If $y^{1 / m} + y^{- 1 / m} = 2 x$ , then $(x^{2} - 1) y_{2} + x y_{1}$ is equal to

m^{2} y

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- m^{2} y

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\pm m^{2} y

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None of these

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Solution

The correct option is A $m^{2} y$
We have $y^{1 / m} + y^{- 1 / m} = 2 x$
$\Rightarrow y^{1 / m} + \frac{1}{y^{1 / m}} = 2 x$

$(y^{1 / m})^{2} - 2 x y^{1 / m} + 1 = 0$

Therefore, $y^{1 / m} = x \pm \sqrt{x^{2} - 1}$

$y = {[x \pm \sqrt{x^{2} - 1}]}^{m}$

Thus $y_{1} = \frac{\pm m {[x \pm \sqrt{x^{2} - 1}]}^{m}}{\sqrt{x^{2} - 1}} = \frac{\pm m y}{\sqrt{x^{2} - 1}}$

$\Rightarrow (x^{2} - 1) y_{1}^{2} = m^{2} y^{2}$

$\Rightarrow (x^{2} - 1) y_{2} + x y_{1} = m^{2} y$

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