To determine the number of roots for all antiderivative of a cubic polynomial

Question

Let $f(x)$ be a cubic polynomial with real coefficients. Suppose that $f(x)$ has exactly one real root which is simple. Which of the following statements holds for all antiderivative $F(x)$ of $f(x)$ ?

(1) $F(x)$ has exactly one real root.

(2) $F(x)$ has exactly four real roots

(3) $F(x)$ has at most two real roots.

(4) $F(x)$ has at most one real root.

My Attempt :-

Let $f(x)= x(x^2 +1)$ has one simple real root . Then $F(x)= \frac {x^2}2 (\frac {x^2}2+1) $ (an anti derivative) has exactly one real root .so this eliminates option 2 .

Now let $F(x)= x(x-1)(x^2+1)$ Then $f(x)=F'(x)= 4x^3 -3x^2 +2x-1$

Thus $f(x)$ has exactly 3 real roots or 1

Claim : $f(x)$ has one real root.

$f'(x)= 12x^2 -6x+2=2(6x^2 -3x+1) $ which has determinant $-15<0$ . So $f'(x)$ has no real root which implies $f(x)$ must have a simple real root.

So correct option is 3 ( only one correct option ) .

Firstly I want you to kindly verify the work and secondly I want any suggestion to finish it quicker as constructing these examples can sometimes be time consuming as this is a question from a competitive exam.

Answer 1

Since $F(x)$ is the antiderivative of $f(x)$ (which has 1 simple root at $x=r$), it follows without loss of generality that $F(x)$ is strictly decreasing on $(-\infty,r]$ and strictly increasing on $(r,\infty)$. From this it follows answer (3).