Question

If $y^{1/m}+y^{-1/m}=2x$, then $\left(x^2-1\right)y_2+x{y}_1$ is equal to

• A. $m^{2}y$
• B. $-m^{2}y$
• C. $\pm m^{2}y$
• D. None of these

Answer 1

Answer: (A)

We have, $y^{\frac{1}{m}}+y^{-\frac{1}{m}}=2x$
or $y^{\frac{1}{m}}+\frac{1}{y^{1/m}}=2x$
or ${\left(y^{1/m}\right)}^2-2x{y}^{1/m}+1=0$
$\therefore y^{1/m}=x\pm \sqrt{\left(x^2-1\right)}$
or $y={\left[x\pm \sqrt{x^2-1}\right]}^m$
$\therefore y_1=m{\left[x\pm \sqrt{\left(x^2-1\right)}\right]}^{m-1}\left[1\pm \frac{2x}{2\sqrt{x^2-1}}\right]$
$\Rightarrow \left(x^2-1\right)y_1^2=m^2{y}^2$
$\Rightarrow \left(x^2-1\right)2y_1{y}_2+2x{y}_1^2=2m^{2}y{y}_1$
$\Rightarrow \left(x^2-1\right)y_2+x{y}_1=m^{2}y$