Pre-U Pre-U 9794/2 2014 June — Question 12 9 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
Year2014
SessionJune
Marks9
TopicParametric curves and Cartesian conversion
TypeFind intersection points
DifficultyStandard +0.8 This question requires finding when x=0 (solving a trigonometric equation with multiple solutions), then determining corresponding y-values, followed by analyzing when y>0 using sign analysis of a product involving trigonometric functions. It demands careful case-by-case reasoning and understanding of parametric curves beyond routine manipulation, but uses standard A-level techniques without requiring novel insights.
Spec1.03g Parametric equations: of curves and conversion to cartesian
12 A curve \(C\) is defined parametrically by $$x = \cos t ( 1 - 2 \sin t ) , \quad y = \sin t ( 1 - 3 \sin t ) , \quad 0 \leqslant t < 2 \pi$$
  1. Show that \(C\) intersects the \(y\)-axis at exactly three points, and state the values of \(t\) and \(y\) at these points.
  2. Find the range of values of \(t\) for which \(C\) lies above the \(x\)-axis.
(i) \(\cos t = 0\) or \(\sin t = \dfrac{1}{2}\)
M1 Attempt to solve at least one of these
\(\cos t = 0 \Rightarrow t = \dfrac{1}{2}\pi, \dfrac{3}{2}\pi\)
\(y = -2\) and \(-4\) respectively
A1 Obtain both values for \(t\)
A1 Obtain both values for \(y\). SR A1 for one correct \(t\), \(y\) pair
\(\sin t = \dfrac{1}{2} \Rightarrow t = \dfrac{1}{6}\pi, \dfrac{5}{6}\pi\)
\(y = -\dfrac{1}{4}\) for both values of \(t\) so there is only one point on the \(y\)-axis associated with both.
A1 Obtain both values for \(t\)
A1 [5] Obtain \(y = \dfrac{-1}{4}\) for both, and comment that same point – allow just listing \((0, \dfrac{-1}{4})\) once. SR A1 for one correct \(t\), \(y\) pair. Max of 4/5 if working in degrees
(ii) \(\sin t < 0\) AND \(\sin t > \dfrac{1}{3}\), but this is not possible
B1
Identify that \(\sin t > 0\) AND \(\sin t < \dfrac{1}{3}\)
M1 If equating to 0 and solving then both inequalities must be used/implied later to get M1
so \(t \in \left(0, \sin^{-1}\dfrac{1}{3}\right) \cup \left(\pi - \sin^{-1}\dfrac{1}{3}, \pi\right)\) oe
A1 Obtain at least \(0 < t < \sin^{-1}(\dfrac{1}{3})\). Allow \(0 < t < 0.34\). Allow \(\geq\) for \(>\)
A1 [4] Allow \(0 < t < 0.34\), \(2.80 < t < 3.14\). Working in degrees can get M1A1 only
(i) $\cos t = 0$ or $\sin t = \dfrac{1}{2}$

M1 Attempt to solve at least one of these

$\cos t = 0 \Rightarrow t = \dfrac{1}{2}\pi, \dfrac{3}{2}\pi$
$y = -2$ and $-4$ respectively

A1 Obtain both values for $t$
A1 Obtain both values for $y$. SR A1 for one correct $t$, $y$ pair

$\sin t = \dfrac{1}{2} \Rightarrow t = \dfrac{1}{6}\pi, \dfrac{5}{6}\pi$
$y = -\dfrac{1}{4}$ for both values of $t$ so there is only one point on the $y$-axis associated with both.

A1 Obtain both values for $t$
A1 [5] Obtain $y = \dfrac{-1}{4}$ for both, and comment that same point – allow just listing $(0, \dfrac{-1}{4})$ once. SR A1 for one correct $t$, $y$ pair. Max of 4/5 if working in degrees

(ii) $\sin t < 0$ AND $\sin t > \dfrac{1}{3}$, but this is not possible

B1

Identify that $\sin t > 0$ AND $\sin t < \dfrac{1}{3}$

M1 If equating to 0 and solving then both inequalities must be used/implied later to get M1

so $t \in \left(0, \sin^{-1}\dfrac{1}{3}\right) \cup \left(\pi - \sin^{-1}\dfrac{1}{3}, \pi\right)$ oe

A1 Obtain at least $0 < t < \sin^{-1}(\dfrac{1}{3})$. Allow $0 < t < 0.34$. Allow $\geq$ for $>$

A1 [4] Allow $0 < t < 0.34$, $2.80 < t < 3.14$. Working in degrees can get M1A1 only
12 A curve $C$ is defined parametrically by

$$x = \cos t ( 1 - 2 \sin t ) , \quad y = \sin t ( 1 - 3 \sin t ) , \quad 0 \leqslant t < 2 \pi$$

(i) Show that $C$ intersects the $y$-axis at exactly three points, and state the values of $t$ and $y$ at these points.\\
(ii) Find the range of values of $t$ for which $C$ lies above the $x$-axis.

\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2014 Q12 [9]}}
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