Question

The function $f\left(x\right)=\frac{\lambda \sin x+2\cos x}{\sin x+\cos x}$ is increasing, if
(a) λ < 1
(b) λ > 1
(c) λ < 2
(d) λ > 2

Answer 1

(d) λ > 2

$$\begin{gathered} f\left(x\right)=\frac{\lambda \sin x+2\cos x}{\sin x+\cos x} \\ f\text{'}\left(x\right)=\frac{\left(\sin x+\cos x\right)\left(\lambda \cos x-2\sin x\right)+\left(\lambda \sin x+2\cos x\right)\left(\cos x-\sin x\right)}{{\left(\sin x+\cos x\right)}^2} \\ =\frac{\lambda \cos x\sin x+\lambda {\cos }^{2}x-2{\sin }^{2}x-2\sin x\cos x-\lambda \sin x\cos x-2{\cos }^{2}x+\lambda {\sin }^{2}x+2\cos x\sin x}{{\left(\sin x+\cos x\right)}^2} \\ =\frac{-2\left({\sin }^{2}x+{\cos }^{2}x\right)+\lambda \left({\sin }^{2}x+{\cos }^{2}x\right)}{{\left(\sin x+\cos x\right)}^2} \\ =\frac{-2+\lambda }{{\left(\sin x+\cos x\right)}^2} \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow \frac{-2+\lambda }{{\left(\sin x+\cos x\right)}^2}>0 \\ \Rightarrow \lambda -2>0\left[\because {\left(\sin x+\cos x\right)}^2>0\right] \\ \Rightarrow \lambda >2 \end{gathered}$$