Question

If 1 < x < $\sqrt{2}$, the number of solutions of the equation $ta{n}^{-1}(x-1)+ta{n}^{-1}x+ta{n}^{-1}(x+1)=ta{n}^{-1}3x$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

Given, ${\tan }^{-1}(x-1)+{\tan }^{-1}(x+1-{\tan }^{-1}(x)={\tan }^{-1}(3x)$

${\tan }^{-1}(x-1)+{\tan }^{-1}(x+1)={\tan }^{-1}\left(3x\right)-{\tan }^{-1}\left(x\right)$

${\tan }^{-1}\frac{2x}{1-x^2+1}={\tan }^{-1}\frac{2x}{1+3x^2}$

${\tan }^{-1}\frac{2x}{2-x^2}={\tan }^{-1}\frac{2x}{1+3x^2}$

$x(1+3x^2)=x(2-x^2)$

$4x^2=1$ and $x=0$

$x=\pm \frac{1}{2}$

but $1<x<\sqrt{2}$

Hence there is no solution.