Question

If ${\sin }^{-1}\left(\frac{4x}{x^2+4}\right)+2{\tan }^{-1}(-\frac{x}{2})$ is independent of x,then the value of x is $-a\le x\le b$.
Find the value of $a+b$

• A. 4
• B. 0
• C. 1
• D. 2

Answer 1

The correct option is A 4

Given, ${\sin }^{-1}\left(\frac{4x}{x^2+4}\right)+2{\tan }^{-1}(-\frac{x}{2})={\sin }^{-1}\left(\frac{2\times \frac{x}{2}}{{\left(\frac{x}{2}\right)}^2+1}\right)-2{\tan }^{-1}\frac{x}{2}$
$=2{\tan }^{-1}\frac{x}{2}-2{\tan }^{-1}\frac{x}{2}=0$
$\Rightarrow$ $\left|\frac{x}{2}\right|\le 1$ or
$\left|x\right|\le 2$ or $-2\le x\le 2$

a = 2 and b =2 so a+b = 2+2 =4