I approached that question as follows: for a point of inflection to occur at \(x\) we require that \(f''(x)=0\) and changes sign at \(x\).
Note from part (d) that:
(1) If \(n \leq 2\) there is no point of inflection at \(x=0\).
(2) If \(n\) is even and greater than 2, then \(x=0\) is a zero of \(f''(x)\) with even multiplicity, hence \(f''(x)\) does not change sign there, so no point of inflection.
(3) If \(n\) is odd and greater than or equal to 3, then \(x=0\) is a zero of \(f''(x)\) with odd multiplicity, hence \(f''(x)\) changes sign there, so one point of inflection.
Now note from part (e)(i) that
(4) If \(n < 0\) then there are no non-zero values of \(x\) at which a point of inflection occurs;
(5) If \(n=1\) then there is a single point of inflection at \(1 + \sqrt{1} = 2\) (note that we already ruled out the point of inflection at \(1 - \sqrt{1} = 0\));
(6) If \(n \geq 2\) then there are two points of inflection at non-zero values of \(x\)
Combining everything, we have:
0 points of inflection requires \(n\) to be non-positive.
1 points of inflection requires \(n=1\).
2 points of inflection requires \(n\) to be even and greater than 2.
3 points of inflection requires \(n\) to be odd and greater than or equal to 3.
It's a reasonable question, but 2 marks is way too little for the amount of time it requires to answer it thoroughly and properly - although, marks were awarded just for correct answers, so if you guessed correctly, or just checked the graph for a few small values of n and extrapolated, then that could save time. I'm sure a lot of students either skipped this one or guessed quickly because they judged (quite correctly) that working it out properly wasn't worth the time.