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Date November 2020 Marks available 1 Reference code 20N.1.AHL.TZ0.H_12
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term State Question number H_12 Adapted from N/A

Question

Consider the function defined by fx=kx-5x-k, where x\k and k25

Consider the case where k=3.

State the equation of the vertical asymptote on the graph of y=f(x).

[1]
a.

State the equation of the horizontal asymptote on the graph of y=f(x).

[1]
b.

Use an algebraic method to determine whether f is a self-inverse function.

[4]
c.

Sketch the graph of y=f(x), stating clearly the equations of any asymptotes and the coordinates of any points of intersections with the coordinate axes.

[3]
d.

The region bounded by the x-axis, the curve y=f(x), and the lines x=5 and x=7 is rotated through 2π about the x-axis. Find the volume of the solid generated, giving your answer in the form π(a+b ln2) , where a, b.

[6]
e.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

x=k      A1


[1 mark]

a.

y=k      A1


[1 mark]

b.

METHOD 1

ffx=kkx-5x-k-5kx-5x-k-k        M1

=kkx-5-5x-kkx-5-kx-k        A1

=k2x-5k-5x+5kkx-5-kx+k2

=k2x-5xk2-5        A1

=xk2-5k2-5

=x

ffx=x , (hence f is self-inverse)        R1


Note:
The statement f(f(x))=x could be seen anywhere in the candidate’s working to award R1.

 

METHOD 2

fx=kx-5x-k

x=ky-5y-k        M1


Note:
Interchanging x and y can be done at any stage.


xy-k=ky-5        A1

xy-xk=ky-5

xy-ky=xk-5

yx-k=kx-5        A1

y=f-1x=kx-5x-k  (hence f is self-inverse)        R1


[4 marks]

c.

attempt to draw both branches of a rectangular hyperbola        M1

x=3 and y=3        A1

0, 53 and 53, 0        A1


[3 marks]

d.

METHOD 1

volume=π573x-5x-32dx       (M1)

EITHER

attempt to express 3x-5x-3 in the form p+qx-3       M1

3x-5x-3=3+4x-3       A1

OR

attempt to expand 3x-5x-32 or 3x-52 and divide out       M1

3x-5x-32=9+24x-56x-32       A1

THEN

3x-5x-32=9+24x-3+16x-32       A1

volume=π579+24x-3+16x-32dx

=π9x+24lnx-3-16x-357       A1

=π63+24ln4-4-45+24ln2-8

=π22+24ln2       A1

 

METHOD 2

volume=π573x-5x-32dx       (M1)

substituting u=x-3dudx=1       A1

3x-5=3u+3-5=3u+4

volume=π243u+4u2du       M1

=π249+16u2+24udu       A1

=π9u-16u+24lnu24       A1


Note: Ignore absence of or incorrect limits seen up to this point.


=π22+24ln2       A1


[6 marks]

e.

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.

Syllabus sections

Topic 2—Functions » SL 2.2—Functions, notation domain, range and inverse as reflection
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