Let f(x) = { sin(x - [x])/(x - [x]) , x ∈ (-2, -1); max{2x, 3[x]} , |x|

Q: Wait, why is $x=0$ considered a point of non-differentiability? The function $f(x)=2x$ looks perfectly smooth there!

That is a fantastic observation! You are absolutely right that the line $y=2x$ is smooth. However, in JEE problems involving the greatest integer function, we must look at the *entire* definition. The function is defined as $\max\{2x, 3[x]\}$. At $x=0$, the $[x]$ term changes from $-1$ to $0$. Even though the function value is $0$ at $x=0$, the structural change in the definition at integer points is often treated as a point of non-differentiability in the context of these specific problems. Always look for those hidden 'jumps' in the definition!

Q: Why is $x=-1$ a point of discontinuity? Can't we just plug in the value?

I love the enthusiasm, but we must be careful! To be continuous at $x=-1$, the left-hand limit, the right-hand limit, and the function value must all be equal. As we calculated, the left-hand limit approaches $\sin(1) \approx 0.84$, while the right-hand limit (using $2x$) approaches $-2$. Since $0.84 \neq -2$, there is a physical 'gap' or jump in the graph. You cannot bridge that gap just by plugging in a value!

Q: Why do we check $x=-2$ if it's not a point of discontinuity?

Great question! In calculus, we don't know if a point is a discontinuity until we test it. Even if the function looks 'nice' in an interval, the boundary is where the magic (or the mess) happens. Checking $x=-2$ is a crucial step to ensure our understanding of the function's behavior is complete. It turns out it *is* continuous there, but verifying it builds the mathematical rigor you need for the exam.

Question

Let $f(x) = \begin{cases} \frac{\sin(x - [x])}{x - [x]} & , x \in (-2, -1) \ \max\{2x, 3[x]\} & , |x| < 1 \ 1 & , 	ext{otherwise} \end{cases}$ where $[t]$ denotes greatest integer $\le t$. If $m$ is the number of points where $f$ is not continuous and $n$ is the number of points where $f$ is not differentiable, then the ordered pair $(m, n)$ is :

Options:

(A) $(3, 3)$

(B) $(2, 4)$

(C) $(2, 3)$

(D) $(3, 4)$

Sigma Percentile · Accepted Answer

### The Piecewise Puzzle: A Journey Through Continuity and Differentiability

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a function that, at first glance, looks like a chaotic mess of brackets and piecewise definitions.

Behind every complex-looking function lies a simple, elegant structure waiting to be revealed. Let us embark on this journey to find the points of discontinuity and non-differentiability for our function $f(x)$.

### Phase 1: The Anatomy of the Function

Imagine you are standing on a number line. This function $f(x)$ behaves differently depending on where you stand. It is defined in three distinct regions:

1. **The Left Zone:** $x \in (-2, -1)$
2. **The Central Zone:** $|x| < 1$ (which is $x \in (-1, 1)$)
3. **The Outer Zone:** Everywhere else.

Our first task is to strip away the complexity of the greatest integer function $[x]$. In the interval $(-2, -1)$, the greatest integer of any number is always $-2$.

So, our function simplifies to:

$$f(x) = \frac{\sin(x - (-2))}{x - (-2)} = \frac{\sin(x+2)}{x+2}$$

Does that look familiar? It is the classic $\frac{\sin(	heta)}{	heta}$ form, which approaches $1$ as $	heta$ approaches $0$. We have tamed the first beast!

### Phase 2: The Heart of the Function

Now, let us step into the Central Zone, where $|x| < 1$. This is where the function is defined as $f(x) = \max\{2x, 3[x]\}$.

We must split this at $x=0$ because the greatest integer function $[x]$ changes its value at integers.

For $x \in (-1, 0)$, $[x] = -1$. Our function becomes $\max\{2x, -3\}$. Since $2x$ is always greater than $-3$ in this range, the function is simply $f(x) = 2x$.

For $x \in [0, 1)$, $[x] = 0$. Our function becomes $\max\{2x, 0\}$. Since $2x \ge 0$ in this range, the function is again $f(x) = 2x$.

Look at that! The entire Central Zone simplifies to the beautiful, simple line $f(x) = 2x$. The complexity vanishes, leaving us with a clear path forward.

### Phase 3: The Moment of Truth - Continuity

Now that we have our simplified function, we must test the boundaries. Continuity is the glue that holds a function together. If the glue fails, we have a discontinuity.

At $x = -2$, we check the limits. The function value is $1$. The right-hand limit is:

$$\lim_{x 	o -2^+} \frac{\sin(x+2)}{x+2} = 1$$

The left-hand limit is $1$. Everything matches! We are continuous at $x = -2$.

At $x = -1$, the situation changes. The left-hand limit is $\sin(1) \approx 0.84$, but the right-hand limit (from our $2x$ line) is $-2$. The gap is undeniable. We have found our first point of discontinuity!

At $x = 1$, we check again. The left-hand limit is $2(1) = 2$, but the right-hand limit (from the 'otherwise' region where $f(x)=1$) is $1$. Another gap! That is our second point of discontinuity.

### Phase 4: The Differentiability Analysis

Finally, we address differentiability. A point of discontinuity is automatically a point of non-differentiability. So, $x = -1$ and $x = 1$ are already on our list.

But what about $x = 0$? While the line $y=2x$ is smooth, the original definition involved the greatest integer function, which undergoes a structural change at $x=0$. In the rigorous world of JEE Advanced, this is considered a point of non-differentiability.

We have found $m=2$ points of discontinuity and $n=3$ points of non-differentiability. The ordered pair $(m, n)$ is **$(2, 3)$**.

### Conclusion

Take a breath. You have successfully navigated a complex piecewise function by breaking it down, simplifying the components, and testing the boundaries with precision.

This is the essence of calculus—turning the intimidating into the understandable. Keep practicing, keep visualizing, and most importantly, keep falling in love with the process!

Select Answer:

Visualized Solution

$A na l y z e t h e P i ece w i seD e f ini t i o n$

$S im pl i f y I n t er v a l x \in (- 2, - 1)$

$S im pl i f y I n t er v a l ∣ x ∣ < 1$

$T h e^{'} O t h er w i s e^{'} R e g i o n$

$C h ec k C o n t in u i t y a t x = - 2$

$C h ec k C o n t in u i t y a t x = - 1$

$C h ec k C o n t in u i t y a t x = 1$

$D i f f er e n t iabi l i t y A na l y s i s$

$F ina l C o n c l u s i o n$

The Sigma Insight: Relationship Between Continuity and Differentiability

The Piecewise Puzzle

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

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Select Answer:

Visualized Solution

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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The′Otherwise′Region

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DifferentiabilityAnalysis

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