JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 3)
Let $$\alpha $$ and $$\beta $$ be two real roots of the equation
(k + 1)tan2x - $$\sqrt 2 $$ . $$\lambda $$tanx = (1 - k), where k($$ \ne $$ - 1) and $$\lambda $$ are real numbers. if tan2 ($$\alpha $$ + $$\beta $$) = 50, then a value of $$\lambda $$ is:
(k + 1)tan2x - $$\sqrt 2 $$ . $$\lambda $$tanx = (1 - k), where k($$ \ne $$ - 1) and $$\lambda $$ are real numbers. if tan2 ($$\alpha $$ + $$\beta $$) = 50, then a value of $$\lambda $$ is:
5$$\sqrt 2 $$
10
5
10$$\sqrt 2 $$
Explanation
Let tan$$\alpha $$ and tan$$\beta $$ are the roots of
(k + 1)tan2x - $$\sqrt 2 $$ . $$\lambda $$tanx - (1 - k) = 0
$$ \therefore $$ tan$$\alpha $$ + tan$$\beta $$ = $${{\sqrt 2 \lambda } \over {k + 1}}$$
and an$$\alpha $$.tan$$\beta $$ = $${{k - 1} \over {k + 1}}$$
Now tan($$\alpha $$ + $$\beta $$)
= $${{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }}$$
= $${{{{\lambda \sqrt 2 } \over {k + 1}}} \over {1 - {{k - 1} \over {k + 1}}}}$$
= $${{{\lambda \sqrt 2 } \over 2}}$$ = $${{\lambda \over {\sqrt 2 }}}$$
Given $${{{{\lambda ^2}} \over 2}}$$ = 50
$$ \Rightarrow $$ $$\lambda $$ = 10
(k + 1)tan2x - $$\sqrt 2 $$ . $$\lambda $$tanx - (1 - k) = 0
$$ \therefore $$ tan$$\alpha $$ + tan$$\beta $$ = $${{\sqrt 2 \lambda } \over {k + 1}}$$
and an$$\alpha $$.tan$$\beta $$ = $${{k - 1} \over {k + 1}}$$
Now tan($$\alpha $$ + $$\beta $$)
= $${{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }}$$
= $${{{{\lambda \sqrt 2 } \over {k + 1}}} \over {1 - {{k - 1} \over {k + 1}}}}$$
= $${{{\lambda \sqrt 2 } \over 2}}$$ = $${{\lambda \over {\sqrt 2 }}}$$
Given $${{{{\lambda ^2}} \over 2}}$$ = 50
$$ \Rightarrow $$ $$\lambda $$ = 10
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