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Date	May 2022	Marks available	1	Reference code	22M.1.AHL.TZ1.10
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Show that and Hence or otherwise	Question number	10	Adapted from	N/A

Question

Consider the series $\ln x + p \ln x + \frac{1}{3} \ln x + \dots$ , where $x \in ℝ, x > 1$ and $p \in ℝ, p \neq 0$ .

Consider the case where the series is geometric.

Now consider the case where the series is arithmetic with common difference $d$ .

Show that $p = \pm \frac{1}{\sqrt{3}}$ .

[2]

a.i.

Hence or otherwise, show that the series is convergent.

[1]

a.ii.

Given that $p > 0$ and $S_{\infty} = 3 + \sqrt{3}$ , find the value of $x$ .

[3]

a.iii.

Show that $p = \frac{2}{3}$ .

[3]

b.i.

Write down $d$ in the form $k \ln x$ , where $k \in ℚ$ .

[1]

b.ii.

The sum of the first $n$ terms of the series is $\ln (\frac{1}{x^{3}})$ .

Find the value of $n$ .

[8]

b.iii.

Markscheme

EITHER

attempt to use a ratio from consecutive terms M1

$\frac{p \ln x}{\ln x} = \frac{\frac{1}{3} \ln x}{p \ln x}$ OR $\frac{1}{3} \ln x = (\ln x) r^{2}$ OR $p \ln x = \ln x (\frac{1}{3 p})$

Note: Candidates may use $\ln x^{1} + \ln x^{p} + \ln x^{\frac{1}{3}} + \dots$ and consider the powers of $x$ in geometric sequence

Award M1 for $\frac{p}{1} = \frac{\frac{1}{3}}{p}$ .

$r = p$ and $r^{2} = \frac{1}{3}$ M1

THEN

$p^{2} = \frac{1}{3}$ OR $r = \pm \frac{1}{\sqrt{3}}$ A1

$p = \pm \frac{1}{\sqrt{3}}$ AG

Note: Award M0A0 for $r^{2} = \frac{1}{3}$ or $p^{2} = \frac{1}{3}$ with no other working seen.

[2 marks]

a.i.

EITHER

since, $|p| = \frac{1}{\sqrt{3}}$ and $\frac{1}{\sqrt{3}} < 1$ R1

since, $|p| = \frac{1}{\sqrt{3}}$ and $- 1 < p < 1$ R1

THEN

$\Rightarrow$ the geometric series converges. AG

Note: Accept $r$ instead of $p$ .
Award R0 if both values of $p$ not considered.

[1 mark]

a.ii.

$\frac{\ln x}{1 - \frac{1}{\sqrt{3}}} (= 3 + \sqrt{3})$ (A1)

$\ln x = 3 - \frac{3}{\sqrt{3}} + \sqrt{3} - \frac{\sqrt{3}}{\sqrt{3}}$ OR $\ln x = 3 - \sqrt{3} + \sqrt{3} - 1 (\Rightarrow \ln x = 2)$ A1

$x = e^{2}$ A1

[3 marks]

a.iii.

METHOD 1

attempt to find a difference from consecutive terms or from $u_{2}$ M1

correct equation A1

$p \ln x - \ln x = \frac{1}{3} \ln x - p \ln x$ OR $\frac{1}{3} \ln x = \ln x + 2 (p \ln x - \ln x)$

Note: Candidates may use $\ln x^{1} + \ln x^{p} + \ln x^{\frac{1}{3}} + \dots$ and consider the powers of $x$ in arithmetic sequence.

Award M1A1 for $p - 1 = \frac{1}{3} - p$

$2 p \ln x = \frac{4}{3} \ln x (\Rightarrow 2 p = \frac{4}{3})$ A1

$p = \frac{2}{3}$ AG

METHOD 2

attempt to use arithmetic mean $u_{2} = \frac{u_{1} + u_{3}}{2}$ M1

$p \ln x = \frac{\ln x + \frac{1}{3} \ln x}{2}$ A1

$2 p \ln x = \frac{4}{3} \ln x (\Rightarrow 2 p = \frac{4}{3})$ A1

$p = \frac{2}{3}$ AG

METHOD 3

attempt to find difference using $u_{3}$ M1

$\frac{1}{3} \ln x = \ln x + 2 d (\Rightarrow d = - \frac{1}{3} \ln x)$

$u_{2} = \ln x + \frac{1}{2} (\frac{1}{3} \ln x - \ln x)$ OR $p \ln x - \ln x = - \frac{1}{3} \ln x$ A1

$p \ln x = \frac{2}{3} \ln x$ A1

$p = \frac{2}{3}$ AG

[3 marks]

b.i.

$d = - \frac{1}{3} \ln x$ A1

[1 mark]

b.ii.

METHOD 1

$S_{n} = \frac{n}{2} ⌊2 \ln x + (n - 1) \times (- \frac{1}{3} \ln x)⌋$

attempt to substitute into $S_{n}$ and equate to $\ln (\frac{1}{x^{3}})$ (M1)

$\frac{n}{2} ⌊2 \ln x + (n - 1) \times (- \frac{1}{3} \ln x)⌋ = \ln (\frac{1}{x^{3}})$

$\ln (\frac{1}{x^{3}}) = - \ln x^{3} (= \ln x^{- 3})$ (A1)

$= - 3 \ln x$ (A1)

correct working with $S_{n}$ (seen anywhere) (A1)

$\frac{n}{2} ⌊2 \ln x - \frac{n}{3} \ln x + \frac{1}{3} \ln x⌋$ OR $n \ln x - \frac{n (n - 1)}{6} \ln x$ OR $\frac{n}{2} (\ln x + (\frac{4 - n}{3}) \ln x)$

correct equation without $\ln x$ A1

$\frac{n}{2} (\frac{7}{3} - \frac{n}{3}) = - 3$ OR $n - \frac{n (n - 1)}{6} = - 3$ or equivalent

Note: Award as above if the series $1 + p + \frac{1}{3} + \dots$ is considered leading to $\frac{n}{2} (\frac{7}{3} - \frac{n}{3}) = - 3$ .

attempt to form a quadratic $= 0$ (M1)

$n^{2} - 7 n - 18 = 0$

attempt to solve their quadratic (M1)

$(n - 9) (n + 2) = 0$

$n = 9$ A1

METHOD 2

$\ln (\frac{1}{x^{3}}) = - \ln x^{3} (= \ln x^{- 3})$ (A1)

$= - 3 \ln x$ (A1)

listing the first $7$ terms of the sequence (A1)

$\ln x + \frac{2}{3} \ln x + \frac{1}{3} \ln x + 0 - \frac{1}{3} \ln x - \frac{2}{3} \ln x - \ln x + \dots$

recognizing first $7$ terms sum to $0$ M1

$8$ ^th term is $- \frac{4}{3} \ln x$ (A1)

$9$ ^th term is $- \frac{5}{3} \ln x$ (A1)

sum of $8$ ^th and $9$ ^th term $= - 3 \ln x$ (A1)

$n = 9$ A1

[8 marks]

b.iii.

Examiners report

Part (a)(i) was well done with few candidates incorrectly using the value of p to verify rather than to 'show' the given result. In part (a)(ii) most did not consider both values of r and some did know the condition for convergence of a geometric series. Part (a)(iii) was generally well done but some had difficulty in simplifying the surd. Part (b) (i) and (ii) was generally well done. Although many completely correct answers to part b (iii) were noted, weaker candidates often made errors in properties of logarithms or algebraic manipulation leading to an incorrect quadratic equation.

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.