The number of points where the function f(x) = { |2x² - 3x - 7| if x ≤ -1; [4x² - 1] if -1

Q: Wait, why isn't $x=0$ a point of discontinuity? The function value is $-1$, but shouldn't the GIF jump there?

That is a brilliant observation! At $x=0$, $4x^2 - 1 = -1$. For $x$ slightly larger or smaller than $0$, $4x^2 - 1$ is slightly *greater* than $-1$. The greatest integer of any number in the interval $[-1, 0)$ is $-1$. Since the limit from both sides is $-1$ and the function value is $-1$, the function is continuous at $x=0$. It's a classic trap—always check the neighborhood, not just the point!

Q: Do I really need to check the boundaries $x=-1$ and $x=1$? They look like they might just connect.

In the world of JEE, 'looking like' is dangerous! Piecewise functions are like bridges between different roads. Even if the roads themselves are smooth, the bridge might not connect perfectly. Always check the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the function value at the transition points. In this problem, $x=1$ is a point of discontinuity because the bridge doesn't connect!

Q: Why do we only consider $k \in \{-1, 0, 1, 2\}$? What about $k=3$?

Great question! We look at the range of the expression inside the bracket. For $x \in (-1, 1)$, $x^2$ is in $[0, 1)$. Multiplying by $4$ gives $[0, 4)$, and subtracting $1$ gives $[-1, 3)$. Since the interval is open at $3$, the expression $4x^2 - 1$ can never actually reach $3$. Therefore, $k=3$ is outside our domain of concern.

Question

The number of points where the function $f(x) = \begin{cases} |2x^2 - 3x - 7| & 	ext{if } x \le -1 \ [4x^2 - 1] & 	ext{if } -1 < x < 1 \ |x + 1| + |x - 2| & 	ext{if } x \ge 1 \end{cases}$, $[t]$ denotes the greatest integer $\le t$, is discontinuous is ______.

Sigma Percentile · Accepted Answer

### The Anatomy of a Piecewise Function

Welcome, fellow explorer of the mathematical landscape! Today, we are going to dissect a function that might look intimidating at first glance, but is actually a beautiful puzzle waiting to be solved.

We are dealing with a piecewise function $f(x)$, defined in three distinct regions. Our mission is to find the points where this function "breaks"—the points of discontinuity.

### Phase 1: The Well-Behaved Regions

Let's start by looking at the outer regions. For $x \le -1$, we have $f(x) = |2x^2 - 3x - 7|$.

This is the absolute value of a quadratic polynomial. Polynomials are continuous everywhere, and the absolute value function is also continuous. Thus, this region is perfectly smooth.

Similarly, for $x \ge 1$, we have $f(x) = |x + 1| + |x - 2|$. This is a sum of two absolute value functions, which are also continuous.

So, we can breathe a sigh of relief: no discontinuities will be found strictly inside these outer intervals. The trouble, as is often the case, lies in the middle.

### Phase 2: The Greatest Integer Trap

In the interval $(-1, 1)$, our function is defined as $f(x) = [4x^2 - 1]$. Here, the Greatest Integer Function (GIF) is the star of the show.

The GIF is notorious for its "staircase" behavior. It remains constant over intervals and then jumps abruptly whenever the input expression reaches an integer value.

To find the discontinuities, we must solve the equation $4x^2 - 1 = k$, where $k$ is an integer.

First, let's determine the range of our expression $4x^2 - 1$ within the interval $x \in (-1, 1)$. Since $x^2$ ranges from $0$ to $1$ (excluding $1$), $4x^2$ ranges from $0$ to $4$.

Subtracting $1$, we find that $4x^2 - 1$ takes values in the interval $[-1, 3)$. The integers in this range are $\{-1, 0, 1, 2\}$. These are the values of $k$ that will cause our function to jump.

### Phase 3: Solving for the Jumps

Let's hunt down these points one by one:

1. **For $k = 0$:** Setting $4x^2 - 1 = 0$, we get $x^2 = \frac{1}{4}$, which gives $x = \pm \frac{1}{2}$. These are two points of discontinuity.

2. **For $k = 1$:** Setting $4x^2 - 1 = 1$, we get $4x^2 = 2$, so $x^2 = \frac{1}{2}$, yielding $x = \pm \frac{1}{\sqrt{2}}$. Two more points!

3. **For $k = 2$:** Setting $4x^2 - 1 = 2$, we get $4x^2 = 3$, so $x^2 = \frac{3}{4}$, giving $x = \pm \frac{\sqrt{3}}{2}$. That's another two points.

Wait, what about $k = -1$? Setting $4x^2 - 1 = -1$ gives $x = 0$.

However, at $x=0$, the function is actually continuous because the limit from both sides approaches $-1$. So, we have $2 + 2 + 2 = 6$ points of discontinuity inside the interval $(-1, 1)$.

### Phase 4: The Boundary Check

We aren't done yet! We must check the transition points $x = -1$ and $x = 1$.

At $x = -1$: The Left-Hand Limit (LHL) is $|2(-1)^2 - 3(-1) - 7| = |2 + 3 - 7| = 2$. The Right-Hand Limit (RHL) is:

$$\lim_{x 	o -1^+} [4x^2 - 1] = [3^-] = 2$$

Since LHL = RHL = $f(-1)$, the function is continuous at $x = -1$.

At $x = 1$: The LHL is:

$$\lim_{x 	o 1^-} [4x^2 - 1] = [3^-] = 2$$

The RHL (and $f(1)$) is $|1 + 1| + |1 - 2| = 2 + 1 = 3$. Since $2 
eq 3$, we have found our seventh point of discontinuity!

### Conclusion

By systematically analyzing the behavior of the GIF and carefully checking the boundaries, we have uncovered all the hidden breaks in the function.

We found $6$ points within the interval and $1$ at the boundary, giving us a total of **$7$ points of discontinuity**. Mathematics is all about this kind of careful, logical investigation.

Enter Numerical Value:

Visualized Solution

Understanding the Piecewise Function

Analyzing the Outer Intervals

Focusing on the Middle Interval

Finding the Range of $4 x^{2} - 1$

Solving for $4 x^{2} - 1 = 0$

Solving for $4 x^{2} - 1 = 1$

Solving for $4 x^{2} - 1 = 2$

Checking the Minimum at $x = 0$

Checking Boundary at $x = - 1$

Checking Boundary at $x = 1$

Final Count of Discontinuities

The Sigma Insight: Continuity at a Point and in an Interval

The Anatomy of a Piecewise Function

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Enter Numerical Value:

Visualized Solution

Understanding the Piecewise Function

Analyzing the Outer Intervals

Focusing on the Middle Interval

Finding the Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4x2−1

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4x2−1=0

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4x2−1=1

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4x2−1=2

Checking the Minimum at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Checking Boundary at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1

Checking Boundary at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1