Write down three equations that express the information given above.

Find the number of each type of ticket sold.

$x+y+z=600$           A1

$15x+10y+12z=7816$           A1

$x=2y$           A1

Note: Condone other labelling if clear, e.g. $a$ (adult), $c$ (child) and $s$ (student). Accept equivalent, distinct equations e.g. $2y+y+z=600$.

[3 marks]

$x=308, y=154, z=138$           A1A1

 
Note: Award A1 for all three correct values seen, A1 for correctly labelled as $x, y$ or $z$.
Accept answers written in words: e.g. $308$ adult tickets.

[2 marks]

Many candidates had at least two of the three equations written down correctly. The interpretation of the phrase “twice as many adult tickets sold as child tickets” was enigmatic. Consequently, $2x=y$ was a popular but erroneous answer.

Too many candidates spent considerable time attempting to solve three equations with three unknowns by hand with pages of working rather than using their GDC.

Find the value of $a$.

Find the value of $b$.

Given $0<M<8$, find the range for $N$.

Find the expected length of time between this earthquake and the next earthquake of at least this magnitude. Give your answer to the nearest year.

${\log }_{10} 100=a-3$        (M1)

$a=5$             A1

[2 marks]

EITHER

$N={10}^{5-M}$        (M1)

$=\frac{{10}^5}{{10}^M}\left(=\frac{100000}{{10}^M}\right)$

OR

$100=\frac{b}{{10}^3}$        (M1)

THEN

$b=100000 \left(={10}^5\right)$             A1

[2 marks]

$0.001<N<100000 \left({10}^{-3}<N<{10}^5\right)$             A1A1

Note: Award A1 for correct endpoints and A1 for correct inequalities/interval notation.

[2 marks]

$N=\frac{{10}^5}{{10}^{7.2}} \left(=0.0063095\dots \right)$             (M1)

length of time $=\frac{1}{0.0063095\dots }={10}^{2.2}$

$=158$ years             A1

[2 marks]

Many candidates did not attempt this question. Of those who did attempt the question, most of these candidates arrived at the correct answer to this part with the most common incorrect answer being 103.

Those that were successful in part (a) answered this well.

This was only answered correctly by the strongest candidates.

This part of the question was a discriminator as correct responses were few and far between.

Find the range of $h$.

Find $h^{-1}\left(10\right)$.

In the context of the question, interpret your answer to part (b)(i).

Write down the range of $h^{-1}$.

$h\left(4\right)=\frac{640}{4^2}+0.5$  OR  $h\left(14\right)=\frac{640}{{14}^2}+0.5$                    (M1)

Note: Award (M1) for substituting $4$ or $14$ into $h$. This can be implicit from seeing $3.77 (3.76530\dots )$ or $40.5$.

$3.77\le h\left(x\right)\le 40.5 \left(3.76530\dots \le h\left(x\right)\le 40.5\right)$              A1A1

Note: Award A1 for both correct endpoints seen, A1 for the endpoints in a correct interval.

[3 marks]

$h\left(x\right)=10$  OR  $h^{-1}\left(x\right)=\sqrt{\frac{640}{x-0.5}}$  OR  $h^{-1}\left(10\right)=\sqrt{\frac{640}{10-0.5}}$                    (M1)

$\left(x=\right) 8.21 \text{cm} \left(8.20782\dots \right)$              A1

[2 marks]

a tin that is $10 \text{cm}$ high will have a diameter of $8.21 \text{cm} (8.20782...)$            A1

Note: Condone a correct answer expressed as the converse.

[1 mark]

$4\le h^{-1}\le 14$           A1

Note: Accept $4\le y\le 14$. Do not FT in this part.

[1 mark]

Part (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as $40.5\ge y\ge 3.77$ while some others wrote $40.5\le y\le 3.77$. A few candidates used integer $x$ values from $4$ to $14$ to find corresponding values for $h\left(x\right)$ and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was $6.9$ with weaker candidates simply finding $h\left(10\right)$. Several candidates equated $h\left(x\right)$ to $10$ but missed out $+0.5$ in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting $x$ by $10$ and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).

Part (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as $40.5\ge y\ge 3.77$ while some others wrote $40.5\le y\le 3.77$. A few candidates used integer $x$ values from $4$ to $14$ to find corresponding values for $h\left(x\right)$ and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was $6.9$ with weaker candidates simply finding $h\left(10\right)$. Several candidates equated $h\left(x\right)$ to $10$ but missed out $+0.5$ in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting $x$ by $10$ and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).

Part (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as $40.5\ge y\ge 3.77$ while some others wrote $40.5\le y\le 3.77$. A few candidates used integer $x$ values from $4$ to $14$ to find corresponding values for $h\left(x\right)$ and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was $6.9$ with weaker candidates simply finding $h\left(10\right)$. Several candidates equated $h\left(x\right)$ to $10$ but missed out $+0.5$ in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting $x$ by $10$ and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).

Part (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as $40.5\ge y\ge 3.77$ while some others wrote $40.5\le y\le 3.77$. A few candidates used integer $x$ values from $4$ to $14$ to find corresponding values for $h\left(x\right)$ and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was $6.9$ with weaker candidates simply finding $h\left(10\right)$. Several candidates equated $h\left(x\right)$ to $10$ but missed out $+0.5$ in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting $x$ by $10$ and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).

Charlie ran on day $20$ of his fitness programme.

Daniella ran on day $20$ of her fitness programme.

