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 find the ordered pair \((a, b)\) such that the function \(f(x)\) is differentiable everywhere, particularly at \(x = 0\). The function is defined as follows:

\[ f(x) = \begin{cases} x^2 - ax + 1 & \text{if } x < 0 \\ b(1 - x)^3 & \text{if } x \geq 0 \end{cases} \]

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

For \(f(x)\) to be differentiable at \(x = 0\), it must first be continuous at that point. This means that the left-hand limit as \(x\) approaches 0 must equal the right-hand limit at \(x = 0\):

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

Calculating the left-hand limit:

\[ \lim_{x \to 0^-} f(x) = f(0) = 0^2 - a(0) + 1 = 1 \]

Calculating the right-hand limit:

\[ \lim_{x \to 0^+} f(x) = f(0) = b(1 - 0)^3 = b \]

Setting these equal for continuity:

\[ 1 = b \quad \text{(1)} \]

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

Next, we need to ensure that the derivatives from the left and right at \(x = 0\) are equal.

Calculating the derivative for \(x < 0\):

\[ f'(x) = 2x - a \]
Thus,
\[ f'(0^-) = 2(0) - a = -a \]

Calculating the derivative for \(x \geq 0\):

\[ f'(x) = b \cdot 3(1 - x)^2 \cdot (-1) = -3b(1 - x)^2 \]
Thus,
\[ f'(0^+) = -3b(1 - 0)^2 = -3b \]

Setting these equal for differentiability:

\[ -a = -3b \quad \text{(2)} \]

### Step 3: Substitute \(b\) from Equation (1) into Equation (2)

From Equation (1), we have \(b = 1\). Substituting this into Equation (2):

\[ -a = -3(1) \]
\[ -a = -3 \implies a = 3 \]

### Conclusion

Thus, the ordered pair \((a, b)\) is:

\[ (a, b) = (3, 1) \]

### Final Answer

The ordered pair \((a, b)\) is \((3, 1)\).

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