The least value of $2^{sin x} + 2^{cos x}$ is

2^{\frac{1}{\sqrt{2}}}

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2^{1 - \frac{1}{\sqrt{2}}}

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2^{1 + 2^{- \frac{1}{2}}}

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2^{1 - \sqrt{2}}

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Solution

The correct option is B $2^{1 - \frac{1}{\sqrt{2}}}$
we know that, $A . M . \geq G . M .$

$⟹ \frac{a + b}{2} = \sqrt{a b}$

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

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

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

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

$⟹ 2^{s i n x} + 2^{c o s x} \geq 2^{1 + \frac{1}{2} (s i n x + c o s x)}$

$2^{1 + \frac{1}{2} (s i n x + c o s x)}$ to be least $s i n x + c o s x$ should be minimum

Therefore, finding the derivative and equating zero.

$\frac{d}{d x} s i n x + c o s x = 0 ⟹ c o s x - s i n x = 0$

$⟹ c o s x = s i n x ⟹ t a n x = 1$

$⟹ x = \frac{π}{4}, \frac{5 π}{4}, . . .$

on substituting we get min at $\frac{5 π}{4}$

therefore, $2^{s i n x} + 2^{c o s x} \geq 2^{1 + \frac{1}{2} (s i n (\frac{5 π}{4}) + c o s (\frac{5 π}{4}))}$

$⟹ 2^{s i n x} + 2^{c o s x} \geq 2^{1 + \frac{1}{2} (- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})}$

$⟹ 2^{s i n x} + 2^{c o s x} \geq 2^{1 + \frac{1}{2} (- \frac{2}{\sqrt{2}})}$

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

therefore the least value of $2^{s i n x} + 2^{c o s x}$ is $2^{1 - \frac{1}{\sqrt{2}}}$

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