If the function f(x) is given by `f(x)={{:(,2^(1//(x-1)),x lt 1),(,ax^(2)+bx,x ge 1):}` is everywhere differentiable, then

Options

A. a=0, b=1
B. a-0, b=0
C. a=1, b=0
D. none of these

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable everywhere, particularly at the point \( x = 1 \). The function is defined as follows:

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

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

For \( f(x) \) to be continuous at \( x = 1 \), the left-hand limit as \( x \) approaches 1 must equal the right-hand limit and the value of the function at that point.

1. **Calculate \( f(1) \)**:
\[ f(1) = a(1)^2 + b(1) = a + b \]

2. **Calculate the left-hand limit**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 2^{\frac{1}{x-1}} \]
As \( x \to 1^- \), \( \frac{1}{x-1} \to -\infty \), hence:
\[ \lim_{x \to 1^-} f(x) = 2^{-\infty} = 0 \]

3. **Calculate the right-hand limit**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (ax^2 + bx) = a(1)^2 + b(1) = a + b \]

Setting the left-hand limit equal to the right-hand limit gives:
\[ 0 = a + b \quad \text{(Equation 1)} \]

### Step 2: Ensure Differentiability at \( x = 1 \)

For \( f(x) \) to be differentiable at \( x = 1 \), the left-hand derivative must equal the right-hand derivative.

1. **Calculate the left-hand derivative**:
\[ f'(x) = \frac{d}{dx} \left( 2^{\frac{1}{x-1}} \right) \]
Using the chain rule:
\[ f'(x) = 2^{\frac{1}{x-1}} \cdot \ln(2) \cdot \left(-\frac{1}{(x-1)^2}\right) \]
Thus, the left-hand derivative at \( x = 1 \) is:
\[ \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} 2^{\frac{1}{x-1}} \cdot \frac{-\ln(2)}{(x-1)^2} \]
As \( x \to 1^- \), this approaches:
\[ -\infty \quad \text{(since } 2^{\frac{1}{x-1}} \to 0 \text{ and } (x-1)^2 \to 0\text{)} \]

2. **Calculate the right-hand derivative**:
\[ f'(x) = \frac{d}{dx} (ax^2 + bx) = 2ax + b \]
Thus, the right-hand derivative at \( x = 1 \) is:
\[ f'(1) = 2a(1) + b = 2a + b \]

Setting the left-hand derivative equal to the right-hand derivative gives:
\[ -\infty = 2a + b \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations

From Equation 1:
\[ a + b = 0 \implies b = -a \]

Substituting \( b = -a \) into Equation 2:
\[ 2a - a = 0 \implies a = 0 \]
Then substituting \( a = 0 \) back into Equation 1:
\[ 0 + b = 0 \implies b = 0 \]

### Conclusion

The values of \( a \) and \( b \) that make the function \( f(x) \) everywhere differentiable are:
\[ a = 0 \quad \text{and} \quad b = 0 \]