If the function `f(x)=(sin3x+a sin 2x+b)/(x^(3)), x ne 0` is continuous at `x=0` and `f(0)=K, AA K in R`, then `b-a` is equal to

Options

A. 4
B. `(5)/(2)`
C. 5
D. `(3)/(2)`

Answer

To solve the problem, we need to ensure that the function \( f(x) = \frac{\sin(3x) + a \sin(2x) + b}{x^3} \) is continuous at \( x = 0 \). This means that the limit of \( f(x) \) as \( x \) approaches 0 must equal \( f(0) = K \).

### Step 1: Find the limit as \( x \to 0 \)

We need to evaluate:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin(3x) + a \sin(2x) + b}{x^3} \]

### Step 2: Use Taylor series expansion

Using the Taylor series expansion for \( \sin(x) \):
- \( \sin(3x) \approx 3x - \frac{(3x)^3}{6} + O(x^5) = 3x - \frac{27x^3}{6} + O(x^5) \)
- \( \sin(2x) \approx 2x - \frac{(2x)^3}{6} + O(x^5) = 2x - \frac{8x^3}{6} + O(x^5) \)

Substituting these expansions into \( f(x) \):
\[ \sin(3x) + a \sin(2x) + b \approx \left(3x - \frac{27x^3}{6}\right) + a\left(2x - \frac{8x^3}{6}\right) + b \]
\[ = (3 + 2a)x - \left(\frac{27}{6} + \frac{8a}{6}\right)x^3 + b \]

### Step 3: Combine terms

Now, we can rewrite \( f(x) \):
\[ f(x) = \frac{(3 + 2a)x + b - \left(\frac{27 + 8a}{6}\right)x^3}{x^3} \]
\[ = \frac{(3 + 2a)}{x^2} + \frac{b}{x^3} - \frac{27 + 8a}{6} \]

### Step 4: Evaluate the limit

For \( f(x) \) to be continuous at \( x = 0 \), the terms involving \( x \) must vanish as \( x \to 0 \). This means:
1. The coefficient of \( x^2 \) must be zero:
\[ 3 + 2a = 0 \implies a = -\frac{3}{2} \]

2. The constant term must equal zero:
\[ b = 0 \]

### Step 5: Calculate \( b - a \)

Now we can find \( b - a \):
\[ b - a = 0 - \left(-\frac{3}{2}\right) = \frac{3}{2} \]

### Final Answer

Thus, \( b - a = \frac{3}{2} \).