Question

Function $f\left(x\right)=\frac{a\sin x+b\cos x}{c\sin x+d\cos x}$ is monotonic decreasing if

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

Answer 1

The correct option is C ab - cd < 0

Given $f\left(x\right)=\frac{asinx+bcosx}{csinx+dcosx}$

For function to be monotonically decreasing, $\frac{dy}{dx}<0$

$\therefore \frac{(csinx+dcosx)\frac{d}{dx}(asinx+bcosx)-(asinx+bcosx)\frac{d}{dx}(csinx+dcosx)}{{(csinx+dcosx)}^2}<0$

$\therefore \frac{(csinx+dcosx)(acosx-bsinx)-(asinx+bcosx)(ccosx-dsinx)}{{(csinx+dcosx)}^2}<0$

Multiply both sides by ${(csinx+dcosx)}^2$, we get,

$\therefore (csinx+dcosx)(acosx-bsinx)-(asinx+bcosx)(ccosx-dsinx)<0$

$\therefore (ac.sinx.cox-bc{sin}^{2}x+ad{cos}^{2}x-bdsinx.cosx)-(ac.sinx.cox-ad{sin}^{2}x+bc{cos}^{2}x-bdsinx.cosx)<0$

$\therefore ac.sinx.cox-bc{sin}^{2}x+ad{cos}^{2}x-bdsinx.cosx-ac.sinx.cox+ad{sin}^{2}x-bc{cos}^{2}x+bdsinx.cosx<0$

$\therefore -bc{sin}^{2}x+ad{cos}^{2}x+ad{sin}^{2}x-bc{cos}^{2}x<0$

$\therefore -bc({sin}^{2}x+{cos}^{2}x)+ad({sin}^{2}x+{cos}^{2}x)<0$

$\therefore -bc\left(1\right)+ad\left(1\right)<0$

$\therefore ad-bc<0$