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if the mean of X and 1 by X is M and then the mean of x square and 1 by x square is
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Given:
mean of x and 1/x = M
so (x + 1/x) / 2 = M -----------(1)
or x + 1/x = 2M ------------(2)
Now,
mean of x3 and 1/x3
= (x3 + 1/x3) / 2
= (x + 1/x)(x2 - x*1/x + 1/x2) / 2
= (x + 1/x) / 2 * (x2 + 1/x2 - 1)
= (x + 1/x) / 2 * [(x + 1/x)2 - 2*x*1/x - 1)
= (x + 1/x) / 2 * [(x + 1/x)2 - 2 - 1)
= (x + 1/x) / 2 * [(x + 1/x)2 - 3)
= M * (M2 - 3) (from (1) and 2)
= M3 - 3M
mean of x and 1/x = M
so (x + 1/x) / 2 = M -----------(1)
or x + 1/x = 2M ------------(2)
Now,
mean of x3 and 1/x3
= (x3 + 1/x3) / 2
= (x + 1/x)(x2 - x*1/x + 1/x2) / 2
= (x + 1/x) / 2 * (x2 + 1/x2 - 1)
= (x + 1/x) / 2 * [(x + 1/x)2 - 2*x*1/x - 1)
= (x + 1/x) / 2 * [(x + 1/x)2 - 2 - 1)
= (x + 1/x) / 2 * [(x + 1/x)2 - 3)
= M * (M2 - 3) (from (1) and 2)
= M3 - 3M
siri978:
Karthik brother you got wrong answer
Answered by
1
Hope it will to help you .
Thank you and stay connected.
Thank you and stay connected.
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