If `f(x)=(x-e^(x)+cos2x)/(x^(2))` where `x!=0`, is continuous at `x=0` then (where, {.} and [x] denotes the fractional part and greatest integer)

Options

A. `f(0)=5/2`
B. `[f(0)]=-2`
C. `{f(0)}=-0.5`
D. `[f(0)].{f(x)}=-1.5`

Answer

To determine if the function \( f(x) = \frac{x - e^x + \cos(2x)}{x^2} \) is continuous at \( x = 0 \), we need to find the limit of \( f(x) \) as \( x \) approaches 0 and check if it equals \( f(0) \).

### Step 1: Evaluate the limit as \( x \) approaches 0
We start by substituting \( x = 0 \) into the function:
\[ f(0) = \frac{0 - e^0 + \cos(0)}{0^2} = \frac{0 - 1 + 1}{0} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule states that for limits of the form \( \frac{0}{0} \), we can differentiate the numerator and denominator:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x - e^x + \cos(2x)}{x^2} \]
Differentiate the numerator:
- The derivative of \( x \) is \( 1 \).
- The derivative of \( -e^x \) is \( -e^x \).
- The derivative of \( \cos(2x) \) is \( -2\sin(2x) \).

Thus, the derivative of the numerator is:
\[ 1 - e^x - 2\sin(2x) \]
The derivative of the denominator \( x^2 \) is \( 2x \).

Now we have:
\[ \lim_{x \to 0} \frac{1 - e^x - 2\sin(2x)}{2x} \]

### Step 3: Evaluate the new limit
Substituting \( x = 0 \) again gives us:
\[ \frac{1 - e^0 - 2\sin(0)}{2 \cdot 0} = \frac{1 - 1 - 0}{0} = \frac{0}{0} \]
We apply L'Hôpital's Rule again.

### Step 4: Differentiate again
Differentiate the numerator again:
- The derivative of \( 1 \) is \( 0 \).
- The derivative of \( -e^x \) is \( -e^x \).
- The derivative of \( -2\sin(2x) \) is \( -4\cos(2x) \).

Thus, the new numerator is:
\[ -e^x - 4\cos(2x) \]
The derivative of the denominator \( 2x \) is \( 2 \).

Now we have:
\[ \lim_{x \to 0} \frac{-e^x - 4\cos(2x)}{2} \]

### Step 5: Evaluate the limit as \( x \) approaches 0
Substituting \( x = 0 \):
\[ \frac{-e^0 - 4\cos(0)}{2} = \frac{-1 - 4}{2} = \frac{-5}{2} \]

### Step 6: Conclusion
Thus, we find:
\[ \lim_{x \to 0} f(x) = -\frac{5}{2} \]
For \( f(x) \) to be continuous at \( x = 0 \), we must have \( f(0) = -\frac{5}{2} \).

### Step 7: Check the options
1. **Greatest Integer Function**: \( \lfloor -\frac{5}{2} \rfloor = -3 \)
2. **Fractional Part**: \( \{-\frac{5}{2}\} = 1 - 0.5 = 0.5 \)

### Final Answer
The correct option is:
\[ \lfloor f(0) \rfloor \cdot \{f(0)\} = -3 \cdot 0.5 = -1.5 \]