The correct option is B ≥ 2√(2) we know that, A.M.≥ G.M. a+b2=√(ab) let a=2^sin^2x and #160; b=2^cos^2x #160;therefore, #160; ...

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Complex Numbers

2sin 2x+2cos ...

Question

$2^{{sin}^{2} x} + 2^{{cos}^{2} x}$ is

> 2

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\geq 2 \sqrt{2}

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\leq 2

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< 1

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Solution

The correct option is B $\geq 2 \sqrt{2}$
we know that, $A . M . \geq G . M .$

$⟹ \frac{a + b}{2} = \sqrt{a b}$

let $a = 2^{s i n^{2} x}$ and $b = 2^{c o s^{2} x}$

therefore, $⟹ \frac{2^{s i n^{2} x} + 2^{c o s^{2} x}}{2} \geq \sqrt{2^{s i n^{2} x} * 2^{c o s^{2} x}}$

$⟹ \frac{2^{s i n^{2} x} + 2^{c o s^{2} x}}{2} \geq \sqrt{2^{s i n^{2} x + c o s^{2} x}}$

$⟹ 2^{s i n^{2} x} + 2^{c o s^{2} x} \geq 2 * \sqrt{2^{1}}$

$⟹ 2^{s i n^{2} x} + 2^{c o s^{2} x} \geq 2 \sqrt{2}$

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