If `f(x)={{:(e^(2x^(3)+x),x gt 0),(ax+b, x le 0):}` is differentiable at x = 0, then

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

\[ f(x) = \begin{cases} e^{2x^3 + x} & \text{if } x > 0 \\ ax + b & \text{if } x \leq 0 \end{cases} \]

is differentiable at \( x = 0 \), we need to ensure that the function is continuous at \( x = 0 \) and that the derivatives from both sides at \( x = 0 \) are equal.

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

For \( f(x) \) to be continuous at \( x = 0 \), we must have:

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

Calculating the left-hand limit:

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

Calculating the right-hand limit:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{2x^3 + x} = e^{0} = 1 \]

Setting these equal for continuity:

\[ b = 1 \]

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

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

Calculating the left-hand derivative:

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

Calculating the right-hand derivative using the chain rule:

\[ f'(x) = \frac{d}{dx} e^{2x^3 + x} = e^{2x^3 + x} \cdot (6x^2 + 1) \]

Evaluating the right-hand derivative at \( x = 0 \):

\[ \lim_{x \to 0^+} f'(x) = e^{0} \cdot (6 \cdot 0^2 + 1) = 1 \]

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

\[ a = 1 \]

### Conclusion

Thus, we find:

\[ a = 1, \quad b = 1 \]

### Final Answer

The values of \( a \) and \( b \) are:

\[ \boxed{1} \text{ and } \boxed{1} \]