Question

The number $k$ is such that $\tan \{arctan\left(2\right)+arctan\left(20k\right)\}=k$. 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

Answer: (A)

$-\frac{19}{40}$

Given, $\tan \left[{\tan }^{-1}\right(2)+{\tan }^{-1}20k]$
$=\tan \left[{\tan }^{-1}\right(\frac{2+20k}{1-40k}\left)\right]$
$=\frac{2+20k}{1-40k}$
$=k$
Hence
$2+20k=k-40k^2$
$40k^2+19k+2=0$
$k^2+\frac{19}{40}k+\frac{2}{40}=0$
Hence sum of roots $=-\frac{b}{a}$

$=\frac{-19}{40}$