2. in the xy - plane, the graph of which of t...
2. in the xy - plane, the graph of which of the following functions has a vertical asymptote at x = π/2? a f(x)=csc(x) b f(x)=csc(2x) c f(x)=sec(x - π/2) d f(x)=sec(1/2x) 4. let h(x)=3sec(4x)+1. which of the following statements about the graph of h is correct? a the graph of h has vertical asymptotes when x = π/8+π/4k, where k is an integer. b the graph of h has vertical asymptotes when x = π/4+π/2k, where k is an integer. c the graph of h has vertical asymptotes when x = π/2+π/4k, where k is an integer. d the graph of h has vertical asymptotes when x = 2π+4πk, where k is an integer.
Answer
# Explanation:
## Step1: Recall the definition of vertical - asymptote for trigonometric functions
Vertical asymptotes of \(y = \csc(x)=\frac{1}{\sin(x)}\) occur when \(\sin(x)=0\), and vertical asymptotes of \(y=\sec(x)=\frac{1}{\cos(x)}\) occur when \(\cos(x) = 0\).
## Step2: Analyze option A for \(y = \csc(x)\)
For \(y=\csc(x)=\frac{1}{\sin(x)}\), \(\sin(x)=0\) when \(x = k\pi,k\in\mathbb{Z}\). When \(x=\frac{\pi}{2}\), \(\sin(\frac{\pi}{2}) = 1\), so there is no vertical - asymptote at \(x=\frac{\pi}{2}\).
## Step3: Analyze option B for \(y=\csc(2x)\)
For \(y = \csc(2x)=\frac{1}{\sin(2x)}\), \(\sin(2x)=0\) when \(2x=k\pi\), or \(x=\frac{k\pi}{2},k\in\mathbb{Z}\). When \(x = \frac{\pi}{2}\), \(\sin(2\times\frac{\pi}{2})=\sin(\pi)=0\), but we want to check if it's the only relevant one.
## Step4: Analyze option C for \(y=\sec(x - \frac{\pi}{2})\)
For \(y=\sec(x-\frac{\pi}{2})=\frac{1}{\cos(x - \frac{\pi}{2})}\), and we know that \(\cos(x-\frac{\pi}{2})=\sin(x)\). When \(x=\frac{\pi}{2}\), \(\cos(\frac{\pi}{2}-\frac{\pi}{2})=\cos(0)=1\), so there is no vertical - asymptote at \(x=\frac{\pi}{2}\).
## Step5: Analyze option D for \(y=\sec(\frac{1}{2}x)\)
For \(y=\sec(\frac{1}{2}x)=\frac{1}{\cos(\frac{1}{2}x)}\), \(\cos(\frac{1}{2}x)=0\) when \(\frac{1}{2}x=(2k + 1)\frac{\pi}{2}\), or \(x=(2k + 1)\pi,k\in\mathbb{Z}\). When \(x=\frac{\pi}{2}\), \(\cos(\frac{1}{2}\times\frac{\pi}{2})=\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\neq0\), so there is no vertical - asymptote at \(x=\frac{\pi}{2}\).
For the second question:
## Step1: Recall the vertical - asymptote formula for \(y = \sec(u)\)
The function \(y=\sec(u)=\frac{1}{\cos(u)}\) has vertical asymptotes when \(\cos(u)=0\). For \(u = 4x\), \(\cos(4x)=0\) when \(4x=(2k + 1)\frac{\pi}{2},k\in\mathbb{Z}\).
## Step2: Solve for \(x\)
Solve the equation \(4x=(2k + 1)\frac{\pi}{2}\) for \(x\). Divide both sides by 4: \(x=\frac{(2k + 1)\pi}{8}=\frac{\pi}{8}+\frac{\pi}{4}k,k\in\mathbb{Z}\).
# Answer:
2. B
4. A