The only value of x for which 2^{sin x} + 2^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) By the $A.M.-G.M.$ inequality,we have $\frac{2^{\sin x} + 2^{\cos x}}{2} \ge \sqrt{2^{\sin x} \cdot 2^{\cos x}}$.This simplifies to $2^{\sin x} +

Vedclass · Accepted Answer

$\frac{5\pi}{4}$

Vedclass · Answer

(A) By the $A.M.-G.M.$ inequality,we have $\frac{2^{\sin x} + 2^{\cos x}}{2} \ge \sqrt{2^{\sin x} \cdot 2^{\cos x}}$.This simplifies to $2^{\sin x} + 2^{\cos x} \ge 2 \cdot 2^{\frac{\sin x + \cos x}{2}} = 2^{1 + \frac{\sin x + \cos x}{2}}$.We know that $\sin x + \cos x = \sqrt{2} \sin(x + \frac{\pi}{4})$,which has a minimum value of $-\sqrt{2}$.Thus,the minimum value of the expression $2^{\sin x} + 2^{\cos x}$ is $2^{1 - \frac{\sqrt{2}}{2}} = 2^{1 - (1/\sqrt{2})}$.The inequality $2^{\sin x} + 2^{\cos x} > 2^{1 - (1/\sqrt{2})}$ holds for all $x$ except where $\sin x + \cos x = -\sqrt{2}$.$\sin x + \cos x = -\sqrt{2}$ occurs when $\sin(x + \frac{\pi}{4}) = -1$,which implies $x + \frac{\pi}{4} = \frac{3\pi}{2}$,so $x = \frac{5\pi}{4}$.

The only value of $x$ for which $2^{\sin x} + 2^{\cos x} > 2^{1 - (1/\sqrt{2})}$ holds,is

Similar Questions

The maximum value of $e^{(2 + \sqrt{3} \cos x + \sin x)}$ is

If $A+B+C=270^{\circ}$,then $\cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C =$

The maximum value of $\sin \left( x + \frac{\pi}{6} \right) + \cos \left( x + \frac{\pi}{6} \right)$ in the interval $\left( 0, \frac{\pi}{2} \right)$ is attained at

If $A + B + C = 180^{\circ}$,then $\tan A + \tan B + \tan C = $

If $f(x) = \sin^6 x + \cos^6 x$ for $x \in R$,then $f(x)$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module