If $y=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2$, then $\left(1-x^2\right) y_2-x y_1=$

If $y=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2$, then $\left(1-x^2\right) y_2-x y_1=$
  1. $1$
  2. $4$
  3. $-4$
  4. $-1$

Solution

$\begin{aligned} & y=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2 \\ \therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 \sin ^{-1} x}{\sqrt{1-x^2}}-\frac{2 \cos ^{-1} x}{\sqrt{1-x^2}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2\left(\sin ^{-1} x-\cos ^{-1} x\right)}{\sqrt{1-x^2}} \\ & \Rightarrow \sqrt{1-x^2} \frac{\mathrm{d} y}{\mathrm{~d} x}=2\left(\sin ^{-1} x-\cos ^{-1} x\right)\end{aligned}$ Differentiating both sides w.r.t. $x$, we get $\begin{aligned} & \quad \sqrt{1-x^2} \cdot \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\frac{\mathrm{d} y}{\mathrm{~d} x} \cdot \frac{1}{2 \sqrt{1-x^2}} \cdot(-2 x) \\ & =2\left(\frac{1}{\sqrt{1-x^2}}-\frac{(-1)}{\sqrt{1-x^2}}\right)=\frac{4}{\sqrt{1-x^2}} \\ & \therefore \quad\left(1-x^2\right) \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-x \frac{\mathrm{d} y}{\mathrm{~d} x}=4 \end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 1)

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