Let `f(x)={{:(x^(2)-ax+1",", x lt 0),(b(1-x)^(3)",", x ge0):}`. If is a differentiable function, then the ordered pair (a, b) is

Options

A. (1, 3)
B. (3, 1)
C. `(-1, -3)`
D. `(-3, 1)`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is both continuous and differentiable at the point where it changes definition, which is at \( x = 0 \).

### Step 1: Ensure Continuity at \( x = 0 \)

For \( f(x) \) to be continuous at \( x = 0 \), the left-hand limit (LHL) as \( x \) approaches 0 from the left must equal the right-hand limit (RHL) as \( x \) approaches 0 from the right.

- For \( x < 0 \):
\[ f(x) = x^2 - ax + 1 \]
Thus,
\[ \text{LHL} = \lim_{x \to 0^-} f(x) = 0^2 - a(0) + 1 = 1 \]

- For \( x \geq 0 \):
\[ f(x) = b(1 - x)^3 \]
Thus,
\[ \text{RHL} = \lim_{x \to 0^+} f(x) = b(1 - 0)^3 = b \]

Setting the two limits equal for continuity:
\[ 1 = b \quad \Rightarrow \quad b = 1 \]

### Step 2: Ensure Differentiability at \( x = 0 \)

For \( f(x) \) to be differentiable at \( x = 0 \), the left-hand derivative (LHD) must equal the right-hand derivative (RHD).

- The derivative for \( x < 0 \):
\[ f'(x) = 2x - a \]
Thus,
\[ \text{LHD} = \lim_{x \to 0^-} f'(x) = 2(0) - a = -a \]

- The derivative for \( x \geq 0 \):
\[ f'(x) = 3b(1 - x)^2(-1) = -3b(1 - x)^2 \]
Thus,
\[ \text{RHD} = \lim_{x \to 0^+} f'(x) = -3b(1 - 0)^2 = -3b \]

Setting the two derivatives equal for differentiability:
\[ -a = -3b \]
Substituting \( b = 1 \):
\[ -a = -3(1) \quad \Rightarrow \quad a = 3 \]

### Conclusion

The ordered pair \( (a, b) \) is:
\[ (a, b) = (3, 1) \]