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Date May Example question Marks available 3 Reference code EXM.3.AHL.TZ0.7
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Find Question number 7 Adapted from N/A

Question

This question explores methods to determine the area bounded by an unknown curve.

The curve y=f(x) is shown in the graph, for 0x4.4.

The curve y=f(x) passes through the following points.

It is required to find the area bounded by the curve, the x-axis, the y-axis and the line x=4.4.

One possible model for the curve y=f(x) is a cubic function.

A second possible model for the curve y=f(x) is an exponential function, y=peqx, where p,qR.

Use the trapezoidal rule to find an estimate for the area.

[3]
a.i.

With reference to the shape of the graph, explain whether your answer to part (a)(i) will be an over-estimate or an underestimate of the area.

[2]
a.ii.

Use all the coordinates in the table to find the equation of the least squares cubic regression curve.

[3]
b.i.

Write down the coefficient of determination.

[1]
b.ii.

Write down an expression for the area enclosed by the cubic function, the x-axis, the y-axis and the line x=4.4.

[2]
c.i.

Find the value of this area.

[2]
c.ii.

Show that lny=qx+lnp.

[2]
d.i.

Hence explain how a straight line graph could be drawn using the coordinates in the table.

[1]
d.ii.

By finding the equation of a suitable regression line, show that p=1.83 and q=0.986.

[5]
d.iii.

Hence find the area enclosed by the exponential function, the x-axis, the y-axis and the line x=4.4.

[2]
d.iv.

Markscheme

Area =1.12(2+2(5+15+47)+148)         M1A1

Area = 156 units2          A1

[3 marks]

a.i.

The graph is concave up,         R1

so the trapezoidal rule will give an overestimate.         A1

[2 marks]

a.ii.

f(x)=3.88x312.8x2+14.1x+1.54         M1A2

[3 marks]

b.i.

R2=0.999        A1

[1 mark]

b.ii.

Area =4.40(3.88x312.8x2+14.1x+1.54)dx        A1A1

[2 marks]

c.i.

Area = 145 units2    (Condone 143–145 units2, using rounded values.)      A2

[2 marks]

c.ii.

lny=ln(peqx)      M1

lny=lnp+ln(eqx)      A1

lny=qx+lnp      AG

[2 marks]

d.i.

Plot lny against p.      R1

[1 mark]

d.ii.

Regression line is lny=0.986x+0.602       M1A1

So q= gradient = 0.986    R1

p=e0.602=1.83       M1A1

[5 marks]

d.iii.

Area =4.401.83e0.986xdx=140 units2     M1A1

[2 marks]

d.iv.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.i.
[N/A]
d.ii.
[N/A]
d.iii.
[N/A]
d.iv.

Syllabus sections

Topic 5—Calculus » SL 5.8—Trapezoid rule
Show 21 related questions
Topic 2—Functions » AHL 2.10—Scaling large numbers, log-log graphs
Topic 4—Statistics and probability » AHL 4.13—Non-linear regression
Topic 2—Functions
Topic 4—Statistics and probability
Topic 5—Calculus

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