Q229 / 270 · NDA 2025 (Phase 2) · Mathematics
For the following three (03) items:
Let $y = f(x) = \dfrac{x\sin^{-1} x}{\sqrt{1 - x^2}} + \ln\sqrt{1 - x^2}$.
What is $\dfrac{d^2y}{dx^2}$ at $x = 0$ equal to?
Let $y = f(x) = \dfrac{x\sin^{-1} x}{\sqrt{1 - x^2}} + \ln\sqrt{1 - x^2}$.
What is $\dfrac{d^2y}{dx^2}$ at $x = 0$ equal to?
- A$0$
- B$0.5$
- Correct answerC$1$
- D$1.5$
Solution
Given $y = \dfrac{x\sin^{-1} x}{\sqrt{1 - x^2}} + \ln\sqrt{1 - x^2}$
We can rewrite the function as:
$y = \dfrac{x\sin^{-1} x}{\sqrt{1 - x^2}} + \dfrac{1}{2}\ln(1 - x^2)$
Differentiating with respect to $x$ using the quotient and chain rules:
$\dfrac{dy}{dx} = \dfrac{\left(\sin^{-1} x + \dfrac{x}{\sqrt{1 - x^2}}\right)\sqrt{1 - x^2} - x\sin^{-1} x \left(\dfrac{-x}{\sqrt{1 - x^2}}\right)}{1 - x^2} + \dfrac{1}{2} \left(\dfrac{-2x}{1 - x^2}\right)$
$\dfrac{dy}{dx} = \dfrac{\sqrt{1 - x^2}\sin^{-1} x + x + \dfrac{x^2\sin^{-1} x}{\sqrt{1 - x^2}}}{1 - x^2} - \dfrac{x}{1 - x^2}$
$\dfrac{dy}{dx} = \dfrac{(1 - x^2)\sin^{-1} x + x\sqrt{1 - x^2} + x^2\sin^{-1} x}{(1 - x^2)^{3/2}} - \dfrac{x}{1 - x^2}$
$\dfrac{dy}{dx} = \dfrac{\sin^{-1} x + x\sqrt{1 - x^2}}{(1 - x^2)^{3/2}} - \dfrac{x}{1 - x^2}$
$\dfrac{dy}{dx} = \dfrac{\sin^{-1} x}{(1 - x^2)^{3/2}} + \dfrac{x}{1 - x^2} - \dfrac{x}{1 - x^2} = \dfrac{\sin^{-1} x}{(1 - x^2)^{3/2}}$
Differentiating again with respect to $x$:
$\dfrac{d^2y}{dx^2} = \dfrac{\dfrac{1}{\sqrt{1 - x^2}} (1 - x^2)^{3/2} - \sin^{-1} x \cdot \dfrac{3}{2}(1 - x^2)^{1/2}(-2x)}{(1 - x^2)^3}$
$\dfrac{d^2y}{dx^2} = \dfrac{(1 - x^2) + 3x\sqrt{1 - x^2}\sin^{-1} x}{(1 - x^2)^3}$
$\dfrac{d^2y}{dx^2} = \dfrac{1}{(1 - x^2)^2} + \dfrac{3x\sin^{-1} x}{(1 - x^2)^{5/2}}$
Substituting $x = 0$ into the second derivative:
$\left. \dfrac{d^2y}{dx^2} \right|_{x=0} = \dfrac{1}{(1 - 0)^2} + 0 = 1$
Answer: $1$
We can rewrite the function as:
$y = \dfrac{x\sin^{-1} x}{\sqrt{1 - x^2}} + \dfrac{1}{2}\ln(1 - x^2)$
Differentiating with respect to $x$ using the quotient and chain rules:
$\dfrac{dy}{dx} = \dfrac{\left(\sin^{-1} x + \dfrac{x}{\sqrt{1 - x^2}}\right)\sqrt{1 - x^2} - x\sin^{-1} x \left(\dfrac{-x}{\sqrt{1 - x^2}}\right)}{1 - x^2} + \dfrac{1}{2} \left(\dfrac{-2x}{1 - x^2}\right)$
$\dfrac{dy}{dx} = \dfrac{\sqrt{1 - x^2}\sin^{-1} x + x + \dfrac{x^2\sin^{-1} x}{\sqrt{1 - x^2}}}{1 - x^2} - \dfrac{x}{1 - x^2}$
$\dfrac{dy}{dx} = \dfrac{(1 - x^2)\sin^{-1} x + x\sqrt{1 - x^2} + x^2\sin^{-1} x}{(1 - x^2)^{3/2}} - \dfrac{x}{1 - x^2}$
$\dfrac{dy}{dx} = \dfrac{\sin^{-1} x + x\sqrt{1 - x^2}}{(1 - x^2)^{3/2}} - \dfrac{x}{1 - x^2}$
$\dfrac{dy}{dx} = \dfrac{\sin^{-1} x}{(1 - x^2)^{3/2}} + \dfrac{x}{1 - x^2} - \dfrac{x}{1 - x^2} = \dfrac{\sin^{-1} x}{(1 - x^2)^{3/2}}$
Differentiating again with respect to $x$:
$\dfrac{d^2y}{dx^2} = \dfrac{\dfrac{1}{\sqrt{1 - x^2}} (1 - x^2)^{3/2} - \sin^{-1} x \cdot \dfrac{3}{2}(1 - x^2)^{1/2}(-2x)}{(1 - x^2)^3}$
$\dfrac{d^2y}{dx^2} = \dfrac{(1 - x^2) + 3x\sqrt{1 - x^2}\sin^{-1} x}{(1 - x^2)^3}$
$\dfrac{d^2y}{dx^2} = \dfrac{1}{(1 - x^2)^2} + \dfrac{3x\sin^{-1} x}{(1 - x^2)^{5/2}}$
Substituting $x = 0$ into the second derivative:
$\left. \dfrac{d^2y}{dx^2} \right|_{x=0} = \dfrac{1}{(1 - 0)^2} + 0 = 1$
Answer: $1$
Hence, the correct answer is option: 'C'.