Let `f, g: R rarr R` be functions defined by
`f(x)= {([x],x lt 0),(|1-x|,x ge 0):} and g(x)= {(e^(x)-1,x lt 0),((x-1)^(2)-1,x ge 0):}`
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

Answer

To find the points of discontinuity of the function \( f \circ g \), we will follow these steps:

### Step 1: Define the functions \( f(x) \) and \( g(x) \)

The functions are defined as follows:
- For \( f(x) \):
\[ f(x) = \begin{cases} [x] & \text{if } x < 0 \\ |1 - x| & \text{if } x \geq 0 \end{cases} \]
- For \( g(x) \):
\[ g(x) = \begin{cases} e^x - 1 & \text{if } x < 0 \\ (x - 1)^2 - 1 & \text{if } x \geq 0 \end{cases} \]

### Step 2: Determine the composition \( f(g(x)) \)

We need to evaluate \( f(g(x)) \) for both cases of \( g(x) \).

**Case 1: \( x < 0 \)**
- Here, \( g(x) = e^x - 1 \).
- Since \( e^x < 1 \) for \( x < 0 \), we have \( g(x) < 0 \).
- Thus, \( f(g(x)) = [g(x)] = [e^x - 1] \).

**Case 2: \( x \geq 0 \)**
- Here, \( g(x) = (x - 1)^2 - 1 \).
- We need to analyze when \( g(x) \) is less than or greater than 0.
- For \( x = 0 \): \( g(0) = (0 - 1)^2 - 1 = 0 \).
- For \( x = 1 \): \( g(1) = (1 - 1)^2 - 1 = -1 \).
- For \( x = 2 \): \( g(2) = (2 - 1)^2 - 1 = 0 \).
- For \( x > 2 \): \( g(x) > 0 \).

Thus, we can summarize:
- For \( 0 \leq x < 1 \): \( g(x) < 0 \) and \( f(g(x)) = [g(x)] \).
- For \( 1 \leq x < 2 \): \( g(x) < 0 \) and \( f(g(x)) = [g(x)] \).
- For \( x \geq 2 \): \( g(x) \geq 0 \) and \( f(g(x)) = |1 - g(x)| = |1 - ((x-1)^2 - 1)| = |2 - (x-1)^2| \).

### Step 3: Analyze discontinuities

**At \( x = 0 \)**:
- Left-hand limit: \( \lim_{x \to 0^-} f(g(x)) = [e^0 - 1] = [0] = 0 \).
- Right-hand limit: \( \lim_{x \to 0^+} f(g(x)) = [g(0)] = [0] = 0 \).
- Since both limits are equal, \( f \circ g \) is continuous at \( x = 0 \).

**At \( x = 1 \)**:
- Left-hand limit: \( \lim_{x \to 1^-} f(g(x)) = [g(1)] = [-1] = -1 \).
- Right-hand limit: \( \lim_{x \to 1^+} f(g(x)) = f(g(1)) = f(-1) = [-1] = -1 \).
- Since both limits are equal, \( f \circ g \) is continuous at \( x = 1 \).

**At \( x = 2 \)**:
- Left-hand limit: \( \lim_{x \to 2^-} f(g(x)) = [g(2)] = [0] = 0 \).
- Right-hand limit: \( \lim_{x \to 2^+} f(g(x)) = f(g(2)) = f(0) = |1 - 0| = 1 \).
- Since the limits are not equal, \( f \circ g \) is discontinuous at \( x = 2 \).

### Conclusion

The function \( f \circ g \) is discontinuous at exactly **one point**, which is \( x = 2 \).