Let the function `f: R to R` be defined by`f(x)=x^(3)-x^(2)+(x-1) sin x and "let" g: R to R` be an arbitrary function Let `f g: R to R` be the function defined by `(fg)(x)=f(x)g(x)`. Then which of the folloiwng statements is/are TRUE ?

Options

A. If g is continuous at x=1, then f g is differentiable at x=1
B. If g is differentiable at x=1, then f g is continuous at x=1
C. if g is differentiable at x=1, then f g is differentiable at x=1
D. If f g is diffrentiable at x=1, then g is diffrentiable at x=1

Answer

To solve the problem, we need to analyze the function \( f(x) = x^3 - x^2 + (x-1) \sin x \) and the product of functions \( fg(x) = f(x)g(x) \). We will determine which statements about the differentiability of \( fg \) and the continuity of \( g \) are true.

### Step-by-Step Solution:

1. **Evaluate \( f(1) \)**:
\[ f(1) = 1^3 - 1^2 + (1-1) \sin(1) = 1 - 1 + 0 = 0 \]
Thus, \( f(1) = 0 \).

2. **Evaluate \( fg(1) \)**:
\[ fg(1) = f(1)g(1) = 0 \cdot g(1) = 0 \]
Therefore, \( fg(1) = 0 \).

3. **Find the right-hand derivative \( fg'(1^+) \)**:
Using the definition of the derivative:
\[ fg'(1^+) = \lim_{h \to 0} \frac{fg(1+h) - fg(1)}{h} = \lim_{h \to 0} \frac{f(1+h)g(1+h) - 0}{h} \]
We can express \( f(1+h) \):
\[ f(1+h) = (1+h)^3 - (1+h)^2 + (1+h-1)\sin(1+h) \]
Expanding this gives:
\[ = (1 + 3h + 3h^2 + h^3) - (1 + 2h + h^2) + h \sin(1+h) \]
Simplifying:
\[ = 3h + 2h^2 + h^3 + h \sin(1+h) \]
Thus,
\[ fg'(1^+) = \lim_{h \to 0} \frac{(3h + 2h^2 + h^3 + h \sin(1+h))g(1+h)}{h} \]
Cancelling \( h \):
\[ = \lim_{h \to 0} (3 + 2h + h^2 + \sin(1+h))g(1+h) \]
As \( h \to 0 \), this becomes:
\[ = 3g(1) + \sin(1)g(1) \]

4. **Find the left-hand derivative \( fg'(1^-) \)**:
Similarly,
\[ fg'(1^-) = \lim_{h \to 0} \frac{fg(1-h) - fg(1)}{-h} = \lim_{h \to 0} \frac{f(1-h)g(1-h)}{-h} \]
Expanding \( f(1-h) \):
\[ f(1-h) = (1-h)^3 - (1-h)^2 + (1-h-1)\sin(1-h) \]
Simplifying this gives:
\[ = -3h + 2h^2 - h^3 - h \sin(1-h) \]
Thus,
\[ fg'(1^-) = \lim_{h \to 0} \frac{(-3h + 2h^2 - h^3 - h \sin(1-h))g(1-h)}{-h} \]
Cancelling \( -h \):
\[ = \lim_{h \to 0} (3 - 2h + h^2 + \sin(1-h))g(1-h) \]
As \( h \to 0 \), this becomes:
\[ = 3g(1) - \sin(1)g(1) \]

5. **Set the derivatives equal for differentiability**:
For \( fg \) to be differentiable at \( x = 1 \):
\[ 3g(1) + \sin(1)g(1) = 3g(1) - \sin(1)g(1) \]
This simplifies to:
\[ 2\sin(1)g(1) = 0 \]
Hence, either \( g(1) = 0 \) or \( \sin(1) = 0 \) (which is not true).

6. **Conclusion about statements**:
- **Statement A**: If \( g \) is continuous at \( x = 1 \), then \( fg \) is differentiable at \( x = 1 \). **True**.
- **Statement B**: If \( fg \) is differentiable, then \( g \) is continuous. **False**.
- **Statement C**: If \( g \) is differentiable, then \( fg \) is differentiable. **True**.
- **Statement D**: If \( fg \) is differentiable, then \( g \) is differentiable. **False**.

### Final Answer:
The true statements are A and C.