Calculate, in CAD, the total amount John pays for the bicycle.

Find the value of the bicycle during the 5th year. Give your answer to two decimal places.

Calculate, in years, when the bicycle value will be less than 50 USD.

Find the total amount John has paid to insure his bicycle for the first 5 years.

John purchased the bicycle in 2008.

Justify why John should not insure his bicycle in 2019.

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

1.042 × 880 × 1.25  OR  (880 + 0.042 × 880) × 1.25      (M1)(M1)

Note: Award (M1) for multiplying 880 by 1.042 and (M1) for multiplying 880 by 1.25.

1150 (CAD)  (1146.20 (CAD))      (A1)(G2)

Note: Accept 1146.2 (CAD)

[3 marks]

$\frac{704}{880}$  OR  $\frac{563.20}{704}$      (M1)

Note: Award (M1) for correctly dividing sequential terms to find the common ratio, or 0.8 seen.

880(0.8)⁵⁻¹      (M1)

Note: Award (M1) for correct substitution into geometric sequence formula.

360.45 (USD)      (A1)(G3)

Note: Do not award the final (A1) if the answer is not correct to 2 decimal places. Award at most (M0)(M1)(A0) if $r=1.25$.

[3 marks]

     (M1)

Note: Award (M1) for correct substitution into geometric sequence formula and (in)equating to 50. Accept weak or strict inequalities. Accept an equation. Follow through from their common ratio in part (b). Accept a sketch of their GP with $y=50$ as a valid method.

OR

$u_{13}=60.473$  AND  $u_{14}=48.379$      (M1)

Note: Award (M1) for their $u_{13}$ and $u_{14}$ both seen. If the student states , without $u_{13}=60.473$ seen, this is not sufficient to award (M1).

14 or “14th year” or “after the 13th year”     (A1)(ft)(G2)

Note: The context of the question requires the final answer to be an integer. Award at most (M1)(A0) for a final answer of 13.9 years. Follow through from their 0.8 in part (b).

[2 marks]

$\frac{5}{2}\left(\left(2\times 120\right)+\left(-3.5\left(5-1\right)\right)\right)$    (M1)(A1)

Note: Award (M1) for substitution into arithmetic series formula, (A1) for correct substitution.

565 (USD)    (A1)(G2)

[3 marks]

2019 is the 12th year/term        (M1)

Note: Award (M1) for 12 seen.

75.59 (value of bicycle)  AND  81.5 (cost of insurance policy)       (A1)(ft)

Note: Award (A1) for both sequences’ 12th term seen. The value of the bicycle will follow through from their common ratio in part (b). Do not award (M0)(A1).

the cost of the insurance policy is greater than the value of the bicycle      (R1)(ft)

Note: Award (R1)(ft) for a reason consistent with their cost of insurance policy and their value of the bicycle. Follow through within this part. Award (R0) if the correct values are not explicitly seen. Accept the following contextualized reasons: “the insurance is not worth it", "the values are too close", "insurance is as much as the value of the bike", but only if their cost of insurance is greater than the value of the bicycle.

OR

75.59 < 81.5       (R1)(ft)

Note: Award (R1)(ft) for a correct numerical comparison showing their cost of insurance policy is greater than their value of the bicycle. Follow through within this part.

[3 marks]

Consider the expansion of $(3+x^2{)}^{n+1}$, where $n\in {\mathrm{\mathbb{Z}}}^{+}$.

Given that the coefficient of $x^4$ is $20 412$, find the value of $n$.

METHOD 1

product of a binomial coefficient, a power of $3$ (and a power of $x^2$) seen         (M1)

evidence of correct term chosen           (A1)

${}^{{}_{n+1}}C_2\times 3^{n+1-2}\times {\left(x^2\right)}^2 \left(=\frac{n\left(n+1\right)}{2}\times 3^{n-1}\times x^4\right)$  OR  $n-r=1$

equating their coefficient to $20412$ or their term to $20412x^4$         (M1)

EITHER

${}^{{}_{n+1}}C_2\times 3^{n-1}=20412$           (A1)

OR

${}^{r+2}C_r\times 3^r=20412\Rightarrow r=6$           (A1)

THEN

$n=7$         A1

METHOD 2

$3^{n+1}{\left(1+\frac{x^2}{3}\right)}^{n+1}$

product of a binomial coefficient, and a power of $\frac{x^2}{3}$  OR  $\frac{1}{3}$ seen         (M1)

evidence of correct term chosen           (A1)

$3^{n+1}\times \frac{n\left(n+1\right)}{2!}\times {\left(\frac{x^2}{3}\right)}^2 \left(=\frac{3^{n-1}}{2}n\left(n+1\right)x^4\right)$ 

equating their coefficient to $20412$ or their term to $20412x^4$         (M1)

$3^{n-1}\times \frac{n\left(n+1\right)}{2}=20412$           (A1)

$n=7$         A1

[5 marks]

Find $h$, the height of the tank.

Show that the volume of the tank is $624 000 {\text{m}}^3$, correct to three significant figures.

Write down the common difference, $d$.

Find the amount of fuel pumped into the tank in the $\text{13th}$ hour.

Find the value of $n$ such that $u_n=0$.

Write down the number of hours that the pump was pumping fuel into the tank.

Find the total amount of fuel pumped into the tank in the first $8$ hours.

Show that the tank will never be completely filled using this pump.

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

$\sin 60°=\frac{h}{40}$  OR  $\tan 60°=\frac{h}{20}$           (M1)

Note: Award (M1) for correct substitutions in trig ratio.

OR

${20}^2+h^2={40}^2 \left(\sqrt{{40}^2-{20}^2}\right)$           (M1)

Note: Award (M1) for correct substitutions in Pythagoras’ theorem.

