a) x = 1/2 + 2πk, where k is an integer c) x = π/2 + πk, where k is an integer 17 the graph of k is…

Explanation:

Step1: Recall tangent and cotangent properties

The derivative of $y = \tan(x)$ is $y'=\sec^{2}(x)>0$ for all $x$ in its domain, and it is increasing. The second - derivative $y'' = 2\sec^{2}(x)\tan(x)$. In the interval $(\frac{\pi}{2},\pi)$, $\tan(x)<0$ and $\sec^{2}(x)>0$, so $y''<0$ and $y = \tan(x)$ is concave - down. The derivative of $y=-\tan(x)$ is $y'=-\sec^{2}(x)<0$, so it is decreasing. The derivative of $y = \cot(x)= \frac{\cos(x)}{\sin(x)}$, using the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u=\cos(x)$ and $v = \sin(x)$. Then $y'=\frac{-\sin(x)\sin(x)-\cos(x)\cos(x)}{\sin^{2}(x)}=-\csc^{2}(x)<0$ for all $x$ in its domain, so it is decreasing. The derivative of $y =-\cot(x)=\frac{1}{\tan(x)}$, and $y'=\csc^{2}(x)>0$. The second - derivative $y''=- 2\csc^{2}(x)\cot(x)$. In the interval $(\frac{\pi}{2},\pi)$, $\cot(x)<0$ and $\csc^{2}(x)>0$, so $y''>0$.

Step2: Determine the function

Since we want a function $k(x)$ that is increasing and concave - up on the interval $(\frac{\pi}{2},\pi)$, the function $k(x)=-\cot(x)$ satisfies these conditions.

Answer:

D. $k(x)=-\cot(x)$