The function f: R rightarrow R is defined by | vedclass

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(C) Given $f(x) = \cos^2 x + \sin^4 x$. Using $\cos^2 x = 1 - \sin^2 x$,we get: $f(x) = 1 - \sin^2 x + \sin^4 x = 1 - \sin^2 x(1 - \sin^2 x) = 1 - \si

Vedclass · Accepted Answer

$\left[\frac{3}{4}, 1\right]$

Vedclass · Answer

(C) Given $f(x) = \cos^2 x + \sin^4 x$. Using $\cos^2 x = 1 - \sin^2 x$,we get: $f(x) = 1 - \sin^2 x + \sin^4 x = 1 - \sin^2 x(1 - \sin^2 x) = 1 - \sin^2 x \cos^2 x$. Multiplying and dividing by $4$,we have: $f(x) = 1 - \frac{4 \sin^2 x \cos^2 x}{4} = 1 - \frac{(2 \sin x \cos x)^2}{4} = 1 - \frac{\sin^2(2x)}{4}$. Since $0 \leq \sin^2(2x) \leq 1$,we have $0 \leq \frac{\sin^2(2x)}{4} \leq \frac{1}{4}$. Subtracting from $1$,we get: $1 - \frac{1}{4} \leq 1 - \frac{\sin^2(2x)}{4} \leq 1 - 0$. Thus,$\frac{3}{4} \leq f(x) \leq 1$. The range is $\left[\frac{3}{4}, 1\right]$.

The function $f: R \rightarrow R$ is defined by $f(x) = \cos^2 x + \sin^4 x$ for $x \in R$. Then the range $f(R)$ is equal to

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