If the function f(x) = frac{Ksin x + 2cos x}{ | vedclass

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(D) Given $f(x) = \frac{K\sin x + 2\cos x}{\sin x + \cos x}$.To find the derivative $f'(x)$,we use the quotient rule $\left( \frac{u}{v} \right)' = \f

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$K > 2$

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(D) Given $f(x) = \frac{K\sin x + 2\cos x}{\sin x + \cos x}$.To find the derivative $f'(x)$,we use the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$.Let $u = K\sin x + 2\cos x$ and $v = \sin x + \cos x$.Then $u' = K\cos x - 2\sin x$ and $v' = \cos x - \sin x$.$f'(x) = \frac{(K\cos x - 2\sin x)(\sin x + \cos x) - (K\sin x + 2\cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2}$.Expanding the numerator:$= (K\sin x \cos x + K\cos^2 x - 2\sin^2 x - 2\sin x \cos x) - (K\sin x \cos x - K\sin^2 x + 2\cos^2 x - 2\sin x \cos x)$.$= K\cos^2 x - 2\sin^2 x + K\sin^2 x - 2\cos^2 x$.$= K(\sin^2 x + \cos^2 x) - 2(\sin^2 x + \cos^2 x) = K - 2$.Thus,$f'(x) = \frac{K - 2}{(\sin x + \cos x)^2}$.Since the function is increasing for all $x$,$f'(x) > 0$.Since $(\sin x + \cos x)^2 > 0$ for all $x$ where the function is defined,we must have $K - 2 > 0$,which implies $K > 2$.

If the function $f(x) = \frac{K\sin x + 2\cos x}{\sin x + \cos x}$ is increasing for all values of $x,$ then

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