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pintukr
If f(x+y) = f(x) + f(y) + f(x)f(y) for all real values of x and y except 0; and f(1) =1; then find the value of f(12).

A. 2047
B. 4095
C. 1884
D. 3768
E. 9420

(adapted from Varsity Tutors)

f(x+y) = f(x) + f(y) + f(x)f(y)
f(2) = f(1+1) = f(1) + f(1) + f(1)×f(1) = 3

f(4) = f(2+2) = f(2) + f(2) + f(2)×f(2) = 15

f(8) = f(4+4) = f(4) + f(4) + f(4)×f(4) = 255

f(12) = f(4+8) = f(4) + f(8) + f(4)×f(8) = 4095
Answer - 4095


Alternative:

f(x+y) = f(x) + f(y) + f(x)f(y)
= 1 + f(x) + f(y) + f(x)f(y) - 1
f(x+y) = (1 + f(x))(1 + f(y)) - 1

f(2) = f(1) + f(1) = (1 + f(1))(1 + f(1)) - 1 = 2² - 1 = 3

f(3) = f(2) + f(1) + f(2)f(1) = 3 + 1 + 3 = 7

f(6) = (1 + f(3))(1 + f(3)) - 1 = 8² - 1 = 63

f(12) = (1 + f(6))(1 + f(6)) - 1 = 64² - 1 = 4095


Alternative:
Observe the values we have obtained earlier:
f(1) = 1 = 2¹-1
f(2) = 3 = 2²-1
f(3) = 7 = 2³-1
...
f(12) = 2^12-1 = 4095

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Given, f(x+y) = f(x) + f(y) + f(x)f(y) for all real values of x and y except 0; and f(1) =1

To find, f(12) = f(6+6) = f(6) + f(6) + f(6)*f(6)
f(6) = f(3+3) = f(3) + f(3) + f(3)*f(3)
f(3) = f(1+2) = f(1)+f(2)+ f(1)*f(2)
f(2) = f(1+1) = f(1) + f(1) + f(1)*f(1) = 1+1+1 = 3

f(3) = 1+3 + 1*3 = 7
f(6) = 7 + 7 + 7*7 = 63

f(12) = 63 + 63 + 63*63 = 126 + (60+3)^2 = 126 + 3600 + 360 + 9 = 4095

Answer: B

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