Question

If $2m-\frac{1}{2m}=2$ then the value of $m^2+\frac{1}{16m^2}$ is

• A. 3/2
• B. 1/2
• C. 5/2
• D. 7/2

Answer 1

The correct option is A 3/2

$2m-\frac{1}{2m}=2$

Divided both side with 2

$m-\frac{1}{4m}=1$

square both sides

${(m-\frac{1}{4m})}^2=(1{)}^2$

$\Rightarrow m^2+\frac{1}{16m^2}-\frac{1}{2}=1$

$\Rightarrow m^2+\frac{1}{16m^2}=1+\frac{1}{2}$

$\Rightarrow m^2+\frac{1}{16m^2}=\frac{3}{2}$