If f(x) = `(a sin x + sin 2x)/(x^(3)) ne 0` and f(x) is continuous at x =0 then

Options

A. a=2
B. f(0) =1
C. f(0) =-1
D. a =1

Answer

To solve the problem, we need to ensure that the function \( f(x) = \frac{a \sin x + \sin 2x}{x^3} \) is continuous at \( x = 0 \). For continuity at that point, we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0 and set it equal to \( f(0) \).

### Step-by-step Solution:

1. **Identify the function and the limit**:
\[ f(x) = \frac{a \sin x + \sin 2x}{x^3} \]
We need to find:
\[ \lim_{x \to 0} f(x) \]

2. **Evaluate the limit directly**:
Substituting \( x = 0 \) directly gives:
\[ f(0) = \frac{a \sin(0) + \sin(0)}{0^3} = \frac{0}{0} \]
This is an indeterminate form \( \frac{0}{0} \).

3. **Apply L'Hôpital's Rule**:
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
Here, \( f(x) = a \sin x + \sin 2x \) and \( g(x) = x^3 \).

4. **Differentiate the numerator and denominator**:
- Derivative of the numerator:
\[ f'(x) = a \cos x + 2 \cos 2x \]
- Derivative of the denominator:
\[ g'(x) = 3x^2 \]

5. **Re-evaluate the limit**:
Now we find:
\[ \lim_{x \to 0} \frac{a \cos x + 2 \cos 2x}{3x^2} \]
Substituting \( x = 0 \) gives:
\[ \frac{a \cdot 1 + 2 \cdot 1}{0} = \frac{a + 2}{0} \]
This is still an indeterminate form \( \frac{0}{0} \).

6. **Apply L'Hôpital's Rule again**:
Differentiate again:
- Second derivative of the numerator:
\[ f''(x) = -a \sin x - 4 \sin 2x \]
- Second derivative of the denominator:
\[ g''(x) = 6x \]

7. **Re-evaluate the limit again**:
Now we find:
\[ \lim_{x \to 0} \frac{-a \sin x - 4 \sin 2x}{6x} \]
Substituting \( x = 0 \) gives:
\[ \frac{-a \cdot 0 - 4 \cdot 0}{0} = \frac{0}{0} \]
We can apply L'Hôpital's Rule one more time.

8. **Differentiate one more time**:
- Third derivative of the numerator:
\[ f'''(x) = -a \cos x - 8 \cos 2x \]
- Third derivative of the denominator:
\[ g'''(x) = 6 \]

9. **Final limit evaluation**:
Now we find:
\[ \lim_{x \to 0} \frac{-a \cos x - 8 \cos 2x}{6} \]
Substituting \( x = 0 \):
\[ \frac{-a \cdot 1 - 8 \cdot 1}{6} = \frac{-a - 8}{6} \]

10. **Setting the limit equal to \( f(0) \)**:
For continuity at \( x = 0 \), we need:
\[ \frac{-a - 8}{6} = 0 \]
This implies:
\[ -a - 8 = 0 \Rightarrow a = -8 \]

### Conclusion:
Thus, the value of \( a \) for which \( f(x) \) is continuous at \( x = 0 \) is:
\[ \boxed{-8} \]