If the area bounded by the curve y=f(x), x-axis and the ordinates x=1 and x=b is (b-1) `sin`(3b+4), then-

Options

A. f(x)=cos(3x+4)+3(x-1)sin(3x+4)
B. f(x)=sin(3x+4)+3(x-1)cos(3x+4)
C. f(x)=sin(3x+4)-3(x-1)cos(3x+4)
D. None of these

Explanation

To solve the problem, we need to find the function \( f(x) \) given that the area bounded by the curve \( y = f(x) \), the x-axis, and the ordinates \( x = 1 \) and \( x = b \) is equal to \( (b - 1) \sin(3b + 4) \).

### Step-by-Step Solution:

1. **Understand the Area Representation**:
The area under the curve from \( x = 1 \) to \( x = b \) can be expressed as:
\[ \int_{1}^{b} f(x) \, dx = (b - 1) \sin(3b + 4) \]

2. **Differentiate Both Sides**:
To find \( f(b) \), we differentiate both sides of the equation with respect to \( b \):
\[ \frac{d}{db} \left( \int_{1}^{b} f(x) \, dx \right) = \frac{d}{db} \left( (b - 1) \sin(3b + 4) \right) \]
By the Fundamental Theorem of Calculus, the left side becomes \( f(b) \).

3. **Apply the Product Rule on the Right Side**:
For the right side, we apply the product rule:
\[ \frac{d}{db} \left( (b - 1) \sin(3b + 4) \right) = \sin(3b + 4) + (b - 1) \cdot \frac{d}{db}(\sin(3b + 4)) \]
The derivative of \( \sin(3b + 4) \) is \( 3 \cos(3b + 4) \), so we have:
\[ \frac{d}{db} \left( (b - 1) \sin(3b + 4) \right) = \sin(3b + 4) + (b - 1)(3 \cos(3b + 4)) \]

4. **Combine the Results**:
Now we can combine the results from both sides:
\[ f(b) = \sin(3b + 4) + (b - 1)(3 \cos(3b + 4)) \]

5. **Simplify the Expression**:
Expanding the right side gives:
\[ f(b) = \sin(3b + 4) + 3(b - 1) \cos(3b + 4) \]

6. **Final Function**:
Replacing \( b \) with \( x \) to express the function in terms of \( x \):
\[ f(x) = \sin(3x + 4) + 3(x - 1) \cos(3x + 4) \]

### Final Answer:
\[ f(x) = \sin(3x + 4) + 3(x - 1) \cos(3x + 4) \]