Question

The complete set of values of 'a' such that the equation ${\left({\tan }^{-1}x\right)}^2+a\left({\tan }^{-1}x\right)-\pi {\cot }^{-1}x=0$ has no real solution is

• A. $(-\frac{\pi }{4},\frac{\pi }{2})$
• B. $(-\frac{3\pi }{2},-\frac{\pi }{2})$
• C. $(\frac{\pi }{2},\frac{3\pi }{2})$
• D. $(-\frac{\pi }{4},\frac{3\pi }{4})$

Answer 1

Answer: (B)

$(-\frac{3\pi }{2},-\frac{\pi }{2})$

Using ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$
We get
${\left({\tan }^{-1}x\right)}^2+a\left({\tan }^{-1}x\right)-\pi (\frac{\pi }{2}-{\tan }^{-1}x)=0\Rightarrow {\left({\tan }^{-1}x\right)}^2+a\left({\tan }^{-1}x\right)+\pi {\tan }^{-1}x-\frac{{\pi }^2}{2}=0\Rightarrow {\tan }^{-1}x=\frac{-(a+\pi )\pm \sqrt{a^2+2a\pi +3{\pi }^2}}{2}$
As $-\frac{\pi }{2}<{\tan }^{-1}x<\frac{\pi }{2}$
${\tan }^{-1}x<\frac{\pi }{2}\Rightarrow \frac{-(a+\pi )+\sqrt{a^2+2a\pi +3{\pi }^2}}{2}<\frac{\pi }{2}\Rightarrow a<-\frac{\pi }{2}$
${\tan }^{-1}x>\frac{\pi }{2}\Rightarrow \frac{-(a+\pi )-\sqrt{a^2+2a\pi +3{\pi }^2}}{2}>\frac{\pi }{2}\Rightarrow a>-\frac{3\pi }{2}$
Therefore, $aϵ(-\frac{3\pi }{2},-\frac{\pi }{2})$