If `f(x)=(4+sin2x+a sinx+Bcosx)/(x^(2))` for `x!=0` is continuous at `x=0`, then `A+B+f(0)` is

Options

A. 4
B. -4
C. 2
D. -2

Answer

To solve the problem, we need to ensure that the function \( f(x) = \frac{4 + \sin(2x) + a \sin(x) + b \cos(x)}{x^2} \) is continuous at \( x = 0 \). This means that the limit of \( f(x) \) as \( x \) approaches 0 must equal \( f(0) \).

### Step 1: Evaluate \( f(0) \)

Since \( f(x) \) is defined for \( x \neq 0 \), we need to find \( f(0) \) by taking the limit as \( x \) approaches 0.

### Step 2: Calculate the limit as \( x \to 0 \)

We need to compute:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{4 + \sin(2x) + a \sin(x) + b \cos(x)}{x^2} \]

### Step 3: Substitute \( \sin(2x) \) and \( \cos(x) \)

Using the small angle approximations:
- \( \sin(2x) \approx 2x \) as \( x \to 0 \)
- \( \sin(x) \approx x \) as \( x \to 0 \)
- \( \cos(x) \approx 1 \) as \( x \to 0 \)

So we can rewrite the limit:
\[ \lim_{x \to 0} \frac{4 + 2x + ax + b}{x^2} \]

### Step 4: Combine terms in the numerator

This simplifies to:
\[ \lim_{x \to 0} \frac{(4 + b) + (2 + a)x}{x^2} \]

### Step 5: Analyze the limit

For the limit to exist and be finite as \( x \to 0 \), the constant term in the numerator must equal zero:
\[ 4 + b = 0 \implies b = -4 \]

### Step 6: Substitute \( b \) back into the limit

Now substituting \( b = -4 \):
\[ \lim_{x \to 0} \frac{(2 + a)x}{x^2} = \lim_{x \to 0} \frac{2 + a}{x} \]

For this limit to be finite, we need \( 2 + a = 0 \):
\[ a = -2 \]

### Step 7: Calculate \( f(0) \)

Now we have \( a = -2 \) and \( b = -4 \). We can now evaluate \( f(0) \):
\[ f(0) = \lim_{x \to 0} \frac{4 + \sin(2x) - 2\sin(x) - 4\cos(x)}{x^2} \]

Substituting \( a \) and \( b \):
\[ \lim_{x \to 0} \frac{(4 - 4) + (2 - 2)x}{x^2} = \lim_{x \to 0} \frac{0}{x^2} = 0 \]

### Step 8: Find \( A + B + f(0) \)

Now we can find:
\[ A + B + f(0) = -2 + (-4) + 0 = -6 \]

Thus, the final answer is:
\[ \boxed{-6} \]