If 2m- \frac{1}{2m}=2 , where m≠0, then the value of m^{2}+ \frac{1}{16m^{2}} is

CTET 2 Math and Science Feb 2016 Paper

Question : 34 of 120

Marks: +1, -0

2m-\;\;{1}/{2m}=2

, where m≠0, then the value of

m^{2}+\;\;{1}/{16m^{2}}

Solution:

2m-\;\;{1}/{2m}=2

\;\;{4m^{2}-1}/{2m}=2\;\;\;∴4m^{2}-1=4m

Now,

m^{2}+\;\;{1}/{16m^{2}}=(m-\;\;{1}/{4m})^{2}+2.m.\;\;{1}/{4m}

=(\;\;{4m^{2}-1}/{4m})^{2}+\;\;{1}/{2}

(\;\;{4m}/{4m})^{2}+\;\;{1}/{2}=1+\;\;{1}/{2}=1\;\;{1}/{2}

Go to Question:

Prev Question

Next Question