# 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(A) 10                      (B) 24(C) 21                      (D) none of these

User
12 Points
5 years ago
According to newton`s interpolation x=0,1,2..... g(x)=1,2,5.....That`s x^2+1Then g(3)=9+1=10Substitute 0,1,2....G function is 1,2,5......For 3 g value is 10
Samyak Jain
333 Points
5 years ago
g(x) g(y) = g(x) + g(y) + g(xy) – 2        …....(1)
Put x = 1 in (1)
So g(1) g(1) = g(1) + g(1) + g(1) – 2
{g(1)}^2 – 3 g(1) + 2 = 0
i.e.  g(1) = 1 or g(1) =  2
Put x = 1 & y = 2 in (1)
g(1) g(2) = g(1) + g(2) + g(2) – 2  i.e.
g(1) g(2) – g(1) – 2 g(2) + 2 = 0  …....(2)
It is given that g(2) = 5
If g(1) = 1, LHS of (2) = 1.5 – 1 – 2.5 + 2 =    4,   RHS of (2) = 0
(2) is not satisfied when g(1) = 1.
But if g(1) = 2, LHS of (2) = 2.5 – 2 – 2.5 + 2 = 0 = RHS of (2)
So g(1) = 2       ….....(3)
Put x = x & y = 1/x in (1)
g(x) g(1/x) = g(x) + g(1/x) + g(1) – 2
So g(x) g(1/x) = g(x) + g(1/x)            [g(1)=2  from (3)]
Hence, g(x) = x^n + 1.  Since g(2) = 2^n +1 = 5, so n = 2
g(x) = x^2 + 1
Therefore, g(3) = 3^2 + 1 = 10