On day $n$ of the fitness programmes Daniella runs more than Charlie for the first time.

Find the value of $n$.

attempt to find $u_{20}$ using an arithmetic sequence         (M1)

e.g.  $u_1=500$ and $d=100$  OR  $u_{20}=500+1900$  OR  $500, 600, 700, \dots$

(Charlie ran) $2400 \text{m}$             A1

[2 marks]

$\left(r=\right) 1.02$         (A1)

attempt to find $u_{20}$ using a geometric sequence         (M1)

e.g.  $u_1=500$ and a value for $r$  OR  $500\times r^{19}$  OR  $500, 510, 520.2, \dots$

(Daniella ran) $728 \text{m }\left(728.405\dots \right)$             A1

[3 marks]

$500\times 1.{02}^{n-1}>500+\left(n-1\right)\times 100$         (M1)

attempt to solve inequality         (M1)

$n>184.215\dots$

$n=185$             A1

[3 marks]

Find the value of $p$.

Find the value of

(i)     $a$.

(ii)    $b$.

(iii)   $c$.

Write down the equation of the axis of symmetry of the graph.

$3$         A1

Note: Accept $(3, 0)$ seen.

[1 mark]

METHOD 1

$0=4a-2b+c, 0=9a+3b+c, -\frac{25}{2}=\frac{1}{4}a+\frac{1}{2}b+c$            (M1)(A1) 

(i)     $2$         A1

(ii)    $-2$         A1

(iii)   $-12$         A1

Note: Award the (M1)(A1) if at least one correct value is seen. Do not apply FT form part (a) if workings are not shown.

METHOD 2

$-12.5=a\left(0.5+2\right)\left(0.5-3\right)$            (M1)

(i)     $a=2$         A1

        $0=2\times {\left(3\right)}^2+3b+c$

        $0=2\times {\left(-2\right)}^2+\left(-2\right)b+c$            (M1)

(ii)    $b=-2$         A1

(iii)   $c=-12$         A1

[5 marks]

$x=0.5$          A1

Note: Do not FT from their part (b), this is a contradiction with the diagram.

[1 mark]

Find an equation which links $C$ and $r$.

Find, to the nearest USD, the cost of disk that has a radius of $1.1$cm.

* 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.

$C=k{r}^3$       (M1)

$375=k\times 0.8^3\Rightarrow k=732 \left(732.421\dots \right)$        (M1)

$C=732r^3$       A1

[3 marks]

$C=732.42\dots \times 1.1^3$        (M1)

$C=\$975 \left(974.853\dots \right)$       A1

Note: accept $\$974$ from use of $C=732r^3$ .

[2 marks]

Write down the height of the ball above the ground at the instant it is hit by the bat.

Find the value of $t$ when the ball hits the ground.

State an appropriate domain for $t$ in this model.

$1.2$ metres        A1

[1 mark]

$-4.8t^2+21t+1.2=0$         (M1)

$\left(t=\right) 4.43 \mathrm{s} \left(4.431415\dots \text{s}\right)$        A1

 
Note: If both values for $t$ are seen do not award the A1 mark unless the negative is explicitly excluded.

[2 marks]

$0\le t\le 4.43$  OR  $\left[0, 4.43\right]$        A1A1

 
Note: Award A1 for correct endpoints and A1 for expressing answer with correct notation. Award at most A1A0 for use of $x$ instead of $t$.

[2 marks]

Probably the best answered question on the paper with many correct answers seen.

Many candidates correctly solved the quadratic equation.

Some cases, the lower bound was given as 1.2 from confusing height with time. Often the variable $x$ was used in the interval notation which lost a mark.

Find the equation of the axis of symmetry of the graph of $f$.

Write down the value of $h$.

Find the value of $k$.

The graph of a second function, $g$, is obtained by a reflection of the graph of $f$ in the $y$-axis, followed by a translation of $\left(\begin{array}{c}-3\\ 6\end{array}\right)$.

Find the coordinates of the vertex of the graph of $g$.

correct working      (A1)

eg   $\frac{-\left(-4\right)}{2\left(1\right)}$,  $\frac{-1+5}{2}$

$x=2$  (must be an equation with $x=$)      A1 N2

[2 marks]

$h$ = 2      A1  N1

[1 mark]

METHOD 1

valid approach      (M1)

eg   $f$(2)

correct substitution      (A1)

eg   (2)² − 4(2) − 5

$k$ = −9     A1 N2

METHOD 2

valid attempt to complete the square      (M1)

eg   $x$² − 4$x$ + 4

correct working      (A1)

eg  ($x$² − 4$x$ + 4) − 4 − 5,  ($x$ − 2)² − 9

$k$ = −9     A1 N2

[3 marks]

METHOD 1 (working with vertex)

vertex of $f$ is at (2, −9)      (A1)

correct horizontal reflection      (A1)

eg  $x$ = −2,  (−2, −9)

valid approach for translation of their $x$ or $y$ value      (M1)

eg   $x$ − 3,  $y$ + 6,  $\left(\begin{array}{c}-2\\ -9\end{array}\right)+\left(\begin{array}{c}-3\\ 6\end{array}\right)$,  one correct coordinate for vertex

vertex of $g$ is (−5, −3) (accept $x$ = −5, $y$ = −3)      A1A1 N1N1

METHOD 2 (working with function)

correct approach for horizontal reflection      (A1)

eg   $f$(−$x$)

correct horizontal reflection      (A1)

eg   (−$x$)² −4(−$x$) − 5,   $x$² + 4$x$ − 5, (−$x$ − 2)² − 9

valid approach for translation of their $x$ or $y$ value      (M1)

eg   ($x$ + 3)² + 4($x$ + 3) − 5 + 6,  $x$² + 10$x$ + 22,  ($x$ + 5)² − 3,  one correct coordinate for vertex

vertex of $g$ is (−5, −3) (accept $x$ = −5, $y$ = −3)      A1A1 N1N1

[5 marks]

Write down what the value of 150 represents in the context of the question.

