JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 21)
If sin4$$\alpha $$ + 4 cos4$$\beta $$ + 2 = 4$$\sqrt 2 $$ sin $$\alpha $$ cos $$\beta $$; $$\alpha $$, $$\beta $$ $$ \in $$ [0, $$\pi $$],
then cos($$\alpha $$ + $$\beta $$) $$-$$ cos($$\alpha $$ $$-$$ $$\beta $$) is equal to :
then cos($$\alpha $$ + $$\beta $$) $$-$$ cos($$\alpha $$ $$-$$ $$\beta $$) is equal to :
$$ - \sqrt 2 $$
0
$$-$$ 1
$$\sqrt 2 $$
Explanation
A.M. $$ \ge $$ G.M.
$${{{{\sin }^4}\alpha + 4{{\cos }^4}\beta + 1 + 1} \over 4} \ge {\left( {{{\sin }^4}\alpha .4{{\cos }^4}\beta .1.1} \right)^{{1 \over 4}}}$$
sin4$$\alpha $$ + 4 cos2$$\beta $$ + 2 $$ \ge $$ 4 $$\sqrt 2 $$ sin $$\alpha $$ cos $$\beta $$
Given that sin4$$\alpha $$ + 4cos4$$\beta $$ + 2 = 4$$\sqrt 2 $$ sin$$\alpha $$ cos$$\beta $$
$$ \Rightarrow $$ A.M.=G.M. $$ \Rightarrow $$ sin4$$\alpha $$ = 1 = 4 cos4 $$\beta $$
sin $$\alpha $$ = 1, cos $$\beta $$ = $$ \pm $$ $${1 \over {\sqrt 2 }}$$
$$ \Rightarrow $$ sin$$\beta $$ = $${1 \over {\sqrt 2 }}$$ as $$\beta $$ $$ \in $$ [0, $$\pi $$]
cos($$\alpha $$ + $$\beta $$) $$-$$ cos ($$\alpha $$ $$-$$ $$\beta $$) = $$-$$ 2 sin $$\alpha $$ $$\beta $$
= $$ - \sqrt 2 $$
$${{{{\sin }^4}\alpha + 4{{\cos }^4}\beta + 1 + 1} \over 4} \ge {\left( {{{\sin }^4}\alpha .4{{\cos }^4}\beta .1.1} \right)^{{1 \over 4}}}$$
sin4$$\alpha $$ + 4 cos2$$\beta $$ + 2 $$ \ge $$ 4 $$\sqrt 2 $$ sin $$\alpha $$ cos $$\beta $$
Given that sin4$$\alpha $$ + 4cos4$$\beta $$ + 2 = 4$$\sqrt 2 $$ sin$$\alpha $$ cos$$\beta $$
$$ \Rightarrow $$ A.M.=G.M. $$ \Rightarrow $$ sin4$$\alpha $$ = 1 = 4 cos4 $$\beta $$
sin $$\alpha $$ = 1, cos $$\beta $$ = $$ \pm $$ $${1 \over {\sqrt 2 }}$$
$$ \Rightarrow $$ sin$$\beta $$ = $${1 \over {\sqrt 2 }}$$ as $$\beta $$ $$ \in $$ [0, $$\pi $$]
cos($$\alpha $$ + $$\beta $$) $$-$$ cos ($$\alpha $$ $$-$$ $$\beta $$) = $$-$$ 2 sin $$\alpha $$ $$\beta $$
= $$ - \sqrt 2 $$
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