JEE Main 2025 Apr 2 Shift 1, Mathematics Q15: Quadratic Equations

JEE Main2025Apr 2, Shift 1Mathematics

Let $P_{n} = α^{n} + β^{n}$ , $n \in N$ . If $P_{10} = 123$ , $P_{9} = 76$ , $P_{8} = 47$ and $P_{1} = 1$ , then the quadratic equation having roots $\frac{1}{α}$ and $\frac{1}{β}$ is :

A
$x^{2} - x + 1 = 0$
B
$x^{2} + x - 1 = 0$
C
$x^{2} - x - 1 = 0$
D
$x^{2} + x + 1 = 0$

Solution

Observe that $P_{10} = P_{9} + P_{8}$ since $123 = 76 + 47$ . This is the Fibonacci-style recurrence $P_{n} = P_{n - 1} + P_{n - 2}$ , which corresponds to roots satisfying:

$x^{2} - x - 1 = 0$

So $α + β = 1$ and $α β = - 1$ (consistent with $P_{1} = α + β = 1$ ).

Quadratic for the reciprocals: let $s = \frac{1}{α} + \frac{1}{β}$ and $p = \frac{1}{α β}$ .

$s = \frac{α + β}{α β} = \frac{1}{- 1} = - 1, p = \frac{1}{- 1} = - 1$

Required quadratic: $x^{2} - s x + p = 0$ , i.e.

$x^{2} - (- 1) x + (- 1) = 0 \Rightarrow x^{2} + x - 1 = 0$

$∴$ Option (2) is correct.

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JEE Main 2025 Apr 2 Shift 1, Mathematics Q15: Quadratic Equations

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