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f(x)={ (sgn x)^(sgn x) }^n ; n is an odd integer . Find whether f(x) is even or odd.
Hi Manuj,
sgnx = |x|/x
As |x|^2 = x^2, we have |x|/x = x/|x|.
Hence f(x) as mentioned above will be odd for odd integers " n", since f(-x) = -f(x).
Hope it helps.
Wish you all the best.
Regards,
Ashwin (IIT Madras).
It is clearly an odd function.
Solution:
Sgn(x) which stands for signum function of x which always return either 1 or -1 for positive and negative values respectively. It is defined for R-{0}
To find whether it is an odd or even function , lets take two values 1 and -1 for testing the function
If f(x)= f(-x) it is even otherwise odd.
for x=1, the function would be like this (11)n where n is an odd no, f(x)= 1
for x=-1, the function would be like this (-1-1)n where n is an odd no, f(-x)= -1
since f(x) is not equal to f(-x). It is an odd function
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