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Date November Example question Marks available 3 Reference code EXN.2.SL.TZ0.6
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Hence and Find Question number 6 Adapted from N/A

Question

A theatre set designer is designing a piece of flat scenery in the shape of a hill. The scenery is formed by a curve between two vertical edges of unequal height. One edge is 2 metres high and the other is 1 metre high. The width of the scenery is 6 metres.

A coordinate system is formed with the origin at the foot of the 2 metres high edge. In this coordinate system the highest point of the cross‐section is at 2, 3.5.

A set designer wishes to work out an approximate value for the area of the scenery (Am2 ).

In order to obtain a more accurate measure for the area the designer decides to model the curved edge with the polynomial hx=ax3+bx2+cx+d  a,b,c,d where h metres is the height of the curved edge a horizontal distance xm from the origin.

Explain why A<21.

[1]
a.

By dividing the area between the curve and the x‐axis into two trapezoids of unequal width show that A>14.5, justifying the direction of the inequality.

[4]
b.

Write down the value of d.

[1]
c.

Use differentiation to show that 12a+4b+c=0.

[2]
d.

Determine two other linear equations in a, b and c.

[3]
e.

Hence find an expression for hx.

[3]
f.

Use the expression found in (f) to calculate a value for A.

[2]
g.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

The area A is less than the rectangle containing the cross-section which is equal to 6×3.5=21        R1

 

Note: 6×3.5=21 is not sufficient for R1.

 

[1 mark]

a.

12×2×2+3.5+12×4×3.5+1        (M1)(A1)

=14.5        A1

This is an underestimate as the trapezoids are enclosed by (are under) the curve.        R1

 

Note: This can be shown in a diagram.

 

[4 marks]

b.

h0=2d=2       A1

 

[1 mark]

c.

h'x=3ax2+2bx+c       A1

h'2=0       M1

hence 12a+4b+c=0       AG

 

[2 marks]

d.

Substitute the points 2, 3.5 and 6, 1        (M1)

8a+4b+2c+2=3.5 8a+4b+2c=1.5

and

216a+36b+6c+2=1 216a+36b+6c=-1        A1A1

 

[3 marks]

e.

Solve on a GDC        (M1)

hx=0.0365x3-0.521x2+1.65x+2        A2

hx=0.0364583x3-0.520833x2+1.64583x+2

 

[3 marks]

f.

060.0364583x3-0.520833x2+1.64583x+2dx

=15.9 15.9374 m2       (M1)A1

 

Note: Accept 16.0 (16.014) from the three significant figure answer to part (g).

 

[2 marks]

g.

Examiners report

[N/A]
a.
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b.
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c.
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d.
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e.
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f.
[N/A]
g.

Syllabus sections

Topic 2—Functions » SL 2.5—Modelling functions
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Topic 2—Functions

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