The function f(x) = frac{lambda sin x + 6 cos | vedclass

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(D) For the function $f(x)$ to be monotonically increasing,we must have $f'(x) > 0$. Applying the quotient rule $\left( \frac{u}{v} \right)' = \frac{u

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$\lambda > 4$

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(D) For the function $f(x)$ to be monotonically increasing,we must have $f'(x) > 0$. Applying the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$,we get: $f'(x) = \frac{(\lambda \cos x - 6 \sin x)(2 \sin x + 3 \cos x) - (\lambda \sin x + 6 \cos x)(2 \cos x - 3 \sin x)}{(2 \sin x + 3 \cos x)^2} > 0$. Expanding the numerator: $(2\lambda \sin x \cos x + 3\lambda \cos^2 x - 12 \sin^2 x - 18 \sin x \cos x) - (2\lambda \sin x \cos x - 3\lambda \sin^2 x + 12 \cos^2 x - 18 \sin x \cos x) > 0$. Simplifying the terms: $3\lambda \cos^2 x - 12 \sin^2 x + 3\lambda \sin^2 x - 12 \cos^2 x > 0$. $3\lambda(\sin^2 x + \cos^2 x) - 12(\sin^2 x + \cos^2 x) > 0$. Since $\sin^2 x + \cos^2 x = 1$,we have: $3\lambda - 12 > 0$. $3\lambda > 12$. $\lambda > 4$.

The function $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$ is monotonically increasing,if:

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