Question

If $x,y,2x+\frac{y}{2}$ are non zero real numbers, then ${(2x+\frac{y}{2})}^{-1}\{{\left(2x\right)}^{-1}+{\left(\frac{y}{2}\right)}^{-1}\}$ is equal to:

• A. $1$
• B. $x,y^{-1}$
• C. $x^{-1}.y$
• D. $x^{-1}.y^{-1}$

Answer 1

Answer: (D)

$x^{-1}.y^{-1}$

The given expression ${(2x+\frac{y}{2})}^{-1}\{{\left(2x\right)}^{-1}+{\left(\frac{y}{2}\right)}^{-1}\}$ can be simplified as follows:

${(2x+\frac{y}{2})}^{-1}\{{\left(2x\right)}^{-1}+{\left(\frac{y}{2}\right)}^{-1}\}=\frac{1}{(2x+\frac{y}{2})}\{\frac{1}{2x}+\frac{1}{\frac{y}{2}}\}(\because \frac{1}{x}=x^{-1})=\frac{1}{(2x+\frac{y}{2})}(\frac{1}{2x}+\frac{2}{y})=\frac{\frac{1}{2x}+\frac{2}{y}}{2x+\frac{y}{2}}$

$=\frac{\frac{y+4x}{2xy}}{\frac{4x+y}{2}}=\frac{y+4x}{2xy}\times \frac{2}{4x+y}=\frac{2}{2xy}=\frac{1}{xy}=\frac{1}{x}\times \frac{1}{y}=x^{-1}\times y^{-1}(\because \frac{1}{x}=x^{-1})$

Hence, the value of ${(2x+\frac{y}{2})}^{-1}\{{\left(2x\right)}^{-1}+{\left(\frac{y}{2}\right)}^{-1}\}$ is $x^{-1}{y}^{-1}$.