Let [t] denote the greatest integer ≤ t. The number of points where the function f(x) = [x] |x² - 1|

Q: Wait, why do we only check integers like -1, 0, and 1? Why not check every point in the interval?

That is a brilliant question! In calculus, we look for 'trouble spots' where the function's definition changes abruptly. The Greatest Integer Function $[x]$ is constant between integers, but it 'jumps' at every integer. Since the rest of our function (polynomials and sine) is continuous everywhere, the only possible candidates for discontinuity are the points where the GIF changes its value. By checking only the integers, we are efficiently targeting the only places where the function could possibly break.

Q: I am confused about the modulus $|x^2 - 1|$. How do I know when to write it as $(x^2 - 1)$ versus $(1 - x^2)$?

Think of the modulus as a gatekeeper. It asks: 'Is the expression inside positive or negative?' If $x^2 - 1 \ge 0$ (which happens when $|x| \ge 1$), the gate opens as $(x^2 - 1)$. If $x^2 - 1 < 0$ (which happens when $|x| < 1$), the gate flips the sign to $(1 - x^2)$ to ensure the result is positive. Always check the interval you are working in first—if $x$ is between $-1$ and $1$, then $x^2$ is less than $1$, so $x^2 - 1$ is negative. That is your cue to flip the sign!

Q: Why does the function value at $x=0$ not match the limit? Is that a hole or a jump?

When the left-hand limit and the right-hand limit are not equal, we call that a 'jump discontinuity'. In this problem, as $x$ approaches $0$ from the left, the function behaves like $x^2$, heading toward $0$. But as soon as $x$ hits $0$ and moves to the right, the GIF term $[x]$ becomes $0$, causing the function to suddenly jump to a constant value of $\frac{\sqrt{3}}{2} - 1$. Because the function 'jumps' from one value to another, it is discontinuous.

Question

Let $[t]$ denote the greatest integer $\le t$. The number of points where the function $f(x) = [x] |x^2 - 1| + \sin\left(\frac{\pi}{[x]+3}\right) - [x + 1]$, $x \in (-2, 2)$ is not continuous is .

Sigma Percentile · Accepted Answer

### The Anatomy of a Discontinuity: A Journey Through the Function Welcome, warriors of JEE Advanced. Today, we are not just solving a math problem; we are performing surgery on a function. We are looking at $f(x) = [x] |x^2 - 1| + \sin\left(\frac{\pi}{[x]+3} ight) - [x + 1]$ within the domain $x \in (-2, 2)$. At first glance, this expression looks like a chaotic jumble of symbols. But remember, in the world of mathematics, chaos is simply order waiting to be discovered. Our mission is to find the points of discontinuity. ### Phase 1: The Strategy of the Integer Boundary The Greatest Integer Function, denoted by $[x]$, is the heartbeat of this problem. It acts like a staircase, remaining constant between integers and jumping at every integer value. Because the other components of our function—the modulus $|x^2 - 1|$, the sine function, and the linear term $[x+1]$—are well-behaved, the only places where this function can possibly break are the integer boundaries within our domain $(-2, 2)$. These critical points are $x = -1$, $x = 0$, and $x = 1$. Our strategy is to dissect the function into intervals and check the continuity at each of these three points. ### Phase 2: The Investigation **The Interval $(-2, -1)$:** Here, $[x] = -2$ and $[x+1] = -1$. Since $x < -1$, $x^2 > 1$, so $|x^2 - 1| = x^2 - 1$. Our function becomes: $$f(x) = -2(x^2 - 1) + \sin\left(\frac{\pi}{-2+3} ight) - (-1) = -2x^2 + 2 + \sin(\pi) + 1 = -2x^2 + 3$$ As we approach $x = -1$ from the left, the limit is $-2(-1)^2 + 3 = 1$. **The Interval $[-1, 0)$:** Now, $[x] = -1$ and $[x+1] = 0$. Since $|x| < 1$, $x^2 < 1$, so $|x^2 - 1| = 1 - x^2$. Our function transforms into: $$f(x) = -1(1 - x^2) + \sin\left(\frac{\pi}{-1+3} ight) - 0 = -1 + x^2 + \sin\left(\frac{\pi}{2} ight) = -1 + x^2 + 1 = x^2$$ As we approach $x = -1$ from the right, the limit is $(-1)^2 = 1$. Since the left-hand limit ($1$) equals the right-hand limit ($1$), the function is **continuous** at $x = -1$. **The Interval $[0, 1)$:** Here, $[x] = 0$ and $[x+1] = 1$. The modulus $|x^2 - 1|$ remains $1 - x^2$. Our function becomes: $$f(x) = 0(1 - x^2) + \sin\left(\frac{\pi}{0+3} ight) - 1 = \sin\left(\frac{\pi}{3} ight) - 1 = \frac{\sqrt{3}}{2} - 1$$ Notice that the variable $x$ has vanished; the function is a constant in this interval. Checking $x = 0$, the left-hand limit is $0^2 = 0$, while the right-hand limit is $\frac{\sqrt{3}}{2} - 1$. Since $0 eq \frac{\sqrt{3}}{2} - 1$, we have found our first point of **discontinuity** at $x = 0$. **The Interval $[1, 2)$:** Finally, $[x] = 1$ and $[x+1] = 2$. Since $x > 1$, $|x^2 - 1| = x^2 - 1$. Our function becomes: $$f(x) = 1(x^2 - 1) + \sin\left(\frac{\pi}{1+3} ight) - 2 = x^2 - 1 + \sin\left(\frac{\pi}{4} ight) - 2 = x^2 - 3 + \frac{1}{\sqrt{2}}$$ Checking $x = 1$, the left-hand limit is $\frac{\sqrt{3}}{2} - 1$. The right-hand limit is $1^2 - 3 + \frac{1}{\sqrt{2}} = -2 + \frac{1}{\sqrt{2}}$. Since these are not equal, $x = 1$ is our second point of **discontinuity**. ### The Final Verdict We have systematically walked through the domain. We found that at $x = -1$, the function holds together. However, at $x = 0$ and $x = 1$, the function suffers a jump. Therefore, there are exactly **2 points of discontinuity**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $[t]$ denote the greatest integer $\leq t$ . The number of points where the function $f (x) = [x] ∣ x^{2} - 1∣ + sin (\frac{π}{[ x ] + 3}) - [x + 1]$ , $x \in (- 2, 2)$ is not continuous is .

Enter Numerical Value:

Visualized Solution

Analyzing the Function $f (x)$

Identifying Critical Points

Interval $x \in (- 2, - 1)$

Continuity at $x = - 1$ (Left Limit)

Interval $x \in [- 1, 0)$

Continuity at $x = - 1$ (Right Limit)

Continuity at $x = 0$ (Left Limit)

Interval $x \in [0, 1)$

Continuity at $x = 0$ (Right Limit)

Continuity at $x = 1$ (Left Limit)

Interval $x \in [1, 2)$

Continuity at $x = 1$ (Right Limit)

Final Count of Discontinuities

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

The Anatomy of a Discontinuity

Phase 1

Phase 2

The Final Verdict

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Enter Numerical Value:

Visualized Solution

Analyzing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Identifying Critical Points

Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈(−2,−1)

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1 (Left Limit)

Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[−1,0)

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1 (Right Limit)

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0 (Left Limit)

Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[0,1)

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0 (Right Limit)

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1 (Left Limit)

Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[1,2)

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1 (Right Limit)