Question

Let $f\left(x\right) = \left(sin\left(tan^{-1}x\right)+sin\left(cot^{-1}x\right)\right)^{2} -1,$ $\left|x\right| > 1.$ If $\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} \left(sin^{-1} \left(f\left(x\right)\right)\right)$ and $y\left(\sqrt{3}\right) = \frac{\pi}{6},$ then $y \left(-\sqrt{3}\right)$ is equal to :

• A. $\frac{\pi}{3}$
• B. $\frac{2\pi}{3}$
• C. $-\frac{\pi}{6}$
• D. $\frac{5\pi}{6}$

Answer 1

Answer: (C)

Let $tan^{-1}x=\theta, \theta\,\in\left(-\frac{\pi}{2}-\frac{\pi}{4}\right)\cup\left(\frac{\pi }{2},\frac{\pi }{4}\right)$
$f \left(x\right)-\left(sin\,\theta+cos\,\theta \right)^{2}-1=sin2\theta =\frac{2x}{1+x^{2}}$
Now, $\frac{dy}{dx}=\frac{1}{2} \frac{d}{dx} sin^{-1}\left(\frac{2x}{1+x^{2}}\right)$
$=\frac{1}{1+x^{2}}, \left|x\right|>1$
Since, we can integrate only in the continuousinterval. So we have to take integral in two casessepartely namely for $x < -1$ and for $x > 1.$
$\Rightarrow $ $y = \begin{cases} -tan^{-1} \,x+c_1\,; & \text{$x>1$ } \\ -tan^{-1}\,x+c_2 & \text{ $x<-1$ } \end{cases}$
so, $c_{1}=\frac{\pi}{2} as y\left(\sqrt{3}\right)=\frac{\pi}{6}$
But we cannot find $c_2$ as we do not have any other additional information for $x < -1.$ So, all of the given options may be correct as $c_2$ is unknown so, it should be bonus.