Question

Function f(x) = $\frac{\lambda \sin x+3\cos x}{2\sin x+6\cos x}$ is monotonic increasing when-

• A. $\lambda <1$
• B. $\lambda >1$
• C. $\lambda <2$
• D. $\lambda >2$

Answer 1

The correct option is B $\lambda >1$

Given $f\left(x\right)=\frac{\lambda \sin x+3\cos x}{2\sin x+6\cos x}$
$f^{\prime }\left(x\right)=\frac{(2\sin x+6\cos x)(\lambda \cos x-3\sin x)-(\lambda \sin x+3\cos x)(2\cos x-6\sin x)}{(2\sin x+6\cos x{)}^2}$
Since, $f\left(x\right)$ is increasing
$\Rightarrow f^{\prime }\left(x\right)>0$
$\frac{(2\sin x+6\cos x)(\lambda \cos x-3\sin x)-(\lambda \sin x+3\cos x)(2\cos x-6\sin x)}{(2\sin x+6\cos x{)}^2}>0$
$\frac{2\lambda \left(\sin x\cos x\right)-6{\sin }^{2}x+6\lambda {\cos }^{2}x-18\sin x\cos x-(2\lambda \sin x\cos x-6\lambda {\sin }^{2}x+6{\cos }^{2}x-18\sin x\cos x)}{(2\sin x+6\cos x{)}^2}>0$
$\frac{2\lambda \sin x\cos x-6{\sin }^{2}x+6\lambda {\cos }^{2}x-18\sin x\cos x-2\lambda \sin x\cos x+6\lambda {\sin }^{2}x+18\sin x\cos x}{(2\sin x+6\cos x{)}^2}>0$
$\frac{-6+6\lambda }{(2\sin x+6\cos x{)}^2}>0$
$6\lambda >6$
$\lambda >1$