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f(x) = {ax - 1}/{xn(ax + 1)} ......................1
f(-x) = {a-x - 1}/{(-x)n(a-x + 1}
f(-x) = {1-ax}/{xn(1 + ax)}(-1)n
f(-x) = {-1/(-1)n} {f(x)} ...................2
f(x) is even function if f(x) = f(-x)
for this condition -1/(-1)n should be 1
n = 1 , 3 , 5 .........
n should be odd integer
......
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