Question

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

• A. $\frac{5\pi}{4}$
• B. $\frac{3\pi}{4}$
• C. $\frac{\pi}{2}$
• D. All values of x.

Answer 1

Answer: (A)

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

Answer 2

Since A.M. $\geq$G.M.

$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathbf{2}^{\mathbf{\sin}\mathbf{x}}\mathbf{+}\mathbf{2}^{\mathbf{\cos}\mathbf{x}}\mathbf{) \geq}\sqrt{\mathbf{2}^{\mathbf{\sin}\mathbf{x}}\mathbf{.}\mathbf{2}^{\mathbf{\cos}\mathbf{x}}}$ $\Rightarrow$ $\mathbf{2}^{\mathbf{\sin}\mathbf{x}}\mathbf{+}\mathbf{2}^{\mathbf{\cos}\mathbf{x}}\mathbf{\geq}\mathbf{2.}\mathbf{2}^{\frac{\mathbf{\sin}\mathbf{x}\mathbf{+}\mathbf{\cos}\mathbf{x}}{\mathbf{2}}}$

$\mathbf{\Rightarrow}$ $\mathbf{2}^{\mathbf{\sin}\mathbf{x}}\mathbf{+}\mathbf{2}^{\mathbf{\cos}\mathbf{x}}\mathbf{\geq}\mathbf{2}^{\mathbf{1 +}\frac{\mathbf{\sin}\mathbf{x}\mathbf{+}\mathbf{\cos}\mathbf{x}}{\mathbf{2}}}$

And, we know that $\sin x + \cos x \geq - \sqrt{2}$

$\therefore 2^{\sin x} + 2^{\cos x} > 2^{1 - (1/\sqrt{2})}$ for $x = \frac{5\pi}{4}.$