If `f(x)={{:( e^(x^(2)+x), x gt0),( ax+b, xle0):}`is differentiable at `x=0` , then (a) `a=1,b=-1` (b) `a=-1,b=1` (c) `a=1,b=1` (d) `a=-1,b=-1`

Options

A. a=1,b=-1
B. a=-1,b=1
C. a=1,b=1
D. a=-1,b=-1

Answer

To determine the values of \( a \) and \( b \) such that the function

\[ f(x) = \begin{cases} e^{x^2 + x} & \text{if } x > 0 \\ ax + b & \text{if } x \leq 0 \end{cases} \]

is differentiable at \( x = 0 \), we need to ensure that the function is continuous at \( x = 0 \) and that the left-hand derivative equals the right-hand derivative at that point.

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

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

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

Calculating the left-hand limit:

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

Calculating the right-hand limit:

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

Setting the two limits equal for continuity:

\[ b = 1 \]

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

Next, we need to ensure that the left-hand derivative equals the right-hand derivative at \( x = 0 \).

Calculating the left-hand derivative:

\[ f'(x) = a \quad \text{for } x < 0 \]

Thus,

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

Calculating the right-hand derivative:

For \( x > 0 \), we need to differentiate \( f(x) = e^{x^2 + x} \):

Using the chain rule:

\[ f'(x) = e^{x^2 + x} \cdot (2x + 1) \]

Evaluating this at \( x = 0 \):

\[ \lim_{x \to 0^+} f'(x) = e^{0^2 + 0} \cdot (2(0) + 1) = 1 \cdot 1 = 1 \]

Setting the left-hand derivative equal to the right-hand derivative:

\[ a = 1 \]

### Conclusion

We have found that:

\[ a = 1 \quad \text{and} \quad b = 1 \]

Thus, the values of \( a \) and \( b \) that make the function differentiable at \( x = 0 \) are:

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

The correct option is (c) \( a = 1, b = 1 \).