Find the value of $a$.

Find the total time that the baking tin is in the oven.

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

the temperature in the oven     (A1)

OR

the maximum possible temperature of the cake mix     (A1)     (C1)

Note:     Award (A0) for “the maximum temperature”.

[1 mark]

$18=150-a(\times {1.1}^{\circ })$     (M1)

Note:     Award (M1) for correct substitution of 18 and 0. Substitution of 0 can be implied.

$(a)=132$     (A1)     (C2)

[2 marks]

$150-132\times {1.1}^{-t}=130$     (M1)

Note:     Award (M1) for substituting their a and equating to 130. Accept an inequality.

Award (M1) for a sketch of the horizontal line on the graph.

$t=19.8(19.7992\dots )$     (A1)(ft)

Note:     Follow through from part (b).

34.8 (minutes) (34.7992…, 34 minutes 48 seconds)     (A1)(ft)     (C3)

Note:     Award the final (A1) for adding 15 minutes to their $t$ value.

In part (c), award (C2) for a final answer of 19.8 with no working.

[3 marks]

Find the $y$-intercept of the graph of $f\left(x\right)+3$.

Find the $y$-intercept of the graph of $f\left(4x\right)$.

Find the $x$-intercept of the graph of $f\left(2x\right)$.

Describe the transformation $f\left(x+1\right)$.

$y$-intercept is 11   (accept (0, 11) )     A1 N1 

[1 mark]

valid approach        (M1)

eg  $f\left(4\times 0\right)=f\left(0\right)$,   recognizing stretch of $\frac{1}{4}$ in $x$-direction

$y$-intercept is 8      (accept (0, 8) )    A1 N2 

[2 marks]

$x$-intercept is $\frac{5}{2}\left(=2.5\right)$   (accept $\left(\frac{5}{2}\text{,}0\right)$ or (2.5, 0) )    A2 N2 

[2 marks]

correct name, correct magnitude and direction        A1A1 N2

eg   name: translation, (horizontal) shift (do not accept move)

eg   magnitude and direction: 1 unit to the left, $\left(\begin{array}{c}-1\\ 0\end{array}\right)$, horizontal by –1

[2 marks]

Find the value of $a$ and of $b$.

$-\frac{b}{2a}=-2$

$a(-2{)}^2-2b+5=3$ or equivalent

$a(-4{)}^2-4b+5=5$ or equivalent

$2a(-2)+b=0$ or equivalent     (M1)

Note:     Award (M1) for two of the above equations.

$a=0.5$     (A1)(ft)

$b=2$     (A1)(ft)     (C3)

Note:     Award at most (M1)(A1)(ft)(A0) if the answers are reversed.

Follow through from parts (a) and (b).

[3 marks]

Use this information to write down an equation involving $x$ and $y$.

Use this ratio to write down $y$ in terms of $x$.

Find the value of $x$ and of $y$.

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

$x^2+y^2={68}^2$ (or 4624 or equivalent)     (A1)     (C1)

[1 mark]

$\frac{y}{x}=\frac{3}{4}$     (M1)

Note:     Award (M1) for a correct equation.

$y=\frac{3}{4}x(y=0.75x)$     (A1)     (C2)

[2 marks]

$x^2+{\left(\frac{3}{4}x\right)}^2={68}^2\left(\text{or }{x}^2+\frac{9}{16}{x}^2=4624\text{ or equivalent}\right)$     (M1)

Note:     Award (M1) for correct substitution of their expression for $y$ into their answer to part (a). Accept correct substitution of $x$ in terms of $y$.

$x=54.4\text{ (cm), }y=40.8\text{ (cm)}$     (A1)(ft)(A1)(ft)     (C3)

Note:     Follow through from parts (a) and (b) as long as $x>0$ and $y>0$.

[3 marks]

The equation of the axis of symmetry is x = p. Find p.

Hence, show that a = 2.

METHOD 1 (using symmetry to find p)

valid approach      (M1)

eg  $\frac{-1+3}{2}$, 

p = 1     A1 N2

Note: Award no marks if they work backwards by substituting a = 2 into $-\frac{b}{2a}$ to find p.

Do not accept $p=\frac{2}{a}$.

METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate a      M1

eg   a + 4 − c = a(3²) − 4(3) − c,  f(−1) = f(3)

correct working      A1

eg   8a = 16

a = 2      AG N0

valid approach to find p      (M1)

eg   $-\frac{b}{2a},\frac{4}{2\left(2\right)}$

p = 1      A1 N2

[2 marks]

METHOD 1

valid approach       M1

eg  $-\frac{b}{2a},\frac{4}{2a}$ (might be seen in (i)), f' (1) = 0

correct equation     A1

eg  $\frac{4}{2a}$ = 1, 2a(1) − 4 = 0

a = 2      AG N0

METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate a      M1

eg   a + 4 − c = a(3²) − 4(3) − c,  f(−1) = f(3)

correct working      A1

eg   8a = 16

a = 2      AG N0

[2 marks]

Write down the amount of money Jashanti saves per month.

