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)

Answer 1

(c) (3/4, 1)

Given:
f(x) = cos²x + sin⁴x
$\Rightarrow f\left(x\right)=1-{\sin }^{2}x+{\sin }^{4}x$
$\Rightarrow f\left(x\right)={\left({\sin }^{2}x-\frac{1}{2}\right)}^2+\frac{3}{4}$
The minimum value of $f\left(x\right)$ is $\frac{3}{4}$.
Also,
$$\begin{gathered} {\sin }^{2}x\le 1 \\ \Rightarrow {\sin }^{2}x-\frac{1}{2}\le \frac{1}{2} \\ \Rightarrow {\left({\sin }^{2}x-\frac{1}{2}\right)}^2\le \frac{1}{4} \\ \Rightarrow {\left({\sin }^{2}x-\frac{1}{2}\right)}^2+\frac{3}{4}\le \frac{1}{4}+\frac{3}{4} \\ \Rightarrow f\left(x\right)\le 1 \end{gathered}$$

The maximum value of $f\left(x\right)$ is 1.

∴ f(R) = (3/4, 1)