If f(x) = frac{k sin x + 2 cos x}{sin x + cos | vedclass

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(D) For $f(x)$ to be strictly increasing,we must have $f'(x) > 0$ for all $x$. Given $f(x) = \frac{k \sin x + 2 \cos x}{\sin x + \cos x}$. Using the q

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$k > 2$

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(D) For $f(x)$ to be strictly increasing,we must have $f'(x) > 0$ for all $x$. Given $f(x) = \frac{k \sin x + 2 \cos x}{\sin x + \cos x}$. Using the quotient rule,$f'(x) = \frac{(k \cos x - 2 \sin x)(\sin x + \cos x) - (k \sin x + 2 \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2}$. Simplifying the numerator: $N = (k \sin x \cos x + k \cos^2 x - 2 \sin^2 x - 2 \sin x \cos x) - (k \sin x \cos x - k \sin^2 x + 2 \cos^2 x - 2 \sin x \cos x)$. $N = k \cos^2 x - 2 \sin^2 x + k \sin^2 x - 2 \cos^2 x$. $N = (k - 2) \cos^2 x + (k - 2) \sin^2 x = (k - 2)(\cos^2 x + \sin^2 x) = k - 2$. Thus,$f'(x) = \frac{k - 2}{(\sin x + \cos x)^2}$. For $f(x)$ to be strictly increasing,$f'(x) > 0$. Since $(\sin x + \cos x)^2 > 0$ for all $x$ where the function is defined,we require $k - 2 > 0$,which implies $k > 2$.

If $f(x) = \frac{k \sin x + 2 \cos x}{\sin x + \cos x}$ is strictly increasing for all real values of $x$,then

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