Question : If $\mathrm{p}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$ and $\mathrm{q}=\frac{\sqrt{2}-1}{\sqrt{2}+1}$ then, find the value of $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}$.
Option 1: 200
Option 2: 196
Option 3: 198
Option 4: 188
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Correct Answer: 198
Solution :
Given:
$\mathrm{p}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$, $\mathrm{q}=\frac{\sqrt{2}-1}{\sqrt{2}+1}$
Rationalising $p$, we get:
$p= 3+2\sqrt{2}$
⇒ $p^2=17+12\sqrt{2}$
$\mathrm{q}=\frac{\sqrt{2}-1}{\sqrt{2}+1}$
Rationalising $q$, we get:
$q= 3-2\sqrt{2}$
⇒ $q^2=17-12\sqrt{2}$
$\therefore$ $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}=(\frac{17+12\sqrt{2}}{3-2\sqrt{2}})+(\frac{17-12\sqrt{2}}{3+2\sqrt{2}})$
⇒ $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}=\frac{(17+12\sqrt{2})(3+2\sqrt{2})+(17-12\sqrt{2})(3-2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})}$
⇒ $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}=\frac{2\times 17\times 3+2\times 12\times2\times 2}{(3+2\sqrt{2})(3-2\sqrt{2})}$
⇒ $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}=198$
Hence, the correct answer is 198.
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