Use your graphic display calculator to find how long it will take for Jashanti to have saved enough money to buy the car.

Calculate how much extra money Jashanti needs.

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

400 (USD)     (A1)     (C1)

[1 mark]

$8500(0.95{)}^t=400\times t+2000$     (M1)

Note:     Award (M1) for equating $8500(0.95{)}^t$ to $400\times t+2000$ or for comparing the difference between the two expressions to zero or for showing a sketch of both functions.

$(t=)8.64\text{ (months) }\left(8.6414\dots \text{ (months)}\right)$     (A1)     (C2)

Note:     Accept 9 months.

[2 marks]

$8500(0.95{)}^2-(400\times 2+2000)$     (M1)(M1)

Note:     Award (M1) for correct substitution of $t=2$ into equation for $P$, (M1) for finding the difference between a value/expression for $P$ and a value/expression for $S$. The first (M1) is implied if 7671.25 seen.

4870 (USD) (4871.25)     (A1)     (C3)

Note:     Accept 4871.3.

[3 marks]

State, in the context of the question, what the value of $34.50$ represents.

State, in the context of the question, what the value of $8.50$ represents.

Write down the minimum number of pizzas that can be ordered.

Kaelani has $450 \text{PGK}$.

Find the maximum number of large cheese pizzas that Kaelani can order from Olava’s Pizza Company.

the cost of each (large cheese) pizza / a pizza / one pizza / per pizza       (A1)   (C1)

Note: Award (A0) for “the cost of (large cheese) pizzas”. Do not accept “the minimum cost of a pizza”.

[1 mark]

the (fixed) delivery cost      (A1)   (C1)

[1 mark]

$2$     (A1)   (C1)

[1 mark]

$450=34.50n+8.50$        (M1)

Note: Award (M1) for equating the cost equation to $450$ (may be stated as an inequality).

$12.8 \left(12.7971\dots \right)$      (A1)

$12$      (A1)(ft)   (C3)

Note: The final answer must be an integer.
The final (A1)(ft) is awarded for rounding their answer down to a whole number, provided their unrounded answer is seen.


[3 marks]

Find the total cost of buying $40$ litres of gas at Leon’s gas station.

Find $L^{-1}\left(70\right)$.

The price of gas at Erica’s gas station is $\$1.30$ per litre. A customer must buy a minimum of $10$ litres of gas. The total cost at Erica’s gas station is cheaper than Leon’s gas station when $x>k$.

Find the minimum value of $k$.

$L\left(40\right)=1.50\times 40-5$                      (M1)

$=\$55$              A1

[2 marks]

$70=1.50x-5$                     (M1)

$\left(x=\right) 50$ litres             A1

[2 marks]

$1.30x$                     (A1)

$1.30x<1.50x-5$                     (M1)

Note: Award M1 for a graph showing two intersecting linear functions, provided one function has a $y$-intercept of $0$ and the other function has a negative $y$-intercept.

(minimum value of $k=$) $25$                     A1

Note: Accept $x>25$.

[3 marks]

Find the value of $a$.

Find the value of $b$.

Find the value of $c$.

Find the value of $t$ at which the population first reaches $2200$.

* 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.

$\frac{2600-800}{2}=900$        (M1)A1

[2 marks]

$\frac{360}{12}=30$        (M1)A1

Note: Accept $\frac{2\pi }{12}=0.524 \left(0.523598\dots \right)$.

[2 marks]

$\frac{2600+800}{2}=1700$         A1

[1 mark]

Solve $900 \sin \left(30t\right)+1700=2200$        (M1)

$t=1.12 \left(1.12496\dots \right)$         A1

[2 marks]

Write down the value of $c$.

Hence form two equations in terms of $a$ and $b$.

Hence find the equation of the quadratic curve.

Find the area of the shaded region in Irina’s design.

$c=10$             A1

[1 mark]

$64a+8b+10=10$            A1

$16a+4b+10=12$            A1

Note: Award A1 for each equivalent expression or A1 for the use of the axis of symmetry formula to find $4=\frac{-b}{2a}$ or from use of derivative.  Award A0A1 for $64a+8b+c=10$ and $16a+4b+c=12$.

[2 marks]

$y=-\frac{1}{8}{x}^2+x+10$            A1A1

Note: Award A1A0 if one term is incorrect, A0A0 if two or more terms are incorrect. Award at most A1A0 if correct $a, b$ and $c$ values are seen but answer not expressed as an equation.

[2 marks]

recognizing the need to integrate their expression                        (M1)

${\int }_0^8-\frac{1}{8}{x}^2+x+10 dx$                        (A1)

Note: Award (A1) for correct integral, including limits. Condone absence of $dx$.

$90.7 {\text{cm}}^2 \left(\frac{272}{3}, 90.6666\dots \right)$               A1

[3 marks]

Generally, the responses were good for this last question on the paper. The main issue here was to not give the two equations in part (a)(ii) with simplified coefficients of $a$ and $b$. Several candidates understood what was required but left their answers with ${\left(8\right)}^{2}a$ and ${\left(4\right)}^{2}a$ un-simplified and lost marks. Some candidates used the coordinates $\left(0,0\right)$ to substitute in the equation with an incorrect equation of $a+b=0$. Candidates were successful at writing the equations in part (a)(iii). In part (b), most candidates realized that they had to use integration to find the area of the shaded region and, for the most part, were able to find a correct value for the area using either the correct equation or their obtained equation from the previous part. A common error was to integrate between $0$ and $10$ instead of $0$ and $8$.

