2. in the xy - plane, the graph of which of the following functions has a vertical asymptote at x = π/2? a…

Explanation:

Step1: Recall csc and sec definitions

Recall that $\csc(x)=\frac{1}{\sin(x)}$ and $\sec(x)=\frac{1}{\cos(x)}$. A vertical - asymptote occurs where the denominator of the function is zero.

Step2: Analyze option A

For $y = \csc(x)=\frac{1}{\sin(x)}$, when $x=\frac{\pi}{2}$, $\sin(\frac{\pi}{2}) = 1$, so $y=\csc(\frac{\pi}{2}) = 1$, no vertical - asymptote.

Step3: Analyze option B

For $y=\csc(2x)=\frac{1}{\sin(2x)}$, when $x = \frac{\pi}{2}$, $\sin(2\times\frac{\pi}{2})=\sin(\pi)=0$. But we want the vertical asymptote at $x=\frac{\pi}{2}$, and for $y = \csc(2x)$, the vertical asymptote occurs when $2x = k\pi,k\in\mathbb{Z}$, or $x=\frac{k\pi}{2}$. When $k = 1$, $x=\frac{\pi}{2}$ is a vertical asymptote.

Step4: Analyze option C

For $y=\sec(x - \frac{\pi}{2})=\frac{1}{\cos(x-\frac{\pi}{2})}$, and $\cos(x-\frac{\pi}{2})=\sin(x)$. When $x=\frac{\pi}{2}$, $\cos(\frac{\pi}{2}-\frac{\pi}{2})=\cos(0) = 1$, no vertical - asymptote.

Step5: Analyze option D

For $y=\sec(\frac{1}{2}x)=\frac{1}{\cos(\frac{1}{2}x)}$, when $x=\frac{\pi}{2}$, $\cos(\frac{1}{2}\times\frac{\pi}{2})=\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$, no vertical - asymptote.

Answer:

B. $f(x)=\csc(2x)$