If `f(x)=(sin 2x+A sinx+B cosx)/(x^(3))` is continuous at x = 0, then the values of A, B and f(0) are

Options

A. `A=-2, B=0 and f(0)=-1`
B. `A=0,B=-2 and f(0)=1`
C. `A=1, B=-1 and f(0)=0`
D. None of the above

Explanation

To determine the values of \( A \), \( B \), and \( f(0) \) for the function

\[ f(x) = \frac{\sin(2x) + A \sin(x) + B \cos(x)}{x^3} \]

such that \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 exists and is finite.

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

We start by calculating the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin(2x) + A \sin(x) + B \cos(x)}{x^3} \]

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

Substituting \( x = 0 \):

\[ \sin(2 \cdot 0) + A \sin(0) + B \cos(0) = 0 + 0 + B = B \]

Thus, the limit becomes:

\[ \frac{B}{0} \]

For the limit to exist and be finite, \( B \) must be 0.

### Step 3: Update the function with \( B = 0 \)

Now, substituting \( B = 0 \) into the function gives us:

\[ f(x) = \frac{\sin(2x) + A \sin(x)}{x^3} \]

### Step 4: Re-evaluate the limit with \( B = 0 \)

Now, we need to find:

\[ \lim_{x \to 0} \frac{\sin(2x) + A \sin(x)}{x^3} \]

Substituting \( x = 0 \) again gives us:

\[ \sin(2 \cdot 0) + A \sin(0) = 0 + 0 = 0 \]

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

### Step 5: Apply L'Hôpital's Rule

Differentiating the numerator and denominator:

\[ \text{Numerator: } \frac{d}{dx}(\sin(2x) + A \sin(x)) = 2 \cos(2x) + A \cos(x) \]
\[ \text{Denominator: } \frac{d}{dx}(x^3) = 3x^2 \]

Thus, we have:

\[ \lim_{x \to 0} \frac{2 \cos(2x) + A \cos(x)}{3x^2} \]

### Step 6: Substitute \( x = 0 \) again

Substituting \( x = 0 \):

\[ \frac{2 \cos(0) + A \cos(0)}{3 \cdot 0^2} = \frac{2 + A}{0} \]

For this limit to exist, \( 2 + A \) must equal 0, which gives us:

\[ A = -2 \]

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

Now we can find \( f(0) \):

\[ f(0) = \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin(2x) - 2 \sin(x)}{x^3} \]

This again gives us the form \( \frac{0}{0} \). We apply L'Hôpital's Rule again.

### Step 8: Differentiate again

Differentiating again:

\[ \text{Numerator: } \frac{d}{dx}(\sin(2x) - 2 \sin(x)) = 2 \cos(2x) - 2 \cos(x) \]
\[ \text{Denominator: } \frac{d}{dx}(x^3) = 3x^2 \]

Thus, we have:

\[ \lim_{x \to 0} \frac{2 \cos(2x) - 2 \cos(x)}{3x^2} \]

### Step 9: Substitute \( x = 0 \) again

Substituting \( x = 0 \):

\[ \frac{2 \cos(0) - 2 \cos(0)}{3 \cdot 0^2} = \frac{0}{0} \]

We apply L'Hôpital's Rule one more time.

### Step 10: Differentiate one last time

Differentiating again gives:

\[ \text{Numerator: } \frac{d}{dx}(2 \cos(2x) - 2 \cos(x)) = -4 \sin(2x) + 2 \sin(x) \]
\[ \text{Denominator: } 6x \]

Thus, we have:

\[ \lim_{x \to 0} \frac{-4 \sin(2x) + 2 \sin(x)}{6x} \]

### Step 11: Substitute \( x = 0 \) again

Substituting \( x = 0 \):

\[ \frac{-4 \sin(0) + 2 \sin(0)}{6 \cdot 0} = \frac{0}{0} \]

We apply L'Hôpital's Rule again.

### Step 12: Final differentiation

Differentiating again gives:

\[ \text{Numerator: } \frac{d}{dx}(-4 \sin(2x) + 2 \sin(x)) = -8 \cos(2x) + 2 \cos(x) \]

Thus, we have:

\[ \lim_{x \to 0} \frac{-8 \cos(2x) + 2 \cos(x)}{6} \]

### Step 13: Substitute \( x = 0 \) one last time

Substituting \( x = 0 \):

\[ \frac{-8 \cos(0) + 2 \cos(0)}{6} = \frac{-8 + 2}{6} = \frac{-6}{6} = -1 \]

### Final Results

Thus, we have:

- \( A = -2 \)
- \( B = 0 \)
- \( f(0) = -1 \)