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(B) Let $t = \sin \frac{x}{4}$. Since $x \in R$,$t \in [-1, 1]$.The expression becomes $f(t) = (1 - t^2) + t = -t^2 + t + 1$.This is a downward-openin

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$\frac{9}{4}$

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(B) Let $t = \sin \frac{x}{4}$. Since $x \in R$,$t \in [-1, 1]$.The expression becomes $f(t) = (1 - t^2) + t = -t^2 + t + 1$.This is a downward-opening parabola with vertex at $t = -\frac{b}{2a} = -\frac{1}{2(-1)} = \frac{1}{2}$.Since $t = \frac{1}{2}$ is within the interval $[-1, 1]$,the maximum value $\alpha = f(\frac{1}{2}) = -(\frac{1}{2})^2 + \frac{1}{2} + 1 = -\frac{1}{4} + \frac{1}{2} + 1 = \frac{5}{4}$.The minimum value $\beta$ occurs at the endpoints of the interval $[-1, 1]$.$f(-1) = -(-1)^2 + (-1) + 1 = -1 - 1 + 1 = -1$.$f(1) = -(1)^2 + (1) + 1 = -1 + 1 + 1 = 1$.Thus,$\beta = -1$.Therefore,$\alpha - \beta = \frac{5}{4} - (-1) = \frac{5}{4} + 1 = \frac{9}{4}$.

If $\alpha$ is the maximum value and $\beta$ is the minimum value of $\cos^2 \frac{x}{4} + \sin \frac{x}{4}$,$x \in R$,then $\alpha - \beta =$

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