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]

Show that $T_0=100$.

Show that $k=\frac{1}{10}\ln \frac{10}{7}$.

Find the temperature of the water when $t=15$.

Sketch the graph of $T$ versus $t$, clearly indicating any asymptotes with their equations and stating the coordinates of any points of intersection with the axes.

Find the time taken for the water to have a temperature of $50 °\text{C}$. Give your answer correct to the nearest second.

The model for the temperature of the water can also be expressed in the form $T=T_0{a}^{\frac{t}{10}}$ for $t\ge 0$ and $a$ is a positive constant.

Find the exact value of $a$.

* 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.

when $t=0, T=100\Rightarrow 100=T_0{\text{e}}^0$         A1

so $T_0=100$         AG

[1 mark]

correct substitution of $t=10, T=70$         M1

$70=100{\text{e}}^{-10k}$  or  ${\text{e}}^{-10k}\mathrm{=}\frac{7}{10}$

EITHER

$-10k=\ln \frac{7}{10}$         A1

$\ln \frac{7}{10}=-\ln \frac{10}{7}$  or  $-\ln \frac{7}{10}=\ln \frac{10}{7}$         A1

OR

${\text{e}}^{10k}\mathrm{=}\frac{10}{7}$         A1

$10k=\ln \frac{10}{7}$         A1

THEN

$k=\frac{1}{10}\ln \frac{10}{7}$         AG

[3 marks]

substitutes $t=15$ into $T$         (M1)

$T=58.6 \left(°\text{C}\right)$         A1

[2 marks]

a decreasing exponential         A1

starting at $\left(0, 100\right)$ labelled on the graph or stated         A1

$T\to 0$ as $t\to \infty$         A1

horizontal asymptote $T=0$ labelled on the graph or stated         A1

Note: Award A0 for stating $y=0$ as the horizontal asymptote.

[4 marks]

$100{\text{e}}^{-kt}=50$  where $k=\frac{1}{10}\ln \frac{10}{7}$        A1

EITHER

uses an appropriate graph to attempt to solve for $t$         (M1)

OR

manipulates logs to attempt to solve for $t$ e.g. $\ln \frac{1}{2}=\left(-\frac{1}{10}\ln \frac{10}{7}\right)t$         (M1)

$t=\frac{\ln 2}{{\displaystyle \frac{1}{10}}\ln {\displaystyle \frac{10}{7}}}=19.433\dots$        A1

THEN

temperature will be $50 °\text{C}$ after $19$ minutes and $26$ seconds        A1

[4 marks]

METHOD 1

substitutes $T_0=100$, $t=10$ and $T=70$ into $T=T_0{a}^{\frac{t}{10}}$         (M1)

$70=100a^{\frac{10}{10}}$        A1

$a=\frac{7}{10}$        A1

METHOD 2

$100a^{\frac{t}{10}}=100{\text{e}}^{-kt}$  where $k=\frac{1}{10}\ln \frac{10}{7}$

EITHER

${\text{e}}^{-k}=a^{\frac{1}{10}}\Rightarrow a={\text{e}}^{-10k}$         (M1)

OR

$a={\left({\text{e}}^{\left(-\frac{1}{10}\ln \frac{10}{7}\right)t}\right)}^{\frac{10}{t}}$         (M1)

THEN

$a={\text{e}}^{-\ln \frac{10}{7}} \left(={\text{e}}^{\ln \frac{7}{10}}\right)$        A1

$a=\frac{7}{10}$        A1

[3 marks]

Find the $x$-coordinates of the points of intersection of the two curves.

Find the area, $A$, of the region enclosed by the two curves.

* 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.

attempts to solve $x^2 \sin x=-1-\sqrt{1+4{\left(x+2\right)}^2}$       (M1)

$x=-2.76, -1.54$        A1A1

Note: Award A1A0 if additional solutions outside the domain are given.

[3 marks]

$A=\underset{-2.762\dots }{\overset{-1.537\dots }{\int }}\left(-1-\sqrt{1+4{\left(x+2\right)}^2}-x^2 \sin x\right) \mathrm{d}x$ (or equivalent)       (M1)(A1)

Note: Award M1 for attempting to form an integrand involving “top curve” − “bottom curve”.

so $A=1.47$          A2

[4 marks]

Expand the expression for $f(x)$.

Find $f^{\prime }(x)$.

Draw the graph of $f$ for $-3⩽x⩽3$ and $-40⩽y⩽20$. Use a scale of 2 cm to represent 1 unit on the $x$-axis and 1 cm to represent 5 units on the $y$-axis.

Write down the coordinates of the point of intersection.

$10x-2x^3+10-2x^2$     (A1)

Notes:     The expansion may be seen in part (b)(ii).

[1 mark]

$10-6x^2-4x$     (A1)(ft)(A1)(ft)(A1)(ft)

Notes:     Follow through from part (b)(i). Award (A1)(ft) for each correct term. Award at most (A1)(ft)(A1)(ft)(A0) if extra terms are seen.

[3 marks]

     (A1)(A1)(ft)(A1)(ft)(A1)

Notes:     Award (A1) for correct scale; axes labelled and drawn with a ruler.