$\left(h=\right) 34.6 \left(\text{m}\right) \left(\sqrt{1200}, 20\sqrt{3}, 34.6410\dots \right)$       (A1)(G2)

[2 marks]

$\frac{1}{2}\left(70+110\right)\left(34.6410\dots \right)\times 200$           (M1)(M1)

Note: Award (M1) for their correctly substituted area of trapezium formula, provided all substitutions are positive. Award (M1) for multiplying by $200$. Follow through from part (a).

OR

$\left(2\times \frac{1}{2}\times 20\times 34.6410\dots +70\times 34.6410\dots \right)\times 200$           (M1)(M1)

Note: Award (M1) for the addition of correct areas for two triangles and one rectangle. Award (M1) for multiplying by $200$. Follow through from part (a).

OR

$70\times 34.6410\dots \times 200+2\times \frac{1}{2}\times 34.6410\dots \times 20\times 200$           (M1)(M1)

Note: Award (M1) for their correct substitution in volume of cuboid formula. Award (M1) for correctly substituted volume of triangular prism(s). Follow through from part (a).

$623538\dots$         (A1)

$624000 \left({\text{m}}^3\right)$            (AG)

Note: Both an unrounded answer that rounds to the given answer and the rounded value must be seen for the (A1) to be awarded.

[3 marks]

$\left(d=\right) -1800$           (A1)

[1 mark]

$\left(u_{13}=\right) 45000+\left(13-1\right)\left(-1800\right)$           (M1)

Note: Award (M1) for correct substitutions in arithmetic sequence formula.
OR
Award (M1) for a correct ${\text{4}}^{\text{th}}$ term seen as part of list.

$23400 \left({\text{m}}^3\right)$        (A1)(ft)(G2)

Note: Follow through from part (c) for their value of $d$.

[2 marks]

$0=45000+\left(n-1\right)\left(-1800\right)$           (M1)

Note: Award (M1) for their correct substitution into arithmetic sequence formula, equated to zero.

$\left(n=\right) 26$        (A1)(ft)(G2)

Note: Follow through from part (c). Award at most (M1)(A0) if their $n$ is not a positive integer.

[2 marks]

$25$           (A1)(ft)

Note: Follow through from part (e)(i), but only if their final answer in (e)(i) is positive. If their $n$ in part (e)(i) is not an integer, award  (A1)(ft) for the nearest lower integer.

[1 mark]

$\left(S_8=\right) \frac{8}{2}\left(2\times 45000+\left(8-1\right)\times \left(-1800\right)\right)$           (M1)

Note: Award (M1) for their correct substitutions in arithmetic series formula. If a list method is used, award (M1) for the addition of their $8$ correct terms.

$310 000 \left({\text{m}}^3\right) \left(309 600\right)$       (A1)(ft)(G2)

Note: Follow through from part (c). Award at most (M1)(A0) if their final answer is greater than $624 000$.

[2 marks]

$\left(S_{25}=\right) \frac{25}{2}\left(2\times 45000+\left(25-1\right)\times \left(-1800\right)\right) , \left(S_{25}=\right) \frac{25}{2}\left(45000+1800\right)$           (M1)

Note: Award (M1) for their correct substitutions into arithmetic series formula.

$S_{25}=585000 \left({\text{m}}^3\right)$       (A1)(ft)(G1)

Note: Award (M1)(A1) for correctly finding $S_{26}=585000 \left({\text{m}}^3\right)$, provided working is shown e.g. $\left(S_{26}=\right) \frac{26}{2}\left(2\times 45000+\left(26-1\right)\times \left(-1800\right)\right)$ , $\left(S_{26}=\right) \frac{26}{2}\left(45000+0\right)$. Follow through from part (c) and either their (e)(i) or (e)(ii). If $d<0$ and their final answer is greater than $624 000$, award at most (M1)(A1)(ft)(R0). If $d>0$, there is no maximum, award at most (M1)(A0)(R0). Award no marks if their number of terms is not a positive integer.

$585000 \left({\text{m}}^3\right)<624000 \left({\text{m}}^3\right)$        (R1)

Hence it will never be filled        (AG)

Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.
For unsupported $\left(S_{25}\right)=585000$ seen, award at most (G1)(R1)(AG). Working must be seen to follow through from parts (c) and (e)(i) or (e)(ii).

OR

$\left(S_n=\right) \frac{n}{2}\left(2\times 45000+\left(n-1\right)\times \left(-1800\right)\right)$           (M1)

Note: Award (M1) for their correct substitution into arithmetic series formula, with $n$.

Maximum of this function $585225 \left({\text{m}}^3\right)$       (A1)

Note: Follow through from part (c). Award at most (M1)(A1)(ft)(R0) if their final answer is greater than $624 000$. Award at most (M1)(A0)(R0) if their common difference is not $–1800$. Award at most (M1)(A0)(R0) if $585 225$ is not explicitly identified as the maximum of the function.

$585225 \left({\text{m}}^3\right)<624000 \left({\text{m}}^3\right)$        (R1)

Hence it will never be filled        (AG)

Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.

OR

sketch with concave down curve and labelled $624000$ horizontal line           (M1)

Note: Accept a label of “tank volume” instead of a numerical value. Award (M0) if the line and the curve intersect.

curve explicitly labelled as $\left(S_n=\right) \frac{n}{2}\left(2\times 45000+\left(n-1\right)\times \left(-1800\right)\right)$ or equivalent       (A1)

Note: Award (A1) for a written explanation interpreting the sketch. Accept a comparison of values, e.g $585225 \left({\text{m}}^3\right)<624000 \left({\text{m}}^3\right)$, where $585225$ is the graphical maximum. Award at most (M1)(A0)(R0) if their common difference is not $–1800$.

the line and the curve do not intersect        (R1)

hence it will never be filled        (AG)

Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.

