Question

the mean of x,1÷x,is m, then the mean of xcube,1÷xcube​

Answer 1

Basic Concept Used :-

$\purple{ \boxed{\bf \: Mean \: = \: \dfrac{Sum \: of \: observations}{Number \: of \: observations}}}$

$\green{\large\underline{\sf{Solution-}}}$

Given that,

$\rm :\longmapsto\:Mean \: of \: x \: and \: \dfrac{1}{x} \: is \: m$

$\rm :\implies\:m = \dfrac{x + \dfrac{1}{x} }{2}$

$\bf\implies \:x + \dfrac{1}{x} = 2m - - - (1)$

Now, we have to find

$\red{\bf :\longmapsto\:Mean \: of \: {x}^{3} \: and \: \dfrac{1}{ {x}^{3}}}$

We know,

$\purple{ \boxed{ \bf \: {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y)}}$

Now,

${\bf :\longmapsto\:Mean \: of \: {x}^{3} \: and \: \dfrac{1}{ {x}^{3}} \: is \: }$

$\rm :\longmapsto\:Mean = \dfrac{ {x}^{3} + \dfrac{1}{ {x}^{3}}}{2}$

$\rm :\longmapsto\:Mean = \dfrac{ {\bigg(x + \dfrac{1}{x} \bigg) }^{3} - 3 \times \cancel x \times \dfrac{1}{ \cancel{x}}\bigg(x +\dfrac{1}{x}\bigg) }{2}$

$\rm :\longmapsto\:Mean = \dfrac{ {(2m)}^{3} - 3 \times 2m }{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \because \: \bf \:x + \dfrac{1}{x} = 2m}}$

$\rm :\longmapsto\:Mean = \dfrac{ {8m}^{3} - 6m}{2}$

$\bf\implies \:Mean = 4 {m}^{3} - 3m$

Additional Information :

1. If the mean of 'n' observation is 'm' and each observation is

• increased by a

• decreased by a

• multiplied by a

then mean is also

• increased by a

• decreased by a

• multiplied by a

2. The sum of deviations taken from mean is always 0.