Let f : R → R be defined as f(x) = { [e^x], x

Q: Why is $[e^x] = 0$ for $x < 0$?

Great question! Remember that $e^x$ for $x < 0$ is always a value between $0$ and $1$. The greatest integer function $[t]$ returns the largest integer less than or equal to $t$. Since $0 < e^x < 1$, the largest integer less than or equal to $e^x$ is $0$.

Q: Wait, why is there always a discontinuity at $x=1$?

This is the trap! At $x=1$, the left-hand limit is $ae-1$ (which is $e-1$ since $a=1$). The function value $f(1)$ is $b$. The right-hand limit is $b-1$. Since $b$ can never equal $b-1$, the function jumps at $x=1$ regardless of the value of $b$.

Q: How do I handle the $\sin(\pi x)$ term inside the GIF?

For $1 < x < 2$, the angle $\pi x$ is in the third and fourth quadrants, where $\sin(\pi x)$ is negative. Specifically, it's between $-1$ and $0$. The greatest integer of any number in $(-1, 0)$ is $-1$. So, the term $[\sin(\pi x)]$ simplifies to $-1$.

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ be defined as $f(x) = \begin{cases} [e^x], & x < 0 \ ae^x + [x - 1], & 0 \le x < 1 \ b + [\sin(\pi x)], & 1 \le x < 2 \ [e^{-x}] - c, & x \ge 2 \end{cases}$ where $a, b, c \in \mathbb{R}$ and $[t]$ denotes greatest integer less than or equal to $t$. Then, which of the following statements is true ?

Options:

(A) There exists $a, b, c \in \mathbb{R}$ such that $f$ is continuous of $\mathbb{R}$.

(B) If $f$ is discontinuous at exactly one point, then $a + b + c = 1$.

(C) If $f$ is discontinuous at exactly one point, then $a + b + c 
eq 1$.

(D) $f$ is discontinuous at atleast two points, for any values of $a, b$ and $c$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a number line, looking at a function that changes its personality at three specific gates: $x=0$, $x=1$, and $x=2$. This is the essence of a piecewise function.

It is not just one equation; it is a collection of rules that dictate behavior in different territories. Our goal is to ensure this function is as smooth as possible, or at least understand where it breaks.

### Phase 1: The Exponential Staircase

Let us start in the region where $x < 0$. Here, $f(x) = [e^x]$.

As $x$ becomes more and more negative, $e^x$ gets closer to $0$ but never reaches it. For all $x < 0$, $e^x$ is a fraction between $0$ and $1$.

The greatest integer function, denoted by $[t]$, is a ruthless floor function; it rounds everything down to the nearest integer. Since $0 < e^x < 1$, the floor of $e^x$ is always $0$. So, for the entire left side of the $y$-axis, our function is just a flat line at $y=0$.

Now, consider the region $0 \le x < 1$. Here, $f(x) = ae^x + [x - 1]$.

If $x$ is between $0$ and $1$, then $x-1$ is between $-1$ and $0$. The greatest integer of any number in this range is $-1$. Thus, this branch simplifies to $f(x) = ae^x - 1$.

### Phase 2: The First Junction at $x=0$

To have continuity at $x=0$, the path from the left must meet the path from the right. From the left, the limit is $0$.

From the right, we evaluate:
$$f(0) = ae^0 - 1 = a - 1$$

For these to meet, we must have $a - 1 = 0$, which forces $a = 1$. This is our first anchor point.

### Phase 3: The Inevitable Break at $x=1$

Now we approach the most treacherous part of the problem: $x=1$. The left-hand limit is:
$$\lim_{x 	o 1^-} (ae^x - 1) = ae - 1$$

Since we know $a=1$, this is $e - 1$. The function value at $x=1$ is $f(1) = b + [\sin(\pi)] = b + 0 = b$.

The right-hand limit is $\lim_{x 	o 1^+} (b + [\sin(\pi x)])$. For $x$ just slightly larger than $1$, $\pi x$ is slightly larger than $\pi$, putting it in the third quadrant where sine is negative.

Specifically, $\sin(\pi x)$ is between $-1$ and $0$, so its greatest integer is $-1$. Thus, the right-hand limit is $b - 1$.

Look at the values: $f(1) = b$ and the right-hand limit is $b - 1$. Can $b$ ever equal $b - 1$? Absolutely not. This is a permanent, unavoidable jump discontinuity.

### Phase 4: The Final Junction at $x=2$

Finally, we look at $x=2$. The left-hand limit is $b - 1$.

The right-hand limit is $f(2) = [e^{-2}] - c$. Since $e^{-2}$ is a tiny positive fraction, its greatest integer is $0$. So, the right-hand limit is $-c$.

For continuity at $x=2$, we need $b - 1 = -c$, which rearranges to $b + c = 1$.

### The Conclusion

We have established that $f(x)$ is always discontinuous at $x=1$. If the problem asks for the condition where $f$ is discontinuous at *exactly* one point, it implies that $x=1$ is that point, and the function must be continuous at $x=0$ and $x=2$.

This requires $a=1$ and $b+c=1$. Adding these together:
$$a + b + c = 1 + 1 = 2$$

Since $2 
eq 1$, we have successfully proven that if the function has exactly one discontinuity, the sum $a + b + c$ cannot be $1$. You have navigated the traps, simplified the greatest integer functions, and emerged with a clear, logical result.

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Visualized Solution

The Piecewise Function Setup

Simplifying the First Branch ( $x < 0$ )

Simplifying the Second Branch ( $0 \leq x < 1$ )

Enforcing Continuity at $x = 0$

Analyzing the Third Branch ( $1 \leq x < 2$ )

Checking Continuity at $x = 1$

Simplifying the Fourth Branch ( $x \geq 2$ )

Enforcing Continuity at $x = 2$

Evaluating the Given Statements

Calculating $a + b + c$

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

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

The Conclusion

Similar Questions

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Visualized Solution

The Piecewise Function Setup

Simplifying the First Branch ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0)

Simplifying the Second Branch ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0≤x<1)

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

Analyzing the Third Branch ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1≤x<2)

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

Simplifying the Fourth Branch ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥2)

Enforcing Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2