JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 21)
If $$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$$,
x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ then $$dy \over dx$$ is equal to:
x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ then $$dy \over dx$$ is equal to:
$$2x - {\pi \over 3}$$
$${\pi \over 6} - x$$
$${\pi \over 3} - x$$
$$x - {\pi \over 6}$$
Explanation
$$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$$
$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}$$
$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\left( {{{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}} \right)} \right)^2}$$
$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$$
$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 2} - {{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$$
As x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ then
$${{{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)}$$ = $${\left( {{\pi \over 3} + x} \right)}$$
$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 2} - \left( {{\pi \over 3} + x} \right)} \right)^2}$$
$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 6} - x} \right)^2}$$
$$ \therefore $$ $$2{{dy} \over {dx}} = 2\left( {{\pi \over 6} - x} \right)\left( { - 1} \right)$$
$$ \Rightarrow $$ $${{dy} \over {dx}} = \left( {x - {\pi \over 6}} \right)$$
$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}$$
$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\left( {{{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}} \right)} \right)^2}$$
$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$$
$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 2} - {{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$$
As x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ then
$${{{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)}$$ = $${\left( {{\pi \over 3} + x} \right)}$$
$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 2} - \left( {{\pi \over 3} + x} \right)} \right)^2}$$
$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 6} - x} \right)^2}$$
$$ \therefore $$ $$2{{dy} \over {dx}} = 2\left( {{\pi \over 6} - x} \right)\left( { - 1} \right)$$
$$ \Rightarrow $$ $${{dy} \over {dx}} = \left( {x - {\pi \over 6}} \right)$$
Comments (0)
