Let f, g : R → R be functions defined by f(x) = { [x] , x

Q: Wait, why do we check $x=2$ specifically? How did we know to look there?

Great question! We look for points where the inner function $g(x)$ 'hits' the discontinuity points of the outer function $f(x)$. We know $f(x)$ is discontinuous at $0, -1, -2, \dots$. So, we set $g(x) = 0$ and $g(x) = -1$ to see if any $x$ values produce these outputs. Solving $(x-1)^2 - 1 = 0$ gives $x=2$, which is why we must test it!

Q: Why does $f(0^-)$ result in $-1$? Shouldn't it be $0$?

Remember the definition of the Greatest Integer Function $[x]$: it is the greatest integer *less than or equal to* $x$. If we have $0^-$, we are talking about a number like $-0.0001$. The greatest integer less than or equal to $-0.0001$ is $-1$. Always visualize the number line!

Q: Can't we just graph both functions and see where they break?

While graphing is a powerful tool for intuition, it can be deceptive with composite functions. The 'relay' nature of $f(g(x))$ means that even if $g(x)$ is continuous, the composite can break if $g(x)$ jumps across a discontinuity of $f(x)$. Algebraic limits are the only way to be 100% sure in a JEE Advanced exam.

Question

Let $f, g : \mathbb{R} 	o \mathbb{R}$ be functions defined by $f(x) = \begin{cases} [x] & , x < 0 \ |1 - x| & , x \ge 0 \end{cases}$ and $g(x) = \begin{cases} e^x - x & , x < 0 \ (x - 1)^2 - 1 & , x \ge 0 \end{cases}$ where $[x]$ denote the greatest integer less than or equal to $x$. Then, the function $fog$ is discontinuous at exactly :

Options:

(A) one point

(B) two points

(C) three points

(D) four points

Sigma Percentile · Accepted Answer

### The Relay Race of Continuity: Mastering Composite Functions

Welcome, future engineer! Today, we are not just solving a math problem; we are dissecting the anatomy of a composite function.

When we look at $f(g(x))$, we are essentially watching a relay race. The input $x$ is handed to the inner runner, $g(x)$, and the result of that race is then handed to the outer runner, $f(x)$. If either runner stumbles, or if the hand-off is messy, the entire race is compromised.

### Phase 1: Identifying the Players

First, let's look at our outer function, $f(x)$. It is defined as:

$$f(x) = 
\begin{cases} 
[x] & x < 0 \ 
|1 - x| & x \ge 0 
\end{cases}$$

This function is a bit of a troublemaker. It has a jump discontinuity at $x=0$ (where the left side approaches $-1$ and the right side starts at $1$) and at every negative integer because of the Greatest Integer Function $[x]$.

Our inner function, $g(x)$, is defined as:

$$g(x) = 
\begin{cases} 
e^x - x & x < 0 \ 
(x - 1)^2 - 1 & x \ge 0 
\end{cases}$$

Notice that $g(x)$ itself has a discontinuity at $x=0$ because $\lim_{x 	o 0^-} g(x) = 1$ while $\lim_{x 	o 0^+} g(x) = 0$. This is our first red flag!

### Phase 2: The Critical Investigation at $x=0$

We must check the point $x=0$ with extreme care.

For the Right Hand Limit (RHL), as $x 	o 0^+$, $g(x) = (x-1)^2 - 1$. As $x$ approaches $0$ from the right, $g(x)$ approaches $0$ from below, which we write as $0^-$.

Now, we pass this $0^-$ to $f(x)$. Since $0^-$ is negative, we use the $[x]$ definition:

$$f(0^-) = [0^-] = -1$$

Now for the Left Hand Limit (LHL): as $x 	o 0^-$, $g(x) = e^x - x$. As $x$ approaches $0$, $g(x)$ approaches $e^0 - 0 = 1$. Since we are approaching from the left, it approaches $1^+$.

We pass this $1^+$ to $f(x)$. Since $1^+$ is positive, we use the $|1-x|$ definition:

$$f(1^+) = |1 - 1^+| = 0$$

Since $-1 
eq 0$, the function $f(g(x))$ is definitely discontinuous at $x=0$.

### Phase 3: Hunting for Hidden Discontinuities

Are there other points? We must check where $g(x)$ hits the discontinuity points of $f(x)$. We know $f(x)$ breaks at $0, -1, -2, \dots$.

For $x \ge 0$, $g(x) = (x-1)^2 - 1$. Setting this equal to $0$ gives $(x-1)^2 = 1$, which means $x=0$ or $x=2$. We already checked $x=0$.

Let's check $x=2$. As $x 	o 2^+$, $g(x) 	o 0^+$, so $f(g(x)) 	o f(0^+) = |1-0| = 1$. As $x 	o 2^-$, $g(x) 	o 0^-$, so $f(g(x)) 	o f(0^-) = [0^-] = -1$.

Again, $1 
eq -1$, so $x=2$ is a point of discontinuity! Finally, we check $x=1$ (where $g(x)=-1$), but the limits match there, so it remains continuous.

We have found our culprits: **$x=0$ and $x=2$**. There are **two points of discontinuity**. You have successfully navigated the relay race!

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Visualized Solution

Introduction to $f (g (x))$

Discontinuity of $f (x)$

Discontinuity of $g (x)$

$f (g (x))$ at $x = 0$ : Right Hand Limit

$f (g (x))$ at $x = 0$ : Left Hand Limit

Discontinuity at $x = 0$

Identifying Potential Points

Checking $x = 2$ : Right Hand Limit

Checking $x = 2$ : Left Hand Limit

Checking $x = 1$

Final Conclusion

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

The Relay Race of Continuity

Phase 1

Phase 2

Phase 3

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Visualized Solution

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

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

Discontinuity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

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

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

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