If p = \(\frac{\sqrt{2}+1}{\sqrt{2}-1}\) and q = \(\frac{\sqrt{2}-1}{\sqrt{2}+1}\) then find the value of \(\frac{p^2}{q}+\frac{q^2}{p}\) .

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This question was previously asked in

SSC CGL 2023 Tier-I Official Paper (Held On: 18 Jul 2023 Shift 4)

Answer 1

Given:

p = \(\frac{√{2}+1}{√{2}-1}\) and q = \(\frac{√{2}-1}{√{2}+1}\)

Concept used:

p = 1 / q

Calculation:

p =\(\frac{√{2}+1}{√{2}-1}\)

= (√2 + 1)²

⇒ 3 + 2√2

Again,

q = \(\frac{√{2}-1}{√{2}+1}\)

= 3 - 2√2

p + q = 3 + 2√2 + 3 - 2√2 = 6

pq = (3 + 2√2)(3 - 2√2) = 1

We know that, p3 + q3 = (p + q)3 - 3pq(p + q)

⇒ p3 + q3 = (6)3 - 3 × 1 × 6 = 216 - 3 × 6 = 216 - 18 = 198

Now, p2/q + q2/p = (p3 + q3)/pq = 198/1 = 198

∴ The correct option is 3

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