The function f : R → R is defined by f(x)= ^2x+ ^4x. Then, f(R) = - Vidyadip

MCQ

The function f : R → 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
$\Big(\frac{3}{4},1\Big]$
C
$\Big[\frac{3}{4},1\Big]$
D
$\Big(\frac{3}{4},1\Big)$

✓

Answer

$\Big[\frac{3}{4},1\Big]$

Solution:

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

$\therefore\ \text{f(R)}=\Big(\frac{3}{4},1\Big)$

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