`f (x)=[x] and g (x)= {{:( 0"," , x in I ),( x ^(2)"," , cancel(in)I):}` where [.] dentoes the greatest integer function. Then

Options

A. gof is continous for all x
B. gof is not continous for all x
C. fog is continous everywhere
D. fog is not continous everywhere

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given in the question:

1. **Define the functions:**
- \( f(x) = [x] \), where \([x]\) is the greatest integer function (also known as the floor function).
- \( g(x) = \begin{cases} 0 & \text{if } x \text{ is an integer} \\ x^2 & \text{if } x \text{ is not an integer} \end{cases} \)

2. **Evaluate \( g(f(x)) \):**
- Since \( f(x) = [x] \), \( f(x) \) is always an integer for any real \( x \).
- Therefore, \( g(f(x)) = g([x]) \).
- Since \([x]\) is an integer, we have \( g([x]) = 0 \).
- Thus, \( g(f(x)) = 0 \) for all \( x \).

3. **Check continuity of \( g(f(x)) \):**
- The function \( g(f(x)) \) is constant (always 0) for all \( x \).
- A constant function is continuous everywhere.
- Therefore, \( g(f(x)) \) is continuous for all \( x \).

4. **Evaluate \( f(g(x)) \):**
- We need to consider two cases based on the definition of \( g(x) \):
- **Case 1:** If \( x \) is an integer, then \( g(x) = 0 \).
- Thus, \( f(g(x)) = f(0) = [0] = 0 \).
- **Case 2:** If \( x \) is not an integer, then \( g(x) = x^2 \).
- Thus, \( f(g(x)) = f(x^2) = [x^2] \).
- The value of \([x^2]\) depends on the range of \( x \):
- If \( 0 \leq x < 1 \), then \( 0 \leq x^2 < 1 \) and \([x^2] = 0\).
- If \( 1 \leq x < \sqrt{2} \), then \( 1 \leq x^2 < 2 \) and \([x^2] = 1\).
- If \( \sqrt{2} \leq x < \sqrt{3} \), then \( 2 \leq x^2 < 3 \) and \([x^2] = 2\).
- If \( \sqrt{3} \leq x < 2 \), then \( 3 \leq x^2 < 4 \) and \([x^2] = 3\).
- The function \( f(g(x)) \) is not continuous at the points where \( x \) transitions from integer to non-integer values, as the output jumps from 0 to 1, etc.

5. **Conclusion:**
- \( g(f(x)) \) is continuous for all \( x \).
- \( f(g(x)) \) is not continuous everywhere.

### Summary of Results:
- \( g(f(x)) \) is continuous for all \( x \) (True).
- \( f(g(x)) \) is not continuous everywhere (True).