Question

Minimum value of 22^sinx + 2^cosx is:

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

AM - GM INEQUALITY

For any two non zero real numbers a , b

$\displaystyle \: \frac{a + b}{2} \geqslant \sqrt{ab}$

The equality occurs when a = b

EVALUATION

Applying above mentioned inequality on

$\displaystyle \: {2}^{ sinx} \: \: and \: \: \: {2}^{ cosx} \: \: we \: \: get$

$\displaystyle \: \frac{ {2}^{ sinx} + {2}^{ cosx} }{2} \: \geqslant \sqrt{ {2}^{ sinx} \times {2}^{ cosx} }$

$\implies \: \displaystyle \: {2}^{ sinx} + {2}^{ cosx} \: \geqslant 2\sqrt{ {2}^{ sinx + cosx} } \: \: \: ...(1)$

Now

$sinx + cosx$

$\displaystyle \: \sqrt{2} ( \frac{1}{ \sqrt{2} } \: sinx + \frac{1}{ \sqrt{2} } cosx)$

$= \displaystyle \: \sqrt{2} ( cos \frac{\pi}{4} \: sinx + sin \frac{\pi}{4} cosx)$

$= \displaystyle \: \sqrt{2} \: sin(x + \frac{\pi}{4} )$

Now we know that minimum value of Sine of an angle is - 1

So

$\displaystyle \: \: sin(x + \frac{\pi}{4} ) \geqslant - 1$

So

$\displaystyle \: \sqrt{2} \: sin(x + \frac{\pi}{4} ) \geqslant - \sqrt{2}$

So

$\implies \: \displaystyle \: {2}^{ sinx} + {2}^{ cosx} \: \geqslant 2\times \sqrt{ {2}^{ - \sqrt{2} } }$

$\implies \: \displaystyle \: {2}^{ sinx} + {2}^{ cosx} \: \geqslant 2\times {2}^{ - \frac{ \sqrt{2} }{2} }$

$\implies \: \displaystyle \: {2}^{ sinx} + {2}^{ cosx} \: \geqslant 2\times {2}^{ - \frac{1 }{ \sqrt{2} } }$

$\implies \: \displaystyle \: {2}^{ sinx} + {2}^{ cosx} \: \geqslant {2}^{(1 - \frac{1 }{ \sqrt{2} } )}$

So the required minimum value is

$\displaystyle \: {2}^{(1 - \frac{1 }{ \sqrt{2} } )}$