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

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Question

(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 - \s

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 get: $f(x) = 1 - \frac{4 \sin^2 x \cos^2 x}{4} = 1 - \frac{(2 \sin x \cos x)^2}{4} = 1 - \frac{\sin^2 2x}{4}$. Since $0 \leq \sin^2 2x \leq 1$, we have: $0 \leq \frac{\sin^2 2x}{4} \leq \frac{1}{4}$. Multiplying by $-1$ and adding $1$: $1 - 0 \geq 1 - \frac{\sin^2 2x}{4} \geq 1 - \frac{1}{4}$. $1 \geq f(x) \geq \frac{3}{4}$. Thus, the range $f(R) = \left[\frac{3}{4}, 1\right]$.

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

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