Question

If $y^{1/m}=[x+\sqrt{1+x^2}]$ , then $(1+x^2)y_2+x{y}_1$ is equal to :

• A. $m^{2}y$
• B. $m^2{y}^2$
• C. $m^2{y}^2$
• D. $my$

Answer 1

The correct option is A $m^{2}y$

$y={(x+\sqrt{1+x^2})}^m$
$\Rightarrow \left(\sqrt{1+x^2}\right)y_1=m{(x+\sqrt{1+x^2})}^m$
$\Rightarrow \left(\sqrt{1+x^2}\right)y_1=my$
$\left(\frac{x}{\sqrt{1+x^2}}\right)y_1+\sqrt{1+x^2}{y}_2=m{y}_1$
$(1+x^2)y_2+x{y}_1=m\left(\sqrt{1+x^2}\right)y_1$
$=m^{2}y$