Christmas Streak 58/88: 2018 CSAT (Korean SAT) #20 of Liberal Arts

Since $f'(0)=0$ and $f(x)$ decreases in the intervals $(-\infty,~0)$ and $(0,~k),$ we infer that it contacts the line $y=f(0)$ as it passes through.

Then $f(x)$ would look like the picture shown on the right. (ignore the Korean text >w>)

since $f'(2)>0,$ we can notice that $k<2,$ and therefore $f'(x)=0$ has one root in the open interval $(0,~2).$ ( A. true)

And obviously $f(x)$ doesn't have a local maximum. ( B. false)

If $f(0)=0,$ then $f(x)=x^3(x-p).$

This leads to $f'(x)=3x^2(x-p)+x^3=x^2(4x-3p),$ and from $f'(2)=16,$ we get $p=\dfrac{4}{3}.$

$f'(x)=4x^2(x-1),$ and therefore, the local minimum (and thus the actual minimum of this function) occurs at $x=1,$ which yields $f(1)=-\dfrac{1}{3}.$ ( C. true)

---

From above, only A and C are true.