The function $f: R \rightarrow R$ is defined by $f(x)=\cos ^2 x+\sin ^4 x$ for $x \in R$, then $f(R)$ is…
The function $f: R \rightarrow R$ is defined by $f(x)=\cos ^2 x+\sin ^4 x$ for $x \in R$, then $f(R)$ is equal to
- $\left(\frac{3}{4}, 1\right]$
- $\left[\frac{3}{4}, 1\right)$
- $\left[\frac{3}{4}, 1\right]$
- $\left(\frac{3}{4}, 1\right)$
Solution
We have,
$
\begin{aligned}
& f(x)=\cos ^2 x+\sin ^4 x=1-\sin ^2 x+\sin ^4 x \\
& =1-\sin ^2 x\left(1-\sin ^2 x\right)=1-\sin ^2 x \cos ^2 x \\
& =1-\frac{1}{4} \sin ^2 2 x
\end{aligned}
$
$
\text { Here, } 0 \leq \sin ^2 2 x \leq 1 \Rightarrow 0 \leq \frac{\sin ^2 2 x}{4} \leq \frac{1}{4}
$
$
\begin{aligned}
& \Rightarrow \quad 0 \geq-\frac{\sin ^2 2 x}{4} \geq-\frac{1}{4} \\
& \Rightarrow \quad 1 \geq 1-\frac{\sin ^2 2 x}{4} \geq-\frac{1}{4} \\
& 1 \geq 1-\frac{\sin ^2 2 x}{4} \geq-\frac{1}{4}+1 \\
& \Rightarrow \quad 1 \geq 1-\frac{\sin ^2 2 x}{4} \geq \frac{3}{4} \\
&
\end{aligned}
$
Range of $f=\left[\frac{3}{4}, 1\right]$
Asked in: AP EAMCET 2002
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