Question

The number of points of discontinuity of the function f:(-2, 2) → R, f(x) = $[2^{-x^2}[2^x]]$ is _______ (where [k] represents greatest integer less than or equal to k).

Answer 1

2

Answer 2

The function is given by $f(x) = [2^{-x^2}[2^x]]$. We need to find the number of points of discontinuity in the interval $(-2, 2)$.

Let's analyze the term $[2^x]$. For $x \in (-2, 2)$, we have $2^{-2} < 2^x < 2^2$, which means $1/4 < 2^x < 4$. The integer values that $2^x$ can take are $1, 2, 3$. These values determine the points where $[2^x]$ might change its value and thus be discontinuous. The points where $2^x$ is an integer are:

1. $2^x = 1 \implies x = 0$.
2. $2^x = 2 \implies x = 1$.
3. $2^x = 3 \implies x = \log_2 3$. (Note that $1 < \log_2 3 < 2$ since $2^1=2$ and $2^2=4$).

Let's divide the interval $(-2, 2)$ into sub-intervals based on the value of $[2^x]$:

Case 1: $x \in (-2, 0)$

For $x \in (-2, 0)$, $1/4 < 2^x < 1$. So, $[2^x] = 0$. $f(x) = [2^{-x^2} \cdot 0] = [0] = 0$. The function is constant and continuous in this interval.

Case 2: $x \in [0, 1)$

For $x \in [0, 1)$, $1 \le 2^x < 2$. So, $[2^x] = 1$. $f(x) = [2^{-x^2} \cdot 1] = [2^{-x^2}]$. For $x \in [0, 1)$, $x^2 \in [0, 1)$. Therefore, $-x^2 \in (-1, 0]$. This implies $2^{-x^2} \in (2^{-1}, 2^0] = (1/2, 1]$. So, $f(x) = [2^{-x^2}]$ will be $0$ if $2^{-x^2} \in (1/2, 1)$ and $1$ if $2^{-x^2} = 1$. $2^{-x^2} = 1 \implies -x^2 = 0 \implies x = 0$. So, for $x=0$, $f(0) = [2^{-0^2} \cdot [2^0]] = [1 \cdot 1] = 1$. For $x \in (0, 1)$, $2^{-x^2} \in (1/2, 1)$, so $f(x) = [2^{-x^2}] = 0$. At $x=0$: $f(0)=1$. $\lim_{x \to 0^-} f(x) = 0$ (from Case 1). $\lim_{x \to 0^+} f(x) = 0$. Since $f(0) \ne \lim_{x \to 0^+} f(x)$, $f(x)$ is discontinuous at $x=0$.

Case 3: $x \in [1, \log_2 3)$

For $x \in [1, \log_2 3)$, $2 \le 2^x < 3$. So, $[2^x] = 2$. $f(x) = [2^{-x^2} \cdot 2] = [2^{1-x^2}]$. For $x \in [1, \log_2 3)$, $x^2 \in [1, (\log_2 3)^2)$. (Note: $\log_2 3 \approx 1.58$, so $(\log_2 3)^2 \approx 2.496$). Therefore, $1-x^2 \in (1-(\log_2 3)^2, 1-1] = (1-2.496, 0] = (-1.496, 0]$. This implies $2^{1-x^2} \in (2^{-1.496}, 2^0] = (2^{-1.496}, 1]$. $2^{-1.496} = 1/2^{1.496}$. Since $2^1=2, 2^2=4$, $2^{1.496}$ is between 2 and 4 (approx $2.82$). So $2^{1-x^2} \in (1/2.82, 1] \approx (0.354, 1]$. So, $f(x) = [2^{1-x^2}]$ will be $0$ if $2^{1-x^2} \in (0.354, 1)$ and $1$ if $2^{1-x^2} = 1$. $2^{1-x^2} = 1 \implies 1-x^2 = 0 \implies x^2 = 1 \implies x = 1$ (since $x \in [1, \log_2 3)$). So, for $x=1$, $f(1) = [2^{-1^2} \cdot [2^1]] = [2^{-1} \cdot 2] = [1/2 \cdot 2] = [1] = 1$. For $x \in (1, \log_2 3)$, $2^{1-x^2} \in (0.354, 1)$, so $f(x) = [2^{1-x^2}] = 0$. At $x=1$: $f(1)=1$. $\lim_{x \to 1^-} f(x) = 0$ (from Case 2). $\lim_{x \to 1^+} f(x) = 0$. Since $f(1) \ne \lim_{x \to 1^+} f(x)$, $f(x)$ is discontinuous at $x=1$.

