JEE Main 2024 — 9 April, Evening Shift. Previous Year Question.
Problem
Let $\alpha, \beta$ (with $\alpha > \beta$) be the roots of the equation $x^2 – \sqrt{2}x – \sqrt{3} = 0$. Let $P_n = \alpha^n – \beta^n$, $n \in \mathbb{N}$. Then
$$(11\sqrt{3} – 10\sqrt{2})P_{10} + (11\sqrt{2}+10)P_{11} – 11P_{12}$$
is equal to:
Key insight. Because $\alpha$ and $\beta$ both satisfy the same quadratic, $P_n$ obeys a recurrence built directly from that equation’s coefficients — turning a scary-looking combination of $P_{10}$, $P_{11}$, $P_{12}$ into something that collapses in two substitutions.
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Approach
Since $\alpha$ and $\beta$ are roots of $x^2 = \sqrt{2}x + \sqrt{3}$, multiplying through by $\alpha^{n-1}$ and $\beta^{n-1}$ respectively and subtracting gives a recurrence relation for $P_n$ directly from the equation’s own coefficients — no need to ever find $\alpha$ or $\beta$ explicitly. The given expression is then just this recurrence applied twice, at two different values of $n$.
Solution
Step 1 — Build the recurrence relation for Pₙ
Since $\alpha^2 = \sqrt{2}\alpha + \sqrt{3}$ and $\beta^2 = \sqrt{2}\beta + \sqrt{3}$, multiply the first by $\alpha^{n-1}$ and the second by $\beta^{n-1}$, then subtract:
$$\alpha^{n+1} – \beta^{n+1} = \sqrt{2}\big(\alpha^n – \beta^n\big) + \sqrt{3}\big(\alpha^{n-1} – \beta^{n-1}\big)$$
$$P_{n+1} = \sqrt{2}\,P_n + \sqrt{3}\,P_{n-1}$$
Step 2 — Apply the recurrence at n = 11
$$P_{12} = \sqrt{2}\,P_{11} + \sqrt{3}\,P_{10}$$
Substitute this into the target expression to eliminate $P_{12}$:
$$(11\sqrt{3}-10\sqrt{2})P_{10} + (11\sqrt{2}+10)P_{11} – 11\big(\sqrt{2}\,P_{11}+\sqrt{3}\,P_{10}\big)$$
Distributing the $-11$ and collecting terms:
$$= \big[11\sqrt{3}-10\sqrt{2}-11\sqrt{3}\big]P_{10} + \big[11\sqrt{2}+10-11\sqrt{2}\big]P_{11}$$
$$= -10\sqrt{2}\,P_{10} + 10\,P_{11} = 10\big(P_{11} – \sqrt{2}\,P_{10}\big)$$
Step 3 — Apply the recurrence again, at n = 10
$$P_{11} = \sqrt{2}\,P_{10} + \sqrt{3}\,P_9 \implies P_{11} – \sqrt{2}\,P_{10} = \sqrt{3}\,P_9$$
Substituting back:
$$10\big(P_{11}-\sqrt{2}\,P_{10}\big) = 10\sqrt{3}\,P_9$$
Answer
$$10\sqrt{3}\,P_9$$
Common mistakes
- Trying to compute $\alpha$ and $\beta$ explicitly and then raise them to the 9th, 10th, 11th and 12th powers — technically possible, but far too slow for exam conditions. The recurrence avoids ever finding the roots themselves.
- Substituting the recurrence only once. The expression needs the relation applied twice — once to eliminate $P_{12}$, and again to simplify the resulting $P_{11}-\sqrt2\,P_{10}$ term into a multiple of $P_9$.
Practise next
- If $\alpha, \beta$ are the roots of $x^2 – \sqrt{3}x – \sqrt{2} = 0$ and $Q_n = \alpha^n – \beta^n$, find the recurrence relation for $Q_n$ and use it to simplify $Q_{n+1} – \sqrt{3}\,Q_n$.
Show answer
$Q_{n+1}-\sqrt3\,Q_n=\sqrt2\,Q_{n-1}$.
Both roots satisfy $t^2=\sqrt3\,t+\sqrt2$. Multiplying by $t^{n-1}$ gives $t^{n+1}=\sqrt3\,t^{n}+\sqrt2\,t^{n-1}$, true for $t=\alpha$ and for $t=\beta$.
Subtracting the two and writing $Q_n=\alpha^n-\beta^n$ gives $Q_{n+1}=\sqrt3\,Q_n+\sqrt2\,Q_{n-1}$, so $Q_{n+1}-\sqrt3\,Q_n=\sqrt2\,Q_{n-1}$. No root is ever computed — the equation itself supplies the recurrence.
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