| Date | May 2009 | Marks available | 18 | Reference code | 09M.2.hl.TZ2.12 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Explain, Find, and Show that | Question number | 12 | Adapted from | N/A |
Question
Two non-intersecting circles C1 , containing points M and S , and C2 , containing points N and R, have centres P and Q where PQ \( = 50\) . The line segments [MN] and [SR] are common tangents to the circles. The size of the reflex angle MPS is \( \alpha\), the size of the obtuse angle NQR is \( \beta\) , and the size of the angle MPQ is \( \theta\) . The arc length MS is \({l_1}\) and the arc length NR is \({l_2}\) . This information is represented in the diagram below.
The radius of C1 is \(x\) , where \(x \geqslant 10\) and the radius of C2 is \(10\).
(a) Explain why \(x < 40\) .
(b) Show that cosθ = x −10 50.
(c) (i) Find an expression for MN in terms of \(x\) .
(ii) Find the value of \(x\) that maximises MN.
(d) Find an expression in terms of \(x\) for
(i) \( \alpha\) ;
(ii) \( \beta\) .
(e) The length of the perimeter is given by \({l_1} + {l_2} + {\text{MN}} + {\text{SR}}\).
(i) Find an expression, \(b (x)\) , for the length of the perimeter in terms of \(x\) .
(ii) Find the maximum value of the length of the perimeter.
(iii) Find the value of \(x\) that gives a perimeter of length \(200\).
Markscheme
(a) PQ \( = 50\) and non-intersecting R1
[1 mark]
(b) a construction QT (where T is on the radius MP), parallel to MN, so that \({\text{Q}}\hat {\text{T}}{\text{M}} = 90^\circ \) (angle between tangent and radius \( = 90^\circ \) ) M1
lengths \(50\), \(x - 10\) and angle \( \theta\) marked on a diagram, or equivalent R1
Note: Other construction lines are possible.
[2 marks]
(c) (i) MN \( = \sqrt {{{50}^2} - {{\left( {x - 10} \right)}^2}} \) A1
(ii) maximum for MN occurs when \(x = 10\) A1
[2 marks]
(d) (i) \(\alpha = 2\pi - 2\theta \) M1
\( = 2\pi - 2\arccos \left( {\frac{{x - 10}}{{50}}} \right)\) A1
(ii) \(\beta = 2\pi - \alpha \) ( \( = 2\theta \) ) A1
\( = 2\left( {{{\cos }^{ - 1}}\left( {\frac{{x - 10}}{{50}}} \right)} \right)\) A1
[4 marks]
(e) (i) \(b(x) = x\alpha + 10\beta + 2\sqrt {{{50}^2} - {{\left( {x - 10} \right)}^2}} \) A1A1A1
\( = x\left( {2\pi - 2\left( {{{\cos }^{ - 1}}\left( {\frac{{x - 10}}{{50}}} \right)} \right)} \right) + 20\left( {\left( {{{\cos }^{ - 1}}\left( {\frac{{x - 10}}{{50}}} \right)} \right)} \right) + 2\sqrt {{{50}^2} - {{\left( {x - 10} \right)}^2}} \) M1A1
(ii) maximum value of perimeter \( = 276\) A2
(iii) perimeter of \(200\) cm \(b(x) = 200\) (M1)
when \(x = 21.2\) A1
[9 marks]
Total [18 marks]
Examiners report
This is not an inherently difficult question, but candidates either made heavy weather of it or avoided it almost entirely. The key to answering the question is in obtaining the displayed answer to part (b), for which a construction line parallel to MN through Q is required. Diagrams seen by examiners on some scripts tend to suggest that the perpendicularity property of a tangent to a circle and the associated radius is not as firmly known as they had expected. Some candidates mixed radians and degrees in their expressions.
