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Date May 2021 Marks available 9 Reference code 21M.1.AHL.TZ2.12
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Prove Question number 12 Adapted from N/A

Question

The following diagram shows the graph of y=arctan2x+1+π4 for x, with asymptotes at y=-π4 and y=3π4.

Describe a sequence of transformations that transforms the graph of y=arctan x to the graph of y=arctan2x+1+π4 for x.

[3]
a.

Show that arctanp+arctanqarctanp+q1-pq where p, q>0 and pq<1.

[4]
b.

Verify that arctan 2x+1=arctan xx+1+π4 for x, x>0.

[3]
c.

Using mathematical induction and the result from part (b), prove that Σr=1narctan12r2=arctannn+1 for n+.

[9]
d.

Markscheme

EITHER
horizontal stretch/scaling with scale factor 12


Note: Do not allow ‘shrink’ or ‘compression’


followed by a horizontal translation/shift 12 units to the left           A2


Note: Do not allow ‘move’


OR

horizontal translation/shift 1 unit to the left

followed by horizontal stretch/scaling with scale factor 12     A2


THEN

vertical translation/shift up by π4 (or translation through 0π4          A1
(may be seen anywhere)

 

[3 marks]

a.

let α=arctanp and β=arctanq        M1

p=tanα and q=tanβ        (A1)

tanα+β=p+q1-pq        A1

α+β=arctanp+q1-pq        A1

so arctanp+arctanqarctanp+q1-pq where p, q>0 and pq<1.       AG

 

[4 marks]

b.

METHOD 1

π4=arctan1 (or equivalent)        A1

arctanxx+1+arctan1=arctanxx+1+11-xx+11        A1

=arctanx+x+1x+1x+1-xx+1        A1

=arctan2x+1       AG

 

METHOD 2

tanπ4=1 (or equivalent)        A1

Consider arctan2x+1-arctanxx+1=π4

tanarctan2x+1-arctanxx+1

=arctan2x+1-xx+11+x2x+1x+1        A1

=arctan2x+1x+1-xx+1+x2x+1        A1

=arctan 1       AG

 

METHOD 3

tan arctan2x+1=tanarctanxx+1+π4

tanπ4=1 (or equivalent)        A1

LHS=2x+1        A1

RHS=xx+1+11-xx+1=2x+1        A1

 

[3 marks]

c.

let Pn be the proposition that  Σr=1narctan12r2=arctannn+1 for n+

consider P1

when n=1, Σr=11arctan12r2=arctan12=RHS and so P1 is true          R1

assume Pk is true, ie. Σr=1karctan12r2=arctankk+1 k+           M1

 

Note: Award M0 for statements such as “let n=k”.
Note: Subsequent marks after this M1 are independent of this mark and can be awarded.

 

consider Pk+1:

Σr=1k+1arctan12r2=Σr=1karctan12r2+arctan12k+12           (M1)

=arctankk+1+arctan12k+12        A1

=arctankk+1+12k+121-kk+112k+12           M1

=arctank+12k2+2k+12k+13-k        A1

 

Note: Award A1 for correct numerator, with (k+1) factored. Denominator does not need to be simplified

 

=arctank+12k2+2k+12k3+6k2+5k+2        A1

 

Note: Award A1 for denominator correctly expanded. Numerator does not need to be simplified. These two A marks may be awarded in any order

 

=arctank+12k2+2k+1k+22k2+2k+1=arctank+1k+2        A1

 

Note: The word ‘arctan’ must be present to be able to award the last three A marks

 

Pk+1 is true whenever Pk is true and P1 is true, so

Pn is true for for n+         R1

 

Note: Award the final R1 mark provided at least four of the previous marks have been awarded.
Note: To award the final R1, the truth of Pk must be mentioned. ‘Pk implies Pk+1’ is insufficient to award the mark.

 

[9 marks]

d.

Examiners report

[N/A]
a.
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b.
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c.
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d.

Syllabus sections

Topic 1—Number and algebra » AHL 1.15—Proof by induction, contradiction, counterexamples
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Topic 1—Number and algebra

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