Question

The number $k$ is such that $\tan \\{ {\tan ^{ - 1}}\left( 2 \right) + {\tan ^{ - 1}}\left( {20k} \right)\\} = k$. Then the sum of all possible values of $k$ is-
A.$- \dfrac{{19}}{{40}}$
B.$- \dfrac{{21}}{{40}}$
C.0
D.$\dfrac{1}{5}$

Answer 1

Hint: We will first simplify the given expression using the formula ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 - xy}}} \right)$ and ${\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta$. We will get an equation in $k$.
Now, find the sum of values of $k$ using the condition that if the equation is $a{x^2} + bx + c = 0$, then the sum of all the roots of the equation is given by $- \dfrac{b}{a}$.

Complete step-by-step answer:
First of all we will simplify the inner bracket of the given expression using the formula,
${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 - xy}}} \right)$
Therefore, we can rewrite, ${\tan ^{ - 1}}\left( 2 \right) + {\tan ^{ - 1}}\left( {20k} \right)$ as ${\tan ^{ - 1}}\left( {\dfrac{{2 + 20k}}{{1 - 2\left( {20k} \right)}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{2 + 20k}}{{1 - 40k}}} \right)$
Also, ${\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta$
Thus,
$\tan \left( {{{\tan }^{ - 1}}\left( {\dfrac{{2 + 20k}}{{1 - 40k}}} \right)} \right) = k \\\ \left( {\dfrac{{2 + 20k}}{{1 - 40k}}} \right) = k \\\$
On simplifying we get,
$2 + 20k = k - 40{k^2} \\\ 40{k^2} + 19k + 2 = 0 \\\$
If the equation is $a{x^2} + bx + c = 0$, then the sum of all the roots of the equation is given by $- \dfrac{b}{a}$
Hence, sum of all the possible values of $k$ is $- \dfrac{{19}}{{40}}$
Hence, option A is correct.

Note: The formula ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 - xy}}} \right)$ will help in simplifying the question.
If $a{x^2} + bx + c = 0$, then the sum of all the roots of the equation is given by $- \dfrac{b}{a}$ and the product of roots is given by $\dfrac{c}{a}$