The function f(x) = frac{lambda sin x + 3 cos | vedclass

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(B) For a function of the form $f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x}$ to be monotonically increasing,its derivative $f'(x)$ must be

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$\lambda > 1$

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(B) For a function of the form $f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x}$ to be monotonically increasing,its derivative $f'(x)$ must be greater than $0$.The derivative of $f(x) = \frac{u}{v}$ is $f'(x) = \frac{u'v - uv'}{v^2}$.Here,$u = \lambda \sin x + 3 \cos x$ and $v = 2 \sin x + 6 \cos x$.$u' = \lambda \cos x - 3 \sin x$ and $v' = 2 \cos x - 6 \sin x$.$u'v - uv' = (\lambda \cos x - 3 \sin x)(2 \sin x + 6 \cos x) - (\lambda \sin x + 3 \cos x)(2 \cos x - 6 \sin x)$.Expanding this,we get $(2\lambda \sin x \cos x + 6\lambda \cos^2 x - 6 \sin^2 x - 18 \sin x \cos x) - (2\lambda \sin x \cos x - 6\lambda \sin^2 x + 6 \cos^2 x - 18 \sin x \cos x)$.Simplifying,the terms $2\lambda \sin x \cos x$ and $-18 \sin x \cos x$ cancel out.We are left with $6\lambda \cos^2 x - 6 \sin^2 x + 6\lambda \sin^2 x - 6 \cos^2 x = 6(\lambda - 1)(\cos^2 x + \sin^2 x) = 6(\lambda - 1)$.For $f'(x) > 0$,we must have $6(\lambda - 1) > 0$,which implies $\lambda - 1 > 0$,or $\lambda > 1$.

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

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