If the function f(x) = frac{2sqrt{2} - (cos x | vedclass

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(A) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$, $f(\frac{\pi}{4}) = \lim_{x 	o \frac{\pi}{4}} f(x)$.Let $t = x - \frac{\pi}{4}$, so as $x 	o

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$3\sqrt{2}$

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(A) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$, $f(\frac{\pi}{4}) = \lim_{x 	o \frac{\pi}{4}} f(x)$.Let $t = x - \frac{\pi}{4}$, so as $x 	o \frac{\pi}{4}$, $t 	o 0$. Then $x = t + \frac{\pi}{4}$.$\cos x + \sin x = \sqrt{2} \sin(x + \frac{\pi}{4}) = \sqrt{2} \sin(t + \frac{\pi}{2}) = \sqrt{2} \cos t$.$1 - \sin 2x = 1 - \sin(2t + \frac{\pi}{2}) = 1 - \cos 2t = 2 \sin^2 t$.$f(x) = \frac{2\sqrt{2} - (\sqrt{2} \cos t)^3}{2 \sin^2 t} = \frac{2\sqrt{2} - 2\sqrt{2} \cos^3 t}{2 \sin^2 t} = \sqrt{2} \frac{1 - \cos^3 t}{\sin^2 t}$.Using $1 - \cos^3 t = (1 - \cos t)(1 + \cos t + \cos^2 t)$ and $\sin^2 t = (1 - \cos t)(1 + \cos t)$:$f(x) = \sqrt{2} \frac{(1 - \cos t)(1 + \cos t + \cos^2 t)}{(1 - \cos t)(1 + \cos t)} = \sqrt{2} \frac{1 + \cos t + \cos^2 t}{1 + \cos t}$.As $t 	o 0$, $\cos t 	o 1$, so $f(\frac{\pi}{4}) = \sqrt{2} \frac{1 + 1 + 1}{1 + 1} = \sqrt{2} \cdot \frac{3}{2} = \frac{3\sqrt{2}}{2}$.Note: The provided options do not contain $\frac{3\sqrt{2}}{2}$. Re-evaluating the limit: $\lim_{t 	o 0} \sqrt{2} \frac{1 + \cos t + \cos^2 t}{1 + \cos t} = \frac{3\sqrt{2}}{2}$. Since this is not in the options, the question or options may be flawed.

If the function $f(x) = \frac{2\sqrt{2} - (\cos x + \sin x)^3}{1 - \sin 2x}$ is continuous at $x = \frac{\pi}{4}$, then the value of $f(\frac{\pi}{4})$ is...

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