If `f (x)= {{:((x-e ^(x)+1-(1-cos 2x))/(x ^(2)), x ne 0),(k,x=0):}` is continous at `x=0` then which of the following statement is false ?

Options

A. `k= (-5)/(2)`
B. `{k}=1/2`
C. `[k]=-2`
D. `[k] {k} =(-3)/(2)`

Answer

To determine the value of \( k \) such that the function

\[ f(x) = \begin{cases} \frac{x - e^x + 1 - (1 - \cos 2x)}{x^2} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to find the limit of \( f(x) \) as \( x \) approaches 0 and set it equal to \( k \).

### Step 1: Find the limit as \( x \to 0 \)

We will calculate the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x - e^x + 1 - (1 - \cos 2x)}{x^2} \]

### Step 2: Simplify the expression

The expression simplifies to:

\[ \lim_{x \to 0} \frac{x - e^x + \cos 2x}{x^2} \]

### Step 3: Evaluate the limit using L'Hôpital's Rule

Since substituting \( x = 0 \) gives a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule. We differentiate the numerator and denominator:

1. 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:

\[ 1 - e^x - 2\sin 2x \]

2. Differentiate the denominator:
- The derivative of \( x^2 \) is \( 2x \).

So we have:

\[ \lim_{x \to 0} \frac{1 - e^x - 2\sin 2x}{2x} \]

### Step 4: Apply L'Hôpital's Rule again

This again results in a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again:

1. Differentiate the numerator:
- 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:

\[ -e^x - 4\cos 2x \]

2. Differentiate the denominator:
- The derivative of \( 2x \) is \( 2 \).

Now we have:

\[ \lim_{x \to 0} \frac{-e^x - 4\cos 2x}{2} \]

### Step 5: Evaluate the limit

Substituting \( x = 0 \):

\[ -e^0 - 4\cos(0) = -1 - 4 = -5 \]

So,

\[ \lim_{x \to 0} f(x) = \frac{-5}{2} \]

### Step 6: Set the limit equal to \( k \)

For continuity at \( x = 0 \):

\[ k = -\frac{5}{2} \]

### Step 7: Analyze the statements

1. \( k = -\frac{5}{2} \) is true.
2. The greatest integer of \( k \) is \( -3 \) (true).
3. The fractional part of \( k \) is \( \frac{1}{2} \) (true).
4. The statement "greatest integer \( k \) and fractional part of \( k \) will be equal to \( -\frac{3}{2} \)" is false.

### Conclusion

The false statement is:

**The greatest integer and fractional part of \( k \) will be equal to \( -\frac{3}{2} \)**.