Multiple choice

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - (1 - 2a^2) x + (1 - 2a^2) = 0$.Under what condition is $\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} < 1.$

  1. $a^2 < \dfrac{1}{2}$
  2. $a^2 > \dfrac{1}{2}$
  3. $a^2 > 1$
  4. $a^2 \varepsilon \left( \dfrac{1}{3}, \dfrac{1}{2} \right)$ only
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A Correct answer
Explanation

Roots alpha, beta satisfy alpha+beta = 1-2a^2 and alpha*beta = 1-2a^2. The expression 1/alpha^2 + 1/beta^2 = (alpha^2 + beta^2) / (alpha*beta)^2 = ((alpha+beta)^2 - 2alpha*beta) / (alpha*beta)^2. Substituting: ((1-2a^2)^2 - 2(1-2a^2)) / (1-2a^2)^2 = 1 - 2/(1-2a^2). For this to be < 1, -2/(1-2a^2) < 0, which implies 1-2a^2 > 0, so a^2 < 1/2.

AI explanation

Using the given equation, the sum of roots is alpha + beta = 1 - 2a^2 and the product is alpha beta = 1 - 2a^2. Substitute these into 1/alpha^2 + 1/beta^2 to get ((1 - 2a^2)^2 - 2(1 - 2a^2)) / (1 - 2a^2)^2. Simplifying the numerator yields -1 + 4a^4, so the inequality is (-1 + 4a^4) / (1 - 2a^2)^2 < 1. Solving this leads to a^2 < 1/2. The result is a^2 < 1/2.

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