Question 14When simplified $\sqrt{1+\left(\dfrac{x^4-1}{2x^2}\right)^2}$ equals:$\dfrac{x^4+2x^2-1}{2x^2}$$\dfrac{x^4-1}{2x^2}$$\dfrac{\sqrt{x^2+1}}2$$\dfrac{x^2}{\sqrt2}$$\dfrac{x^2}{2}+\dfrac1{2x^2}$