Year: 2023
Paper: 2
Question Number: 2
Course: LFM Pure
Section: Trigonometry 2
No solution available for this problem.
Many candidates were able to express their reasoning clearly and presented good solutions to the questions that they attempted. There were excellent solutions seen for all of the questions. An area where candidates struggled in several questions was in the direction of the logic that was required in a solution. Some candidates failed to appreciate that separate arguments may be needed for the "if" and "only if" parts of a question and, in some cases, candidates produced correct arguments, but for the wrong direction. In several questions it was clear that candidates who used sketches or diagrams generally performed much better that those who did not. Sketches often also helped to make the solution clearer and easier to understand. Several questions on the STEP papers ask candidates to show a given result. Candidates should be aware that there is a need to present sufficient detail in their solutions so that it is clear that the reasoning is well understood.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item The real numbers $x$, $y$ and $z$ satisfy the equations
\[y = \frac{2x}{1-x^2}\,,\qquad z = \frac{2y}{1-y^2}\,,\qquad x = \frac{2z}{1-z^2}\,.\]
Let $x = \tan\alpha$. Deduce that $y = \tan 2\alpha$ and show that $\tan\alpha = \tan 8\alpha$.
Find all solutions of the equations, giving each value of $x$, $y$ and $z$ in the form $\tan\theta$ where $-\frac{1}{2}\pi < \theta < \frac{1}{2}\pi$.
\item Determine the number of real solutions of the simultaneous equations
\[y = \frac{3x - x^3}{1-3x^2}\,,\qquad z = \frac{3y - y^3}{1-3y^2}\,,\qquad x = \frac{3z - z^3}{1-3z^2}\,.\]
\item Consider the simultaneous equations
\[y = 2x^2 - 1\,,\qquad z = 2y^2 - 1\,,\qquad x = 2z^2 - 1\,.\]
\begin{enumerate}
\item[(a)] Determine the number of real solutions of these simultaneous equations with $|x| \leqslant 1$, $|y| \leqslant 1$, $|z| \leqslant 1$.
\item[(b)] By finding the degree of a single polynomial equation which is satisfied by $x$, show that all solutions of these simultaneous equations have $|x| \leqslant 1$, $|y| \leqslant 1$, $|z| \leqslant 1$.
\end{enumerate}
\end{questionparts}
In part (i) most attempts to show tan α = tan 8α were successful and included sufficient detail to earn the marks. Some candidates attempted to use the half-angle formula instead of the double-angle formula. This does not work, as the logic goes in the wrong direction, and leads to a quadratic with two solutions following which candidates simply asserted that the half angle is the correct solution. A large proportion of students made no further progress on this question. Of the students that did progress further on this part, many became confused by the restrictions on range. They realised solutions were of the form (tan α, tan 2α, tan 4α), but then tried to simultaneously have α, 2α, and 4α between −π/2 and π/2. These solutions erroneously discarded (or did not even find) solutions other than (0, 0, 0). However, some students did realise that they could subtract or add multiples of π to some of the arguments until all were in the required range. These attempts often obtained full marks for this part. Some attempts used an alternative method, rather than using periodicity of tan to solve tan α = tan 8α, they rewrote in terms of sin and cos and used addition formulae to obtain sin 7α = 0. However, these attempts often only checked the logic in one direction and did not comment that cos α and cos 8α were non-zero in these cases. In part (ii) a good number of students attempted the substitution x = tan α, and many of these either quoted or proved the triple angle formula for tan. Again, some made no further progress (often scripts that attempted both (i) and (ii) made similar amounts of progress on both parts). Since this part asked for the number of solutions, rather than finding all solutions, many students only found solutions for x. This lost credit unless it was accompanied by a check that each value of x led to a value of y and z. Several candidates failed to discard x = ±π/2, for which tan(x) is undefined, leading to an answer of 27, rather than 25. In part (iii)(a) many students attempted a useful trigonometric substitution in this part (either cos or sin). Most scripts that attempted a useful substitution made correct use of the double angle formula to arrive at cos α = cos 8α or a similar equation using a sin substitution. Solving the equation cos(α) = cos(8α) gave many students significant difficulty. Many attempts used only the periodicity of cos, and not the evenness, thus only obtaining half of the solutions. Others failed to restrict to a range where cos is single-valued, thus finding the same solution for x for different values of α and erroneously counting these as different solutions. Those who chose to draw a sketch of the graph to aid their thinking generally produced better solutions in this part. In part (iii)(b) most attempts found the correct octic. Attempts that had found fewer than 8 solutions in (a) often made no further progress. Candidates who had found 8 solutions in (a) often obtained full marks in this part.