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
(B) 24
(C) 21
(D) 12
(A) 10
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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
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