Let `f: R to R` be defined by
`f(x) =x/(x+1), x ne -1`
If `a,b in R` and `ab ne 0, f (a/b) + f(b/a)` and `a+b ne 0` is equal to

Options

A. 1
B. 0
C. `(a-b)/(a+b)`
D. `(b-a)/(b+a)`

Answer

To solve the problem, we need to evaluate the expression \( f\left(\frac{a}{b}\right) + f\left(\frac{b}{a}\right) \) given the function \( f(x) = \frac{x}{x+1} \) where \( x \neq -1 \).

### Step 1: Calculate \( f\left(\frac{a}{b}\right) \)

Using the definition of the function:
\[ f\left(\frac{a}{b}\right) = \frac{\frac{a}{b}}{\frac{a}{b} + 1} \]

To simplify this, we can rewrite the denominator:
\[ \frac{a}{b} + 1 = \frac{a}{b} + \frac{b}{b} = \frac{a + b}{b} \]

Thus, we have:
\[ f\left(\frac{a}{b}\right) = \frac{\frac{a}{b}}{\frac{a+b}{b}} = \frac{a}{a+b} \]

### Step 2: Calculate \( f\left(\frac{b}{a}\right) \)

Now we calculate \( f\left(\frac{b}{a}\right) \):
\[ f\left(\frac{b}{a}\right) = \frac{\frac{b}{a}}{\frac{b}{a} + 1} \]

Again, simplifying the denominator:
\[ \frac{b}{a} + 1 = \frac{b}{a} + \frac{a}{a} = \frac{b + a}{a} \]

Thus, we have:
\[ f\left(\frac{b}{a}\right) = \frac{\frac{b}{a}}{\frac{b+a}{a}} = \frac{b}{b+a} \]

### Step 3: Combine the results

Now we combine the results from Step 1 and Step 2:
\[ f\left(\frac{a}{b}\right) + f\left(\frac{b}{a}\right) = \frac{a}{a+b} + \frac{b}{b+a} \]

Since \( a+b = b+a \), we can combine the fractions:
\[ f\left(\frac{a}{b}\right) + f\left(\frac{b}{a}\right) = \frac{a + b}{a + b} = 1 \]

### Final Result

Thus, the value of \( f\left(\frac{a}{b}\right) + f\left(\frac{b}{a}\right) \) is:
\[ \boxed{1} \]