Under what condition is \;\;{1}{α²}+\;\;{1}{β²} [NDA II 2016]

Correct Answer is : a2<12a² < 1/2a2<21

Correct Answer is : a2<12a² < 1/2a2<21

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NDA Math Quadratic Equations and Inequations

Consider the following for the next two (02) items that follow:

Let α and β be the roots of the equation

x^{2}-(1-2 a^{2}) x+(1-2 a^{2})=0

[NDA II 2016]

Question : 52 of 67

Marks: +1, -0

Under what condition is

\;\;\frac{1}{\alpha^{2}}+\;\;\frac{1}{\beta^{2}}<1

Solution:

\;\;\frac{1}{\alpha^{2}}+\;\;\frac{1}{\beta^{2}}<1\Rightarrow\;\;\frac{\alpha^{2}+\beta^{2}}{(\alpha\beta)^{2}}<1

\Rightarrow\;\;\frac{(\alpha+\beta)^{2}-2\alpha\beta}{(\alpha\beta)^{2}}<1

\alpha+\beta=\;\;\frac{-b}{a}=(1-2a^{2})\&\alpha\beta=\;\;\frac{c}{a}=(1-2a^{2})

On solving

\;\;\frac{4a^{4}-1}{4a^{4}-4a^{2}+1}<1

\Rightarrow4a^{4}-1<4a^{4}-4a^{2}+1

4a^{2}<2\Rightarrow a^{2} < \;\;\frac{1}{2}

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