Determine whether the following functions are even or odd.
`{:((i)f(x)=log(x+sqrt(1+x^(2))),(ii)f(x)=x((a^(x)+1)/(a^(x)-1))),((iii)f(x)=sinx+cosx,(iv)f(x)=x^(2)-abs(x)),((v)f(x)=log((1-x)/(1+x)),(vi)f(x)={(sgn x)^(sgnx)}^(n)," n is an odd integer"):}`
`{:((vii) f(x)=sgn(x)+x^(2),""),((viii)f(x+y)+f(x-y)=2f(x)*f(y)," where " f(0) ne 0 and x","y ne R.,""):}`

Answer

To determine whether the given functions are even, odd, or neither, we will apply the definitions of even and odd functions:

1. A function \( f(x) \) is **even** if \( f(-x) = f(x) \).
2. A function \( f(x) \) is **odd** if \( f(-x) = -f(x) \).

Let's analyze each function step by step:

### (i) \( f(x) = \log(x + \sqrt{1 + x^2}) \)

**Step 1:** Calculate \( f(-x) \):
\[ f(-x) = \log(-x + \sqrt{1 + (-x)^2}) = \log(-x + \sqrt{1 + x^2}) \]

**Step 2:** Simplify \( f(-x) \):
Using the property of logarithms:
\[ f(-x) = \log\left(\frac{1}{x + \sqrt{1 + x^2}}\right) = -\log(x + \sqrt{1 + x^2}) = -f(x) \]

**Conclusion:** Since \( f(-x) = -f(x) \), the function is **odd**.

### (ii) \( f(x) = x \cdot \frac{a^x + 1}{a^x - 1} \)

**Step 1:** Calculate \( f(-x) \):
\[ f(-x) = -x \cdot \frac{a^{-x} + 1}{a^{-x} - 1} = -x \cdot \frac{\frac{1}{a^x} + 1}{\frac{1}{a^x} - 1} = -x \cdot \frac{1 + a^x}{1 - a^x} \]

**Step 2:** Simplify \( f(-x) \):
\[ f(-x) = -x \cdot \frac{1 + a^x}{1 - a^x} = -f(x) \]

**Conclusion:** Since \( f(-x) = -f(x) \), the function is **odd**.

### (iii) \( f(x) = \sin x + \cos x \)

**Step 1:** Calculate \( f(-x) \):
\[ f(-x) = \sin(-x) + \cos(-x) = -\sin x + \cos x \]

**Conclusion:** Since \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), the function is **neither even nor odd**.

### (iv) \( f(x) = x^2 - |x| \)

**Step 1:** Calculate \( f(-x) \):
\[ f(-x) = (-x)^2 - |-x| = x^2 - |x| = f(x) \]

**Conclusion:** Since \( f(-x) = f(x) \), the function is **even**.

### (v) \( f(x) = \log\left(\frac{1-x}{1+x}\right) \)

**Step 1:** Calculate \( f(-x) \):
\[ f(-x) = \log\left(\frac{1+x}{1-x}\right) \]

**Step 2:** Simplify \( f(-x) \):
\[ f(-x) = -\log\left(\frac{1-x}{1+x}\right) = -f(x) \]

**Conclusion:** Since \( f(-x) = -f(x) \), the function is **odd**.

### (vi) \( f(x) = (\text{sgn } x)^{\text{sgn } x}^n \) where \( n \) is an odd integer

**Step 1:** Calculate \( f(-x) \):
\[ f(-x) = (\text{sgn } (-x))^{\text{sgn } (-x)}^n = (-\text{sgn } x)^n \]

**Step 2:** Since \( n \) is odd:
\[ f(-x) = -(\text{sgn } x)^{\text{sgn } x}^n = -f(x) \]

**Conclusion:** Since \( f(-x) = -f(x) \), the function is **odd**.

### (vii) \( f(x) = \text{sgn}(x) + x^2 \)

**Step 1:** Calculate \( f(-x) \):
\[ f(-x) = \text{sgn}(-x) + (-x)^2 = -\text{sgn}(x) + x^2 \]

**Conclusion:** Since \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), the function is **neither even nor odd**.

### (viii) \( f(x+y) + f(x-y) = 2f(x)f(y) \)

**Step 1:** Substitute \( -x \):
\[ f(-x+y) + f(-x-y) = 2f(-x)f(y) \]

**Step 2:** Since the equation remains unchanged:
\[ f(-x+y) + f(-x-y) = 2f(-x)f(y) \]

**Conclusion:** The function is **even**.

### Summary of Results:
1. (i) Odd
2. (ii) Odd
3. (iii) Neither
4. (iv) Even
5. (v) Odd
6. (vi) Odd
7. (vii) Neither
8. (viii) Even