Question

The number k is such that $\tan \left\{{\tan }^{-1}\right(2)+{\tan }^{-1}(20k\left)\right\}=k$. The sum of all possible values of k is -

• A. $-\frac{19}{40}$
• B. $-\frac{21}{40}$
• C. $0$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $-\frac{19}{40}$

$\tan \left\{{\tan }^{-1}\right(2)+{\tan }^{-1}(20k\left)\right\}=k$
$\tan \left({\tan }^{-1}\left(\frac{2+20k}{1-40k}\right)\right)=k$
$\Rightarrow \frac{2+20k}{1-40k}=k$
$\Rightarrow 40k^2+19k+2=0$
Sum of roots of the eqn in $k$ is $\frac{-19}{40}$