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Date November Example questions Marks available 4 Reference code EXN.1.AHL.TZ0.11
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Sketch Question number 11 Adapted from N/A

Question

A function f is defined by fx=3x2+2, x.

The region R is bounded by the curve y=fx, the x-axis and the lines x=0 and x=6. Let A be the area of R.

The line x=k divides R into two regions of equal area.

Let m be the gradient of a tangent to the curve y=fx.

Sketch the curve y=fx, clearly indicating any asymptotes with their equations and stating the coordinates of any points of intersection with the axes.

[4]
a.

Show that A=2π2.

[4]
b.

Find the value of k.

[4]
c.

Show that m=-6xx2+22.

[2]
d.

Show that the maximum value of m is 273223.

[7]
e.

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.

a curve symmetrical about the y-axis with correct concavity that has a local maximum point on the positive y-axis        A1

a curve clearly showing that y0 as x±        A1

0,32        A1

horizontal asymptote y=0 (x-axis)        A1

 

[4 marks]

a.

attempts to find 3x2+2dx        (M1)

=32arctanx2        A1

 

Note: Award M1A0 for obtaining k arctanx2 where k32.

Note: Condone the absence of or use of incorrect limits to this stage.

 

=32arctan3-arctan0        (M1)

=32×π3=π2        A1

A=2π2        AG

 

[4 marks]

b.

METHOD 1

EITHER

0k3x2+2dx=2π4

32arctank2=2π4        (M1)

 

OR

k63x2+2dx=2π4

32arctan3-arctank2=2π4        (M1)

arctan3-arctank2=π6

 

THEN

arctank2=π6        A1

k2=tanπ6=13        A1

k=63=23        A1

 

METHOD 2

0k3x2+2dx=k63x2+2dx

32arctank2=32arctan3-arctank2        (M1)

arctank2=π6        A1

k2=tanπ6=13        A1

k=63=23        A1

 

[4 marks]

c.

attempts to find ddx3x2+2        (M1)

=3-12xx2+2-2        A1

so m=-6xx2+22        AG

 

[2 marks]

d.

attempts product rule or quotient rule differentiation        M1

EITHER

dmdx=-6x-22xx2+2-3+x2+2-2-6        A1

OR

dmdx=x2+22-6--6x22xx2+2x2+24        A1

 

Note: Award A0 if the denominator is incorrect. Subsequent marks can be awarded.

 

THEN

attempts to express their dmdx as a rational fraction with a factorized numerator        M1

dmdx=6x2+23x2-2x2+24=63x2-2x2+23

attempts to solve their dmdx=0 for x        M1

x=±23        A1

from the curve, the maximum value of m occurs at x=-23        R1

(the minimum value of m occurs at x=23)

 

Note: Award R1 for any equivalent valid reasoning.

 

maximum value of m is -6-23-232+22        A1

leading to a maximum value of 273223        AG

 

[7 marks]

e.

Examiners report

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b.
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d.
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e.

Syllabus sections

Topic 2—Functions » SL 2.4—Key features of graphs, intersections using technology
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Topic 2—Functions

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