Question

The function $\frac{a\sin \left(x\right)+b\cos \left(x\right)}{c\sin \left(x\right)+d\cos \left(x\right)}$ is decreasing, if

• A. $ad-bc>0$
• B. $ad-bc<0$
• C. $ab-ad>0$
• D. $ab-cd<0$

Answer 1

The correct option is B

$ad-bc<0$

Explanation for the correct option

The given function, $f\left(x\right)=\frac{a\sin \left(x\right)+b\cos \left(x\right)}{c\sin \left(x\right)+d\cos \left(x\right)}$.

Differentiate the given function with respect to $x$.

$$\begin{gathered} \frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left(\frac{a\sin \left(x\right)+b\cos \left(x\right)}{c\sin \left(x\right)+d\cos \left(x\right)}\right) \\ \Rightarrow f\text{'}\left(x\right)=\frac{\left(c\sin \left(x\right)+d\cos \left(x\right)\right){\displaystyle \frac{d}{dx}}\left(a\sin \left(x\right)+b\cos \left(x\right)\right)-\left(a\sin \left(x\right)+b\cos \left(x\right)\right){\displaystyle \frac{d}{dx}}\left(c\sin \left(x\right)+d\cos \left(x\right)\right)}{{\left(c\sin \left(x\right)+d\cos \left(x\right)\right)}^2}\left[\because \frac{d}{d\mathrm{x}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v}{\displaystyle \frac{d\mathrm{u}}{d\mathrm{x}}}-\mathrm{u}{\displaystyle \frac{d\mathrm{v}}{d\mathrm{x}}}}{{\mathrm{v}}^2}\right] \\ \Rightarrow f\text{'}\left(x\right)=\frac{\left(c\sin \left(x\right)+d\cos \left(x\right)\right)\times \left(a\cos \left(x\right)-b\sin \left(x\right)\right)-\left(a\sin \left(x\right)+b\cos \left(x\right)\right)\times \left(c\cos \left(x\right)-d\sin \left(x\right)\right)}{{\left(c\sin \left(x\right)+d\cos \left(x\right)\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{ac\sin \left(x\right)\cos \left(x\right)-bc{\sin }^2\left(x\right)+ad{\cos }^2\left(x\right)-bd\sin \left(x\right)\cos \left(x\right)-ac\sin \left(x\right)\cos \left(x\right)+ad{\sin }^2\left(x\right)-bc{\cos }^2\left(x\right)+bd\sin \left(x\right)\cos \left(x\right)}{{\left(c\sin \left(x\right)+d\cos \left(x\right)\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{ad{\cos }^2\left(x\right)+ad{\sin }^2\left(x\right)-bc{\sin }^2\left(x\right)-bc{\cos }^2\left(x\right)}{{\left(c\sin \left(x\right)+d\cos \left(x\right)\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{ad\left[{\cos }^2\left(x\right)+{\sin }^2\left(x\right)\right]-bc\left[{\sin }^2\left(x\right)+{\cos }^2\left(x\right)\right]}{{\left(c\sin \left(x\right)+d\cos \left(x\right)\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{\left(ad\times 1\right)-\left(bc\times 1\right)}{{\left(c\sin \left(x\right)+d\cos \left(x\right)\right)}^2}\left[\because {\sin }^2\left(\mathrm{x}\right)+{\cos }^2\left(\mathrm{x}\right)=1\right] \\ \Rightarrow f\text{'}\left(x\right)=\frac{ad-bc}{{\left(c\sin \left(x\right)+d\cos \left(x\right)\right)}^2} \end{gathered}$$

Now, for decreasing intervals $f\text{'}\left(\mathrm{x}\right)<0$.

$$\begin{gathered} \Rightarrow \frac{ad-bc}{{\left(c\sin \left(x\right)+d\cos \left(x\right)\right)}^2}<0 \\ \Rightarrow ad-bc<0\times {\left(c\sin \left(x\right)+d\cos \left(x\right)\right)}^2\left[\because {\left(c\sin \left(x\right)+d\cos \left(x\right)\right)}^2\ge 0\right] \\ \Rightarrow ad-bc<0 \end{gathered}$$

So, the function $\frac{a\sin \left(x\right)+b\cos \left(x\right)}{c\sin \left(x\right)+d\cos \left(x\right)}$ is decreasing if $ad-bc<0$.

Hence, the correct option is (B).