Generally, the responses were good for this last question on the paper. The main issue here was to not give the two equations in part (a)(ii) with simplified coefficients of $a$ and $b$. Several candidates understood what was required but left their answers with ${\left(8\right)}^{2}a$ and ${\left(4\right)}^{2}a$ un-simplified and lost marks. Some candidates used the coordinates $\left(0,0\right)$ to substitute in the equation with an incorrect equation of $a+b=0$. Candidates were successful at writing the equations in part (a)(iii). In part (b), most candidates realized that they had to use integration to find the area of the shaded region and, for the most part, were able to find a correct value for the area using either the correct equation or their obtained equation from the previous part. A common error was to integrate between $0$ and $10$ instead of $0$ and $8$.

Generally, the responses were good for this last question on the paper. The main issue here was to not give the two equations in part (a)(ii) with simplified coefficients of $a$ and $b$. Several candidates understood what was required but left their answers with ${\left(8\right)}^{2}a$ and ${\left(4\right)}^{2}a$ un-simplified and lost marks. Some candidates used the coordinates $\left(0,0\right)$ to substitute in the equation with an incorrect equation of $a+b=0$. Candidates were successful at writing the equations in part (a)(iii). In part (b), most candidates realized that they had to use integration to find the area of the shaded region and, for the most part, were able to find a correct value for the area using either the correct equation or their obtained equation from the previous part. A common error was to integrate between $0$ and $10$ instead of $0$ and $8$.

Generally, the responses were good for this last question on the paper. The main issue here was to not give the two equations in part (a)(ii) with simplified coefficients of $a$ and $b$. Several candidates understood what was required but left their answers with ${\left(8\right)}^{2}a$ and ${\left(4\right)}^{2}a$ un-simplified and lost marks. Some candidates used the coordinates $\left(0,0\right)$ to substitute in the equation with an incorrect equation of $a+b=0$. Candidates were successful at writing the equations in part (a)(iii). In part (b), most candidates realized that they had to use integration to find the area of the shaded region and, for the most part, were able to find a correct value for the area using either the correct equation or their obtained equation from the previous part. A common error was to integrate between $0$ and $10$ instead of $0$ and $8$.

Find $\frac{\text{d}y}{\text{d}x}$.

Find the gradient of this tangent at point P.

Find the equation of this tangent. Give your answer in the form y = mx + c.

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

15x² − 3      (A1)(A1) (C2)

Note: Award (A1) for 15x², (A1) for −3. Award at most (A1)(A0) if additional terms are seen.

[2 marks]

15 (−1)² − 3      (M1)

Note: Award  (M1) for substituting −1 into their $\frac{\text{d}y}{\text{d}x}$.

= 12     (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

(y − (−2)) = 12 (x − (−1))     (M1)

OR

−2 = 12(−1) + c     (M1)

Note: Award  (M1) for point and their gradient substituted into the equation of a line.

y = 12x + 10     (A1)(ft) (C2)

Note: Follow through from part (b).

[2 marks]

Find f'(x)

Find the gradient of the graph of f at $x=-\frac{1}{2}$.

Find the x-coordinate of the point at which the normal to the graph of f has gradient $-\frac{1}{8}$.

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

x³     (A1) (C1)

Note: Award (A0) for $\frac{4x^3}{4}$ and not simplified to x³.

[1 mark]

${\left(-\frac{1}{2}\right)}^3$     (M1)

Note: Award (M1) for correct substitution of $-\frac{1}{2}$ into their derivative.

$-\frac{1}{8}$  (−0.125)     (A1)(ft) (C2)

Note: Follow through from their part (a).

[2 marks]

x³ = 8     (A1)(M1)

Note: Award (A1) for 8 seen maybe seen as part of an equation y = 8x + c, (M1) for equating their derivative to 8.

(x =) 2     (A1) (C3)

Note: Do not accept (2, 4).

[3 marks]

Write down the number of Elvis impersonators in $1977$.

Calculate the time taken for the number of Elvis impersonators to reach $130 000$.

Calculate the number of Elvis impersonators when $t=70$.

The world population in $2047$ is projected to be $9 500 000 000$ people.

Use this information to explain why the model for the number of Elvis impersonators is unrealistic.

$170$      (A1)   (C1)

[1 mark]

$130 000=170\times 1.{31}^t$      (M1)

Note: Award (M1) for equating $130 000$ to the exponential function.

$\left(t=\right) 24.6$  ($24.5882\dots$ (years))        (A1)   (C2)

[2 marks]

$170\times 1.{31}^{70}$      (M1)

Note: Award (M1) for correct substitution in the function $N\left(t\right)$.

$2.75\times {10}^{10} \left(2.75067\dots \times {10}^{10}, 27 500 000 000, 27 506 771 343\right)$        (A1)   (C2)

[2 marks]

The number of Elvis impersonators in $2047$, is greater than the world population.       (R1) (C1)

OR

$2.75\times {10}^{10}>9 500 000 000$       (R1) (C1)

Note: Award (R1) for a correct comparison of their number of impersonators with the world population. Follow through from part (c) if a reasonable argument can be made that the model is unrealistic.
Award (R0) if the number of impersonators is not explicitly seen in part (c) or in part (d).

