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 $\sin ^{-1}\left(x_0\right)+\tan ^{-1}\left(\frac{2 x_0}{2-\left(x_0\right)^2}\right)+\cot ^{-1}(2)$ 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 $x ^{\log _2 x }= t$
$t ^2+2 t -3=0 \Rightarrow ( t +3)( t -1)=0 \Rightarrow t =1$
$\Rightarrow x ^{\log _2 x }=1 $
$(\log _2 x )\left(\log _2 x \right)=0 \Rightarrow x _0=1$
$\therefore \sin ^{-1}\left( x _0\right)+\tan ^{-1}\left(\frac{2 x _0}{2-\left( x _0\right)^2}\right)+\cot ^{-1}(2)=\frac{\pi}{2}+\tan ^{-1} 2+\cot ^{-1} 2=\pi$