If g(x) is a polynomial satisfying g(x)g (y)=g(x) + g(y) + g(xy) - 2 for all real x, y and g (2) = 5, then g (3) is equal to

Options

A. 10
B. 24
C. 21
D. None of these

Answer

To solve the problem, we need to find the polynomial \( g(x) \) that satisfies the equation:

\[ g(x)g(y) = g(x) + g(y) + g(xy) - 2 \]

for all real \( x \) and \( y \), and we know that \( g(2) = 5 \). We will also find \( g(3) \).

### Step 1: Analyze the functional equation

Let's substitute \( y = 1 \) into the functional equation:

\[ g(x)g(1) = g(x) + g(1) + g(x \cdot 1) - 2 \]

This simplifies to:

\[ g(x)g(1) = g(x) + g(1) + g(x) - 2 \]

Rearranging gives:

\[ g(x)g(1) = 2g(x) + g(1) - 2 \]

### Step 2: Set \( g(1) = c \)

Let \( g(1) = c \):

\[ g(x)c = 2g(x) + c - 2 \]

Rearranging this, we have:

\[ g(x)(c - 2) = c - 2 \]

### Step 3: Determine the value of \( c \)

If \( c - 2 \neq 0 \), we can divide both sides by \( c - 2 \):

\[ g(x) = 1 \]

However, this contradicts \( g(2) = 5 \). Thus, we must have \( c - 2 = 0 \), which means:

\[ c = 2 \quad \Rightarrow \quad g(1) = 2 \]

### Step 4: Substitute \( x = 2 \) and \( y = 2 \)

Now, substituting \( x = 2 \) and \( y = 2 \) into the original equation gives:

\[ g(2)g(2) = g(2) + g(2) + g(4) - 2 \]

Substituting \( g(2) = 5 \):

\[ 5 \cdot 5 = 5 + 5 + g(4) - 2 \]

This simplifies to:

\[ 25 = 10 + g(4) - 2 \]

Thus:

\[ 25 = 8 + g(4) \quad \Rightarrow \quad g(4) = 17 \]

### Step 5: Assume a polynomial form

Assuming \( g(x) \) is a quadratic polynomial of the form:

\[ g(x) = ax^2 + bx + c \]

Using the known values \( g(1) = 2 \) and \( g(2) = 5 \):

1. \( g(1) = a(1)^2 + b(1) + c = a + b + c = 2 \)
2. \( g(2) = a(2)^2 + b(2) + c = 4a + 2b + c = 5 \)

### Step 6: Solve the system of equations

We have the equations:

1. \( a + b + c = 2 \)
2. \( 4a + 2b + c = 5 \)

From the first equation, we can express \( c \):

\[ c = 2 - a - b \]

Substituting into the second equation:

\[ 4a + 2b + (2 - a - b) = 5 \]

This simplifies to:

\[ 3a + b + 2 = 5 \quad \Rightarrow \quad 3a + b = 3 \]

### Step 7: Solve for \( a \) and \( b \)

Now we have:

1. \( a + b + c = 2 \)
2. \( 3a + b = 3 \)

Substituting \( b = 3 - 3a \) into the first equation:

\[ a + (3 - 3a) + c = 2 \]

This simplifies to:

\[ -2a + 3 + c = 2 \quad \Rightarrow \quad c = 2a - 1 \]

### Step 8: Use \( g(2) = 5 \)

Substituting \( c \) back into \( g(2) = 5 \):

\[ 4a + 2(3 - 3a) + (2a - 1) = 5 \]

This simplifies to:

\[ 4a + 6 - 6a + 2a - 1 = 5 \quad \Rightarrow \quad 0 = 0 \]

This means we can choose any \( a \). Let's try \( a = 1 \):

Then \( b = 0 \) and \( c = 1 \). Thus:

\[ g(x) = x^2 + 1 \]

### Step 9: Calculate \( g(3) \)

Now, we can find \( g(3) \):

\[ g(3) = 3^2 + 1 = 9 + 1 = 10 \]

### Final Answer

Thus, \( g(3) = 10 \).