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

ManipalManipal 2016

Solution:

We have, $y^{\frac{1}{m}}+y^{-\frac{1}{m}}=2 x$
or $y^{\frac{1}{m}}+\frac{1}{y^{1 / m}}=2 x$
or $\left(y^{1 / m}\right)^{2}-2 x y^{1 / m}+1=0$
$\therefore y^{1 / m}=x \pm \sqrt{\left(x^{2}-1\right)}$
or $y=\left[x \pm \sqrt{x^{2}-1}\right]^{m}$
$\therefore y_{1}=m\left[x \pm \sqrt{\left(x^{2}-1\right)}\right]^{m-1}\left[1 \pm \frac{2 x}{2 \sqrt{x^{2}-1}}\right]$
$\Rightarrow\left(x^{2}-1\right) y_{1}^{2}=m^{2} \,y^{2}$
$\Rightarrow\left(x^{2}-1\right) 2 y_{1} \,y_{2}+2 x \,y_{1}^{2}=2 m^{2} \,y\, y_{1}$
$\Rightarrow\left(x^{2}-1\right) y_{2}+x y_{1}=m^{2} \,y$