Let `f(x)={:{((sin(x-[x]))/(x-[x]),","x in (-2,-1)),(max{2x,3[|x|]},","|x|lt1),(1,",otherwise"):}`
where [t] denotes greatest `le t` .If m is the number of points where f is not continuous and n is the number of points where f is not differentiable , then the ordered pair (m,n) is :

Options

A. (3,3)
B. (2,4)
C. (2,3)
D. (3,4)

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined piecewise:

\[ f(x) = \begin{cases} \frac{\sin(x - [x])}{x - [x]} & \text{for } x \in (-2, -1) \\ \max(2x, 3[|x|]) & \text{for } |x| < 1 \\ 1 & \text{otherwise} \end{cases} \]

where \([t]\) denotes the greatest integer less than or equal to \(t\).

### Step 1: Analyze the function in the interval \((-2, -1)\)

In this interval, \([x] = -2\). Therefore, we can rewrite the function as:

\[ f(x) = \frac{\sin(x + 2)}{x + 2} \]

This function is continuous for all \(x \neq -2\). Since \(-2\) is not in the interval \((-2, -1)\), \(f(x)\) is continuous in this interval.

**Hint:** Check the continuity of the sine function and the denominator.

### Step 2: Analyze the function in the interval \((-1, 1)\)

In this interval, we need to consider two cases based on the definition of \(|x|\):

1. For \(-1 < x < 0\):
- Here, \(|x| = -x\) and \([|x|] = 0\).
- Thus, \(f(x) = \max(2x, 0)\).
- Since \(2x\) is a linear function, \(f(x)\) will be \(0\) when \(x \leq 0\) and \(2x\) when \(x > 0\).

2. For \(0 < x < 1\):
- Here, \(|x| = x\) and \([|x|] = 0\).
- Thus, \(f(x) = 2x\).

**Hint:** Identify the points where the function changes its definition.

### Step 3: Check continuity at the transition points

- At \(x = -1\):
- Left-hand limit: \(f(-1^-) = \max(2(-1), 0) = 0\).
- Right-hand limit: \(f(-1^+) = 0\).
- Since both limits are equal, \(f(x)\) is continuous at \(x = -1\).

- At \(x = 0\):
- Left-hand limit: \(f(0^-) = 0\).
- Right-hand limit: \(f(0^+) = 0\).
- Since both limits are equal, \(f(x)\) is continuous at \(x = 0\).

- At \(x = 1\):
- Left-hand limit: \(f(1^-) = 2(1) = 2\).
- Right-hand limit: \(f(1^+) = 1\).
- Since both limits are not equal, \(f(x)\) is discontinuous at \(x = 1\).

**Hint:** Evaluate limits from both sides at transition points.

### Step 4: Analyze the function for \(x \geq 1\) and \(x \leq -2\)

For \(x \geq 1\) and \(x \leq -2\), \(f(x) = 1\). This part of the function is continuous everywhere in its domain.

### Step 5: Check differentiability at the critical points

- At \(x = -1\):
- Left-hand derivative: \(f'(-1^-) = 2\).
- Right-hand derivative: \(f'(-1^+) = 0\).
- Since both derivatives are not equal, \(f(x)\) is not differentiable at \(x = -1\).

- At \(x = 0\):
- Left-hand derivative: \(f'(0^-) = 0\).
- Right-hand derivative: \(f'(0^+) = 2\).
- Since both derivatives are not equal, \(f(x)\) is not differentiable at \(x = 0\).

- At \(x = 1\):
- Left-hand derivative: \(f'(1^-) = 2\).
- Right-hand derivative: \(f'(1^+) = 0\).
- Since both derivatives are not equal, \(f(x)\) is not differentiable at \(x = 1\).

### Conclusion

- The function \(f(x)\) is not continuous at \(x = 1\) (1 point).
- The function \(f(x)\) is not differentiable at \(x = -1\), \(x = 0\), and \(x = 1\) (3 points).

Thus, the ordered pair \((m, n)\) is \((1, 3)\).

### Final Answer

\((m, n) = (1, 3)\)