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 find 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, we set the first derivative equal to zero:
\[ 4x^3 - 8ax = 0 \]
Factoring out \( 4x \):
\[ 4x(x^2 - 2a) = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad x^2 - 2a = 0 \]

### Step 3: Solve for critical points
From \( x^2 - 2a = 0 \):
\[ x^2 = 2a \quad \Rightarrow \quad x = \pm \sqrt{2a} \]
Thus, the critical points are:
\[ x = 0, \quad x = \sqrt{2a}, \quad x = -\sqrt{2a} \]

### Step 4: Determine the conditions for more than two local extrema
For the function to have more than two local extrema, we need at least three distinct critical points. The critical points are \( 0, \sqrt{2a}, -\sqrt{2a} \). For these points to be distinct, \( \sqrt{2a} \) must be a real number and not equal to zero:
- \( \sqrt{2a} \) is real if \( 2a > 0 \) (which is always true for positive \( a \)).
- To ensure \( \sqrt{2a} \) is not equal to zero, we need \( a > 0 \).

### Step 5: Find the smallest positive integral value of \( a \)
Since we need \( a \) to be a positive integer, the smallest positive integral value of \( a \) is:
\[ a = 1 \]

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