Question

Number of real values of ' $x$ ' satisfying the equation ${\tan }^{-1}\left(\frac{x-1}{x+1}\right)+\frac{\pi }{2}={\tan }^{-1}\left(4-2x-2x^2\right)+{\cot }^{-1}\left(\frac{x-1}{x+3}\right)$ is

• A. 0
• B. 1
• C. 2
• D. more than 2

Answer 1

Answer: (C)

${\tan }^{-1}\left(\frac{x-1}{x+1}\right)+{\tan }^{-1}\left(\frac{x-1}{x+3}\right)={\tan }^{-1}\left(4-2x-2x^2\right)$
$\Rightarrow \frac{\left(\frac{x-1}{x+1}\right)+\left(\frac{x-1}{x+3}\right)}{1-\frac{(x-1{)}^2}{(x+1)(x+3)}}=-2(x+2)(x-1)$
$\therefore x=1$ is a solution.
If $x\ne 1$, then $\frac{2(x+2)}{2(3x+1)}=-2(x+2)$
$\therefore x=-2\text{ or }x=\frac{-1}{2}\text{ (not is domain). }$