Question

If the function $f\left(x\right)=\frac{\left[k\sin \left(x\right)+2\cos \left(x\right)\right]}{\left[\sin \left(x\right)+\cos \left(x\right)\right]}$ is increasing on $x$, then:

• A. $k<1$
• B. $k>1$
• C. $k<2$
• D. $k>2$

Answer 1

The correct option is D

$k>2$

Explanation for the correct option

Given function, $f\left(x\right)=\frac{\left[k\sin x+2\cos x\right]}{\left[\sin x+\cos x\right]}$ is increasing on all $x$ so $f\text{'}\left(x\right)$ will be greater than zero for all $x$.

Differentiate by applying quotient rule, ${\left(\frac{u}{v}\right)}^{\text{'}}=\frac{u\text{'}v-uv\text{'}}{v^2}$,

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=\frac{\left[k\cos \left(x\right)-2\sin \left(x\right)\right]\left[\sin \left(x\right)+\cos \left(x\right)\right]-\left[k\sin \left(x\right)+2\cos \left(x\right)\right]\left[\cos \left(x\right)-\sin \left(x\right)\right]}{{\left[\sin x+\cos x\right]}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{\cancel{k\sin \left(x\right)\cos \left(x\right)}+k{\cos }^2\left(x\right)-2{\sin }^2\left(x\right)-\cancel{2\sin \left(x\right)\cos \left(x\right)}-\cancel{k\sin \left(x\right)\cos \left(x\right)}+k{\sin }^2\left(x\right)-2{\cos }^2\left(x\right)+\cancel{2\sin \left(x\right)\cos \left(x\right)}}{{\left[\sin x+\cos x\right]}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{k\left({\sin }^2\left(x\right)+{\cos }^2\left(x\right)\right)-2({\sin }^2\left(x\right)+{\cos }^2\left(x\right))}{{\left[\sin \left(x\right)+\cos \left(x\right)\right]}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{k\left(1\right)-2\left(1\right)}{{\left[\sin \left(x\right)+\cos \left(x\right)\right]}^2}\mathbf{}\mathbf{}[\because {\sin }^2\left(\mathrm{x}\right)+{\cos }^2\left(\mathrm{x}\right)=1] \end{gathered}$$

Putting $f\text{'}\left(x\right)>0$, we get,

$$\begin{gathered} \Rightarrow k-2>0 \\ \Rightarrow k>2 \end{gathered}$$

Hence, option (D), i.e. $k>2$ is the correct answer.