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 \[\dfrac{c}{a}\]