If `R to R` is defined by
`f(x)={((2 sinx-sin2x)/(2x cos x)",","if "x ne 0),(a",","if " x =0):}`
then the value of a so that f is continuous at x = 0 is

Options

A. 2
B. 1
C. `-1`
D. 0

Explanation

To find the value of \( a \) such that the function \( f \) is continuous at \( x = 0 \), we need to ensure that:

\[ f(0) = \lim_{x \to 0} f(x) \]

Given the function:

\[ f(x) = \begin{cases} \frac{2 \sin x - \sin 2x}{2x \cos x} & \text{if } x \neq 0 \\ a & \text{if } x = 0 \end{cases} \]

### Step 1: Calculate the limit as \( x \) approaches 0

We need to evaluate:

\[ \lim_{x \to 0} \frac{2 \sin x - \sin 2x}{2x \cos x} \]

### Step 2: Substitute \( x = 0 \)

Substituting \( x = 0 \) directly into the expression gives:

\[ 2 \sin(0) - \sin(2 \cdot 0) = 0 - 0 = 0 \]

The denominator also becomes:

\[ 2 \cdot 0 \cdot \cos(0) = 0 \]

This results in the indeterminate form \( \frac{0}{0} \).

### Step 3: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

where \( f(x) = 2 \sin x - \sin 2x \) and \( g(x) = 2x \cos x \).

### Step 4: Differentiate the numerator and denominator

1. Differentiate the numerator:

\[ f'(x) = 2 \cos x - 2 \cos 2x \cdot 2 = 2 \cos x - 4 \cos 2x \]

2. Differentiate the denominator:

\[ g'(x) = 2 \cos x - 2x \sin x \]

### Step 5: Evaluate the limit again

Now we evaluate:

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

Substituting \( x = 0 \):

- The numerator becomes:

\[ 2 \cos(0) - 4 \cos(0) = 2 - 4 = -2 \]

- The denominator becomes:

\[ 2 \cos(0) - 2 \cdot 0 \cdot \sin(0) = 2 - 0 = 2 \]

Thus, we have:

\[ \lim_{x \to 0} \frac{-2}{2} = -1 \]

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

For continuity at \( x = 0 \):

\[ f(0) = a = \lim_{x \to 0} f(x) = -1 \]

### Conclusion

The value of \( a \) such that \( f \) is continuous at \( x = 0 \) is:

\[ \boxed{-1} \]