If $2a +\frac{1}{a} = 4$, then the value of $a^2 + \frac{1}{4a^2}$ is :

3

If $K+\frac{1}{K}=n$

then, $K^2+\frac{1}{K^2}$ = n² – 2 × k × \(\frac{1}{k}\)

If $2a +\frac{1}{a} = 4$,

Then $4a^2 +\frac{1}{a^2}$ = 4² – 2 × 2a × \(\frac{1}{a}\)

$4a^2 +\frac{1}{a^2}$ = 16 – 2 × 2 = 12

Divide the equation by 4 for getting the desired type of result,

then the value of $a^2 + \frac{1}{4a^2}$ is = \(\frac{12}{4}\) = 3