Question

If $f\left(x\right)=2^{\sin x}+2^{\cos x},x\in [-\pi ,\pi ],$ then which of the following is/are always true ?

• A. $f\left(x\right)\le 2^{1+\frac{1}{\sqrt{2}}}$
• B. $f\left(x\right)\ge 2^{1-\frac{1}{\sqrt{2}}}$
• C. $f\left(x\right)\ge 2^{1+\sqrt{2}}$
• D. $f\left(x\right)\le 2^{1-\sqrt{2}}$

Answer 1

Answer: (A), (B)

The correct options are
A $f\left(x\right)\le 2^{1+\frac{1}{\sqrt{2}}}$
B $f\left(x\right)\ge 2^{1-\frac{1}{\sqrt{2}}}$

Let $g\left(x\right)=\sin x+\cos x$
$\Rightarrow g\left(x\right)=\sqrt{2}\sin (\frac{\pi }{4}+x)$
So, $g(x{)}_{\text{max}}=\sqrt{2}\text{at}x=\frac{\pi }{4}$
$g(x{)}_{\text{mini}}=-\sqrt{2}\text{at}x=-\frac{3\pi }{4}$

Now $f\left(x\right)$ will be maximum when $g\left(x\right)$ will be maximum
and $f\left(x\right)$ will be minimum when $g\left(x\right)$ will be minimum

So, $f(x{)}_{\text{max}}=2^{\sin \pi /4}+2^{\cos \pi /4}=2^{1/\sqrt{2}}+2^{1/\sqrt{2}}={2.2}^{1/\sqrt{2}}=2^{1+\frac{1}{\sqrt{2}}}$
Similarly $f(x{)}_{\text{mini}}=2^{1-\frac{1}{\sqrt{2}}}$