JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 11)
If
$${e^{\left( {{{\cos }^2}x + {{\cos }^4}x + {{\cos }^6}x + ...\infty } \right){{\log }_e}2}}$$
satisfies the equation t2 - 9t + 8 = 0, then the value of
$${{2\sin x} \over {\sin x + \sqrt 3 \cos x}}\left( {0 < x < {\pi \over 2}} \right)$$ is :
$${{2\sin x} \over {\sin x + \sqrt 3 \cos x}}\left( {0 < x < {\pi \over 2}} \right)$$ is :
$$\sqrt 3 $$
$${3 \over 2}$$
2$$\sqrt 3 $$
$${1 \over 2}$$
Explanation
$${e^{({{\cos }^2}x + {{\cos }^4}x + ...........\infty )\ln 2}} = {2^{{{\cos }^2}x + {{\cos }^4}x + ...........\infty }}$$
= $${2^{{{{{\cos }^2}x} \over {1 - {{\cos }^2}x}}}}$$
$$ = {2^{{{\cot }^2}x}}$$
Given, $${t^2} - 9t + 8 = 0 \Rightarrow t = 1,8$$
$$ \Rightarrow {2^{{{\cot }^2}x}} = 1,8 \Rightarrow co{t^2}x = 0,3$$
$$0 < x < {\pi \over 2} \Rightarrow \cot x = \sqrt 3 $$
$$ \therefore $$ $$ {{2\sin x} \over {\sin x + \sqrt 3 \cos x}} = {2 \over {1 + \sqrt 3 \cot x}} = {2 \over 4} = {1 \over 2}$$
= $${2^{{{{{\cos }^2}x} \over {1 - {{\cos }^2}x}}}}$$
$$ = {2^{{{\cot }^2}x}}$$
Given, $${t^2} - 9t + 8 = 0 \Rightarrow t = 1,8$$
$$ \Rightarrow {2^{{{\cot }^2}x}} = 1,8 \Rightarrow co{t^2}x = 0,3$$
$$0 < x < {\pi \over 2} \Rightarrow \cot x = \sqrt 3 $$
$$ \therefore $$ $$ {{2\sin x} \over {\sin x + \sqrt 3 \cos x}} = {2 \over {1 + \sqrt 3 \cot x}} = {2 \over 4} = {1 \over 2}$$
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