The smallest positive integral value of a, such that the function `f(x)=x^(4)-4ax^(2)+10` has more two local extrema, is

Options

A. 1
B. 2
C. 4
D. 16

Answer

To determine the smallest positive integral value of \( a \) such that the function \( f(x) = x^4 - 4ax^2 + 10 \) has more than two local extrema, we will follow these steps:

### Step 1: Find the first derivative of the function
We start by differentiating the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^4 - 4ax^2 + 10) = 4x^3 - 8ax \]

### Step 2: Set the first derivative to zero
To find the critical points (where local extrema can occur), we set the first derivative equal to zero:
\[ 4x^3 - 8ax = 0 \]

### Step 3: Factor the equation
We can factor out \( 4x \):
\[ 4x(x^2 - 2a) = 0 \]
This gives us two cases:
1. \( 4x = 0 \) which implies \( x = 0 \)
2. \( x^2 - 2a = 0 \) which implies \( x^2 = 2a \)

### Step 4: Solve for critical points
From \( x^2 = 2a \), we find:
\[ x = \sqrt{2a} \quad \text{and} \quad x = -\sqrt{2a} \]
Thus, the critical points are \( x = 0, \sqrt{2a}, -\sqrt{2a} \).

### Step 5: Count the number of local extrema
The function has three critical points: \( x = 0 \), \( x = \sqrt{2a} \), and \( x = -\sqrt{2a} \). For the function to have more than two local extrema, we need \( \sqrt{2a} \) to be a real number, which requires:
\[ 2a \geq 0 \quad \Rightarrow \quad a \geq 0 \]
Since we are looking for the smallest positive integral value of \( a \), we check for \( a = 1 \):
\[ \sqrt{2 \cdot 1} = \sqrt{2} \quad \text{(which is real)} \]
Thus, \( a = 1 \) gives us three critical points.

### Step 6: Conclusion
The smallest positive integral value of \( a \) such that the function \( f(x) \) has more than two local extrema is:
\[ \boxed{1} \]