The following functions are defined for any two distinct, non zero integers a and b
`f_(1)(a, b)=|a|xxb^(2)`
`f_(2)(a, b)=a^(2)xx|b|`
`f_(3)(a, b)=((a^(2)+b^(2)))/(2)`
`f_(4)(a, b)=((a)/(2)+(b)/(2))`
Which of the following is equal to `(a+b)^(2)`?

Options

A. `f_(3)(a,b)div f_(4)(a, b)`
B. `2xxf_(3)(a, b)xxf_(4)(axxb)`
C. `4xxf_(3)(a, b)xxf_(4)(a, b)`
D. none of these

Answer

To solve the problem, we need to determine which of the given functions is equal to \((a + b)^2\). We will analyze each function step by step.

### Step 1: Expand \((a + b)^2\)
The first step is to recall the formula for the square of a binomial:
\[ (a + b)^2 = a^2 + 2ab + b^2 \]

### Step 2: Analyze each function
Now, let's analyze each function one by one to see if any of them can be manipulated to match the expression \(a^2 + 2ab + b^2\).

#### Function 1: \(f_1(a, b) = |a| \cdot b^2\)
This function does not contain \(a^2\) or \(2ab\) terms. It only involves \(b^2\). Therefore, it cannot equal \((a + b)^2\).

#### Function 2: \(f_2(a, b) = a^2 \cdot |b|\)
This function only contains \(a^2\) and \(|b|\). It also does not contain the term \(2ab\). Thus, it cannot equal \((a + b)^2\).

#### Function 3: \(f_3(a, b) = \frac{a^2 + b^2}{2}\)
This function gives us the average of \(a^2\) and \(b^2\). It lacks the \(2ab\) term and cannot equal \((a + b)^2\).

#### Function 4: \(f_4(a, b) = \frac{a}{2} + \frac{b}{2}\)
This function simplifies to \(\frac{a + b}{2}\). Squaring this gives:
\[ \left(\frac{a + b}{2}\right)^2 = \frac{(a + b)^2}{4} = \frac{a^2 + 2ab + b^2}{4} \]
This is not equal to \((a + b)^2\).

### Conclusion
None of the functions \(f_1\), \(f_2\), \(f_3\), or \(f_4\) are equal to \((a + b)^2\). Therefore, the answer is:
\[ \text{None of these} \]