[1 mark]

Write down the other solution of $f\left(x\right)=k$.

Complete the table below placing a tick (✔) to show whether the unknown parameters $a$ and $b$ are positive, zero or negative. The row for $c$ has been completed as an example.

State the values of $x$ for which $f\left(x\right)$ is decreasing.

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

$\left(x=\right) \left(-2\right)-4$  OR  $\left(x=\right) \left(-2\right)-\left(2-\left(-2\right)\right)$      (M1)

Note: Award (M1) for correct calculation of the left symmetrical point.

$\left(x=\right) -6$      (A1)   (C2)

[2 marks]

      (A1)(A1)   (C2)

Note: Award (A1) for each correct row.

[2 marks]

$x>-2$  OR  $x\ge -2$      (A1)(A1)   (C2)

Note: Award (A1) for $-2$ seen as part of an inequality, (A1) for completely correct notation. Award (A1)(A1) for correct equivalent statement in words, for example “decreasing when $x$ is greater than negative $2$”.

[2 marks]

Show that $I=\frac{9}{d^2}$.

Sketch the curve of $I$ on the axes below showing clearly the point $(1.5, 4)$.

Whilst swimming, Scarlett can hear the siren only if the sound intensity at her location is greater than $1.5\times {10}^{-6}$ ${\text{W\hspace{0.05em}m}}^{-2}$.

Find the values of $d$ where Scarlett cannot hear the siren.

$I=\frac{k}{d^2}$              (M1)

$4=\frac{k}{1.5^2}$              M1

$I=\frac{9}{d^2}$                AG

Note: The AG line must be seen for the second M1 to be awarded.
          Award no marks for substituting $1.5$ and $4$ into $I=\frac{9}{d^2}$ (i.e., working backwards).

[2 marks]

        A1A1

Note: Award A1 for correct general shape (concave up) with no $I$-intercept, passing through the marked point $\left(1.5, 4\right)$; the point must be labelled with either the coordinates or the values $1.5$ and $4$ on the $x$ and $y$ axes. Award A1 for the curve showing asymptotic behavior (i.e. $I$ tends to $0$, as $d$ tends to infinity), extending to at least $d=6$; the curve must not cross nor veer away from the horizontal asymptote.

[2 marks]

$1.5\times {10}^{-6}\ge \frac{9}{d^2}$               (M1)

Note: Award (M1) for a correct inequality.

$d\ge 2450 \left(\text{m}\right) \left(2449.48\dots \right)$            A1

Note: Award A0 for $d=2450$.

[2 marks]

Write down the derivative of $f$.

Find the point on the graph of $f$ at which the gradient of the tangent is equal to 6.

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

$12x^2-\frac{6}{x^3}$ or equivalent     (A1)(A1)(A1)     (C3)

Note:     Award (A1) for $12x^2$, (A1) for $-6$ and (A1) for $\frac{1}{x^3}$ or $x^{-3}$. Award at most (A1)(A1)(A0) if additional terms seen.

[3 marks]

$12x^2-\frac{6}{x^3}=6$     (M1)

Note:     Award (M1) for equating their derivative to 6.

$(1,4)$$$OR$$$x=1,y=4$     (A1)(ft)(A1)(ft)     (C3)

Note:     A frequent wrong answer seen in scripts is $(1,6)$ for this answer with correct working award (M1)(A0)(A1) and if there is no working award (C1).

[3 marks]

Write down how many kilograms of cheese Maria sells in one week if the price of a kilogram of cheese is 8 EUR.

Find how much Maria earns in one week, from selling cheese, if the price of a kilogram of cheese is 8 EUR.

Write down an expression for $W$ in terms of $p$.

Find the price, $p$, that will give Maria the highest weekly profit.

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

522 (kg)     (A1)     (C1)

[1 mark]

$522(8-6.80)$ or equivalent     (M1)

Note:     Award (M1) for multiplying their answer to part (a) by $(8-6.80)$.

626 (EUR) (626.40)     (A1)(ft)     (C2)

Note:     Follow through from part (a).

[2 marks]

$(W=)(882-45p)(p-6.80)$     (A1)

OR

$(W=)-45p^2+1188p-5997.6$     (A1)     (C1)

[1 mark]

sketch of $W$ with some indication of the maximum     (M1)

OR

$-90p+1188=0$     (M1)

Note:     Award (M1) for equating the correct derivative of their part (c) to zero.

OR

$(p=)\frac{-1188}{2\times (-45)}$     (M1)

Note:     Award (M1) for correct substitution into the formula for axis of symmetry.

$(p=)13.2\text{ (EUR)}$     (A1)(ft)     (C2)

Note:     Follow through from their part (c), if the value of $p$ is such that .

[2 marks]

Find the value of A.

Interpret what A represents in this context.

Find the time since the experiment began for the bacteria population to be equal to 40A.

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

6400 = A × 2⁴     (M1)

Note: Award (M1) for correct substitution of 4 and 6400 in equation.

(A =) 400     (A1) (C2)

[2 marks]

the initial population OR the population at the start of experiment      (A1) (C1)

[1 mark]

40A = A × 2^t  OR  40 × 400 = 400 × 2^t       (M1)

Note: Award (M1) for correct substitution into equation. Follow through with their A from part (a).

40 = 2^t     (M1)

Note: Award (M1) for simplifying.

5.32  (5.32192…) (hours)  OR  5 hours 19.3 (19.3156…) minutes     (A1) (C3)

[3 marks]

Determine an equation for the axis of symmetry of the parabola that models the archway.

