Question

The inequality $2^{\sin \theta }+2^{\cos \theta }\ge 2^{(1-\frac{1}{\sqrt{2}})}$ holds for

• A. $0\le \theta <\pi$
• B. $\pi \le \theta <2\pi$
• C. for all real $\theta$
• D. None of these

Answer 1

The correct option is C for all real $\theta$

We have

$\frac{1}{2}[2^{\sin \theta }+2^{\cos \theta }]\ge \sqrt{2^{\sin \theta }.2^{\cos \theta }}[\because A.M.\ge G.M.]$

$\Rightarrow 2^{\sin \theta }+2^{\cos \theta }\ge 2.2^{(sin\theta +\cos \theta )/2}$ ...(1)

Now $(\sin \theta +\cos \theta )=\sqrt{2}\sin (\theta +\frac{\pi }{4})\ge -\sqrt{2}$ for all real $\theta$

$\therefore 2^{\sin \theta }+2^{\cos \theta }\ge 2.2^{(\sin \theta +\cos \theta )/2}>2.2^{-\frac{\sqrt{2}}{2}}=2^{1-\frac{1}{\sqrt{2}}}$

Thus, $2^{\sin \theta }+2^{\cos \theta }\ge 2^{1-\frac{1}{\sqrt{2}}}$ for all real $\theta$