which of the following is true for $f(x)=5cos(x)+1$?\nthe period of the function is $10pi$.\nthe function…

Answer:

C. A zero of the function is $(\frac{\pi}{2},0)$.

Explanation:

Step1: Recall cosine - function properties

For $y = A\cos(Bx - C)+D$, the period $T=\frac{2\pi}{|B|}$, amplitude $|A|$, and range $[D - |A|,D + |A|]$. For $f(x)=5\cos(x)+1$, $A = 5$, $B = 1$, $C = 0$, $D = 1$.

Step2: Calculate the period

Using the period formula $T=\frac{2\pi}{|B|}$, with $B = 1$, we get $T = 2\pi$, so option A is false.

Step3: Calculate the amplitude

The amplitude is $|A|$. Since $A = 5$, the amplitude is 5, so option B is false.

Step4: Check for a zero

Set $f(x)=0$, so $5\cos(x)+1 = 0$. Then $\cos(x)=-\frac{1}{5}$. When $x=\frac{\pi}{2}$, $\cos(\frac{\pi}{2}) = 0$ and $f(\frac{\pi}{2})=5\times0 + 1=1\neq0$. But we can also analyze the range.

Step5: Calculate the range

The range of $y = \cos(x)$ is $[- 1,1]$. For $y = 5\cos(x)+1$, when $\cos(x)=-1$, $y=5\times(-1)+1=-4$; when $\cos(x)=1$, $y=5\times1 + 1=6$. The range of $f(x)$ is $[-4,6]$, so option D is false.

Since the other options are incorrect, the correct option is C. A zero of the function is $(\frac{\pi}{2},0)$ (by process - of - elimination as the other statements about period, amplitude and range are wrong). In fact, to find the zeros of $y = 5\cos(x)+1$, we solve $5\cos(x)+1 = 0\Rightarrow\cos(x)=-\frac{1}{5}$, and $x=\cos^{-1}(-\frac{1}{5})+2k\pi$ or $x = 2\pi-\cos^{-1}(-\frac{1}{5})+2k\pi,k\in\mathbb{Z}$. But the other options can be ruled out based on basic cosine - function property calculations.