JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 8)
The minimum value of 2sinx + 2cosx is :
$${2^{-1 + \sqrt 2 }}$$
$${2^{1 - {1 \over {\sqrt 2 }}}}$$
$${2^{1 - \sqrt 2 }}$$
$${2^{-1 + {1 \over {\sqrt 2 }}}}$$
Explanation
Using AM $$ \ge $$ GM
$$ \Rightarrow {{{2^{\sin \,x}} + {2^{\cos \,x}}} \over 2} \ge \sqrt {{2^{\sin x}}{{.2}^{\cos x}}} $$
$$ \Rightarrow {2^{\sin x}} + {2^{\cos x}} \ge {2^{1 + \left( {{{\sin x + \cos x} \over 2}} \right)}}$$
$$ \Rightarrow \min ({2^{\sin x}} + {2^{\cos x}}) = {2^{1 - {1 \over {\sqrt 2 }}}}$$
As we know range of sin x + cos x is :
$$ - \sqrt 2 $$ $$ \le $$ sin x + cos x $$ \le $$ $$\sqrt 2 $$.
So Minimum value of sin x + cos x = $$ - \sqrt 2 $$
$$ \Rightarrow {{{2^{\sin \,x}} + {2^{\cos \,x}}} \over 2} \ge \sqrt {{2^{\sin x}}{{.2}^{\cos x}}} $$
$$ \Rightarrow {2^{\sin x}} + {2^{\cos x}} \ge {2^{1 + \left( {{{\sin x + \cos x} \over 2}} \right)}}$$
$$ \Rightarrow \min ({2^{\sin x}} + {2^{\cos x}}) = {2^{1 - {1 \over {\sqrt 2 }}}}$$
As we know range of sin x + cos x is :
$$ - \sqrt 2 $$ $$ \le $$ sin x + cos x $$ \le $$ $$\sqrt 2 $$.
So Minimum value of sin x + cos x = $$ - \sqrt 2 $$
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