Question:

Step 1: Simplify $f(x)$. Let $t = 2^{x/2}$, so $2^x = t^2$ and $4^x = t^4$. $$f(x) = \left[\frac{t^4 + t^2 + 1}{t^2 - t + 1}\right] = \left[\frac{(t^2+t+1)(t^2-t+1)}{t^2-t+1}\right] = [t^2 + t + 1] = [2^x + 2^{x/2} + 1]$$ Step 2: Analyze $f(x) = [2^x + 2^{x/2} + 1]$ on $(-\infty, 0]$. Let $h(x) = 2^x + 2^{x/2} + 1$. As $x \to -\infty$, $h(x) \to 1$. At $x=0$, $h(0) = 1+1+1 = 3$. The function $h(x)$ is continuous and increasing on $(-\infty, 0]$, ranging in $(1, 3]$. The greatest integer function $[h(x)]$ is discontinuous when $h(x)$ passes through integers 2 and 3. So there are 2 points of discontinuity in $(-\infty, 0]$. Option (1) is correct. Step 3: Analyze $g(x) = \left[\frac{9}{x^2+5}\right]$. The expression $\frac{9}{x^2+5}$ has maximum value $\frac{9}{5} = 1.8$ at $x=0$ and approaches 0 as $x \to \pm\infty$. So $\frac{9}{x^2+5} \in (0, 9/5]$. The GIF takes values 0 or 1. It equals 1 when $\frac{9}{x^2+5} \geq 1$, i.e., $x^2 \leq 4$, i.e., $x \in [-2, 2]$. Discontinuities occur at $x = -2$ and $x = 2$. So 2 points of discontinuity. Option (2) is correct. Step 4: Analyze $f(x) \cdot g(x)$ on $(-\infty, \infty)$. $g(x) = 0$ for $|x| > 2$ and $g(x) = 1$ for $|x| < 2$. $f(x)$ has discontinuities wherever $2^x + 2^{x/2} + 1$ is an integer. On $(-\infty, \infty)$, $f(x)$ ranges from 1 (as $x\to-\infty$) to $\infty$. The product $f(x)\cdot g(x)$ has discontinuities from: 2 points from $g(x)$ at $x = \pm 2$, plus discontinuities of $f(x)$ within $(-2, 2)$ where $g(x)=1$. Within $(-2,2)$, $h(x) = 2^x + 2^{x/2}+1$ ranges from values near $h(-2)$ to $h(2)$. $h(-2) = 2^{-2}+2^{-1}+1 = 0.25+0.5+1=1.75$ and $h(2)=4+2+1=7$. So $h(x)$ passes through integers 2,3,4,5,6,7 giving 6 discontinuities, but at $x=2$, $h(2)=7$ exactly (boundary). Counting carefully, there are 5 interior discontinuities of $f$ in $(-2,2)$, plus 2 from $g$, totaling 7. Option (3) is correct.