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Step-by-Step Solution
1. Understand the Given Functions
• We have a function f(x) defined by
f(x) = [\,2x^2 + 1\,],
where [\,t\,] denotes the greatest integer less than or equal to t .
• Another function g(x) is given by
g(x) =
\begin{cases}
2x - 3, & x < 0,\\
2x + 3, & x \ge 0.
\end{cases}
• We need to form the composite function (f \circ g)(x) = f\bigl(g(x)\bigr) and find the number of points of discontinuity in the open interval (-1, 1) .
2. Form the Composite Function (f \circ g)(x)
Using f(y) = [\,2y^2\,] + 1 if we rewrite it slightly, we get:
f(y) = [\,2y^2\,] + 1.
Here, y = g(x) , so
(f \circ g)(x) = [\,2(g(x))^2\,] + 1.
Substitute g(x) from its piecewise definition:
For x < 0 : g(x) = 2x - 3 \implies (f \circ g)(x) = [\,2\,(2x - 3)^2\,] + 1.
For x \ge 0 : g(x) = 2x + 3 \implies (f \circ g)(x) = [\,2\,(2x + 3)^2\,] + 1.
3. Identify Where Discontinuities Occur
A greatest integer function [\,h(x)\,] is discontinuous at all points where h(x) is an integer. In our composite function,
h(x) = 2\,(g(x))^2.
Hence, (f \circ g)(x) is discontinuous whenever 2 (g(x))^2 is an integer. We must solve for those x values in (-1,1) that make 2 (g(x))^2 an integer.
4. Case 1: x \in (-1,0)
In this interval, g(x) = 2x - 3 . So we look at
2(2x - 3)^2 \quad \text{being an integer.}
Let
2(2x - 3)^2 = k,
where k is an integer. Rewrite:
(2x - 3)^2 = \frac{k}{2}.
Since x < 0 , we note 2x - 3 is negative for -1 < x < 0 . So we take
2x - 3 = -\sqrt{\frac{k}{2}} \quad \Longrightarrow \quad x = \frac{3 - \sqrt{\frac{k}{2}}}{2}.
We also need -1 < x < 0 . By analyzing the range of (2x - 3)^2 over x \in (-1,0) :
When x=-1 , 2(-1)-3 = -5 \implies (2x -3)^2 = 25 \implies 2(2x -3)^2 = 50.
When x \to 0^- , 2x -3 \to -3 \implies (2x -3)^2 \to 9 \implies 2(2x -3)^2 \to 18.
Hence, 2(2x -3)^2 runs in the interval (18,50) as x goes from -1 to 0 . The integers in that range are 19, 20, \dots, 49 . There are 31 integer values (from 19 up to 49 inclusive).
For each integer k from 19 to 49 , there is exactly one corresponding x in (-1, 0) (since we use the negative square root branch to keep 2x-3 negative). Hence, there are 31 points in (-1,0) where the composite function is discontinuous.
5. Case 2: x \in (0,1)
In this interval, g(x) = 2x + 3 . Thus we look at
2(2x + 3)^2 \quad \text{being an integer.}
Let
2(2x + 3)^2 = k,
where k is an integer. Then
(2x + 3)^2 = \frac{k}{2}.
Now, for x \in (0,1) , 2x + 3 is positive. So
2x + 3 = \sqrt{\frac{k}{2}} \quad \Longrightarrow \quad x = \frac{\sqrt{\frac{k}{2}} - 3}{2}.
Check the range of (2x + 3)^2 for x \in (0,1) :
When x=0 , 2(0)+3 = 3 \implies (2x +3)^2 = 9 \implies 2(2x +3)^2 = 18.
When x \to 1^- , 2(1)+3 = 5 \implies (2x +3)^2 = 25 \implies 2(2x +3)^2 = 50.
Thus, 2(2x + 3)^2 also spans (18,50) over x \in (0,1) , giving the same set of integer values k = 19, 20, \dots, 49 . Again, there are 31 integers in that set, and for each integer k , there is exactly one corresponding x in (0,1) .
Hence, there are another 31 discontinuities in (0,1) .
6. Combine Both Intervals
• From x \in (-1,0) , we found 31 discontinuities.
• From x \in (0,1) , we found 31 discontinuities.
• Note that x=0 itself is excluded as per the problem statement (the open interval (-1,1) ) and also a separate mention that we do not count x=0 for these jumps.
7. Final Count of Discontinuities
The total number of points of discontinuity in (-1,1) is
31 \;+\; 31 \;=\; 62.
