Question

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

• A. 9
• B. 10
• C. 25
• D. 20

Hint:

Polynomials are continuous everywhere, so \(\lim_{x\to a} g(x)=g(a)\).

Answer 1

The Correct Option is B

Step 1: Put \(x=1, y=1\).
\[ g(1)g(1)=g(1)+g(1)+g(1)-2 \]
\[ [g(1)]^2 = 3g(1)-2 \]
\[ [g(1)]^2-3g(1)+2=0 \Rightarrow (g(1)-1)(g(1)-2)=0 \]
So:
\[ g(1)=1 \ \text{or} \ 2 \]
Step 2: Put \(y=1\) in the given equation.
\[ g(x)g(1)=g(x)+g(1)+g(x)-2 \]
\[ g(x)g(1)=2g(x)+g(1)-2 \]
Step 3: Case 1: If \(g(1)=1\).
\[ g(x)\cdot 1 = 2g(x)+1-2 \Rightarrow g(x)=2g(x)-1 \Rightarrow g(x)=1 \]
This gives constant polynomial \(g(x)=1\).
But then \(g(2)=1\), which contradicts \(g(2)=5\).
So this case is rejected.
Step 4: Hence \(g(1)=2\).
Now substitute:
\[ g(x)\cdot 2 = 2g(x)+2-2 \Rightarrow 2g(x)=2g(x) \]
This is true for all \(x\). So no contradiction.
Step 5: Put \(x=0, y=0\).
\[ g(0)g(0)=g(0)+g(0)+g(0)-2 \Rightarrow [g(0)]^2=3g(0)-2 \]
So again:
\[ g(0)=1 \ \text{or} \ 2 \]
Step 6: Put \(y=0\).
\[ g(x)g(0)=g(x)+g(0)+g(0)-2 \Rightarrow g(x)g(0)=g(x)+2g(0)-2 \]
If \(g(0)=1\):
\[ g(x)=g(x)+0 \]
True for all \(x\).
If \(g(0)=2\):
\[ 2g(x)=g(x)+2 \Rightarrow g(x)=2 \]
Then \(g(2)=2\) contradicts \(5\).
So:
\[ g(0)=1 \]
Step 7: Guess polynomial form and use condition \(g(2)=5\).
A known polynomial solution to this functional equation is:
\[ g(x)=x^2+1 \]
Check:
\[ g(2)=2^2+1=5 \]
Matches given value.
Step 8: Find \(\lim_{x\to 3} g(x)\).
Since polynomial is continuous:
\[ \lim_{x\to 3} g(x)=g(3)=3^2+1=10 \]
Final Answer:
\[ \boxed{10} \]