Question

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

• A. $a^2<\frac{1}{2}$
• B. $a^2>\frac{1}{2}$
• C. $a^2>1$
• D. $a^2\epsilon (\frac{1}{3},\frac{1}{2})$ only

Answer 1

The correct option is A $a^2<\frac{1}{2}$

$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}<1$ .... $\left(i\right)$

$\alpha ,$ $\beta$ are roots of the equation $x^2-(1-2a^2)x+(1-2a^2)=0$

Then $\alpha +\beta =1-2a^2$ and $\alpha \beta =1-2a^2$

We have $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{{\alpha }^2+{\beta }^2}{(\alpha \beta {)}^2}$

Now, $\frac{(\alpha +\beta {)}^2-2\alpha \beta }{(\alpha \beta {)}^2}<1$ ....... From $\left(i\right)$

$\Rightarrow \frac{\left(\right(1-2a^2{)}^{{}^2}-2(1-2a^2))}{(1-2a^2{)}^{{}^2}}<1$

$\Rightarrow \frac{-2(1-2a^2)}{(1-2a^2{)}^2}<0$

$\Rightarrow \frac{2(1-2a^2)}{(1-2a^2{)}^2}>0$

Now $(1-2a^2{)}^2>0$ for $a^2\ne 0.5$, So $(1-2a^2)>0$ we have $a^2<\frac{1}{2}$

Hence, A is correct.