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Date	May 2021	Marks available	3	Reference code	21M.1.AHL.TZ2.12
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Describe	Question number	12	Adapted from	N/A

Question

The following diagram shows the graph of $y = arctan (2 x + 1) + \frac{π}{4}$ for $x \in ℝ$ , with asymptotes at $y = - \frac{π}{4}$ and $y = \frac{3 π}{4}$ .

Describe a sequence of transformations that transforms the graph of $y = arctan x$ to the graph of $y = arctan (2 x + 1) + \frac{π}{4}$ for $x \in ℝ$ .

[3]

Show that $arctan p + arctan q \equiv arctan (\frac{p + q}{1 - p q})$ where $p, q > 0$ and $p q < 1$ .

[4]

Verify that $arctan (2 x + 1) = arctan (\frac{x}{x + 1}) + \frac{π}{4}$ for $x \in ℝ, x > 0$ .

[3]

Using mathematical induction and the result from part (b), prove that $Σ_{r = 1}^{n} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{n}{n + 1})$ for $n \in ℤ^{+}$ .

[9]

Markscheme

EITHER
horizontal stretch/scaling with scale factor $\frac{1}{2}$

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

followed by a horizontal translation/shift $\frac{1}{2}$ units to the left A2

Note: Do not allow ‘move’

horizontal translation/shift $1$ unit to the left

followed by horizontal stretch/scaling with scale factor $\frac{1}{2}$ A2

THEN

vertical translation/shift up by $\frac{π}{4}$ (or translation through $(\begin{matrix} 0 \\ \frac{π}{4} \end{matrix})$ A1
(may be seen anywhere)

[3 marks]

let $α = arctan p$ and $β = arctan q$ M1

$p = tan α$ and $q = tan β$ (A1)

$\tan (α + β) = \frac{p + q}{1 - p q}$ A1

$α + β = arctan (\frac{p + q}{1 - p q})$ A1

so $arctan p + arctan q \equiv arctan (\frac{p + q}{1 - p q})$ where $p, q > 0$ and $p q < 1$ . AG

[4 marks]

METHOD 1

$\frac{π}{4} = arctan 1$ (or equivalent) A1

$arctan (\frac{x}{x + 1}) + arctan 1 = arctan (\frac{\frac{x}{x + 1} + 1}{1 - \frac{x}{x + 1} (1)})$ A1

$= arctan (\frac{\frac{x + x + 1}{x + 1}}{\frac{x + 1 - x}{x + 1}})$ A1

$= arctan (2 x + 1)$ AG

METHOD 2

$\tan \frac{π}{4} = 1$ (or equivalent) A1

Consider $arctan (2 x + 1) - arctan (\frac{x}{x + 1}) = \frac{π}{4}$

$\tan (arctan (2 x + 1) - arctan (\frac{x}{x + 1}))$

$= arctan (\frac{2 x +1 - \frac{x}{x + 1}}{1 + \frac{x (2 x + 1)}{x + 1}})$ A1

$= arctan (\frac{(2 x +1) (x + 1) - x}{x + 1 + x (2 x + 1)})$ A1

$= arctan 1$ AG

METHOD 3

$tan (arctan (2 x + 1)) = \tan (arctan (\frac{x}{x + 1}) + \frac{π}{4})$

$\tan \frac{π}{4} = 1$ (or equivalent) A1

$LHS = 2 x + 1$ A1

$RHS = \frac{\frac{x}{x + 1} + 1}{1 - \frac{x}{x + 1}} (= 2 x + 1)$ A1

[3 marks]

let $P (n)$ be the proposition that $Σ_{r = 1}^{n} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{n}{n + 1})$ for $n \in ℤ^{+}$

consider $P (1)$

when $n = 1, Σ_{r = 1}^{1} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{1}{2}) = RHS$ and so $P (1)$ is true R1

assume $P (k)$ is true, ie. $Σ_{r = 1}^{k} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{k}{k + 1}) (k \in ℤ^{+})$ 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 $P (k + 1)$ :

$Σ_{r = 1}^{k + 1} arctan (\frac{1}{2 r^{2}}) = Σ_{r = 1}^{k} arctan (\frac{1}{2 r^{2}}) + arctan (\frac{1}{2 {(k + 1)}^{2}})$ (M1)

$= arctan (\frac{k}{k + 1}) + arctan (\frac{1}{2 {(k + 1)}^{2}})$ A1

$= arctan (\frac{\frac{k}{k + 1} + \frac{1}{2 {(k + 1)}^{2}}}{1 - (\frac{k}{k + 1}) (\frac{1}{2 {(k + 1)}^{2}})})$ M1

$= arctan (\frac{(k + 1) (2 k^{2} + 2 k + 1)}{2 {(k + 1)}^{3} - k})$ A1

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

$= arctan (\frac{(k + 1) (2 k^{2} + 2 k + 1)}{2 k^{3} + 6 k^{2} + 5 k + 2})$ 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

$= arctan (\frac{(k + 1) (2 k^{2} + 2 k + 1)}{(k + 2) (2 k^{2} + 2 k + 1)}) = arctan (\frac{k + 1}{k + 2})$ A1

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

$P (k + 1)$ is true whenever $P (k)$ is true and $P (1)$ is true, so

$P (n)$ is true for for $n \in ℤ^{+}$ 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 $P (k)$ must be mentioned. ‘ $P (k)$ implies $P (k + 1)$ ’ is insufficient to award the mark.

[9 marks]

Examiners report

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.11—Transformation of functions

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19M.2.AHL.TZ1.H_10c:
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19M.2.AHL.TZ1.H_10d:
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16N.2.SL.TZ0.S_4b:
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21M.1.SL.TZ1.1b:
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20N.1.SL.TZ0.S_10c:
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- points $D$ and $F$ lie on the horizontal asymptote of $g$
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Line $L_{2}$ is the tangent to the graph of $g$ at $Q$ , and passes through $E$ and $F$ .

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