Year: 2020
Paper: 2
Question Number: 8
Course: LFM Pure
Section: Integration
No solution available for this problem.
There were just over 800 entries for this paper, and good solutions were seen to all of the questions. Candidates should be aware of the need to provide clear explanations of their reasoning throughout the paper (and particularly in questions where the result to be shown is given in the question). Short explanatory comments at key points in solutions can be very helpful in this regard, as can clearly drawn diagrams of the situation described in the question. The paper included a few questions where a statement of the form "A if and only if B" needed to be proven – candidates should be aware of the meaning of such statements and make sure that both directions of the implication are covered clearly. In general, candidates who performed better on the questions in this paper recognised the relationships between the different parts of each question and were able to adapt methods used in earlier parts when working on the later sections of the question.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, $\mathrm{f}(x)$ is a quartic polynomial where the coefficient of $x^4$ is equal to $1$, and which has four real roots, $0$, $a$, $b$ and $c$, where $0 < a < b < c$.
$\mathrm{F}(x)$ is defined by $\mathrm{F}(x) = \displaystyle\int_0^x \mathrm{f}(t)\,\mathrm{d}t$.
The area enclosed by the curve $y = \mathrm{f}(x)$ and the $x$-axis between $0$ and $a$ is equal to that between $b$ and $c$, and half that between $a$ and $b$.
\begin{questionparts}
\item Sketch the curve $y = \mathrm{F}(x)$, showing the $x$ co-ordinates of its turning points.
Explain why $\mathrm{F}(x)$ must have the form $\mathrm{F}(x) = \frac{1}{5}x^2(x-c)^2(x-h)$, where $0 < h < c$.
Find, in factorised form, an expression for $\mathrm{F}(x) + \mathrm{F}(c-x)$ in terms of $c$, $h$ and $x$.
\item If $0 \leqslant x \leqslant c$, explain why $\mathrm{F}(b) + \mathrm{F}(x) \geqslant 0$ and why $\mathrm{F}(b) + \mathrm{F}(x) > 0$ if $x \neq a$.
Hence show that $c - b = a$ or $c > 2h$.
By considering also $\mathrm{F}(a) + \mathrm{F}(x)$, show that $c = a + b$ and that $c = 2h$.
\item Find an expression for $\mathrm{f}(x)$ in terms of $c$ and $x$ only.
Show that the points of inflection on $y = \mathrm{f}(x)$ lie on the $x$-axis.
\end{questionparts}
For the first part candidates were asked to sketch a curve y = F(x) based on some information about the function f(x). A not insignificant number of candidates instead sketched y = f(x) but those who sketched the correct curve often earned most of the marks. When justifying the given form of F(x) some good explanations were provided, but in many cases the repeated roots at x = 0 and x = c were not explained. The final section of part (i) was generally completed well by those who reached it. The next part was found to be more difficult with many candidates mistakenly using the local maximum of F(x) at x = b to justify the first inequality instead of the local minimum at x = a. It was common to see justification such as |F(x)| < F(b) without showing first that F(b) = −F(a). Candidates who spotted the connection with part (i) and substituted x = b into their expression for F(x) + F(c − x) were usually able to show c − b = a or c > 2h. For the last section of part (ii), those who realised the connection with the first paragraph had no issues. Candidates who reached the final part of the question were often able to obtain expression for f(x) and most realised that they needed to calculate f″(x) in order to find the inflection points. However, the final mark for spotting that the roots of f″(x) = 0 are necessarily roots of f(x) = 0 without explicitly calculating them (and thereby wasting time) eluded the majority of candidates who reached this part.