If f(x) = sgn (x) and g (x) = `(1-x^2)`,then the number of points of discontinuity of function f (g (x)) is -

Options

A. exact two
B. exactly three
C. finite and more than 3
D. infinitely many

Answer

To find the number of points of discontinuity of the function \( f(g(x)) \), where \( f(x) = \text{sgn}(x) \) and \( g(x) = 1 - x^2 \), we will follow these steps:

### Step 1: Define the Functions
The signum function \( f(x) \) is defined as:
- \( f(x) = 1 \) for \( x > 0 \)
- \( f(x) = 0 \) for \( x = 0 \)
- \( f(x) = -1 \) for \( x < 0 \)

The function \( g(x) \) is defined as:
\[ g(x) = 1 - x^2 \]

### Step 2: Find the Composite Function
We need to evaluate \( f(g(x)) \):
\[ f(g(x)) = f(1 - x^2) \]

### Step 3: Determine the Sign of \( g(x) \)
Next, we analyze the expression \( 1 - x^2 \):
- \( 1 - x^2 > 0 \) when \( x^2 < 1 \) (i.e., \( -1 < x < 1 \))
- \( 1 - x^2 = 0 \) when \( x^2 = 1 \) (i.e., \( x = -1 \) or \( x = 1 \))
- \( 1 - x^2 < 0 \) when \( x^2 > 1 \) (i.e., \( x < -1 \) or \( x > 1 \))

### Step 4: Apply the Signum Function
Now we can write \( f(g(x)) \) based on the intervals determined:
- For \( -1 < x < 1 \): \( g(x) > 0 \) so \( f(g(x)) = 1 \)
- For \( x = -1 \) or \( x = 1 \): \( g(x) = 0 \) so \( f(g(x)) = 0 \)
- For \( x < -1 \) or \( x > 1 \): \( g(x) < 0 \) so \( f(g(x)) = -1 \)

### Step 5: Identify Points of Discontinuity
Now we can summarize the behavior of \( f(g(x)) \):
- \( f(g(x)) = 1 \) for \( -1 < x < 1 \)
- \( f(g(x)) = 0 \) at \( x = -1 \) and \( x = 1 \)
- \( f(g(x)) = -1 \) for \( x < -1 \) and \( x > 1 \)

The points where the function changes its value (discontinuities) are:
1. At \( x = -1 \) (from \( -1 \) to \( 1 \) to \( 0 \))
2. At \( x = 1 \) (from \( 1 \) to \( -1 \) to \( 0 \))

### Conclusion
Thus, the number of points of discontinuity of the function \( f(g(x)) \) is \( 2 \).