If the average of x and frac{1}{x} (x neq 0) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that the average of $x$ and $\frac{1}{x}$ is $M$.$\frac{x + \frac{1}{x}}{2} = M$Multiplying both sides by $2$,we get:$x + \frac{1}{x} = 2M$N

Vedclass · Accepted Answer

$2M^{2}-1$

Vedclass · Answer

(C) Given that the average of $x$ and $\frac{1}{x}$ is $M$.$\frac{x + \frac{1}{x}}{2} = M$Multiplying both sides by $2$,we get:$x + \frac{1}{x} = 2M$Now,we need to find the average of $x^{2}$ and $\frac{1}{x^{2}}$,which is given by:$	ext{Average} = \frac{x^{2} + \frac{1}{x^{2}}}{2}$Using the algebraic identity $(a + b)^{2} = a^{2} + b^{2} + 2ab$,we can write:$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2(x)(\frac{1}{x}) = (x + \frac{1}{x})^{2} - 2$Substituting the value of $(x + \frac{1}{x}) = 2M$:$x^{2} + \frac{1}{x^{2}} = (2M)^{2} - 2 = 4M^{2} - 2$Therefore,the required average is:$\frac{4M^{2} - 2}{2} = 2M^{2} - 1$

If the average of $x$ and $\frac{1}{x} (x \neq 0)$ is $M$,then the average of $x^{2}$ and $\frac{1}{x^{2}}$ is

Similar Questions

The average of five numbers is $27$. If one number is excluded,the average becomes $25$. The excluded number is

If the average of $m$ numbers is $n^{2}$ and that of $n$ numbers is $m^{2},$ then the average of $(m + n)$ numbers is

The sum of three numbers is $98$. If the ratio between the first and second is $2:3$ and that between the second and the third is $5:8$,then the second number is

Three $years$ ago,the average age of a family of $5$ members was $17$ $years$. With the birth of a new baby,the average age of the family remains the same even today. Find the age of the baby (in $years$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module