Determine whether the crate will fit through the archway. Justify your answer.

$\left(x=\right) -\frac{4.48}{2\left(-1.6\right)}$   OR   coordinates of maximum point $(1.4, 3.136)$          (M1)

$x=1.4$           A1

[2 marks]

METHOD 1

the cart is centred in the archway when it is between

$x=0.6$ and $x=2.2$,            A1

where $y\ge 2.112 \left(\text{m}\right)$  (which is greater than $2$)           R1

the archway is tall enough for the crate           A1

Note: Do not award R0A1.

METHOD 2

the height of the archway is greater or equal to $2.0$ between

$x=0.557385\dots$  and  $x=2.24261\dots$            A1

width of this section of archway =

$\left(2.24261\dots -0.557385\dots =\right) 1.68522 \left(\text{m}\right)$ (which is greater than $1.6$)           R1

the archway is wide enough for the crate           A1

Note: Do not award R0A1.

[3 marks]

Most candidates were able to substitute into the formula for axis of symmetry or find the vertex of the parabola correctly, both being appropriate methods, but neglected to write an equation from that, even though the question specifically asked for an equation.

Determining a process to see if the crate would fit through the archway proved to be difficult for many candidates. It was common to see the maximum heights compared, the maximum widths compared, or the area of the front surface of the crate compared to the area of the archway opening. Other candidates merely calculated the height at $x=1.6$, positioning the corner of the crate at O, and made their conclusion based on this value, without consideration of how the crate would be moving through the archway.

Find the value of $a$.

Find the time, in years, for the population to decrease to 20 turtles.

There is a number $m$ beyond which the turtle population will not decrease.

Find the value of $m$. Justify your answer.

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

$55=a\times 2^0+10$   (M1)

Note: Award (M1) for correct substitution of zero and 55 into the function.

45      (A1)   (C2)  

[2 marks]

$45\times 2^{-t}+10⩽20$       (M1)

Note: Award (M1) for comparing correct expression involving 20 and their 45. Accept an equation.

$t=2.17$  (2.16992…)      (A1)(ft)   (C2)

Note: Follow through from their part (a), but only if positive.
Answer must be in years; do not accept months for the final (A1).  

[2 marks]

$m=$10       (A1)

because as the number of years increases the number of turtles approaches 10      (R1)   (C2)

Note: Award (R1) for a sketch with an asymptote at approximately $y=10$,
OR for table with values such as 10.003 and 10.001 for $t=14$ and $t=15$, for example,
OR when $t$ approaches large numbers $y$ approaches 10. Do not award (A1)(R0). 

[2 marks]

Consider the following graphs of quadratic functions.

The equation of each of the quadratic functions can be written in the form $y=a{x}^2+bx+c$, where $a\ne 0$.

Each of the sets of conditions for the constants $a$, $b$ and $c$, in the table below, corresponds to one of the graphs above.

Write down the number of the corresponding graph next to each set of conditions.

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

     (A1)(A1)(A1)(A1)(A1)(A1)     (C6)

Note:     Award (A1) for each correct entry.

[6 marks]

Find the cost of producing 70 shirts.

Find the value of s.

Find the number of shirts produced when the cost of production is lowest.

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

(70 − 75)² + 100     (M1)

Note: Award (M1) for substituting in x = 70.

125     (A1) (C2)

[2 marks]

(s − 75)² + 100 = 500     (M1)

Note: Award (M1) for equating C(x) to 500. Accept an inequality instead of =.

OR

     (M1)

Note: Award (M1) for sketching correct graph(s).

(s =) 95    (A1) (C2)

[2 marks]

     (M1)

Note: Award (M1) for an attempt at finding the minimum point using graph.

OR

$\frac{95+55}{2}$     (M1)

Note: Award (M1) for attempting to find the mid-point between their part (b) and 55.

OR

(C'(x) =) 2x − 150 = 0     (M1)

Note: Award (M1) for an attempt at differentiation that is correctly equated to zero.

75     (A1) (C2)

[2 marks]

Write down one feature of this graph which suggests a cubic function might be appropriate to model this scenario.

Write down the value of $d$.

Write down three simultaneous equations for $a, b$ and $c$.

Hence, or otherwise, find the values of $a, b$ and $c$.

Use the model to determine the total length of time, in years, for which a graduate is expected to be in debt after graduating from university.

Accept any one of the following (or equivalent):

one minimum and one maximum point
three $x$-intercepts or three roots (or zeroes)
one point of inflexion           R1

Note: Do not accept “S shape” as a justification.

[1 mark]

$\left(d=\right)-5$         A1

[1 mark]

$8=a+b+c$

$4=8a+4b+2c$

$0=27a+9b+3c$            A2

Note: Award A2 if all three equations are correct.
Award A1 if at least one is correct. Award A1 for three correct equations that include the letter “$d$”.

[2 marks]

$a=2, b=-12, c=18$            A1

[1 mark]

equating found expression to zero            (M1)

$0=2t^3-12t^2+18t-5$

$t=0.358216\dots , 1.83174\dots , 3.81003\dots$            (A1)

(so total time in debt is $3.81003\dots -1.83174\dots +0.358216\approx$)

$2.34 \left(2.33650\dots \right)$ years            A1

[3 marks]

Proved to be difficult with several referring to the shape of the graph, the graph increasing and decreasing, or positive and negative values fitting the context.

