Question

Let $\alpha $ and $\beta $ be two real roots of the equation $\left( {k + 1} \right){\tan ^2}x - \sqrt 2 \lambda \tan x = \left( {1 - k} \right),$ where k $ \ne $1 and $\lambda $ are real numbers. If ${\tan ^2}\left( {\alpha + \beta } \right) = 50$. Then a value of $\lambda $ is :

Answer 1

Hint: We have given $\alpha $ and $\beta $ as two real roots that means when we put $\alpha $ and $\beta $ in the place of x then the equation will be satisfied. Given a quadratic equation is quadratic in tan and hence the roots of the equation will be $\tan \alpha $ and $\tan \beta $. Now apply the relation of sum of roots and product of roots to proceed further.

Complete step-by-step answer:
We have given
$\left( {k + 1} \right){\tan ^2}x - \sqrt 2 \lambda \tan x = \left( {1 - k} \right)$
We can rewrite it as :
$\left( {k + 1} \right){\tan ^2}x - \sqrt 2 \lambda \tan x + \left( {k - 1} \right) = 0$
Roots of this quadratic equation are $\tan \alpha $ and $\tan \beta $.
From the property of sum of roots and product of roots we can write
$\tan \alpha + \tan \beta = \dfrac{{\sqrt 2 \lambda }}{{\left( {k + 1} \right)}}$
$\tan \alpha \times \tan \beta = \dfrac{{\left( {k - 1} \right)}}{{\left( {k + 1} \right)}}$
And we know the formula
$\tan \left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha .\tan \beta }}$
On putting values from above we get,
$\tan \left( {\alpha + \beta } \right) = \dfrac{{\dfrac{{\sqrt 2 \lambda }}{{\left( {k + 1} \right)}}}}{{1 - \dfrac{{\left( {k - 1} \right)}}{{\left( {k + 1} \right)}}}}$
On further solving we get,
$\tan \left( {\alpha + \beta } \right) = \dfrac{{\dfrac{{\sqrt 2 \lambda }}{{\left( {k + 1} \right)}}}}{{\dfrac{{\left( {k + 1} \right) - \left( {k - 1} \right)}}{{\left( {k + 1} \right)}}}}$
On cancel out we get,
$\tan \left( {\alpha + \beta } \right) = \dfrac{{\sqrt 2 \lambda }}{2} = \dfrac{\lambda }{{\sqrt 2 }}$
And hence on squaring we get,
${\tan ^2}\left( {\alpha + \beta } \right) = \dfrac{{{\lambda ^2}}}{2}$
And we have given in the question ${\tan ^2}\left( {\alpha + \beta } \right) = 50$
On comparing both equation we get
$
  \dfrac{{{\lambda ^2}}}{2} = 50 \\
  {\lambda ^2} = 100 \\
  \therefore \lambda = \pm 10 \\
 $

Note: Whenever we get this type of question the key concept of solving is we have to remember the formula of quadratic equations like properties on sum of roots and product of roots. And also remember $\tan \left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha .\tan \beta }}$ this type of formula are very helpful in solving this type of question.