If f(x) = frac{lambda sin x + 6 cos x}{2 sin | vedclass

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(D) Given $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$.Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$

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$\lambda > 4$

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(D) Given $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$.Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$,we find $f'(x)$:$f'(x) = \frac{(\lambda \cos x - 6 \sin x)(2 \sin x + 3 \cos x) - (\lambda \sin x + 6 \cos x)(2 \cos x - 3 \sin x)}{(2 \sin x + 3 \cos x)^2}$Expanding the numerator:$= \frac{(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)}{(2 \sin x + 3 \cos x)^2}$$= \frac{3\lambda \cos^2 x - 12 \sin^2 x + 3\lambda \sin^2 x - 12 \cos^2 x}{(2 \sin x + 3 \cos x)^2}$$= \frac{3\lambda(\cos^2 x + \sin^2 x) - 12(\sin^2 x + \cos^2 x)}{(2 \sin x + 3 \cos x)^2} = \frac{3\lambda - 12}{(2 \sin x + 3 \cos x)^2}$Since $f(x)$ is strictly increasing,$f'(x) > 0$ for all $x$ in the domain.Since the denominator $(2 \sin x + 3 \cos x)^2$ is always positive,we must have $3\lambda - 12 > 0$.$3\lambda > 12 \implies \lambda > 4$.

If $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$ is a strictly increasing function,then .......

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