Question

If the function $f\left(x\right)=\frac{k\sin x+2\cos x}{\sin x+\cos x}$ is strictly increasing for all values of $x$, then

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

Answer 1

The correct option is D $K>2$

The given function is:

$f\left(x\right)=\frac{K\sin x+2\cos x}{\sin x+\cos x}$

Differentiating once w.r.t x and and making it greater than 0 to make it monotonically increasing we get,

$\Rightarrow f^{\prime }\left(x\right)=\frac{(K\cos x-2\sin x)(\sin x+\cos x)-(\cos x-\sin x)(K\sin x+2\cos x)}{(\sin x+\cos x{)}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{K\sin x\cos x+K{\cos }^{2}x-2{\sin }^{2}x-2\sin x\cos x-K\sin x\cos x-2{\cos }^{2}x+K{\sin }^{2}x+2\sin x\cos x}{(\sin x+\cos x{)}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{K{\cos }^{2}x+K{\sin }^{2}x-2{\cos }^{2}x-2{\sin }^{2}x}{(\sin x+\cos x{)}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{K-2}{(\sin x+\cos x{)}^2}$

Now, $f^{\prime }\left(x\right)>0$

$\Rightarrow \frac{K-2}{(\sin x+\cos x{)}^2}>0$

$\Rightarrow K-2>0$

$\Rightarrow K>2$ .....Answer