Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2805 Upvotes
Verified Answer
The correct answer is:
$\left[\frac{3}{4}, 1\right]$
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 $f=\left[\frac{3}{4}, 1\right]$
$$
\begin{aligned}
& f(x)=\cos ^2 x+\sin ^4 x=1-\sin ^2 x+\sin ^4 x \\
& =1-\sin ^2 x\left(1-\sin ^2 x\right)=1-\sin ^2 x \cos ^2 x \\
& =1-\frac{1}{4} \sin ^2 2 x
\end{aligned}
$$
$$
\text { Here, } 0 \leq \sin ^2 2 x \leq 1 \Rightarrow 0 \leq \frac{\sin ^2 2 x}{4} \leq \frac{1}{4}
$$
$$
\begin{aligned}
& \Rightarrow \quad 0 \geq-\frac{\sin ^2 2 x}{4} \geq-\frac{1}{4} \\
& \Rightarrow \quad 1 \geq 1-\frac{\sin ^2 2 x}{4} \geq-\frac{1}{4} \\
& 1 \geq 1-\frac{\sin ^2 2 x}{4} \geq-\frac{1}{4}+1 \\
& \Rightarrow \quad 1 \geq 1-\frac{\sin ^2 2 x}{4} \geq \frac{3}{4} \\
&
\end{aligned}
$$
Range of $f=\left[\frac{3}{4}, 1\right]$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.