Question

$y=\sin ^{-1} 2 x \sqrt{1-x^{2}}= \begin{cases}2 \sin ^{-1} x & |x| \leq \frac{1}{\sqrt{2}} \\ \pi-2 \sin ^{-1} x & x>\frac{1}{\sqrt{2}} \\\ -\left(\pi+2 \sin ^{-1} x\right) & x<\frac{-1}{\sqrt{2}}\end{cases}$ \(\mathrm{x}=0, \mathrm{y}=0\) is only integral point where the function has a unique tangent

Short answer

Answer: The equation of the unique tangent line at \((0,0)\) is \(y = 2x\).

Step-by-step solution

Find the derivative of each part of the function with respect to x

We need to find the derivative of each part of the piecewise-defined function, For the first part, \(y = 2\sin^{-1}(x)\) for \(|x| \leq \frac{1}{\sqrt{2}}\) The derivative with respect to x is: \(y' = 2 \cdot \frac{1}{\sqrt{1 - x^2}}\) For the Second part, \(y = \pi - 2\sin^{-1}(x)\) for \(x > \frac{1}{\sqrt{2}}\) The derivative with respect to x is: \(y' = -2 \cdot \frac{1}{\sqrt{1 - x^2}}\) For the Third part, \(y = -(\pi + 2\sin^{-1}(x))\) for \(x < \frac{-1}{\sqrt{2}}\) The derivative with respect to x is: \(y' = \text{-}2 \cdot \frac{1}{\sqrt{1 - x^2}}\)

Evaluate the derivative at x = 0

For the first part, when \(x = 0\), \(y'(0) = 2 \cdot \frac{1}{\sqrt{1 - 0^2}} = 2\) For the second part, \(x = 0\) is not in the domain, so we don't need to evaluate it. For the third part, \(x = 0\) is not in the domain, so we don't need to evaluate it either.

Find the equation of the tangent line

Since the only part of the piecewise function that has a unique tangent at \((0,0)\) is the first part, we can use the slope \(y'(0) = 2\) to find the equation of the tangent line. The equation of the tangent line is \(y = mx + c\), where \(m\) is the slope and \((0,0)\) is the point where the tangent line passes through. So the equation is \(y = 2x + 0\) or \(y = 2x\).