If f(x) = frac{a sin x + b cos x}{c sin x + d | vedclass

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(B) Given $f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x}$.For $f(x)$ to be decreasing,we must have $f'(x) < 0$.Using the quotient rule,$f'(x)

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$ad - bc < 0$

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(B) Given $f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x}$.For $f(x)$ to be decreasing,we must have $f'(x) Using the quotient rule,$f'(x) = \frac{(c \sin x + d \cos x)(a \cos x - b \sin x) - (a \sin x + b \cos x)(c \cos x - d \sin x)}{(c \sin x + d \cos x)^2}$.Expanding the numerator:$= (ac \sin x \cos x - bc \sin^2 x + ad \cos^2 x - bd \sin x \cos x) - (ac \sin x \cos x - ad \sin^2 x + bc \cos^2 x - bd \sin x \cos x)$.$= ac \sin x \cos x - bc \sin^2 x + ad \cos^2 x - bd \sin x \cos x - ac \sin x \cos x + ad \sin^2 x - bc \cos^2 x + bd \sin x \cos x$.$= ad(\sin^2 x + \cos^2 x) - bc(\sin^2 x + \cos^2 x)$.$= ad - bc$.Since the denominator $(c \sin x + d \cos x)^2$ is always positive,for $f'(x) < 0$,we require $ad - bc < 0$.

If $f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x}$ is decreasing for all $x$,then

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