Question

Question was provided via image upload.

Explanation:

planation:

Step1: Analyze the general form of cotangent function

The general form of the cotangent function is \(y = A\cot(Bx - C)+D\). For the function \(y=\frac{6}{5}\cot x\), we have \(A = \frac{6}{5}\), \(B = 1\), \(C = 0\), \(D = 0\). The period of the cotangent function \(y = A\cot(Bx)\) is \(\frac{\pi}{|B|}\), so the period of \(y=\frac{6}{5}\cot x\) is \(\pi\) since \(B = 1\).

Step2: Determine the key points

The vertical asymptotes of the basic cotangent function \(y=\cot x\) are at \(x = n\pi\), \(n\in\mathbb{Z}\). For \(y=\frac{6}{5}\cot x\), the vertical asymptotes are also at \(x=n\pi\), \(n\in\mathbb{Z}\). When \(x=\frac{\pi}{2}\), \(y=\frac{6}{5}\cot(\frac{\pi}{2})=0\). The function \(y = \cot x\) is positive in \((0,\pi)\) and negative in \((\pi,2\pi)\). Since \(A=\frac{6}{5}>0\), the sign of \(y=\frac{6}{5}\cot x\) follows the sign of \(y = \cot x\).

Now, let's analyze each option:

• Option A: Incorrect, because the basic shape of the cotangent function is a curve that decreases from \(+\infty\) to \(-\infty\) in one period, and this option does not match the correct trend.
• Option B: Incorrect, the cotangent function \(y=\frac{6}{5}\cot x\) is positive in \((0,\pi)\) and negative in \((\pi,2\pi)\), this option has the wrong sign in some intervals.
• Option C: Correct. It has the vertical asymptotes at \(x = n\pi\), passes through \((\frac{\pi}{2},0)\) and has the correct decreasing trend from \(+\infty\) to \(-\infty\) in the interval \((0,\pi)\) and from \(+\infty\) to \(-\infty\) in other periods (since the period is \(\pi\)).
• Option D: Incorrect, because the cotangent function \(y=\frac{6}{5}\cot x\) is not an odd - function in the sense of passing through the origin (it has a zero at \(x=\frac{\pi}{2}+n\pi\), \(n\in\mathbb{Z}\) and vertical asymptotes at \(x=n\pi\), \(n\in\mathbb{Z}\))

Answer:

C.