If $f(x)=\sin ^6 x+\cos ^6 x$ for $x \in R$, then $f(x)$ lies in the interval

1. $\left[\frac{7}{8}, \frac{5}{4}\right]$
2. $\left[\frac{1}{2}, \frac{5}{8}\right]$
3. $\left[\frac{1}{4}, 1\right]$
4. $\left[\frac{1}{4}, \frac{1}{2}\right]$

Solution

Given, $f(x)=\sin ^6 x+\cos ^6 x$ and $x \in R$ $\begin{aligned} & f(x)=\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3 \\ & \Rightarrow \quad f(x)=\left(\sin ^2 x+\cos ^2 x\right) \\ & \left\{\sin ^4 x+\cos ^4 x-\sin ^2 x \cdot \cos ^2 x\right\} \\ & \Rightarrow \\ & f(x)=1\left\{\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cdot \cos ^2 x\right. \\ & \left.-\sin ^2 x \cdot \cos ^2 x\right\} \\ & \Rightarrow \quad f(x)=\left\{1-3 \sin ^2 x \cdot \cos ^2 x\right\} \\ & \end{aligned}$
$\because \quad 0 \leq \sin ^2 x \leq 1$ $\begin{array}{lc}\Rightarrow & 0 \leq(\sin 2 x)^2 \leq 1 \\ \Rightarrow & 0 \geq \frac{-3(\sin 2 x)^2}{4} \geq-3 / 4 \\ \Rightarrow & 1 \geq 1-\frac{3}{4}(\sin 2 x)^2 \geq 1-\frac{3}{4} \\ \Rightarrow & 1 \geq f(x) \geq 1 / 4 \\ \Rightarrow & f(x)[1 / 4,1] \forall x \in R\end{array}$

Asked in: AP EAMCET 2011

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