OR

$624000=\frac{n}{2}\left(2\times 45000+\left(n-1\right)\times \left(-1800\right)\right)$           (M1)

Note: Award (M1) for their correctly substituted arithmetic series formula equated to $624000 \left(623538\right)$.

Demonstrates there is no solution       (A1)

Note: Award (A1) for a correct working that the discriminant is less than zero OR correct working indicating there is no real solution in the quadratic formula.

There is no (real) solution (to this equation)       (R1)

hence it will never be filled        (AG)

Note: At most (M1)(A0)(R0) for their correctly substituted arithmetic series formula $=624000, 623538$ or $622800$ with a statement "no solution". Follow through from their part (b).

[3 marks]

Find the value of Amelia’s investment after $5$ years to the nearest hundred dollars.

Determine the number of years required for Amelia’s investment to reach the target.

Bill invests his $\$9000$ in an account that offers an interest rate of $r\%$ per annum compounded monthly, where $r$ is set to two decimal places.

Find the minimum value of $r$ needed for Bill to reach the target after $10$ years.

Show that Chris will never reach the target if his initial deposit is $\$9000$.

Find the amount Chris needs to deposit initially in order to reach the target after $5$ years. Give your answer to the nearest dollar.

EITHER

$9000\times {\left(1+\frac{7}{100}\right)}^5$           (A1)

$12622.965\dots$           (A1)

OR

$n=5$
$\text{I}\%=7$
$\text{PV}=\mp 9000$
$\text{P/Y}=1$
$\text{C/Y}=1$           (A1)
$\pm 12622.965\dots$           (A1)

THEN

$(\$) 12600$           A1

[3 marks]

EITHER

$9000{\left(1+\frac{7}{100}\right)}^x=20000$           (A1)

OR

$\text{I}\%=7$
$\text{PV}=\mp 9000$
$\text{FV}=\pm 20000$
$\text{P/Y}=1$
$\text{C/Y}=1$           (A1)

THEN

$=12$ (years)           A1

[2 marks]

METHOD 1

attempt to substitute into compound interest formula (condone absence of compounding periods)           (M1)

$9000{\left(1+\frac{r}{100\times 12}\right)}^{12\times 10}=20000$

$8.01170\dots$           (A1)

$r=8.02 \left(\%\right)$           A1

METHOD 2

$n=10$
$\text{PV}=\pm 9000$
$\text{FV}=\mp 20000$
$\text{P/Y}=1$
$\text{C/Y}=12$
$r=8.01170\dots$           (M1)(A1)

Note: Award M1 for an attempt to use a financial app in their technology, award A1 for $( r=) 8.01170\dots$

$r=8.02 \left(\%\right)$           A1

[3 marks]

recognising geometric series (seen anywhere)           (M1)

$r=\frac{4500}{9000} \left(=\frac{1}{2}\right)$           (A1)

EITHER

considering $S_{\infty }$           (M1)

$\frac{9000}{1-0.5}\left(=18000\right)$           A1

correct reasoning that $18000<20000$           R1

Note: Accept $S_{\infty }<20000$ only if $S_{\infty }$ has been calculated.

OR

considering $S_n$ for a large value of $n, n\ge 80$           (M1)

Note: Award M1 only if the candidate gives a valid reason for choosing a value of $n$, where $50\le n<80$.

correct value of $S_n$ for their $n$           A1

valid reason why Chris will not reach the target, which involves their choice of $n$, their value of $S_n$ and Chris’ age OR using two large values of $n$ to recognize asymptotic behaviour of $S_n$ as $n\to \infty$.           R1

Note: Do not award the R mark without the preceding A mark.

THEN

Therefore, Chris will never reach the target.           AG

[5 marks]

recognising geometric sum           M1

$\frac{u_1\left(1-0.5^5\right)}{0.5}=20000$           (A1)

$10322.58\dots$

$(\$) 10323$           A1

[3 marks]

Determine the amount that he will have in his account after 3 years. Give your answer correct to two decimal places.

Find the difference between the cost of the bicycle and the amount of money in Tommaso’s account after 3 years. Give your answer correct to two decimal places.

After $m$ complete months Tommaso will, for the first time, have enough money in his account to buy the bicycle.

Find the value of $m$.

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

$950\times {\left(1+\frac{5}{12\times 100}\right)}^{12\times 3}$    (M1)(A1)

Note: Award (M1) for substitution in the compound interest formula: (A1) for correct substitution.

OR

N = 3
I% = 5
PV = 950
P/Y = 1
C/Y = 12    (A1)(M1)

Note: Award (A1) for C/Y = 12 seen, (M1) for other correct entries.

OR

N = 36
I% = 5
PV = 950
P/Y = 12
C/Y = 12    (A1)(M1)

Note: Award (A1) for C/Y = 12 seen, (M1) for other correct entries.

1103.40 (EUR)    (A1)(G3)

Note: Answer must be given to 2 decimal places.

[3 marks]

(20 × 3 + 1100) − 1103.40    (M1)(M1)

Note: Award (M1) for correct substitution into cost of bike function, (M1) for subtracting their answer to part (a). This subtraction may be implied by their final answer (follow through from their part (a) for this implied subtraction).

55.60 (EUR)    (A1)(ft)(G3)

Note: Follow through from part (a). The answer must be two decimal places.

[3 marks]

METHOD 1

$950\times {\left(1+\frac{5}{12\times 100}\right)}^{12x}=20x+1100$     (M1)(M1)

Note: Award (M1) for their correct substitution in the compound interest formula with a variable in the exponent; (M1) for comparing their expressions provided variables are the same (not an expression with $x$ for years and another with $x$ representing months). Award at most (M0)(M1)(A0)(M1)(A0) for substitution of an integer in both expressions and comparison of the results. Accept inequality.

