Let `f:[-1/2,2] rarr R` and `g:[-1/2,2] rarr R` be functions defined by `f(x)=[x^2-3]` and `g(x)=|x|f(x)+|4x-7|f(x)`, where [y] denotes the greatest integer less than or equal to y for `yinR`. Then,

Options

A. f is discontinuous exactly at three points in `[-1/2,2]`
B. f is discontinuous exactly at four points in `[-1/2,2]`
C. g is NOT differentiable exactly at four points in `(-1/2,2)`
D. g is NOT differentiable exactly at five points in `(-1/2,2)`

Answer

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

### Step 1: Analyze the function \( f(x) \)

The function \( f(x) \) is defined as:
\[ f(x) = \lfloor x^2 - 3 \rfloor \]
where \( \lfloor y \rfloor \) denotes the greatest integer less than or equal to \( y \).

#### Finding the points of discontinuity for \( f(x) \)

1. **Identify when \( x^2 - 3 \) is an integer**:
- Set \( x^2 - 3 = n \) where \( n \) is an integer.
- This gives us \( x^2 = n + 3 \) or \( x = \pm\sqrt{n + 3} \).

2. **Determine the integer values of \( n \)**:
- For \( f(x) \) to be discontinuous, \( n \) must be such that \( n + 3 \) is a perfect square.
- The possible integer values of \( n \) can be calculated by considering the range of \( x \) from \(-\frac{1}{2}\) to \(2\):
- \( x^2 \) ranges from \(0\) to \(4\).
- Thus, \( x^2 - 3 \) ranges from \(-3\) to \(1\).

3. **Calculate the integer values of \( n \)**:
- The integers in the range \([-3, 1]\) are \(-3, -2, -1, 0, 1\).
- This gives us the following equations:
- For \( n = -3 \): \( x^2 = 0 \) → \( x = 0 \)
- For \( n = -2 \): \( x^2 = 1 \) → \( x = \pm 1 \)
- For \( n = -1 \): \( x^2 = 2 \) → \( x = \pm \sqrt{2} \)
- For \( n = 0 \): \( x^2 = 3 \) → \( x = \pm \sqrt{3} \)
- For \( n = 1 \): \( x^2 = 4 \) → \( x = 2 \)

4. **List the points of discontinuity**:
- The points of discontinuity are \( x = 0, 1, \sqrt{2}, \sqrt{3}, 2 \).
- However, since \( x \) is restricted to \([-1/2, 2]\), we only consider \( 0, 1, \sqrt{2}, \sqrt{3} \) (note that \(\sqrt{3} \approx 1.732\) is within the interval).

Thus, the function \( f(x) \) is discontinuous at **4 points**.

### Step 2: Analyze the function \( g(x) \)

The function \( g(x) \) is defined as:
\[ g(x) = |x| f(x) + |4x - 7| f(x) \]

1. **Identify the intervals for \( g(x) \)**:
- We need to consider the absolute values:
- \( |x| \) changes at \( x = 0 \).
- \( |4x - 7| \) changes at \( x = \frac{7}{4} = 1.75 \).

2. **Evaluate \( g(x) \) in different intervals**:
- For \( x \in [-\frac{1}{2}, 0) \):
- \( g(x) = -x f(x) + (7 - 4x) f(x) = (7 - 3x) f(x) \)
- For \( x \in [0, 1) \):
- \( g(x) = x f(x) + (7 - 4x) f(x) = (7 - 3x) f(x) \)
- For \( x \in [1, \sqrt{2}) \):
- \( g(x) = x f(x) + (7 - 4x) f(x) = (7 - 3x) f(x) \)
- For \( x \in [\sqrt{2}, \sqrt{3}) \):
- \( g(x) = x f(x) + (4x - 7) f(x) = (3x - 7) f(x) \)
- For \( x \in [\sqrt{3}, 2] \):
- \( g(x) = x f(x) + (4x - 7) f(x) = (3x - 7) f(x) \)

3. **Determine points of discontinuity for \( g(x) \)**:
- Since \( g(x) \) depends on \( f(x) \), the points of discontinuity for \( g(x) \) will occur at the same points where \( f(x) \) is discontinuous.
- Therefore, \( g(x) \) is also discontinuous at the same 4 points \( 0, 1, \sqrt{2}, \sqrt{3} \).

### Conclusion

- The function \( f(x) \) is discontinuous at **4 points**.
- The function \( g(x) \) is also discontinuous at **4 points**.