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

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(C) Given the function $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$.For the function to be increasing,we must have $f'(x) \geq 0$.Us

Vedclass · Accepted Answer

$\lambda \geq 4$

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(C) Given the function $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$.For the function to be increasing,we must have $f'(x) \geq 0$.Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$,we calculate $f'(x)$:$f'(x) = \frac{(\lambda \cos x - 6 \sin x)(2 \sin x + 3 \cos x) - (\lambda \sin x + 6 \cos x)(2 \cos x - 3 \sin x)}{(2 \sin x + 3 \cos x)^2}$.Expanding the numerator:Numerator $= (2\lambda \sin x \cos x + 3\lambda \cos^2 x - 12 \sin^2 x - 18 \sin x \cos x) - (2\lambda \sin x \cos x - 3\lambda \sin^2 x + 12 \cos^2 x - 18 \sin x \cos x)$.Simplifying the numerator:Numerator $= 3\lambda \cos^2 x - 12 \sin^2 x + 3\lambda \sin^2 x - 12 \cos^2 x$.Numerator $= 3\lambda(\sin^2 x + \cos^2 x) - 12(\sin^2 x + \cos^2 x) = 3\lambda - 12$.Since the denominator $(2 \sin x + 3 \cos x)^2$ is always positive,$f'(x) \geq 0$ implies $3\lambda - 12 \geq 0$.Therefore,$3\lambda \geq 12$,which gives $\lambda \geq 4$.

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

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