Ques
Asked by rbhatt16
9th June 2018,
11:21 PM
Answered by Expert
Answer:
f(x)f(y) + 2 = f(x) + f(y) + f(xy).....(i)
Put x = y = 0
f2(0) + 2 = 3f(0)
f2(0) + 2 - 3f(0) = 0
f(0) we get 1 or 2
Put x = 0 and y = 1 Consider f(0) = 1
f(0) f(1) + 2 = f(0) + f(1) + f(0)
f(1) + 2 = 2 + f(1)
Consider, f(0) = 2
f(0) f(1) + 2 = f(0) + f(1) + f(0)
2f(1) + 2 = 4 + f(1)
f(1) = 2
Put y = 1/x in (i)
f(x) f(1/x) + 2 = f(x) + f(1/x) + f(1)
f(x) f(1/x) + 2 = f(x) + f(1/x) + 2
If f(x) f(1/x) = f(x) + f(1/x)
then f(x) = 1 ± xn
f'(x) = nxn-1
f'(0) = 0 and f'(1) = 2
f'(1) = 2 → n = 2
f(x) = 1 + x2
If you solve it for option A then it satisfies the given equation.
Answered by Expert
11th June 2018,
10:43 AM
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