Question

If the mean of x and 1/x is M then the mean of $x^3$ and $1/x^3$ is

• A. $\frac{M(M^2-3)}{2}$
• B. $M(4M^2-3)$
• C. $M^3$
• D. $M^3+3$

Answer 1

The correct option is B $M(4M^2-3)$

$\Rightarrow$ $\frac{x+\frac{1}{x}}{2}=M$ [ Given ]

$\therefore$ $x+\frac{1}{x}=2M$ ----- ( 1 )

$\Rightarrow$ $(x+\frac{1}{x}{)}^3=(2M{)}^3$

$\Rightarrow$ $x^3+\frac{1}{x^3}+3\times x\times \frac{1}{x}\times (x+\frac{1}{x})=8M^3$

$\Rightarrow$ $x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=8M^3$

$\Rightarrow$ $x^3+\frac{1}{x^3}+3\left(2M\right)=8M^3$ [ From ( 1 ) ]

$\Rightarrow$ $x^3+\frac{1}{x^3}+6M=8M^3$

$\Rightarrow$ $x^3+\frac{1}{x^3}=8M^3-6M$

$\therefore$ $\frac{x^3+\frac{1}{x^3}}{2}=\frac{8M^3-6M}{2}$

$\therefore$ $\frac{x^3+\frac{1}{x^3}}{2}=4M^3-3M$

$\therefore$ $\frac{x^3+\frac{1}{x^3}}{2}=M(4M^2-3)$