Verified Solution • 2025–26 Session moderate

QJEEM20Mad373: JEE Main 2020 Mathematics - Class 11 JEEM20Mad373 Solution: Let and be two real roots of the equation (k + 1)tan 2 x - .…

Let α\alpha and β\beta be two real roots of the equation
(k + 1)tan2x - 2\sqrt 2 . λ\lambda tanx = (1 - k), where k( ≠\ne - 1) and λ\lambda are real numbers. if tan2 ( α\alpha + β\beta ) = 50, then a value of λ\lambda is:

Multiple Choice Options

  • A 5 2\sqrt 2
  • B 10
  • C 5
  • D 10 2\sqrt 2

Step-by-Step Verified Solution

Let tan α\alpha and tan β\beta are the roots of

(k + 1)tan2x - 2\sqrt 2 . λ\lambda tanx - (1 - k) = 0

∴\therefore tan α\alpha + tan β\beta = 2λk+1{{\sqrt 2 \lambda } \over {k + 1}}

and an α\alpha .tan β\beta = k−1k+1{{k - 1} \over {k + 1}}

Now tan( α\alpha + β\beta )

= tan⁡α+tan⁡β1−tan⁡αtan⁡β{{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }}

= λ2k+11−k−1k+1{{{{\lambda \sqrt 2 } \over {k + 1}}} \over {1 - {{k - 1} \over {k + 1}}}}

= λ22{{{\lambda \sqrt 2 } \over 2}} = λ2{{\lambda \over {\sqrt 2 }}}

Given λ22{{{{\lambda ^2}} \over 2}} = 50

⇒\Rightarrow λ\lambda = 10

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