Question 35

The least value of $$4^{\sin x} + 4^{\cos x}$$ for $$x \in R$$, is

Solution

Using $$A.M.\ge\ G.M.$$

$$\frac{\left(4^{\sin x}+4^{\cos x}\right)}{2}\ge\ \sqrt{\ 4^{\sin x+\cos x}}$$

For minimum value $$\left(4^{\sin x}+4^{\cos x}\right)=2\ \sqrt{\ 4^{\sin x+\cos x}}$$.

The minimum value of sinx +cosx when x lies in R is $$-\sqrt{\ 2}$$

Thus, the minimum value of the expression is $$2^{1-\sqrt{\ 2}}$$

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: