Let f be a one-one function such that `f(x).f(y) + 2 = f(x) + f(y) + f(xy), AA x, y in R - {0} and f(0) = 1, f'(1) = 2`. Prove that `3(int f(x) dx) - x(f(x) + 2)` is constant.

Answer

To prove that \(3 \int f(x) \, dx - x(f(x) + 2)\) is constant, we start from the given functional equation:

\[ f(x)f(y) + 2 = f(x) + f(y) + f(xy) \]

for all \(x, y \in \mathbb{R} - \{0\}\), along with the conditions \(f(0) = 1\) and \(f'(1) = 2\).

### Step 1: Differentiate the functional equation

We differentiate both sides of the equation with respect to \(x\), treating \(y\) as a constant:

\[ \frac{d}{dx}(f(x)f(y) + 2) = \frac{d}{dx}(f(x) + f(y) + f(xy)) \]

Using the product rule on the left side, we get:

\[ f'(x)f(y) = f'(x) + yf'(xy) \]

### Step 2: Substitute \(x = 1\)

Now, we substitute \(x = 1\) into the differentiated equation:

\[ f'(1)f(y) = f'(1) + yf'(y) \]

Given that \(f'(1) = 2\), we have:

\[ 2f(y) = 2 + yf'(y) \]

### Step 3: Rearranging the equation

Rearranging gives us:

\[ 2f(y) - yf'(y) = 2 \]

### Step 4: Integrate with respect to \(y\)

Now, we integrate both sides with respect to \(y\):

\[ \int (2f(y) - yf'(y)) \, dy = \int 2 \, dy \]

This simplifies to:

\[ 2 \int f(y) \, dy - \int yf'(y) \, dy = 2y + C \]

where \(C\) is a constant of integration.

### Step 5: Integrate \(yf'(y)\) using integration by parts

Using integration by parts on \(\int yf'(y) \, dy\):

Let \(u = y\) and \(dv = f'(y) \, dy\), then \(du = dy\) and \(v = f(y)\):

\[ \int yf'(y) \, dy = yf(y) - \int f(y) \, dy \]

Substituting this back into our equation gives:

\[ 2 \int f(y) \, dy - \left( yf(y) - \int f(y) \, dy \right) = 2y + C \]

### Step 6: Simplifying the equation

Combining terms results in:

\[ 3 \int f(y) \, dy - yf(y) = 2y + C \]

### Step 7: Replace \(y\) with \(x\)

Finally, we replace \(y\) with \(x\):

\[ 3 \int f(x) \, dx - x f(x) = 2x + C \]

Rearranging gives us:

\[ 3 \int f(x) \, dx - x(f(x) + 2) = C \]

This shows that \(3 \int f(x) \, dx - x(f(x) + 2)\) is constant.

### Conclusion

Thus, we have proved that \(3 \int f(x) \, dx - x(f(x) + 2)\) is constant.