Question

If f(x) = $\frac{COS X}{[\frac{x}{\pi}] + \frac{1}{2}}$ where; [.] is G.I.F and x ≠ ηπ (n E integer) Then f(x) is:

• A. Bounded and discontinuous at $x=n\pi$ for all integers $n$.
• B. Continuous everywhere.
• C. Unbounded.
• D. Periodic with period $\pi$.

Answer 1

Answer: (A)

Bounded and discontinuous at $x=n\pi$ for all integers $n$.

Answer 2

The function is given by $f(x) = \frac{\cos x}{[\frac{x}{\pi}] + \frac{1}{2}}$, where $[\cdot]$ denotes the greatest integer function (GIF).

1. Discontinuity: Let $k = [\frac{x}{\pi}]$. For $x$ in the interval $[n\pi, (n+1)\pi)$, $[\frac{x}{\pi}] = n$. As $x \to n\pi^+$ (from the right), $[\frac{x}{\pi}] = n$. The limit is $\lim_{x \to n\pi^+} f(x) = \frac{\cos(n\pi)}{n + \frac{1}{2}} = \frac{(-1)^n}{n + \frac{1}{2}}$. As $x \to n\pi^-$ (from the left), $[\frac{x}{\pi}] = n-1$. The limit is $\lim_{x \to n\pi^-} f(x) = \frac{\cos(n\pi)}{(n-1) + \frac{1}{2}} = \frac{(-1)^n}{n - \frac{1}{2}}$. Since $n + \frac{1}{2} \neq n - \frac{1}{2}$ for any integer $n$, the left and right limits are different, implying that $f(x)$ is discontinuous at $x = n\pi$ for all integers $n$.

2. Boundedness: For $x \in (n\pi, (n+1)\pi)$, $[\frac{x}{\pi}] = n$. So, $f(x) = \frac{\cos x}{n + \frac{1}{2}}$. Since $-1 \le \cos x \le 1$, the range of $f(x)$ in this interval is $(\frac{-1}{n + \frac{1}{2}}, \frac{1}{n + \frac{1}{2}}) = (\frac{-2}{2n+1}, \frac{2}{2n+1})$. The union of these intervals for all integers $n$ is $(-2, 2)$, which means the function is bounded. For instance, when $n=0$ (interval $(0, \pi)$), the range is $(-2, 2)$. As $|n|$ increases, the range of $f(x)$ in the corresponding interval shrinks.

3. Periodicity: $f(x+\pi) = \frac{\cos(x+\pi)}{[\frac{x+\pi}{\pi}] + \frac{1}{2}} = \frac{-\cos x}{[\frac{x}{\pi} + 1] + \frac{1}{2}} = \frac{-\cos x}{[\frac{x}{\pi}] + 1 + \frac{1}{2}}$. This is not equal to $f(x)$, so the function is not periodic with period $\pi$. It is also not periodic with period $2\pi$.

Therefore, the function $f(x)$ is bounded and discontinuous at $x=n\pi$ for all integers $n$.