($x$ =) 4.52157… (years)    (A1)(ft)

4.52157… × 12 (= 54.2588…)     (M1)

Note: Award (M1) for multiplying their value for $x$ by 12. This may be implied.

$m$ = 55 (months)    (A1)(ft)(G4)

METHOD 2

$950\times {\left(1+\frac{5}{12\times 100}\right)}^m=20\times \frac{m}{12}+1100$     (M1)(M1)(M1)

Note: Award (M1) for their correct substitution in the compound interest formula with a variable in the exponent to solve; (M1) for comparing their expressions provided variables are the same; (M1) for converting years to months in these expressions. Award at most (M0)(M1)(A0)(M1)(A0) for substitution of an integer in both expressions and comparison of the results. Accept inequality.

$m$ = 54.2588… (months)    (A1)(ft)

$m$ = 55 (months)    (A1)(ft)(G4)

METHOD 3

     (M1)(M1)

Note: Award (M1) for each graph drawn.

($x$ =) 4.52157… (years)    (A1)(ft)

4.52157… × 12 (= 54.2588…)     (M1)

Note: Award (M1) for multiplying their value for $x$ by 12. This may be implied.

      If the graphs drawn are in terms of months, leading to a value of 54.2588…, award (M1)(M1)(M1)(A1), consistent with METHOD 2.

$m$ = 55 (months)    (A1)(ft)(G4)

Note: Follow through for a compound interest formula consistent with their part (a). The final (A1)(ft) can only be awarded for correct answer, or their correct answer following through from previous parts and only if value is rounded up. For example, do not award (M0)(M0)(A0)(M1)(A1)(ft) for an unsupported “5 years × 12 = 60” or similar.

[5 marks]

Given that the $k$th term of the sequence is zero, find the value of $k$.

Let $S_n$ denote the sum of the first $n$ terms of the sequence.

Find the maximum value of $S_n$.

attempt to use $u_1+\left(n-1\right)d=0$            (M1)

$60-2.5\left(k-1\right)=0$

$k=25$                          A1

[2 marks]

METHOD 1

attempting to express $S_n$ in terms of $n$            (M1)

use of a graph or a table to attempt to find the maximum sum            (M1)

$=750$                A1

METHOD 2

EITHER

recognizing maximum occurs at $n=25$           (M1)

$S_{25}=\frac{25}{2}\left(60+0\right), S_{25}=\frac{25}{2}\left(2\times 60+24\times -2.5\right)$          (A1)

OR

attempting to calculate $S_{24}$          (M1)

$S_{24}=\frac{24}{2}\left(2\times 60+23\times -2.5\right)$          (A1)

THEN

$=750$                A1

[3 marks]

Calculate the café’s total profit for the first 12 weeks.

Calculate the tea-shop’s total profit for the first 12 weeks.

$\frac{12}{2}\left(2\times 60+11\times 10\right)$     (M1)(A1)(ft)

Note: Award (M1) for substituting the arithmetic series formula, (A1)(ft) for correct substitution. Follow through from their first term and common difference in part (a).

= ($) 1380     (A1)(ft)(G2)

[3 marks]

$\frac{60\left({1.1}^{12}-1\right)}{1.1-1}$     (M1)(A1)(ft)

Note: Award (M1) for substituting the geometric series formula, (A1)(ft) for correct substitution. Follow through from part (c) for their first term and common ratio.

= ($)1280  (1283.05…)     (A1)(ft)(G2)

[3 marks]

Find the amount that Sam will have in his account after $10$ years.

Find the value of $r$ required so that the amount in David’s account after $10$ years will be equal to the amount in Sam’s account.

Find the interest David will earn over the $10$ years.

Note: The first time an answer is not given to two decimal places, the final A1 in that part is not awarded.

EITHER

$N=10$              OR             $N=20$

$I\%=2.74$                           $I\%=2.74$

$PV=\left(\mp \right)1700$                     $PV=\left(\mp \right)1700$

$P/Y=1$                              $P/Y=2$

$C/Y=2$                              $C/Y=2$              (M1)(A1)

Note: Award (M1) for an attempt to use a financial app in their technology with at least two entries seen, and award (A1) for all entries correct. Accept a positive or negative value for $PV$.

OR

$1700{\left(1+\frac{0.0274}{2}\right)}^{2\times 10}$         (M1)(A1)

Note: Award (M1) for substitution into compound interest formula.
Award (A1) for correct substitution.

THEN

$\$2231.71$        A1

[3 marks]

Note: The first time an answer is not given to two decimal places, the final A1 in that part is not awarded.

EITHER

$N=10$

$PV=\mp 1700$

$FV=\pm 2231.71\dots$

$P/Y=1$

$C/Y=1$             (M1)

Note: Award (M1) for an attempt to use a financial app in their technology with at least two entries seen.

OR

$1700{\left(1+\frac{r}{100}\right)}^{10}=2231.71\dots$         (M1)

THEN

$r=2.75876\dots$

$r=2.76$        A1

Note: Ignore omission of opposite signs for $PV$ and $FV$ if $r=2.76$ is obtained.

[2 marks]

Note: The first time an answer is not given to two decimal places, the final A1 in that part is not awarded.

$\$531.71$         A1 

[1 mark]

This was the only short question which was not common to both HL and SL papers.

Candidates found this question a straightforward start, with the majority being able to attempt all three parts. While most candidates initially gave answers to two decimal places as required, many subsequently rounded their answers to three significant figures. In parts (a) and (b), most used the compound interest formula rather than the finance app on their GDC. Difficulty using the formula arose in determining the value of r (0.0274 was often seen, rather than 2.74) and the value of k. Many candidates struggled to correctly interpret 'half-yearly', and attempted to use either $k=6$ or $k=\frac{1}{2}$ in the formula.

