Question

The equation of the tangent to the curve

$y={\sin }^{-1}\frac{2x}{1+x^2}$ at $x=\sqrt{3}$

• A. $y=-\frac{x}{2}$
• B. $y-\frac{\pi }{3}=-\frac{x}{2}$
• C. $y-\frac{\pi }{3}=\frac{1}{2}(x-\sqrt{3})$
• D. $y-\frac{\pi }{3}=\frac{-1}{2}(x-\sqrt{3})$

Answer 1

The correct option is D $y-\frac{\pi }{3}=\frac{-1}{2}(x-\sqrt{3})$

$y=si{n}^{-1}\frac{2x}{1+x^2}=\pi -2ta{n}^{-1}x$, for $x>1$

$\Rightarrow \frac{dy}{dx}=-\frac{2}{1+x^2}$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\sqrt{3}}=-\frac{2}{1+3}=-\frac{1}{2}$

Also when $x=\sqrt{3}$, $y=\pi -2\frac{\pi }{3}=\frac{\pi }{3}$

Hence, equation of tangent is

$y-\frac{\pi }{3}=-\frac{1}{2}(x-\sqrt{3})$