Bihar Police Constable Model Paper 5

Given:
x = $\sqrt{2}+\frac{1}{\sqrt{2}}$,
Y = $\sqrt{2}\mathit{–}\frac{1}{\sqrt{2}}$
Formula Used:
${\mathit{x}}^2\mathit{–}{\mathit{y}}^2$ = (x – y)(x + y)
Calculation:
x/y – y/x
⇒ $\left({\mathit{x}}^2\mathit{–}{\mathit{y}}^2\right)∕\mathit{x}\mathit{y}$
⇒ (x – y)(x + y)/xy ---- equation (1)
x – y = $\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right)\mathit{–}\left(\sqrt{2}\mathit{–}\frac{1}{\sqrt{2}}\right)$
⇒ $\sqrt{2}+\frac{1}{\sqrt{2}}-\sqrt{2}+\frac{1}{\sqrt{2}}$
⇒ $\frac{2}{\sqrt{2}}$
⇒ $\sqrt{2}$
x + y = $\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right)+\left(\sqrt{2}\mathit{–}\frac{1}{\sqrt{2}}\right)$
⇒ $\sqrt{2}+\frac{1}{\sqrt{2}}+\sqrt{2}\mathit{–}\frac{1}{\sqrt{2}}$
⇒ $2\sqrt{2}$
xy = $\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right)\left(\sqrt{2}\mathit{–}\frac{1}{\sqrt{2}}\right)$
⇒ ${\left(\sqrt{2}\right)}^2\mathit{–}{\left(\frac{1}{\sqrt{2}}\right)}^2$
⇒ 2 – 1/2
⇒ 3/2
put the value of (x – y), (x + y), and xy in equation (1)
⇒ (x – y)(x + y)/xy
⇒ (√2 × 2√2)/(3/2)
⇒ (4 × 2)/3
⇒ 8/3
∴ The value of x/y – y/x is 8/3.