The function f :
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)

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