Question

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

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

Answer 1

Answer: (A)

$2^{1+{\left({\log }_{2}x\right)}^2}+{\left(x^{{\log }_{2}x}\right)}^2-3=0$
$2\cdot {\left(2^{{\log }_{2}x}\right)}^{{\log }_{2}x}+{\left(x^{{\log }_{2}x}\right)}^2-3=0$
$2\cdot x^{{\log }_{2}x}+{\left(x^{{\log }_{2}x}\right)}^2-3=0$
Now put $x^{{\log }_{2}x}=t$
$t^2+2t-3=0\Rightarrow (t+3)(t-1)=0\Rightarrow t=1$
$\Rightarrow x^{{\log }_{2}x}=1$
$({\log }_{2}x)\left({\log }_{2}x\right)=0\Rightarrow x_0=1$
$\therefore {\sin }^{-1}\left(x_0\right)+{\tan }^{-1}\left(\frac{2x_0}{2-{\left(x_0\right)}^2}\right)+{\cot }^{-1}(2)=\frac{\pi }{2}+{\tan }^{-1}2+{\cot }^{-1}2=\pi$