Q182 / 270 · NDA 2016 (Phase 2) · Mathematics

Paragraph: Let $\alpha$ and $\beta$ be the roots of the equation $\mathrm{x}^2-\left(1-2 \mathrm{a}^2\right) \mathrm{x}+\left(1-2 \mathrm{a}^2\right)=0$
Question: Under what condition is $\frac{1}{\alpha^2}+\frac{1}{\beta^2} \lt 1$ ?

  • Correct answer
    A
    $\mathrm{a}^2 \lt \frac{1}{2}$
  • B
    $a^2 \gt \frac{1}{2}$
  • C
    $a^2 \gt 1$
  • D
    $\mathrm{a}^2 \in\left(\frac{1}{3}, \frac{1}{2}\right)$ only

Solution

$\frac{1}{\alpha^2}+\frac{1}{\beta^2} \lt 1 \Rightarrow \frac{\alpha^2+\beta^2}{(\alpha \beta)^2} \lt 1$ $\begin{aligned} & \Rightarrow \quad \frac{(\alpha+\beta)^2-2 \alpha \beta}{(\alpha \beta)^2} \lt 1 \\ & \alpha+\beta=\frac{-b}{a}=\left(1-2 \mathrm{a}^2\right) \& \alpha \beta=\frac{c}{a}=\left(1-2 \mathrm{a}^2\right) \\ & \text { On solving: } \quad \lt 1 \end{aligned}$ $\begin{aligned} & \frac{4 a^4-1}{4 a^4-4 a^2+1} \\ & \Rightarrow \quad 4 a^4-1 \lt 4 a^4-4 a^2+1 \\ & 4 a^2 \lt 2 \Rightarrow a^2 \lt \frac{1}{2} \end{aligned}$

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

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