The function f:R to R is defined by f(x) = co | vedclass

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(C) Given $f(x) = \cos^2 x + \sin^4 x$. We can rewrite this as $f(x) = \cos^2 x + \sin^2 x \cdot \sin^2 x$. Since $\sin^2 x = 1 - \cos^2 x$,we have $f

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

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(C) Given $f(x) = \cos^2 x + \sin^4 x$. We can rewrite this as $f(x) = \cos^2 x + \sin^2 x \cdot \sin^2 x$. Since $\sin^2 x = 1 - \cos^2 x$,we have $f(x) = \cos^2 x + \sin^2 x(1 - \cos^2 x)$. $f(x) = \cos^2 x + \sin^2 x - \sin^2 x \cos^2 x$. Since $\cos^2 x + \sin^2 x = 1$,we get $f(x) = 1 - \sin^2 x \cos^2 x$. Multiplying and dividing by $4$,we get $f(x) = 1 - \frac{1}{4}(4 \sin^2 x \cos^2 x) = 1 - \frac{1}{4} \sin^2(2x)$. Since $0 \le \sin^2(2x) \le 1$,we have $1 - \frac{1}{4}(1) \le f(x) \le 1 - \frac{1}{4}(0)$. Thus,$\frac{3}{4} \le f(x) \le 1$. Therefore,$f(R) \in \left[ \frac{3}{4}, 1 \right]$.

The function $f:R \to R$ is defined by $f(x) = \cos^2 x + \sin^4 x$ for $x \in R$,then $f(R) \in $

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