Question

Let $f\left(x\right)={\left(sin\left(ta{n}^{-1}x\right)+sin\left(co{t}^{-1}x\right)\right)}^2-1,$ $\left|x\right|>1.$ If $\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left(si{n}^{-1}\left(f\left(x\right)\right)\right)$ 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 $ta{n}^{-1}x=\theta ,\theta \in \left(-\frac{\pi }{2}-\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{\pi }{4}\right)$
$f\left(x\right)-{\left(sin\theta +cos\theta \right)}^2-1=sin2\theta =\frac{2x}{1+x^2}$
Now, $\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}si{n}^{-1}\left(\frac{2x}{1+x^2}\right)$
$=\frac{1}{1+x^2},\left|x\right|>1$
Since, we can integrate only in the continuousinterval. So we have to take integral in two casessepartely namely for $x<-1$ and for $x>1.$
$\Rightarrow$ $y=\{\begin{array}{ll}-ta{n}^{-1}x+c_1;& \text{x>1 }\\ -ta{n}^{-1}x+c_2& \text{ x<−1 }\end{array}$
so, $c_1=\frac{\pi }{2}asy\left(\sqrt{3}\right)=\frac{\pi }{6}$
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.