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Let f be an injective function such that f(x) f(y) + 2 = f(x) + f(y) + f(xy)∀ x, y ∈R. If f(4) = 65 and f(0) ≠ 2, then show that f(x) – 1= x3 x ∈ R.

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Given that f(x) f(y) + 2 = f(x) + f(y) + f(xy) ....(i)

Putting x = y = 0 in equation (i), we get f(0) f(0) + 2 = f(0) + f(0) + f(0)

or (f(0))2 + 2 = 3f(0) or (f(0) – 2) (f(0) – 1) = 0 or f(0)

 ( f(0) 2) ....(ii)

Again putting x = y = 1 in equation (i) and repeating the above steps, we get (f(1) – 2) (f(1) – 1) = 0

But f(1)  1 as f(x) is injective. 

 f(1) = 2 ....(iii)

Now putting y = 1/x in equation (i), we get

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