Question:
The function f : R → R is defined by f(x) = cos2 x + sin4 x. Then, f(R) =
(a) [3/4, 1)
(b) (3/4, 1]
(c) [3/4, 1]
(d) (3/4, 1)
Solution:
(c) (3/4, 1)
Given:
f(x) = cos2x + sin4x
$\Rightarrow f(x)=1-\sin ^{2} x+\sin ^{4} x$
$\Rightarrow f(x)=\left(\sin ^{2} x-\frac{1}{2}\right)^{2}+\frac{3}{4}$
The minimum value of $f(x)$ is $\frac{3}{4}$.
Also,
$\sin ^{2} x \leq 1$
$\Rightarrow \sin ^{2} x-\frac{1}{2} \leq \frac{1}{2}$
$\Rightarrow\left(\sin ^{2} x-\frac{1}{2}\right)^{2} \leq \frac{1}{4}$
$\Rightarrow\left(\sin ^{2} x-\frac{1}{2}\right)^{2}+\frac{3}{4} \leq \frac{1}{4}+\frac{3}{4}$
$\Rightarrow f(x) \leq 1$
The maximum value of $f(x)$ is 1 .
∴ f(R) = (3/4, 1)