Question

lf the square root of $\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}+\frac{1}{2i}(\frac{x}{y}+\frac{y}{x})+\frac{31}{16}\right)$ is $\pm (\frac{x}{y}+\frac{y}{x}-\frac{i}{m})$, then the value of $m$ is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (C)

4

${(\frac{x}{y}+\frac{y}{x}+\frac{-i}{m})}^2$
$=\frac{x^2}{y^2}+\frac{y^2}{x^2}-\frac{1}{m^2}+2-\frac{2i}{m}(\frac{x}{y}+\frac{y}{x})$
Comparing the terms with $\frac{x^2}{y^2}+\frac{y^2}{x^2}+\frac{1}{2i}(\frac{x}{y}+\frac{y}{x})+\frac{31}{16}$,
$\frac{-2i}{m}=\frac{1}{2i}$ and $2-\frac{1}{m^2}=\frac{31}{16}$.
Hence, $m=4$.
Hence, option C is the correct answer.