>What is the area enclosed by the function f(x) = sin²x - x²(sin³x)(cos²x) and the x-axis — between -π/4 and π/4?
Well, the second part is an odd function so we know its areas are 0 over symmetric regions, assuming "area" means "signed area". So we're left with sin²x which has a trick: the integral of sin²x and cos²x are equal over any of the quarter cycles of the graph -- same shape and everything. So we have that over our domain, pi/2 = Int(1) = Int(cos²x+sin²x) = Int(sin²x+sin²x) = 2Int(sin²x)... so the answer would be pi/4 (which doesn't appear).
I also suspect that some of the answers must be wrong, possibly in the counting or probability questions where it is extremely easy to make a logical error.
In my opinion the author should have a third party scrutinize his/her answers. I'm quite certain I didn't get 40% of the questions wrong. (I have a math PhD, taught calculus for 4 years, scored 170 on the quant section of the GRE, and went through this whole "test" very carefully).