In part (b), a correct equation was usually seen, but lengthy analytical methods to solving it were favoured, with varying degrees of success, over an approach using the GDC. A common error was an attempt to use logarithms rather than the tenth root. It was not uncommon to see a final answer of either 2.76% or 0.0276, rather than 2.76. Many did not know what was meant by 'interest' in part (c). Those that did often recalculated David's amount using their value of r, rather than subtract $1700 from Sam's amount.

The candidates who attempted to use a finance app on their GDC in this question were generally able to give accurate and concise answers. In the case where they obtained an incorrect answer, most gave sufficient detail of the values they were using in the app, to be awarded the method mark. The most common error seen was the use of an incorrect value for the number of payments per year or the number of compounding periods per year.

Find the value of a and of b.

Use the regression equation to estimate the value of y when x = 3.57.

The relationship between x and y can be modelled using the formula y = kxⁿ, where k ≠ 0 , n ≠ 0 , n ≠ 1.

By expressing ln y in terms of ln x, find the value of n and of k.

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

valid approach      (M1)

eg  one correct value

−0.453620, 6.14210

a = −0.454, b = 6.14      A1A1 N3

[3 marks]

correct substitution     (A1)

eg   −0.454 ln 3.57 + 6.14

correct working     (A1)

eg  ln y = 5.56484

261.083 (260.409 from 3 sf)

y = 261, (y = 260 from 3sf)       A1 N3

Note: If no working shown, award N1 for 5.56484.
If no working shown, award N2 for ln y = 5.56484.

[3 marks]

METHOD 1

valid approach for expressing ln y in terms of ln x      (M1)

eg  $\text{ln}y=\text{ln}\left(k{x}^n\right),\text{ln}\left(k{x}^n\right)=a\text{ln}x+b$

correct application of addition rule for logs      (A1)

eg  $\text{ln}k+\text{ln}\left(x^n\right)$

correct application of exponent rule for logs       A1

eg  $\text{ln}k+n\text{ln}x$

comparing one term with regression equation (check FT)      (M1)

eg  $n=a,b=\text{ln}k$

correct working for k      (A1)

eg  $\text{ln}k=6.14210,k=e^{6.14210}$

465.030

$n=-0.454,k=465$ (464 from 3sf)     A1A1 N2N2

METHOD 2

valid approach      (M1)

eg  $e^{\text{ln}y}=e^{a\text{ln}x+b}$

correct use of exponent laws for $e^{a\text{ln}x+b}$     (A1)

eg  $e^{a\text{ln}x}\times e^b$

correct application of exponent rule for $a\text{ln}x$     (A1)

eg  $\text{ln}{x}^a$

correct equation in y      A1

eg  $y=x^a\times e^b$

comparing one term with equation of model (check FT)      (M1)

eg  $k=e^b,n=a$

465.030

$n=-0.454,k=465$ (464 from 3sf)     A1A1 N2N2

METHOD 3

valid approach for expressing ln y in terms of ln x (seen anywhere)      (M1)

eg  $\text{ln}y=\text{ln}\left(k{x}^n\right),\text{ln}\left(k{x}^n\right)=a\text{ln}x+b$

correct application of exponent rule for logs (seen anywhere)      (A1)

eg  $\text{ln}\left(x^a\right)+b$

correct working for b (seen anywhere)      (A1)

eg  $b=\text{ln}\left(e^b\right)$

correct application of addition rule for logs      A1

eg  $\text{ln}\left(e^b{x}^a\right)$

comparing one term with equation of model (check FT)     (M1)

eg  $k=e^b,n=a$

465.030

$n=-0.454,k=465$ (464 from 3sf)     A1A1 N2N2

[7 marks]

Find the first term of the sequence, $u_1$.

Find $S_{\infty }$.

Find the least value of $n$ such that $S_{\infty }-S_n<0.001$.

$u_1=S_1=\frac{2}{3}\times \frac{7}{8}$                 (M1)

$=\frac{14}{24}\left(=\frac{7}{12}=0.583333\dots \right)$                 A1

[2 marks]

$r=\frac{7}{8}\left(=0.875\right)$                 (A1)

substituting their values for $u_1$ and $r$ into $S_{\infty }=\frac{u_1}{1-r}$                 (M1)

$=\frac{14}{3}\left(=4.66666\dots \right)$                 A1

[3 marks]

attempt to substitute their values into the inequality or formula for $S_n$                 (M1)

$\frac{14}{3}-\underset{r=1}{\overset{n}{\text{Σ}}}\frac{2}{3}{\left(\frac{7}{8}\right)}^r<0.001$  OR  $S_n=\frac{{\displaystyle \frac{7}{12}}\left(1-{\left({\displaystyle \frac{7}{8}}\right)}^n\right)}{\left(1-{\displaystyle \frac{7}{8}}\right)}$

attempt to solve their inequality using a table, graph or logarithms

(must be exponential)                 (M1)

Note: Award (M0) if the candidate attempts to solve $S_{\infty }-u_n<0.001$.

correct critical value or at least one correct crossover value                 (A1)

$63.2675\dots$  OR  $S_{\infty }-S_{63}=0.001036\dots$  OR  $S_{\infty }-S_{64}=0.000906\dots$

OR  $S_{\infty }-S_{63}-0.001=0.0000363683\dots$  OR  $S_{\infty }-S_{64}-0.001=0.0000931777\dots$

least value is $n=64$                 A1

[4 marks]

Write down the distance Rosa runs in the third training session;

Write down the distance Rosa runs in the $n$th training session.

Find the value of $k$.

Calculate the total distance, in kilometres, Rosa runs in the first 50 training sessions.

Find the distance Carlos runs in the fifth month of training.

Calculate the total distance Carlos runs in the first year.

