Step-by-step Analysis

1. Analyze the function f(x) at x = 1

The given function is $f(x) = x^3 - x^2 + (x - 1)sin(x)$. Let's analyze its properties at x = 1.

• Value of f(1): $f(1) = 1^3 - 1^2 + (1 - 1)sin(1) = 1 - 1 + 0 = 0$.

• Differentiability and f'(1): The function f(x) is a sum and product of polynomial and trigonometric functions, which are differentiable everywhere on R. Thus, f(x) is differentiable for all `x

R. Let's find its derivative f'(x)using the product rule for(x - 1)sin(x): $f'(x) = d/dx (x^3 - x^2) + d/dx ((x - 1)sin(x))$ f'(x) = 3x^2 - 2x + [1

sin(x) + (x - 1)cos(x)]` $f'(x) = 3x^2 - 2x + sin(x) + (x - 1)cos(x)$

Now, let's evaluate the derivative at `x = 1`:$f'(1) = 3(1)^2 - 2(1) + sin(1) + (1 - 1)cos(1)$`f'(1) = 3 - 2 + sin(1) + 0 = 1 + sin(1)`
Since `1` radian is approximately `57.3^

, and 0 < 1 <

/2, we have sin(1) > 0. Therefore, f'(1) = 1 + sin(1) > 1, which means f'(1)

0`.

2. Analyze the differentiability of the product function fg(x) at x = 1

The differentiability of fg(x) at x = 1 is determined by the existence of the following limit: `$(fg)'(1) = lim_{x

1} rac{(fg)(x) - (fg)(1)}{x - 1} = lim_{x

1} rac{f(x)g(x) - f(1)g(1)}{x - 1}$`

Since f(1) = 0, the expression simplifies to: `$(fg)'(1) = lim_{x

1} rac{f(x)g(x)}{x - 1}$`

We can rewrite this limit as: `$(fg)'(1) = lim_{x

1} rac{f(x) - f(1)}{x - 1}

g(x)(sincef(1)=0`)

We know that ` lim_{x

1} rac{f(x) - f(1)}{x - 1} = f'(1) = 1 + sin(1), which is a non-zero finite value. Let L_f = lim_{x

1} rac{f(x) - f(1)}{x - 1} = f'(1). Let L_g = lim_{x

1} g(x)`.

Then, `$(fg)'(1) = lim_{x

1} ( rac{f(x) - f(1)}{x - 1})

lim_{x

1} g(x) = f'(1)

( lim_{x

1} g(x))$` This separation of limits is valid if both limits exist. Since$L_f$ exists and is non-zero, the limit for (fg)'(1) exists if and only if ` lim_{x

1} g(x)` exists.

Key finding: fg is differentiable at x=1 if and only if ` lim_{x

1} g(x)` exists.

Now we evaluate each statement based on this finding.

A: If g is continuous at x = 1, then fg is differentiable at x = 1 If g is continuous at x = 1, then by definition, ` lim_{x

1} g(x)exists and is equal tog(1). Since lim_{x

1} g(x)exists, our key finding implies thatfgis differentiable atx = 1`. Therefore, statement A is TRUE.

B: If fg is differentiable at x = 1, then g is continuous at x = 1 If fg is differentiable at x = 1, our key finding implies that ` lim_{x

1} g(x)exists. Forgto be continuous atx = 1, we need lim_{x

1} g(x) = g(1). The differentiability of fg` does not guarantee this equality.

Counterexample: Let g(x) be defined as: g(x) = 2 if `x

1 g(x) = 3ifx = 1Here, lim_{x

1} g(x) = 2, so the limit exists. This means fgis differentiable atx=1. However, g(1) = 3, so lim_{x

1} g(x)

g(1). Thus, gis not continuous atx=1`. Therefore, statement B is FALSE.

C: If g is differentiable at x = 1, then fg is differentiable at x = 1 If g is differentiable at x = 1, it must also be continuous at x = 1. As established in the analysis of statement A, if g is continuous at x = 1, then fg is differentiable at x = 1. Alternatively, since both f and g are differentiable at x = 1, their product fg is also differentiable at x = 1 by the product rule of differentiation. `$(fg)'(1) = f'(1)g(1) + f(1)g'(1) = (1 + sin(1))g(1) + 0

g'(1) = (1 + sin(1))g(1)`. The derivative exists. Therefore, statement C is TRUE.

D: If fg is differentiable at x = 1, then g is differentiable at x = 1 If fg is differentiable at x = 1, we only know that ` lim_{x

1} g(x)exists. This does not imply thatgis differentiable atx = 1`.

Counterexample: Let g(x) = |x - 1|. This function is continuous at x = 1 (since ` lim_{x

1} |x-1| = 0 = g(1)), but it is not differentiable at x = 1. Since g(x) = |x - 1|is continuous atx = 1, from statement A we know that fgis differentiable atx = 1. So we have a case where fgis differentiable atx=1butgis not. Let's verify the derivative offgfor this case:$(fg)'(1) = f'(1)

( lim_{x

1} g(x)) = (1 + sin(1))

( lim_{x

1} |x-1|) = (1 + sin(1))

0 = 0`. The derivative exists. Therefore, statement D is FALSE.

Conclusion

The correct statements are A and C.