Question

$\bold{If\;the\;mean\;of\;x\;and\;\dfrac{1}{x}\;is\;M.Then\;the\;mean\;of\;x^2\;and\dfrac{1}{x^2}\;is\;KM^2-1\;find\;K?\;}$

Answer 1

Given that mean of x and 1/x is M, i.e.,

$\small\text{ \longrightarrow\dfrac{\left(x+\dfrac{1}{x}\right)}{2}=M }$

$\small\longrightarrow x+\dfrac{1}{x}=2M\quad\dots(1)$

Now given that mean of x² and 1/x² is KM² - 1.

$\small\text{ \longrightarrow KM^2-1=\dfrac{\left(x^2+\dfrac{1}{x^2}\right)}{2}\quad\dots(2) }$

We need to find value of K.

We see that,

$\small\longrightarrow x^2+\dfrac{1}{x^2}=x^2+\dfrac{1}{x^2}+2-2$

$\small\longrightarrow x^2+\dfrac{1}{x^2}=x^2+2\cdot x\cdot\dfrac{1}{x}+\left(\dfrac{1}{x}\right)^2-2$

$\small\longrightarrow x^2+\dfrac{1}{x^2}=\left(x+\dfrac{1}{x}\right)^2-2$

Then (2) becomes,

$\small\text{ \longrightarrow KM^2-1=\dfrac{\left(x+\dfrac{1}{x}\right)^2-2}{2} }$

From (1),

$\small\longrightarrow KM^2-1=\dfrac{\left(2M\right)^2-2}{2}$

$\small\longrightarrow KM^2-1=\dfrac{4M^2-2}{2}$

$\small\longrightarrow KM^2-1=2M^2-1$

$\small\text{ \Longrightarrow\underline{\underline{K=2}} }$

Hence value of K is 2.

Answer 2

⇝ Question :-

If the mean of $\large\rm x \: \: and \: \: \dfrac{1}{x}$ is M.

And the mean of $\large\rm x^2 \: \: and \: \: \dfrac{1}{x^2}$ is KM²

then Find the Value of K.

⇝ Answer :-

$\large\pink{\pmb{\frak{ \text The \: \text Value \: of \: \text K \: is \: 2}}}$

⇝ Step by step Explanation :-

❒ Given in the Question that the mean of $\large\rm x \: \: and \: \: \dfrac{1}{x}$ is M,

$:\longmapsto \rm \frac{x + \dfrac{1}{x} }{2} = M$

$:\longmapsto \rm x + \frac{1}{x} = 2M$

$\large \bigstar$ Squaring Both Sides :

$:\longmapsto \rm \bigg(x + \frac{1}{x} \bigg)^{2} = 4 {M}^{2}$

$:\longmapsto \rm {x}^{2} + \frac{1}{x {}^{2} } + 2.x. \frac{1}{x} = 2M {}^{2}$

$:\longmapsto \rm {x}^{2} + \frac{1}{{x}^{2} } + 2 = 4M {}^{2}$

$\purple{ :\longmapsto  \underline {\boxed{{\bf  {x}^{2} + \frac{1}{ {x}^{2} } = 4M {}^{2} - 2} }}}$

❒ Also According To Question the mean of $\large\rm x^2 \: \: and \: \: \dfrac{1}{x^2}$ is KM² ,

$:\longmapsto \rm \frac{ {x}^{2} + \dfrac{1}{ {x}^{2} } }{2} = {KM}^{2} - 1$

$:\longmapsto \rm {x}^{2} + \frac{1}{ {x}^{2} } = 2(KM {}^{2} - 1)$

$:\longmapsto \rm {x}^{2} + \frac{1}{ {x}^{2} } = 2KM {}^{2} - 2$

$\large \red\bigstar$ Substituting Value of $\large\rm x^2 + \dfrac{1}{x^2}$ :

$:\longmapsto \rm 4M {}^{2} - \cancel{2}= 2KM {}^{2} - \cancel{2}$

$:\longmapsto \rm 4 \: \cancel{M {}^{2} } = 2 K \: \cancel{M^{2} }$

$:\longmapsto \rm2 K =4$

$:\longmapsto \rm K = \cancel \frac{4}{2}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf K = 2} }}}$

Therefore,

$\large \bf \blue\maltese \: \: \orange{ \underbrace{ \underline{The\:Value \: of \: K \: is \: 2}}}$