3800 m     (A1)

[1 mark]

$3000+(n-1)400\text{ m}$$$OR$$$2600+400n\text{ m}$     (M1)(A1)

Note:     Award (M1) for substitution into arithmetic sequence formula, (A1) for correct substitution.

[2 marks]

$3000+(k-1)400>42195$     (M1)

Notes:     Award (M1) for their correct inequality. Accept $3+(k-1)0.4>42.195$.

Accept $=$ OR $⩾$. Award (M0) for $3000+(k-1)400>42.195$.

$(k=)99$     (A1)(ft)(G2)

Note:     Follow through from part (a)(ii), but only if $k$ is a positive integer.

[2 marks]

$\frac{50}{2}\left(2\times 3000+(50-1)(400)\right)$     (M1)(A1)(ft)

Note:     Award (M1) for substitution into sum of an arithmetic series formula, (A1)(ft) for correct substitution.

$640000\text{ m}$     (A1)

Note:     Award (A1) for their $640000$ seen.

$=640\text{ km}$     (A1)(ft)(G3)

Note:     Award (A1)(ft) for correctly converting their answer in metres to km; this can be awarded independently from previous marks.

OR

$\frac{50}{2}\left(2\times 3+(50-1)(0.4)\right)$     (M1)(A1)(ft)(A1)

Note:     Award (M1) for substitution into sum of an arithmetic series formula, (A1)(ft) for correct substitution, (A1) for correctly converting 3000 m and 400 m into km.

$=640\text{ km}$     (A1)(G3)

[4 marks]

$7500\times {1.2}^{5-1}$     (M1)(A1)

Note:     Award (M1) for substitution into geometric series formula, (A1) for correct substitutions.

$=15600\text{ m }(15552\text{ m})$     (A1)(G3)

OR

$7.5\times {1.2}^{5-1}$     (M1)(A1)

Note:     Award (M1) for substitution into geometric series formula, (A1) for correct substitutions.

$=15.6\text{ km}$     (A1)(G3)

[3 marks]

$\frac{7500({1.2}^{12}-1)}{1.2-1}$     (M1)(A1)

Notes:     Award (M1) for substitution into sum of a geometric series formula, (A1) for correct substitutions. Follow through from their ratio ($r$) in part (d). If (distance does not increase) or the final answer is unrealistic (eg $r=20$), do not award the final (A1).

$=297000\text{ m }(296853\dots \text{ m},297\text{ km})$     (A1)(G2)

[3 marks]

Show that $H_n=2400n+67 600$.

Given that $J_n$ follows a geometric sequence, state the value of the common ratio, $r$.

Find the value of $N$.

For the value of $N$ found in part (c) (i), state Helen’s annual salary and Jane’s annual salary, correct to the nearest dollar.

Find Jane’s total earnings at the start of her $10$th year of employment. Give your answer correct to the nearest dollar.

* 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.

uses $H_n=H_1+\left(n-1\right)d$ with $H_1=70 000$ and $d=2400$       (M1)

$H_n=70 000+2400\left(n-1\right)$        A1

so  $H_n=2400n+67 600$         AG

[2 marks]

$r=1.03$        A1

[1 mark]

evidence of use of an appropriate table or graph or GDC numerical solve feature to find the value of $N$ such that $J_n>H_n$         (M1)

EITHER

for example, an excerpt from an appropriate table

        (A1)

OR

for example, use of a GDC numerical solve feature to obtain $N=10.800\dots$         (A1)

Note: Award A1 for an appropriate graph. Condone use of a continuous graph.

THEN

$N=11$        A1

[3 marks]

$H_{11}=94 000 \left(\$\right)$        A1

$J_{11}=94 074 \left(\$\right)$        A1

Helen’s annual salary is $\$94 000$ and Jane’s annual salary is $\$94 074$

Note: Award A1 for a correct $H_{11}$ value and A1 for a correct $J_{11}$ value seen in part (c) (i).

[2 marks]

at the start of the $10$th year, Jane will have worked for $9$ years so the value of $S_9$ is required         R1

Note: Award R1 if $S_9$ is seen anywhere.

uses $S_n=\frac{J_1\left(r^n-1\right)}{r-1}$ with $J_1=70 000$, $r=1.03$ and $n=9$        (M1)

Note: Award M1 if $n=10$ is used.

$S_9=\frac{70 000\left({\left(1.03\right)}^9-1\right)}{1.03-1}=711 137.42\dots$        (A1)

$=711 137 \left(\$\right)$

Jane’s total earnings are $\$711 137$ (correct to the nearest dollar)

[4 marks]

Find the common ratio in terms of $a$.

Find the values of $a$ for which the sum to infinity of the series exists.

Find the value of $a$ when $S_{\infty }=76$.

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

evidence of dividing terms (in any order)      (M1)

eg       $\frac{u_1}{u_2}, \frac{{\displaystyle \frac{1}{4}{a}^2-3a}}{a}$

$r=\frac{1}{4}a-3$      A1   N2

[2 marks]

recognizing $\left| r \right|<1$ (must be in terms of $a$)      (M1)

eg       $\left|\frac{1}{4}a-3\right|<1, -1\le \frac{1}{4}a-3\le 1, -4<a-12<4$

$8<a<16$      A2   N3

[3 marks]

correct equation     (A1)

eg       $\frac{a}{1-\left({\displaystyle \frac{1}{4}}a-3\right)}=76 , a=76\left(4-\frac{1}{4}a\right)$

$a=\frac{76}{5} \left(=15.2\right)$ (exact)      A2   N3

[3 marks]

Let $f(x)=(x^2+3{)}^7$. Find the term in $x^5$ in the expansion of the derivative, $f^{\prime }(x)$.

