Q89 / 180 · COMEDK 2025 (10 May Shift 2) · Mathematics
If $y=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2$, then $\left(1-x^2\right) \dfrac{d^2 y}{d x^2}-x \dfrac{d y}{d x}=$
- Correct answerA4
- B2
- C0
- D3
Solution
Given $y = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$.
Using the identity $\cos^{-1} x = \dfrac{\pi}{2} - \sin^{-1} x$, we have $y = (\sin^{-1} x)^2 + (\dfrac{\pi}{2} - \sin^{-1} x)^2$.
Let $u = \sin^{-1} x$. Then $y = u^2 + (\dfrac{\pi}{2} - u)^2 = u^2 + \dfrac{\pi^2}{4} - \pi u + u^2 = 2u^2 - \pi u + \dfrac{\pi^2}{4}$.
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} = (4u - \pi) \cdot \dfrac{1}{\sqrt{1-x^2}}$.
Thus, $\sqrt{1-x^2} \dfrac{dy}{dx} = 4\sin^{-1} x - \pi$.
Differentiating both sides with respect to $x$ again:
$\sqrt{1-x^2} \dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} \cdot \dfrac{-2x}{2\sqrt{1-x^2}} = 4 \cdot \dfrac{1}{\sqrt{1-x^2}}$.
Multiplying the entire equation by $\sqrt{1-x^2}$:
$(1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} = 4$.
Answer: 4
Using the identity $\cos^{-1} x = \dfrac{\pi}{2} - \sin^{-1} x$, we have $y = (\sin^{-1} x)^2 + (\dfrac{\pi}{2} - \sin^{-1} x)^2$.
Let $u = \sin^{-1} x$. Then $y = u^2 + (\dfrac{\pi}{2} - u)^2 = u^2 + \dfrac{\pi^2}{4} - \pi u + u^2 = 2u^2 - \pi u + \dfrac{\pi^2}{4}$.
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} = (4u - \pi) \cdot \dfrac{1}{\sqrt{1-x^2}}$.
Thus, $\sqrt{1-x^2} \dfrac{dy}{dx} = 4\sin^{-1} x - \pi$.
Differentiating both sides with respect to $x$ again:
$\sqrt{1-x^2} \dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} \cdot \dfrac{-2x}{2\sqrt{1-x^2}} = 4 \cdot \dfrac{1}{\sqrt{1-x^2}}$.
Multiplying the entire equation by $\sqrt{1-x^2}$:
$(1-x^2) \dfrac{d^2y}{dx^2} - x \dfrac{dy}{dx} = 4$.
Answer: 4
Hence, the correct answer is option: 'A'.