Q44 / 75 · JEE Main 2025 (2 Apr Shift 1) · Mathematics

Let $P_n=\alpha^n+\beta^n, n \in \mathbf{N}$. If $P_{10}=123, P_9=76$, $P_8=47$ and $P_1=1$, then the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is :

  • A
    $x^2-x+1=0$
  • Correct answer
    B
    $x^2+x-1=0$
  • C
    $x^2-x-1=0$
  • D
    $x^2+x+1=0$

Solution

$\begin{aligned} & \alpha^{10}+\beta^{10}=123 \\ & \alpha+\beta=1 \\ & \alpha^9+\beta^9=76 \\ & \alpha^8+\beta^8=47 \\ & P_{10}=P_9+P_8 \\ & x^2=x+1 \Rightarrow x^2-x-1=0 \\ & \alpha+\beta=1, \alpha \beta=-1 \\ & \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{1}{-1}=-1, \frac{1}{\alpha \beta}=-1\end{aligned}$

Hence, the correct answer is option: 'B'.

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