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

METHOD 1 

derivative of $f(x)$     A2

$7(x^2+3{)}^6(x2)$

recognizing need to find $x^4$ term in $(x^2+3{)}^6$ (seen anywhere)     R1

eg$$$14x\text{ (term in }{x}^4)$

valid approach to find the terms in $(x^2+3{)}^6$     (M1)

eg$$$\left(\begin{array}{c}6\\ r\end{array}\right)(x^2{)}^{6-r}(3{)}^r,(x^2{)}^6(3{)}^0+(x^2{)}^5(3{)}^1+\dots$, Pascal’s triangle to 6th row

identifying correct term (may be indicated in expansion)     (A1)

eg$$$\text{5th term, }r=2,\left(\begin{array}{c}6\\ 4\end{array}\right),(x^2{)}^2(3{)}^4$

correct working (may be seen in expansion)     (A1)

eg$$$\left(\begin{array}{c}6\\ 4\end{array}\right)(x^2{)}^2(3{)}^4,15\times 3^4,14x\times 15\times 81(x^2{)}^2$

$17010x^5$     A1     N3

METHOD 2

recognition of need to find $x^6$ in $(x^2+3{)}^7$ (seen anywhere) R1 

valid approach to find the terms in $(x^2+3{)}^7$     (M1)

eg$$$\left(\begin{array}{c}7\\ r\end{array}\right)(x^2{)}^{7-r}(3{)}^r,(x^2{)}^7(3{)}^0+(x^2{)}^6(3{)}^1+\dots$, Pascal’s triangle to 7th row

identifying correct term (may be indicated in expansion)     (A1)

eg$$6th term, $r=3,\left(\begin{array}{c}7\\ 3\end{array}\right),\text{ (}{\text{x}}^2{)}^3(3{)}^4$

correct working (may be seen in expansion)     (A1)

eg$$$\left(\begin{array}{c}7\\ 4\end{array}\right)\text{(}{\text{x}}^2{)}^3(3{)}^4,35\times 3^4$

correct term     (A1)

$2835x^6$

differentiating their term in $x^6$     (M1)

eg$$$(2835x^6{)}^{\prime },\text{ (6)(2835}{\text{x}}^5)$

$17010x^5$     A1     N3

[7 marks]

Given that x_{k + 1} = x_k + a, find a.

Hence find the value of n such that $\sum _{k=1}^n{x}_k=861$.

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

valid approach to find maxima     (M1)

eg  one correct value of x_k, sketch of f

any two correct consecutive values of x_k      (A1)(A1)

eg  x₁ = 1, x₂ = 5

a = 4      A1 N3

[4 marks]

recognizing the sequence x_1,  x_2,  x_{3, …,} x_n is arithmetic  (M1)

eg  d = 4

correct expression for sum       (A1)

eg  $\frac{n}{2}\left(2\left(1\right)+4\left(n-1\right)\right)$

valid attempt to solve for n      (M1)

eg  graph, 2n² − n − 861 = 0

n = 21       A1 N2

[4 marks]

Write down the first three non-zero terms of $w_n$.

Find the value of $r$.

Find the value of $m$.

attempt to add corresponding terms      (M1)

eg   $2+2\text{,}6+\left(-6\right)\text{,}2{\left(3\right)}^{n-1}+2{\left(-3\right)}^{n-1}$

correct value for $w_5$        (A1)

eg   324

4, 36, 324 (accept 4 + 36 + 324)      A1 N3

[3 marks]

valid approach     (M1)

eg  $4\times r^1=36$,  $4\times 9^{n-1}$

$r=9$  (accept $\sum _{k=0}^{m}4\times 9^k$; $m$ may be incorrect)      A1 N2

[2 marks]

recognition that 225 terms of $w_n$ consists of 113 non-zero terms    (M1)

eg  $\sum _1^{113}$,  $\sum _0^{112}$,  113

$m=112$  (accept $\sum _{k=0}^{1}124\times r^k$; $r$ may be incorrect)      A1 N2

[2 marks]

Consider the expansion of ${\left(3x^2-\frac{k}{x}\right)}^9$, where $k>0$.

The coefficient of the term in $x^6$ is $6048$. Find the value of $k$.

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

valid approach for expansion (must have correct substitution for parameters, but accept an incorrect value for $r$).       (M1)

eg     $\left(\begin{array}{c}9\\ r\end{array}\right){\left(3x^2\right)}^{9-r}{\left(-\frac{k}{x}\right)}^r , {\left(3x^2\right)}^9+\left(\begin{array}{c}9\\ 1\end{array}\right){\left(3x^2\right)}^8{\left(-\frac{k}{x}\right)}^1+\left(\begin{array}{c}9\\ 2\end{array}\right){\left(3x^2\right)}^7{\left(-\frac{k}{x}\right)}^2+\dots$

valid attempt to identify correct term       (M1)

eg     $2\left(9-r\right)-r=6 , {\left(x^2\right)}^r{\left(x^{-1}\right)}^{9-r}=x^6$

identifying correct term (may be indicated in expansion)       (A1)

eg     $r=4, r=5$

correct term or coefficient in binominal expansion       (A1)

eg     $\left(\begin{array}{c}9\\ 4\end{array}\right){\left(3x^2\right)}^5{\left(-\frac{k}{x}\right)}^4 , 126\left(243x^{10}\right)\left(\frac{k^4}{x^4}\right), 30618k^4$

correct equation in $k$       (A1)

eg     $\left(\begin{array}{c}9\\ 4\end{array}\right)\left(243\right)\left(k^4\right)x^6=6048x^6 , 30618k^4=6048$

$k=\frac{2}{3}$ (exact)  $0.667$       A1  N3 

Note: Do not award A1 if additional answers given.

[6 marks]

Find the common ratio.

Find the sum of the first 8 terms.

Find the least value of n for which S_n > 163.

