If the function f(x) = frac{1-sin 2x + cos 2x | vedclass

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(D) For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,we must have $\lim_{x \rightarrow \frac{\pi}{2}} f(x) = f(\frac{\pi}{2}) = k$. First,simplify t

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$-1$

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(D) For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,we must have $\lim_{x \rightarrow \frac{\pi}{2}} f(x) = f(\frac{\pi}{2}) = k$. First,simplify the expression for $f(x)$: $f(x) = \frac{(1+\cos 2x) - \sin 2x}{(1+\cos 2x) + \sin 2x} = \frac{2\cos^2 x - 2\sin x \cos x}{2\cos^2 x + 2\sin x \cos x}$. Factoring out $2\cos x$ from the numerator and denominator: $f(x) = \frac{2\cos x(\cos x - \sin x)}{2\cos x(\cos x + \sin x)} = \frac{\cos x - \sin x}{\cos x + \sin x}$. Now,evaluate the limit as $x \rightarrow \frac{\pi}{2}$: $\lim_{x \rightarrow \frac{\pi}{2}} \frac{\cos x - \sin x}{\cos x + \sin x} = \frac{\cos(\frac{\pi}{2}) - \sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2})} = \frac{0 - 1}{0 + 1} = -1$. Thus,$k = -1$.

If the function $f(x) = \frac{1-\sin 2x + \cos 2x}{1+\sin 2x + \cos 2x}$ for $x \neq \frac{\pi}{2}$ and $f(x) = k$ for $x = \frac{\pi}{2}$ is continuous at $x = \frac{\pi}{2}$,then $k = $

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