Question : If $2x-\frac{1}{2x}=5,x\neq0$, then the value of $(x^2+\frac{1}{16x^2}-2)$is:
Option 1: $\frac{19}{4}$
Option 2: $\frac{23}{4}$
Option 3: $\frac{27}{4}$
Option 4: $\frac{31}{4}$
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Correct Answer: $\frac{19}{4}$
Solution :
Given;
$2x-\frac{1}{2x}=5,x\neq0$
Dividing both sides by 2, we get
$x-\frac{1}{4x}=\frac{5}{2}$
Now, squaring both sides, we get
$(x^2+\frac{1}{16x^2}-2×x×\frac{1}{4x})=\frac{25}{4}$
⇒$(x^2+\frac{1}{16x^2})=\frac{25}{4}+\frac{1}{2}$
⇒$(x^2+\frac{1}{16x^2})=\frac{27}{4}$
⇒$(x^2+\frac{1}{16x^2}-2)=\frac{27}{4}-2$
$\therefore (x^2+\frac{1}{16x^2}-2)=\frac{19}{4}$
Hence, the correct answer is $\frac{19}{4}$.
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