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

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a

\frac{1}{2 \sqrt{2}}

b

\frac{1}{4}

c

- \frac{1}{2}

d

\frac{1}{2}

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detailed solution

Correct option is D

We have,

$\begin{array}{l} f (x) = \sin^{4} x + \cos^{4} x, 0 \leq x \leq \frac{π}{2} \\ \Rightarrow f (x) = {(\sin^{2} x + \cos^{2} x)}^{2} - \frac{1}{2} \sin^{2} 2 x \\ \Rightarrow f (x) = 1 - \frac{1}{4} (1 - \cos 4 x) \\ \Rightarrow f (x) = \frac{3}{4} + \frac{1}{4} \cos 4 x \end{array}$

$∵ - 1 \leq \cos 4 x \leq 1 for x \in [0, π / 2]$

$∴ - \frac{1}{4} \leq \frac{1}{4} \cos 4 x \leq \frac{1}{4}$ for all $x \in [0, π / 2]$

$\Rightarrow \frac{1}{2} \leq \frac{3}{4} + \frac{1}{4} \cos 4 x \leq 1$ for all $x \in [0, π / 2]$

$\Rightarrow \frac{1}{2} \leq f (x) \leq 1$ for all $x \in [0, π / 2]$

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