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If $α, β$ are real and $α^{2}, - β^{2}$ are the roots of $a^{2} x^{2} + x + 1 - a^{2} = 0, a > 1$ , then $β^{2} =$

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a^{2}

1 - a^{2}

1 + a^{2}

answer is B.

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Detailed Solution

Given, the roots of

a^{2} x^{2} + x + 1 - a^{2} = 0, a > 1

are

α^{2}, - β^{2} .

We need to find

β^{2} .

We know that,

α^{2}, - β^{2}

are the roots of

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

Solving,

\Rightarrow α^{2} - β^{2} = - \frac{1}{a^{2}}

\Rightarrow α^{2} (- β^{2}) = \frac{1 - a^{2}}{a^{2}}

\Rightarrow α^{2} + β^{2} = {(α^{2} - β^{2})}^{2} + 4 α^{2} β^{2}

\Rightarrow α^{2} + β^{2} = \frac{1}{a^{4}} + 4 (\frac{a^{2} - 1}{a^{2}})

\Rightarrow α^{2} + β^{2} = \frac{1}{a^{4}} + 4 - \frac{4}{a^{2}}

\Rightarrow α^{2} + β^{2} = {(2 - \frac{1}{a^{2}})}^{2}

\Rightarrow α^{2} + β^{2} = 2 - \frac{1}{a^{2}}

\Rightarrow β^{2} = \frac{1}{2} [(α^{2} + β^{2}) - (α^{2} - β^{2})]

\Rightarrow β^{2} = \frac{1}{2} [2 - \frac{1}{a^{2}} + \frac{1}{a^{2}}]

\Rightarrow β^{2} = \frac{1}{2} [2]

\Rightarrow β^{2} = 1

Therefore,

β^{2} = 1 .

Hence, the correct option is 2.

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