Case 4: $x \in [\log_2 3, 2)$

For $x \in [\log_2 3, 2)$, $3 \le 2^x < 4$. So, $[2^x] = 3$. $f(x) = [2^{-x^2} \cdot 3]$. For $x \in [\log_2 3, 2)$, $x^2 \in [(\log_2 3)^2, 4)$. Therefore, $-x^2 \in (-4, -(\log_2 3)^2]$. This implies $2^{-x^2} \in (2^{-4}, 2^{-(\log_2 3)^2}] = (1/16, 1/2^{(\log_2 3)^2}]$. $1/16 = 0.0625$. $2^{-(\log_2 3)^2} \approx 2^{-2.496} \approx 0.178$. So, $2^{-x^2} \in (0.0625, 0.178]$. Then $f(x) = [3 \cdot 2^{-x^2}]$. The values of $3 \cdot 2^{-x^2}$ are in $(3 \cdot 0.0625, 3 \cdot 0.178] = (0.1875, 0.534]$. For all $x$ in this interval, $3 \cdot 2^{-x^2}$ is between $0.1875$ and $0.534$. So, $f(x) = [3 \cdot 2^{-x^2}] = 0$ for all $x \in [\log_2 3, 2)$. At $x=\log_2 3$: $f(\log_2 3) = [2^{-(\log_2 3)^2} \cdot [2^{\log_2 3}]] = [2^{-(\log_2 3)^2} \cdot 3]$. As calculated above, this value is $0$. $\lim_{x \to (\log_2 3)^-} f(x) = 0$ (from Case 3). $\lim_{x \to (\log_2 3)^+} f(x) = 0$. Since $f(\log_2 3) = \lim_{x \to (\log_2 3)^-} f(x) = \lim_{x \to (\log_2 3)^+} f(x) = 0$, $f(x)$ is continuous at $x=\log_2 3$.

Summary of $f(x)$:

• For $x \in (-2, 0)$, $f(x) = 0$.
• For $x=0$, $f(0) = 1$.
• For $x \in (0, 1)$, $f(x) = 0$.
• For $x=1$, $f(1) = 1$.
• For $x \in (1, 2)$, $f(x) = 0$. (This covers $(1, \log_2 3)$ and $[\log_2 3, 2)$).

We found discontinuities at $x=0$ and $x=1$. Let's check for any other discontinuities. The function $2^{-x^2}$ is continuous everywhere. The function $[2^x]$ is discontinuous at $x=0, 1, \log_2 3$. We have already analyzed these points. The function $[g(x)]$ can also be discontinuous if $g(x)$ (which is $2^{-x^2}[2^x]$) takes an integer value and its graph "jumps" over it. However, in all the intervals where $[2^x]$ is constant, $f(x)$ either remains constant at 0 or is $[2^{-x^2}]$ or $[2 \cdot 2^{-x^2}]$ or $[3 \cdot 2^{-x^2}]$.

• For $x \in (0, 1)$, $f(x) = [2^{-x^2}]$. $2^{-x^2} \in (1/2, 1)$. So $f(x)=0$. Continuous.
• For $x \in (1, \log_2 3)$, $f(x) = [2^{1-x^2}]$. $2^{1-x^2} \in (0.354, 1)$. So $f(x)=0$. Continuous.
• For $x \in (\log_2 3, 2)$, $f(x) = [3 \cdot 2^{-x^2}]$. $3 \cdot 2^{-x^2} \in (0.1875, 0.534)$. So $f(x)=0$. Continuous.

Thus, the only points of discontinuity are $x=0$ and $x=1$. There are 2 points of discontinuity.

The final answer is $\boxed{2}$.