Year: 2021
Paper: 2
Question Number: 5
Course: LFM Pure
Section: Differential equations
No solution available for this problem.
Candidates were generally well prepared for many of the questions on this paper, with the questions requiring more standard operations seeing the greatest levels of success. Candidates need to ensure that solutions to the questions are supported by sufficient evidence of the mathematical steps, for example when proving a given result or deducing the properties of graphs that are to be sketched. In a significant number of steps there were marks lost through simple errors such as mistakes in arithmetic or confusion of sine and cosine functions, so it is important for candidates to maintain accuracy in their solutions to these questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Use the substitution $y = (x - a)u$, where $u$ is a function of $x$, to solve the differential equation
\[
(x - a)\frac{dy}{dx} = y - x,
\]
where $a$ is a constant.
\item The curve $C$ with equation $y = f(x)$ has the property that, for all values of $t$ except $t = 1$, the tangent at the point $\bigl(t,\, f(t)\bigr)$ passes through the point $(1, t)$.
\begin{enumerate}[label=(\alph*)]
\item Given that $f(0) = 0$, find $f(x)$ for $x < 1$.
Sketch $C$ for $x < 1$. You should find the coordinates of any stationary points and consider the gradient of $C$ as $x \to 1$. You may assume that $z\ln|z| \to 0$ as $z \to 0$.
\item Given that $f(2) = 2$, sketch $C$ for $x > 1$, giving the coordinates of any stationary points.
\end{enumerate}
\end{questionparts}
This proved to be a popular question. In part (i), many candidates were able to use the substitution to reduce the differential equation into a form where the variables could be separated, but a surprising number struggled with the integral that resulted from this process. A variety of approaches were successfully employed by those who were able to complete the integration, but candidates often forgot the modulus function inside the logarithm, which caused problems later in the question. A small number of candidates forgot that the constant of integration would also be multiplied by (x − a) in the final step of this part of the question. In part (ii) some candidates were unsure how to use the information given about the tangent. Those who set a = 1 were generally able to make good progress and many correct sketches were produced. A number of candidates assumed, without justification, that the form of f(x) would remain unchanged from part (a) to part (b).