If $$2x - \frac{1}{2x} = \frac{10}{3}$$​ and x > 1, then find the value of $$x^2+\frac1{16x^2 }$$​​

Correct option is B

Given:
​$$2x - \frac{1}{2x} = \frac{10}{3}$$​​
x > 1
Formula Used:
​$$(a - b)^2 = a^2 + b^2 - 2ab$$​​
Solution:
Square both sides of the given equation.
​$$\left(2x - \frac{1}{2x}\right)^2 = \left(\frac{10}{3}\right)^2$$​​
Apply the algebraic identity.

​$$4x^2 + \frac{1}{4x^2} - 2 × 2x × \frac{1}{2x} = \frac{100}{9}$$​​

​$$4x^2 + \frac{1}{4x^2} - 2 = \frac{100}{9}$$​​

​$$4x^2 + \frac{1}{4x^2} = \frac{100}{9} + 2$$​​

​$$4x^2 + \frac{1}{4x^2} = \frac{118}{9}$$​​

​$$\frac{4x^2}{4} + \frac{1}{16x^2} = \frac{118}{9 × 4}$$​​

​$$x^2 + \frac{1}{16x^2} = \frac{118}{36}$$​​

​$$x^2 + \frac{1}{16x^2} = \frac{59}{18}$$​​
Final Answer
So the correct answer is (b)