Minimum integral value of k for which the equation `e^(x)=kx^(2)` has exactly three real distinct solution,

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To find the minimum integral value of \( k \) for which the equation \( e^x = kx^2 \) has exactly three distinct real solutions, we can follow these steps:

### Step 1: Define the Functions
We start by rewriting the equation as a function:
\[ f(x) = e^x - kx^2 \]
We want to find the values of \( k \) such that \( f(x) = 0 \) has exactly three distinct solutions.

### Step 2: Find the Derivative
To analyze the behavior of the function, we need to find its derivative:
\[ f'(x) = e^x - 2kx \]

### Step 3: Set the Derivative to Zero
To find the critical points, we set the derivative equal to zero:
\[ e^x - 2kx = 0 \]
This implies:
\[ e^x = 2kx \]

### Step 4: Analyze the Critical Points
From the equation \( e^x = 2kx \), we can see that the left side \( e^x \) is always positive, while the right side \( 2kx \) depends on \( k \) and \( x \).

### Step 5: Find the Condition for Three Distinct Solutions
For the function \( f(x) \) to have exactly three distinct real solutions, the derivative \( f'(x) \) must change sign at three points. This occurs when there are two critical points where \( f'(x) = 0 \) and one of them is a local maximum and the other is a local minimum.

### Step 6: Analyze the Behavior of \( f'(x) \)
The equation \( e^x = 2kx \) has solutions that depend on the value of \( k \). We need to find conditions on \( k \) such that there are exactly two solutions to this equation.

### Step 7: Use the Second Derivative Test
To ensure that we have a local maximum and a local minimum, we can use the second derivative:
\[ f''(x) = e^x - 2k \]
Setting \( f''(x) = 0 \) gives:
\[ e^x = 2k \]
This means that for a given \( k \), we can find the corresponding \( x \) values where the concavity changes.

### Step 8: Determine the Minimum Value of \( k \)
To have three distinct solutions, the minimum value of \( k \) must be such that the function \( f(x) \) touches the x-axis at one point and crosses it at two others. This occurs when the local maximum is equal to zero.

### Step 9: Calculate the Minimum Integral Value of \( k \)
To find the minimum integral value of \( k \), we can evaluate:
\[ k \geq \frac{e^2}{4} \]
Calculating \( \frac{e^2}{4} \):
\[ e \approx 2.718 \implies e^2 \approx 7.389 \implies \frac{e^2}{4} \approx 1.847 \]
Thus, the minimum integral value of \( k \) is:
\[ k = 2 \]

### Conclusion
The minimum integral value of \( k \) for which the equation \( e^x = kx^2 \) has exactly three distinct real solutions is:
\[ \boxed{2} \]