MCQ
The function $f : R \rightarrow R$ is defined by $\text{f(x)}=\cos^2\text{x}+\sin^4\text{x}.$ Then, $f(R) =$
- A$\Big[\frac{3}{4},1\Big]$
- B$\Big(\frac{3}{4},1\Big]$
- ✓$\Big[\frac{3}{4},1\Big]$
- D$\Big(\frac{3}{4},1\Big)$
✓
Answer
Correct option: C.
$\Big[\frac{3}{4},1\Big]$
Given,
$\text{f(x)}=\cos^2\text{x}+\sin^4\text{x}$
$\Rightarrow\text{f(x)}=1-\sin^2\text{x}+\sin^4\text{x}$
$\Rightarrow\text{f(x)}=\Big(\sin^2\text{x}-\frac{1}{2}\Big)^2+\frac{3}{4}$
The minimum value of $\text{f(x)}$ is $\frac{3}{4}$
Also,
$\sin^2\text{x}\leq1$
$\Rightarrow\ \sin^2\text{x}-\frac{1}{2}\leq\frac{1}{2}$
$\Rightarrow\ \Big(\sin^2\text{x}-\frac{1}{2}\Big)^2\leq\frac{1}{4}$
$\Rightarrow\ \Big(\sin^2\text{x}-\frac{1}{2}\Big)^2+\frac{3}{4}\leq\frac{1}{4}+\frac{3}{4}$
$\Rightarrow\ \text{f(x)}\leq1$
The maximum value of $f(x)$ is $1$
$\therefore\ \text{f(R)}=\Big(\frac{3}{4},1\Big)$
$\text{f(x)}=\cos^2\text{x}+\sin^4\text{x}$
$\Rightarrow\text{f(x)}=1-\sin^2\text{x}+\sin^4\text{x}$
$\Rightarrow\text{f(x)}=\Big(\sin^2\text{x}-\frac{1}{2}\Big)^2+\frac{3}{4}$
The minimum value of $\text{f(x)}$ is $\frac{3}{4}$
Also,
$\sin^2\text{x}\leq1$
$\Rightarrow\ \sin^2\text{x}-\frac{1}{2}\leq\frac{1}{2}$
$\Rightarrow\ \Big(\sin^2\text{x}-\frac{1}{2}\Big)^2\leq\frac{1}{4}$
$\Rightarrow\ \Big(\sin^2\text{x}-\frac{1}{2}\Big)^2+\frac{3}{4}\leq\frac{1}{4}+\frac{3}{4}$
$\Rightarrow\ \text{f(x)}\leq1$
The maximum value of $f(x)$ is $1$
$\therefore\ \text{f(R)}=\Big(\frac{3}{4},1\Big)$
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