User interface language: English | Español

Date May 2011 Marks available 4 Reference code 11M.2.sl.TZ2.10
Level SL only Paper 2 Time zone TZ2
Command term Find Question number 10 Adapted from N/A

Question

The diagram below shows a plan for a window in the shape of a trapezium.


Three sides of the window are \(2{\text{ m}}\) long. The angle between the sloping sides of the window and the base is \(\theta \) , where \(0 < \theta  < \frac{\pi }{2}\) .

Show that the area of the window is given by \(y = 4\sin \theta + 2\sin 2\theta \) .

[5]
a.

Zoe wants a window to have an area of \(5{\text{ }}{{\text{m}}^2}\). Find the two possible values of \(\theta \) .

[4]
b.

John wants two windows which have the same area A but different values of \(\theta \) .

Find all possible values for A .

[7]
c.

Markscheme

evidence of finding height, h     (A1)

e.g. \(\sin \theta = \frac{h}{2}\) , \(2\sin \theta \)

evidence of finding base of triangle, b     (A1)

e.g. \(\cos \theta = \frac{b}{2}\) , \(2\cos \theta \)

attempt to substitute valid values into a formula for the area of the window     (M1)

e.g. two triangles plus rectangle, trapezium area formula

correct expression (must be in terms of \(\theta \) )     A1

e.g. \(2\left( {\frac{1}{2} \times 2\cos \theta \times 2\sin \theta } \right) + 2 \times 2\sin \theta \) , \(\frac{1}{2}(2\sin \theta )(2 + 2 + 4\cos \theta )\)

attempt to replace \(2\sin \theta \cos \theta \) by \(\sin 2\theta \)     M1

e.g. \(4\sin \theta + 2(2\sin \theta \cos \theta )\)

\(y = 4\sin \theta + 2\sin 2\theta \)     AG     N0

[5 marks]

a.

correct equation     A1

e.g. \(y = 5\) , \(4\sin \theta + 2\sin 2\theta = 5\)

evidence of attempt to solve     (M1)

e.g. a sketch, \(4\sin \theta + 2\sin \theta - 5 = 0\)

\(\theta = 0.856\) \(({49.0^ \circ })\) , \(\theta = 1.25\) \(({71.4^ \circ })\)     A1A1     N3

[4 marks]

b.

recognition that lower area value occurs at \(\theta = \frac{\pi }{2}\)     (M1)

finding value of area at \(\theta  = \frac{\pi }{2}\)     (M1)

e.g. \(4\sin \left( {\frac{\pi }{2}} \right) + 2\sin \left( {2 \times \frac{\pi }{2}} \right)\) , draw square

\(A = 4\)     (A1)

recognition that maximum value of y is needed     (M1)

\(A = 5.19615 \ldots \)     (A1)

\(4 < A < 5.20\) (accept \(4 < A < 5.19\) )     A2      N5

[7 marks]

c.

Examiners report

As the final question of the paper, this question was understandably challenging for the majority of the candidates. Part (a) was generally attempted, but often with a lack of method or correct reasoning. Many candidates had difficulty presenting their ideas in a clear and organized manner. Some tried a "working backwards" approach, earning no marks.

a.

In part (b), most candidates understood what was required and set up an equation, but many did not make use of the GDC and instead attempted to solve this equation algebraically which did not result in the correct solution. A common error was finding a second solution outside the domain.

b.

A pleasing number of stronger candidates made progress on part (c), recognizing the need for the end point of the domain and/or the maximum value of the area function (found graphically, analytically, or on occasion, geometrically). However, it was evident from candidate work and teacher comments that some candidates did not understand the wording of the question. This has been taken into consideration for future paper writing.

c.

Syllabus sections

Topic 3 - Circular functions and trigonometry » 3.5 » Solving trigonometric equations in a finite interval, both graphically and analytically.
Show 41 related questions

View options