Question

If $\alpha ,\beta (\alpha >\beta )$ are two solutions of equation ${\tan }^{-1}x+{\cot }^{-1}(-\left|x\right|)=2{\tan }^{-1}6x,$ then $2\alpha +3\beta$ is equal to

• A. 1
• B. 2
• C. 4
• D. 5

Answer 1

The correct option is B 2

Case 1. when $x>0$
${\tan }^{-1}x+{\cot }^{-1}(-x)=2{\tan }^{-1}6x$
$\Rightarrow$ ${\tan }^{-1}x+\pi -{\cot }^{-1}x=2{\tan }^{-1}6x$
$\Rightarrow$ $\frac{\pi }{2}+2{\tan }^{-1}x=2{\tan }^{-1}6x$
$\Rightarrow$ ${\tan }^{-1}6x-{\tan }^{-1}x=\frac{\pi }{4}$
$\frac{5x}{1+6x^2}=1\Rightarrow 6x^2-5x+1=0$
$\Rightarrow$ $\alpha =\frac{1}{2},\beta =\frac{1}{3}$
Case 2. when, $x<0$
${\tan }^{-1}x+{\cot }^{-1}x=2{\tan }^{-1}6x$
${\tan }^{-1}6x=\frac{\pi }{4}$
Not possible $(\because x<0)$