SAT Question 635f54ee: Nonlinear Functions answer and explanation with Desmos solution

Choice A is incorrect because if the perimeter of one face of the cube is $\frac{a}{4}$, then the total surface area of the cube is $6{\left(\frac{a/4}{4}\right)}^2=6{\left(\frac{a}{16}\right)}^2$, which is not $6{\left(\frac{a}{4}\right)}^2$.

Choice C is incorrect because if the perimeter of one face of the cube is $4a$, then the total surface area of the cube is $6{\left(\frac{4a}{4}\right)}^2=6a^2$, which is not $6{\left(\frac{a}{4}\right)}^2$.

Choice D is incorrect because if the perimeter of one face of the cube is $6a$, then the total surface area of the cube is $6{\left(\frac{6a}{4}\right)}^2=6{\left(\frac{3a}{2}\right)}^2$, which is not $6{\left(\frac{a}{4}\right)}^2$.