Year: 2021
Paper: 2
Question Number: 3
Course: LFM Pure
Section: Simultaneous 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
In this question, $x$, $y$ and $z$ are real numbers.
Let $\lfloor x \rfloor$ denote the largest integer that satisfies $\lfloor x \rfloor \leqslant x$ and let $\{x\}$ denote the fractional part of~$x$, so that $x = \lfloor x \rfloor + \{x\}$ and $0 \leqslant \{x\} < 1$. For example, if $x = 4.2$, then $\lfloor x \rfloor = 4$ and $\{x\} = 0.2$ and if $x = -4.2$, then $\lfloor x \rfloor = -5$ and $\{x\} = 0.8$.
\begin{questionparts}
\item Solve the simultaneous equations
\begin{align*}
\lfloor x \rfloor + \{y\} &= 4.9, \\
\{x\} + \lfloor y \rfloor &= -1.4.
\end{align*}
\item Given that $x$, $y$ and $z$ satisfy the simultaneous equations
\begin{align*}
\lfloor x \rfloor + y + \{z\} &= 3.9, \\
\{x\} + \lfloor y \rfloor + z &= 5.3, \\
x + \{y\} + \lfloor z \rfloor &= 5,
\end{align*}
show that $\{y\} + z = 3.2$ and solve the equations.
\item Solve the simultaneous equations
\begin{align*}
\lfloor x \rfloor + 2y + \{z\} &= 3.9, \\
\{x\} + 2\lfloor y \rfloor + z &= 5.3, \\
x + 2\{y\} + \lfloor z \rfloor &= 5.
\end{align*}
\end{questionparts}
This was a popular question, attempted by a large proportion of the candidates. Candidates who were able to appreciate the method by which the integer and fractional parts could be interpreted to find the original values were able to make good progress and gain high marks with relatively short solutions. Those who did not see this could produce many pages of work without making significant progress towards a solution. In part (i) candidates were often able to deduce the values of x and y successfully, but some did not remember that the fractional part was defined as positive in the explanation at the start of the question, meaning that they found options for the final answer. A variety of successful methods were seen for part (ii) and there were a high proportion of perfect answers. The most successful approach was to combine the simultaneous equations to reach the given two-variable equation. Another method was to analyse the set of 8 different cases to identify the unique solution. The most common problem encountered with this approach was to fail to identify all of the possible cases. Part (iii) was found to be difficult by many of the candidates. While many were able to find the "obvious" solution of halving the value from (ii), the complication presented by the coefficient of 2 was not appreciated by all. Those who did were often then able to earn most of the marks for this part.