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

correct substitution into infinite sum      (A1)
eg  $200=\frac{4}{1-r}$

r = 0.98 (exact)     A1 N2

[2 marks]

correct substitution     (A1)

$\frac{4\left(1-{0.98}^8\right)}{1-0.98}$

29.8473

29.8    A1 N2

[2 marks]

attempt to set up inequality (accept equation)      (M1)
eg  $\frac{4\left(1-{0.98}^n\right)}{1-0.98}>163,\frac{4\left(1-{0.98}^n\right)}{1-0.98}=163$

correct inequality for n (accept equation) or crossover values      (A1)
eg  n > 83.5234, n = 83.5234, S₈₃ = 162.606 and S₈₄ = 163.354

n = 84     A1 N1

[3 marks]

Find the value of $r$.

Find the value of $u_{10}$.

Find the least value of $n$ such that $S_n>5543$.

valid approach        (M1)

eg       $\frac{u_1}{u_2}$,  $\frac{2.226}{2.1}$,  $2.226=2.1r$

$r=1.06$  (exact)       A1  N2

[2 marks]

correct substitution        (A1)

eg       $2.1\times {1.06}^9$

$3.54790$       A1  N2

$u_{10}=3.55$

[2 marks]

correct substitution into $S_n$ formula        (A1)

eg      $\frac{2.1\left({1.06}^n-1\right)}{1.06-1}$,  $\frac{2.1\left({1.06}^n-1\right)}{1.06-1}>5543$,  $2.1\left({1.06}^n-1\right)=332.58$,  sketch of $S_n$ and $y=5543$

correct inequality for $n$ or crossover values       A1

eg       $n>87.0316$,  $S_{87}=5532.73$  and  $S_{88}=5866.79$

$n=88$       A1  N2

[3 marks]

Consider a geometric sequence where the first term is 768 and the second term is 576.

Find the least value of $n$ such that the $n$th term of the sequence is less than 7.

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

attempt to find $r$     (M1)

eg$$$\frac{576}{768},\frac{768}{576},0.75$

correct expression for $u_n$     (A1)

eg$$$768(0.75{)}^{n-1}$

EITHER (solving inequality)

valid approach (accept equation)     (M1)

eg$$

valid approach to find $n$     M1

eg$$$768(0.75{)}^{n-1}=7,n-1>{\log }_{0.75}\left(\frac{7}{768}\right)$, sketch

correct value

eg$$$n=17.3301$     (A1)

$n=18$ (must be an integer)     A1     N2

OR (table of values)

valid approach     (M1)

eg$$$u_n>7$, one correct crossover value

both crossover values, $u_{17}=7.69735$ and $u_{18}=5.77301$     A2

$n=18$ (must be an integer)     A1     N2

OR (sketch of functions)

valid approach     M1

eg$$sketch of appropriate functions

valid approach     (M1) 

eg$$finding intersections or roots (depending on function sketched)

correct value

eg$$$n=17.3301$     (A1)

$n=18$ (must be an integer)     A1     N2

[6 marks]

Find the value of r, giving your answer to four significant figures.

Laurie makes no further deposits to or withdrawals from the account.

Find the year in which the amount of money in Laurie’s account will become double the amount she invested.

${\left(1+\frac{5.5}{4\times 100}\right)}^4$      (M1)(A1)

1.056      A1

[3 marks]

EITHER

$2P=P\times {\left(1+\frac{5.5}{100\times 4}\right)}^{4n}$  OR $2P=P\times {\left(\text{their}\left(a\right)\right)}^m$       (M1)(A1)

Note: Award (M1) for substitution into loan payment formula. Award (A1) for correct substitution.

OR

PV = ±1
FV = $\mp$1
I% = 5.5
P/Y = 4
C/Y = 4
n = 50.756…       (M1)(A1)

OR

PV = ±1
FV = $\mp$2
I% = 100(their (a) − 1)
P/Y = 1
C/Y = 1        (M1)(A1)

THEN

⇒ 12.7 years

Laurie will have double the amount she invested during 2032      A1

[3 marks]

Find the exact value of $S_k$.

Show that $F=3240$.

An infinite geometric series is given as $S_{\mathrm{\infty }}=a+\frac{a}{\sqrt{2}}+\frac{a}{2}+\dots$, $a\in {\mathbb{Z}}^{+}$.

Find the largest value of $a$ such that .

correct substitution      (A1)

eg   $\frac{300}{2}\left(1.3+31.2\right)$ ,  $\frac{300}{2}\left[2\left(1.3\right)+\left(300-1\right)\left(0.1\right)\right]$ ,  $\frac{300}{2}\left[2.6+299\left(0.1\right)\right]$ 

$S_k=4875$        A1  N2

[2 marks]

recognizing need to find the sequence of multiples of 3 (seen anywhere)       (M1)

eg   first term is $u_3$ (= 1.5)   (accept notation $u_1=1.5$) ,

$d=0.1\times 3$  (= 0.3) , 100 terms (accept $n=100$), last term is 31.2

(accept notation $u_{100}=31.2$) ,  $u_3+u_6+u_9+\dots$  (accept $F=u_3+u_6+u_9+\dots$)

correct working for sum of sequence where n is a multiple of 3      A2

$\frac{100}{2}\left(1.5+31.2\right)$ ,  $50\left(2\times 1.5+99\times 0.3\right)$ ,  1635

valid approach (seen anywhere)       (M1)

eg    $S_k-\left(u_3+u_6+\dots \right)$ ,  $S_k-\frac{100}{2}\left(1.5+31.2\right)$ ,  $S_k-$ (their sum for $\left(u_3+u_6+\dots \right)$)

correct working (seen anywhere)       A1

eg   $S_k-1635$ , 4875 − 1635

$F=3240$       AG  N0

[5 marks]

attempt to find $r$       (M1)

eg    dividing consecutive terms

correct value of $r$ (seen anywhere, including in formula)