It seemed easy to find the d-value in the function. Most candidates could derive at least one correct equation, but not always three. Many candidates did not write their equations in proper mathematical form, leaving exponents and like terms in their equations. Even those candidates who did not write correct equations in part (ii) were able to correctly find the values of a, b, and c in part (iii) using cubic regression (an off-syllabus method, but still valid and credited full marks). There were some candidates who attempted an analytic method to solve the system of equations which did not usually prove successful.

Some candidates realized they had to find the roots, but then did not know what to do with them. Several candidates selected one of the roots as the answer to the question, usually the largest root, clearly not understanding the relationship between the roots and the length of time in debt. Others found only one root and stated that as the answer.

Find the equation of the axis of symmetry of the graph of $f$.

(i)     Write down the value of $h$.

(ii)     Find the value of $k$.

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

correct approach     (A1)

eg $\frac{-(-4)}{2},f^{\prime }(x)=2x-4=0,\text{ (}{x}^2-4x+4)+5-4$

$x=2$ (must be an equation)     A1     N2

[2 marks]

(i)     $h=2$     A1     N1

(ii)     METHOD 1

valid attempt to find $k$     (M1)

eg$$$f(2)$

correct substitution into their function     (A1)

eg$$$(2{)}^2-4(2)+5$

$k=1$     A1     N2

METHOD 2

valid attempt to complete the square     (M1)

eg$$$x^2-4x+4$

correct working     (A1)

eg$$$(x^2-4x+4)-4+5,(x-2{)}^2+1$

$k=1$     A1     N2

[4 marks]

Write down the domain of the function.

Draw the line $y=-6$ on the axes.

Write down the number of solutions to $f(x)=-6$.

Find the range of values of $k$ for which $f(x)=k$ has no solution.

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

$(x\in \mathbb{R}),x\ne 0$     (A2)     (C2)

Note:     Accept equivalent notation. Award (A1)(A0) for $y\ne 0$.

Award (A1) for a clear statement that demonstrates understanding of the meaning of domain. For example, $\text{D}:(-\mathrm{\infty },0)\cup (1,\mathrm{\infty })$ should be awarded (A1)(A0).

[2 marks]

     (A1)     (C1)

Note:     The command term “Draw” states: “A ruler (straight edge) should be used for straight lines”; do not accept a freehand $y=-6$ line.

[1 mark]

2     (A1)(ft)     (C1)

Note:     Follow through from part (b)(i).

[1 mark]

    (A1)(A1)     (C2)

Note:     Award (A1) for both end points correct and (A1) for correct strict inequalities.

Award at most (A1)(A0) if the stated variable is different from $k$ or $y$ for example is (A1)(A0).

[2 marks]

Find the value of $p$.

Use this model to find the percentage of information retained by his students $36$ hours after Professor Wei’s lecture.

State a mathematical reason why Professor Wei might believe this.

Write down one possible limitation of the domain of the model.

EITHER

$50=100{\text{e}}^{-1\times p}$  OR  $0.5={\text{e}}^{-1\times p}$              (M1)

OR

THEN

$0.693 (0.693147\dots , \ln 2)$              A1

[2 marks]

$R\left(1.5\right)=100{\text{e}}^{-0.693147\dots \times 1.5}$               (M1)

$35.4\left(\%\right) \left(35.3553\dots \right)$             A1

[2 marks]

$R\left(t\right)>0$  OR  $R\left(t\right)$ has a horizontal asymptote             R1

[1 mark]

Award A1 for one reasonable limitation of the domain:         A1

small values of $t$ produce unrealistic results

$R\left(0\right)=100\%$

large values of $t$ are not possible

people do not live forever

model is not valid at small or large values of $t$

The reason should focus on the domain $t\ge 0$. Do not accept answers such as:

recollection varies for different people

memories are discrete not continuous

the nature of the information will change how easily it is recalled

emotional/physical stress can affect recollection/concentration

Note: Do not accept $t\ge 0$ as this is a limitation that has been given in the question.

[1 mark]

Find the population of the Bulbul birds at the start of the migration season.

Find the population of the Bulbul birds after 5 days.

Calculate the time taken for the population to decrease below 1400.

According to this model, find the smallest possible population of Bulbul birds during the migration season.

1750     A1

[1 mark]

$1350+400{\left(1.25\right)}^{-5}$    (M1)

= 1480    A1

Note: Accept 1481.

[2 marks]

$1400=1350+400{\left(1.25\right)}^{-t}$       (M1)

9.32 (days (9.31885…) (days))   A1

[2 marks]

1350   A1

Note: Accept 1351 as a valid interpretation of the model as $P$ = 1350 is an asymptote.

[1 mark]

the initial dose of the drug.

the percentage of the drug that leaves the body each hour.

Calculate the amount of the drug remaining in the body $10$ hours after the injection.

$23 \text{mg}$                   A1

[1 mark]

$1-0.85$  OR  $\frac{23-19.55}{23}$  OR  $0.15$                  (M1)

$15 \left(\%\right)$                   A1

[2 marks]

$23{\left(0.85\right)}^{10}$                  (M1)

$4.53 \text{mg} \left(4.52811\dots \right)$                   A1

[2 marks]

Find the number of fruit flies which were placed in the container.

Find the number of fruit flies that are in the container after 6 days.

The maximum capacity of the container is 8000 fruit flies.

Find the number of days until the container reaches its maximum capacity.

$12\times 3^{0.498\times 0}$     (M1)