If `f(x)=(sin3x+Asin2x+Bsinx)/(x^(5))` for`x!=0` is continuous at `x=0`, then `A+B+f(0)` is

Options

A. 0
B. 4
C. 2
D. 3

Answer

To solve the problem, we need to ensure that the function \( f(x) = \frac{\sin(3x) + A \sin(2x) + B \sin(x)}{x^5} \) is continuous at \( x = 0 \). This requires that the limit of \( f(x) \) as \( x \) approaches 0 equals \( f(0) \).

### Step-by-Step Solution:

1. **Find \( f(0) \)**:
Since \( f(x) \) is defined for \( x \neq 0 \), we need to find \( f(0) \) by evaluating the limit:
\[ f(0) = \lim_{x \to 0} f(x) \]

2. **Use Taylor Series Expansion**:
We will use the Taylor series expansions for \( \sin(3x) \), \( \sin(2x) \), and \( \sin(x) \) around \( x = 0 \):
- \( \sin(3x) = 3x - \frac{(3x)^3}{6} + \frac{(3x)^5}{120} + O(x^7) = 3x - \frac{27x^3}{6} + \frac{243x^5}{120} + O(x^7) \)
- \( \sin(2x) = 2x - \frac{(2x)^3}{6} + \frac{(2x)^5}{120} + O(x^7) = 2x - \frac{8x^3}{6} + \frac{32x^5}{120} + O(x^7) \)
- \( \sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7) \)

3. **Combine the Expansions**:
Substitute the expansions into \( f(x) \):
\[ f(x) = \frac{(3x - \frac{27x^3}{6} + \frac{243x^5}{120}) + A(2x - \frac{8x^3}{6} + \frac{32x^5}{120}) + B(x - \frac{x^3}{6} + \frac{x^5}{120})}{x^5} \]

4. **Simplify the Numerator**:
Combine like terms in the numerator:
\[ = \frac{(3 + 2A + B)x + \left(-\frac{27}{6} - \frac{8A}{6} - \frac{B}{6}\right)x^3 + \left(\frac{243}{120} + \frac{32A}{120} + \frac{B}{120}\right)x^5}{x^5} \]
\[ = \frac{(3 + 2A + B)}{x^4} + \frac{-\frac{27 + 8A + B}{6}}{x^2} + \left(\frac{243 + 32A + B}{120}\right) \]

5. **Take the Limit as \( x \to 0 \)**:
For \( f(x) \) to be continuous at \( x = 0 \), the coefficients of \( x^4 \) and \( x^2 \) must be zero:
- \( 3 + 2A + B = 0 \) (1)
- \( -\frac{27 + 8A + B}{6} = 0 \) (2)

6. **Solve the System of Equations**:
From equation (1):
\[ B = -3 - 2A \]
Substitute into equation (2):
\[ -\frac{27 + 8A - 3 - 2A}{6} = 0 \]
Simplifying gives:
\[ -\frac{24 + 6A}{6} = 0 \implies 24 + 6A = 0 \implies A = -4 \]
Substitute \( A = -4 \) back into equation (1):
\[ B = -3 - 2(-4) = -3 + 8 = 5 \]

7. **Calculate \( A + B + f(0) \)**:
Now we have \( A = -4 \), \( B = 5 \), and since the limit gives \( f(0) = \frac{243 + 32(-4) + 5}{120} = 1 \):
\[ A + B + f(0) = -4 + 5 + 1 = 2 \]

### Final Answer:
\[ A + B + f(0) = 2 \]