For \(-1 \lt x \lt 1\), let \(L = (1 - x^{2})\displaystyle\frac{d^{2}y}{dx^{2}} - x\frac{dy}{dx}\). Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Function y | Value of L | ||
| (A) | \(y = \sin^{-1}x\) | (I) | \(4y\) |
| (B) | \(y = (\sin^{-1}x)^{2}\) | (II) | 0 |
| (C) | \(y = e^{2\sin^{-1}x}\) | (III) | \(-9y\) |
| (D) | \(y = \cos(3\sin^{-1}x)\) | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Each function has the form \(y = F(\theta)\) with \(\theta = \sin^{-1}x\). A general result handles all four at once.
\( \displaystyle \frac{d\theta}{dx} = \frac{1}{\sqrt{1 - x^{2}}}, \quad \frac{dy}{dx} = \frac{F'(\theta)}{\sqrt{1 - x^{2}}} \)
\( \displaystyle \sqrt{1 - x^{2}}\,\frac{dy}{dx} = F'(\theta) \)
Differentiating again with respect to x:
\( \displaystyle \sqrt{1 - x^{2}}\,\frac{d^{2}y}{dx^{2}} - \frac{x}{\sqrt{1 - x^{2}}}\frac{dy}{dx} = \frac{F''(\theta)}{\sqrt{1 - x^{2}}} \)
Multiplying by \(\sqrt{1 - x^{2}}\):
\(L = F''(\theta)\)
(A) \(y = \sin^{-1}x\)
This matches (II) 0. Here \(F(\theta) = \theta\), so \(F''(\theta) = 0\).
(B) \(y = (\sin^{-1}x)^{2}\)
This matches (IV) 2. Here \(F(\theta) = \theta^{2}\), so \(F''(\theta) = 2\).
(C) \(y = e^{2\sin^{-1}x}\)
This matches (I) 4y. Here \(F(\theta) = e^{2\theta}\), so \(F''(\theta) = 4e^{2\theta} = 4y\).
(D) \(y = \cos(3\sin^{-1}x)\)
This matches (III) −9y. Here \(F(\theta) = \cos 3\theta\), so \(F''(\theta) = -9\cos 3\theta = -9y\).