Let `f(x)={{:((e^(alphax)-e^x)/x^2, x ne 0),(3//2,x=0):}` The value of `alpha` so that f is a continuous function is

Options

A. 1
B. 0
C. 4
D. 2

Answer

To find the value of \( \alpha \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 equals \( f(0) \).

### Step 1: Define the Function
The function is given as:
\[ f(x) = \begin{cases} \frac{e^{\alpha x} - e^x}{x^2} & \text{if } x \neq 0 \\ \frac{3}{2} & \text{if } x = 0 \end{cases} \]

### Step 2: Find \( \lim_{x \to 0} f(x) \)
We need to calculate:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{e^{\alpha x} - e^x}{x^2} \]
Since substituting \( x = 0 \) directly gives us \( \frac{0}{0} \), we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
- The derivative of the numerator \( e^{\alpha x} - e^x \) is:
\[ \alpha e^{\alpha x} - e^x \]
- The derivative of the denominator \( x^2 \) is:
\[ 2x \]

So we have:
\[ \lim_{x \to 0} \frac{e^{\alpha x} - e^x}{x^2} = \lim_{x \to 0} \frac{\alpha e^{\alpha x} - e^x}{2x} \]

### Step 4: Apply L'Hôpital's Rule Again
Substituting \( x = 0 \) again gives \( \frac{0}{0} \), so we apply L'Hôpital's Rule again:
- The derivative of the numerator \( \alpha e^{\alpha x} - e^x \) is:
\[ \alpha^2 e^{\alpha x} - e^x \]
- The derivative of the denominator \( 2x \) is:
\[ 2 \]

Now we have:
\[ \lim_{x \to 0} \frac{\alpha^2 e^{\alpha x} - e^x}{2} \]

### Step 5: Evaluate the Limit
Substituting \( x = 0 \) gives:
\[ \frac{\alpha^2 e^0 - e^0}{2} = \frac{\alpha^2 - 1}{2} \]

### Step 6: Set the Limit Equal to \( f(0) \)
For continuity at \( x = 0 \), we need:
\[ \frac{\alpha^2 - 1}{2} = \frac{3}{2} \]

### Step 7: Solve for \( \alpha \)
Multiplying both sides by 2:
\[ \alpha^2 - 1 = 3 \]
\[ \alpha^2 = 4 \]
Taking the square root:
\[ \alpha = 2 \quad \text{or} \quad \alpha = -2 \]

### Conclusion
The values of \( \alpha \) that make \( f(x) \) continuous at \( x = 0 \) are \( \alpha = 2 \) and \( \alpha = -2 \).