$f (x) = \frac{a sin x + b cos x}{c sin x + d cos x}$ is an increasing funtion if

a d - b c = 0

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a d - b c < 0

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a d - b c > 0

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a b + c d = 0

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Solution

The correct option is C $a d - b c > 0$
$f^{'} (x) = \frac{(a c o s x - b s i n x) (c s i n x + d c o s x) - (a s i n x + b c o s x) (c c o s x - d s i n x)}{(c s i n x + d c o s x)^{2}}$

$= \frac{a d c o s^{2} x - b c s i n^{2} x - b c c o s^{2} x + a d s i n^{2} x}{(c s i n x + d c o s x)^{2}}$

$= \frac{a d - b c}{(c s i n x + d c o s x)^{2}}$

for $f (x)$ to be increasing function the $f^{'} (x)$ will have to be greater than $0$

$f^{'} (x) > 0$ so $a d - b c > 0$

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Similar questions

f (x) = \frac{a sin x + b cos x}{c sin x + d cos x}

a d - b c \neq 0

A = [\begin{matrix} a & b c & d \end{matrix}]

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In a trapezium ABCD, if AB || CD then $(A C^{2} + B D^{2})$ = ?

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(d) $B C^{2} + A D^{2} + 2 A B . C D$

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