Question

The minimum value of $f(x)={\sin }^{4}x+{\cos }^{4}x,0\le x\le \frac{\pi }{2}$

• (A) $\frac{1}{2\sqrt{2}}$
• (B) $\frac{1}{4}$$\frac{1}{4}$
• (C) $\frac{-1}{2}$
• (D) $\frac{1}{2}$

Answer 1

The Correct Option is D

Explanation:
$$\begin{array}{l}\because f(x)={\sin }^{4}x+{\cos }^{4}x\\ ={({\sin }^{2}x+{\cos }^{2}x)}^2-2{\sin }^{2}x\cdot {\cos }^{2}x\\ \Rightarrow f(x)=1-\frac{1}{2}{\sin }^{2}2x\\ \text{ Also }0\le {\sin }^{2}2x\le 1\end{array}$$$\therefore$ Minimum value of $f(x)$$$1-\frac{1}{2}=\frac{1}{2}$$