Question

If $x_0$ is the solution of the equation $2^{1+{\left({\log }_{2}x\right)}^2}+(x^{{\log }_{2}x}{)}^2=3,$ then the value of ${\sin }^{-1}\left(x_0\right)+{\tan }^{-1}\left(\frac{2x_0}{2-(x_0{)}^2}\right)+{\cot }^{-1}\left(2\right)$ equals

• A. $\pi$
• B. $\frac{5\pi }{4}$
• C. $\frac{3\pi }{2}$
• D. $\frac{3\pi }{4}$

Answer 1

The correct option is A $\pi$

$2^{1+({\log }_{2}x{)}^2}+(x^{{\log }_{2}x}{)}^2=3$
$\Rightarrow 2\cdot {\left(2^{{\log }_{2}x}\right)}^{{\log }_{2}x}+{\left(x^{{\log }_{2}x}\right)}^2-3=0$
$\Rightarrow 2\cdot x^{{\log }_{2}x}+{\left(x^{{\log }_{2}x}\right)}^2-3=0$
Put $x^{{\log }_{2}x}=t$
$\therefore t^2+2t-3=0$
$\Rightarrow (t+3)(t-1)=0$
$\Rightarrow x^{{\log }_{2}x}=1$ or $x^{{\log }_{2}x}=-3$ (rejected)
$\Rightarrow \left({\log }_{2}x\right)\cdot \left({\log }_{2}x\right)=0$
$\Rightarrow x=1=x_0$

Now, ${\sin }^{-1}\left(x_0\right)+{\tan }^{-1}\left(\frac{2x_0}{2-(x_0{)}^2}\right)+{\cot }^{-1}\left(2\right)$
$={\sin }^{-1}\left(1\right)+{\tan }^{-1}\left(2\right)+{\cot }^{-1}\left(2\right)$
$=\frac{\pi }{2}+\frac{\pi }{2}=\pi$