If m and M are the minimum and the maximum va | vedclass

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(A) Let $f(x) = 4 + \frac{1}{2} \sin^2 2x - 2 \cos^4 x$.Using $\sin^2 2x = (2 \sin x \cos x)^2 = 4 \sin^2 x \cos^2 x = 4(1 - \cos^2 x) \cos^2 x$,we ge

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$\frac{9}{4}$

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(A) Let $f(x) = 4 + \frac{1}{2} \sin^2 2x - 2 \cos^4 x$.Using $\sin^2 2x = (2 \sin x \cos x)^2 = 4 \sin^2 x \cos^2 x = 4(1 - \cos^2 x) \cos^2 x$,we get:$f(x) = 4 + \frac{1}{2} [4(1 - \cos^2 x) \cos^2 x] - 2 \cos^4 x$$f(x) = 4 + 2 \cos^2 x - 2 \cos^4 x - 2 \cos^4 x$$f(x) = 4 + 2 \cos^2 x - 4 \cos^4 x$.Let $t = \cos^2 x$,where $t \in [0, 1]$.Then $g(t) = -4t^2 + 2t + 4$.This is a downward parabola with vertex at $t = -\frac{b}{2a} = -\frac{2}{2(-4)} = \frac{1}{4}$.Since $\frac{1}{4} \in [0, 1]$,the maximum value $M$ occurs at $t = \frac{1}{4}$:$M = g(\frac{1}{4}) = -4(\frac{1}{16}) + 2(\frac{1}{4}) + 4 = -\frac{1}{4} + \frac{1}{2} + 4 = \frac{1}{4} + 4 = \frac{17}{4}$.The minimum value $m$ occurs at the boundaries $t = 0$ or $t = 1$:$g(0) = 4$$g(1) = -4(1)^2 + 2(1) + 4 = 2$.Thus,$m = 2$.$M - m = \frac{17}{4} - 2 = \frac{17 - 8}{4} = \frac{9}{4}$.

If $m$ and $M$ are the minimum and the maximum values of $4 + \frac{1}{2} \sin^2 2x - 2 \cos^4 x$ for $x \in R$,then $M - m$ is equal to

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