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if $f(3)=\frac{a sin x+b cosx}{ c sinx +dcos x}$ is monotonically increasing, then
Option: 1
$ab\geq bc$

Option: 2
ab < bc

Option: 3
$ab\leq bc$

Option: 4
ab>bc

Answers (1)

Quotient Rule of differentiation -

$\frac{d}{dx}\:\:\frac{f(x)}{g(x)}=\frac{g(x).f'(x)-f(x).g'(x)}{[g(x)]^{2}}$

$\frac{d}{dx}\:\:\frac{u}{v}=\frac{v.\frac{du}{dx}-u.\frac{dv}{dx}}{v^{2}}$

-

Condition for increasing functions -

For increasing function tangents drawn at any point on it makes an acute slope with positive x-axis.

$M_{T}=tan\theta\geq 0$

$\therefore \:\:\:\frac{dy}{dx}=f'(x)\geq 0\:\:for\:\:x\epsilon (a,b)$

- wherein

Where f(x) is continuous and differentiable for (a,b)

Here $f(x)=\frac{a\: \sin \: x+b\: \cos x}{c\: \sin \: x+d\: \cos \: x}$ is monotonically increasing

then $f'(x)\geq 0$

Now $f'(x)$

$=\frac{(c\sin x+d\cos x)(a\cos x-b\sin x)-(a\sin x+b\cos x)(c\cos x-d\sin x)}{(c\sin x+d\cos x)^{2}}$

$ac\sin \: x\cos x-bc\: \sin ^{2}x+ad\cos ^{2}x-bd\sin x\cos x$

$=\frac{-ac\sin x\cos x+ad\sin ^{2}x-bc\cos ^{2}x+bd\sin x\cos x}{(c\sin x+d\cos x)^{2}}$

$=\frac{ad(\sin ^{2}x+\cos ^{2}x)-bc(\sin^{2}x+\cos ^{2}x)}{(c\sin x+d\cos x)^{2}}$

$=\frac{ad-bc}{(c\sin x+d\cos x)^{2}}$

Here $f'(x)\geq 0\: \forall\: x\in R$

$\Rightarrow ad-bc\geq 0$

$ad\geq bc$

Posted by

Sumit Saini

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