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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
f (x) f (y) = f (2xy + 3) + 3f (x + y) - 3f (y) + 6y
Putting x = y = 0,
f (0) f (0) = f (2.0.0 + 3) + 3f (0 + 0) - 3f (0) + 6.0
f (0)2= f (3) + 3f (0) - 3f (0) + 0
f (3) = f (0)2
Putting y = 0,
f (x) f (0) = f (2.x.0 + 3) + 3f (x + 0) - 3f (0) + 6.0
f (x) f (0) = f (3) + 3f (x) - 3f (0)
f (x) f (0) = f (0)2 + 3f (x) - 3f (0)
f (x) f (0) - f (0)2 = 3f (x) - 3f (0)
f (0) (f (x)- f (0)) = 3 (f (x)- f (0))
Either f(x) = f(0) or f(0) = 3
Since all functions can’t have the same value, f(0)=3
Putting x = 0, y = 3,
f (0) f (3) = f (2.0.3 + 3) + 3f (0 + 3) - 3f (3) + 6.3
f(0) f(3) = f(3) + 3f(3) -3f(3) +18 = f(3) + 18
3 f(3) = f(3) + 18
Therefore, f(3) =9
f(3) = 9.
Putting x = 0, y = 8
f (0) f (8) = f (2.0.8 + 3) + 3f (0+ 8) - 3f (8) + 6.8
3.f(8) = f(3) + f(8) – f(8) + 48
3. f(8) = 9 + 48
f(8) = 19