
This matches (III) \(\frac{dy}{dx} = 0\), for 0 < x < 1. By the chain rule, the derivative of the second term is
\(\displaystyle \frac{d}{dx}\sin^{-1}\sqrt{1-x^2} = \frac{1}{\sqrt{1 - (1 - x^2)}}\cdot\frac{-x}{\sqrt{1-x^2}}\)
\(\displaystyle = \frac{-x}{|x|\sqrt{1-x^2}}\)
since \(\sqrt{x^2} = |x|\). For 0 < x < 1, |x| = x, so
\(\displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0\)
(Equivalently, for 0 < x < 1, \(\sin^{-1}\sqrt{1-x^2} = \cos^{-1}x\), so \(y = \frac{\pi}{2}\), a constant.) For −1 < x < 0, however, |x| = −x and \(\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}\), not 0.
(B) \(y=\sqrt{x+y}\)
This matches (I). Squaring, \(y^2 = x + y\). Differentiating implicitly:
\(\displaystyle 2y\,\frac{dy}{dx} = 1 + \frac{dy}{dx}\)
\(\displaystyle \frac{dy}{dx}\,(2y - 1) = 1\)
\(\displaystyle \frac{dy}{dx} = \frac{1}{2y - 1}\) (as \(y \ne \frac{1}{2}\))
(C) \(y = \log_{10} x\)
This matches (IV). By the change of base:
\(\displaystyle y = \frac{\log_e x}{\log_e 10}\)
\(\displaystyle \frac{dy}{dx} = \frac{1}{x\log_e 10}\)
(D) \(y = 10^x\)
This matches (II), since \(\frac{d}{dx}a^x = a^x\log_e a\):
\(\displaystyle \frac{dy}{dx} = 10^x\log_e 10\)
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).
Note: It seems that there is a mistake in the original NTA paper, which prints the condition in (A) as \(|x| \lt 1\) when it should read \(0 \lt x \lt 1\), since for \(-1 \lt x \lt 0\) the derivative is not 0. The question can still be solved because (B), (C) and (D) match only (I), (IV) and (II), which leaves (III) for (A), and Option (1) is the only option with (B)-(I), (C)-(IV), (D)-(II).