Let `f : R rightarrow R` be defined as `f(x) = {(frac{a-b cos 2x}{x^2} ,x < 0),(x^2 + cx + 2, 0 le x le 1), (2x + 1, x gt 1):}`. If fis continuous everywhere in R and m is the number of points where f(x) is NOT differential then `m +a+ b + c` equals :

Options

A. 1
B. 4
C. 3
D. 2

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous everywhere and then find the values of \( a \), \( b \), and \( c \) such that the function is also differentiable at the points of interest.

### Step 1: Check continuity at \( x = 0 \)

The function is defined as follows:
- For \( x < 0 \): \( f(x) = \frac{a - b \cos(2x)}{x^2} \)
- For \( 0 \leq x \leq 1 \): \( f(x) = x^2 + cx + 2 \)
- For \( x > 1 \): \( f(x) = 2x + 1 \)

To check continuity at \( x = 0 \), we need to ensure:
\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \]

**Left-hand limit**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{a - b \cos(2x)}{x^2} \]
Using L'Hôpital's Rule since it is of the form \( \frac{0}{0} \):
\[ = \lim_{x \to 0^-} \frac{2b \sin(2x)}{2x} = \lim_{x \to 0^-} \frac{b \sin(2x)}{x} = b \cdot 2 \quad (\text{since } \lim_{x \to 0} \frac{\sin(2x)}{x} = 2) \]
Thus, the left-hand limit is \( 2b \).

**Right-hand limit**:
\[ f(0) = f(0) = 0^2 + c \cdot 0 + 2 = 2 \]

Setting the left-hand limit equal to the right-hand limit:
\[ 2b = 2 \implies b = 1 \]

### Step 2: Check continuity at \( x = 1 \)

Next, we check continuity at \( x = 1 \):
\[ \lim_{x \to 1^-} f(x) = f(1) = \lim_{x \to 1^+} f(x) \]

**Left-hand limit**:
\[ \lim_{x \to 1^-} f(x) = 1^2 + c \cdot 1 + 2 = 1 + c + 2 = c + 3 \]

**Right-hand limit**:
\[ f(1) = 2 \cdot 1 + 1 = 3 \]

Setting the left-hand limit equal to the right-hand limit:
\[ c + 3 = 3 \implies c = 0 \]

### Step 3: Find \( a \)

From the continuity condition at \( x = 0 \), we already found \( b = 1 \). Now we need to find \( a \).

Using the left-hand limit:
\[ \lim_{x \to 0^-} f(x) = 2b = 2 \cdot 1 = 2 \]
So, we set:
\[ \lim_{x \to 0^-} f(x) = \frac{a - 1}{0} \text{ (as } \cos(0) = 1 \text{)} \implies a - 1 = 2 \implies a = 3 \]

### Step 4: Check differentiability

Now we need to check differentiability at \( x = 0 \) and \( x = 1 \).

**At \( x = 0 \)**:
The left-hand derivative:
\[ \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} \frac{d}{dx} \left( \frac{3 - \cos(2x)}{x^2} \right) = \text{Apply L'Hôpital's Rule again} \]
Using the derivative:
\[ = \lim_{x \to 0^-} \frac{2\sin(2x)}{2x} = 1 \]

The right-hand derivative:
\[ \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} (2x + 0) = 0 \]

Since the left-hand and right-hand derivatives at \( x = 0 \) are not equal, \( f(x) \) is not differentiable at \( x = 0 \).

**At \( x = 1 \)**:
Both left-hand and right-hand derivatives will yield the same result since they are both linear functions.

### Conclusion

The only point where \( f(x) \) is not differentiable is at \( x = 0 \), so \( m = 1 \).

Now we can calculate:
\[ m + a + b + c = 1 + 3 + 1 + 0 = 5 \]

### Final Answer
Thus, the final result is:
\[ \boxed{5} \]