Award (A1)(ft) for their correct $x$-intercepts in approximately correct location.

Award (A1) for correct minimum and maximum points in approximately correct location.

Award (A1) for a smooth continuous curve with approximate correct shape. The curve should be in the given domain.

Follow through from part (a) for the $x$-intercepts.

[4 marks]

$(1.49,13.9)\left((1.48702\dots ,13.8714\dots )\right)$     (G1)(ft)(G1)(ft)

Notes:     Award (G1) for 1.49 and (G1) for 13.9 written as a coordinate pair. Award at most (G0)(G1) if parentheses are missing. Accept $x=1.49$ and $y=13.9$. Follow through from part (b)(i).

[2 marks]

Write down the $y$-intercept of the graph of $y=f\left(x\right)$.

Sketch the graph of $y=f\left(x\right)$ for −3 ≤ $x$ ≤ 3 and −4 ≤ $y$ ≤ 12.

Determine the range of $f\left(x\right)$ for $p$ ≤ $x$ ≤ $q$.

−1    (A1)

Note: Accept (0, −1).

[1 mark]

  (A1)(A1)(A1)(A1)

Note: Award (A1) for correct window and axes labels, −3 to 3 should be indicated on the $x$-axis and −4 to 12 on the $y$-axis.
    (A1)) for smooth curve with correct cubic shape;
    (A1) for $x$-intercepts: one close to −3, the second between −1 and 0, and third between 1 and 2; and $y$-intercept at approximately −1;
    (A1) for local minimum in the 4th quadrant and maximum in the 2nd quadrant, in approximately correct positions.
Graph paper does not need to be used. If window not given award at most (A0)(A1)(A0)(A1).

[4 marks]

$-1.27⩽f\left(x\right)⩽1.33\left(-1.27083\dots ⩽f\left(x\right)⩽1.33333\dots \text{,}-\frac{61}{48}⩽f\left(x\right)⩽\frac{4}{3}\right)$     (A1)(ft)(A1)(ft)(A1)

Note: Award (A1) for −1.27 seen, (A1) for 1.33 seen, and (A1) for correct weak inequalities with their endpoints in the correct order. For example, award (A0)(A0)(A0) for answers like $5⩽f\left(x\right)⩽2$. Accept $y$ in place of $f\left(x\right)$. Accept alternative correct notation such as [−1.27, 1.33].

Follow through from their $p$ and $q$ values from part (g) only if their $f\left(p\right)$ and $f\left(q\right)$ values are between −4 and 12. Award (A0)(A0)(A0) if their values from (g) are given as the endpoints.

[3 marks]

Show that arc $\text{APB}$ has length $6\mathrm{\pi }-3\theta$.

Show that $L=\sqrt{18-18 \cos \theta }$.

Arc $\text{APB}$ is twice the length of chord $\left[\text{AB}\right]$.

Find the value of $\theta$.

* 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.

EITHER

uses the arc length formula        (M1)

arc length is $3\left(2\mathrm{\pi }-\theta \right)$        A1

OR

length of arc $\text{AB}$ is $3\theta$        A1

the sum of the lengths of arc $\text{AB}$ and arc $\text{APB}$ is $6\mathrm{\pi }$        A1

THEN

so arc $\text{APB}$ has length $6\mathrm{\pi }-3\theta$        AG

[2 marks]

uses the cosine rule       (M1)

$L^2=3^2+3^2-2\left(3\right)\left(3\right) \cos \theta$        A1

so $L=\sqrt{18-18 \cos \theta }$        AG

[2 marks]

$6\mathrm{\pi }-3\theta =2\sqrt{18-18 \cos \theta }$        A1

attempts to solve for $\theta$       (M1)

$\theta =2.49$        A1

[3 marks]

the boat is taken from $\text{A}$ to $\text{P}$, and the bus from $\text{P}$ to $\text{B}$.

the boat travels directly to $\text{B}$.

Find an expression, in terms of $x$ for the travel time $T$, from $\text{A}$ to $\text{B}$, passing through $\text{D}$.

Find the value of $x$ so that $T$ is a minimum.

Write down the minimum value of $T$.

Find the new value of $x$ so that the total cost $C$ to travel from $\text{A}$ to $\text{B}$ via $\text{D}$ is a minimum.

Write down the minimum total cost for this journey.

$\frac{\text{AP}}{42}$  OR  $\frac{215}{84}$  OR  $\frac{65}{42}+\frac{215}{84}$                 (M1)

time $=4.10714\dots$ (hours)

time $=4.11$ (hours)                 A1

[2 marks]

$\text{AB}=\sqrt{{215}^2+{65}^2}\left(=224.610\dots \right)$                 (A1)

time $=5.34787\dots$ (hours)

time $=5.35$ (hours)                 A1

[2 marks]

$\text{AD}=\sqrt{{\left(215-x\right)}^2+{65}^2}$                 (A1)

$t=\frac{\sqrt{{\left(215-x\right)}^2+{65}^2}}{42}$                 (A1)

$T=\frac{\sqrt{{\left(215-x\right)}^2+{65}^2}}{42}+\frac{x}{84}\left(=\frac{\sqrt{x^2-430x+50450}}{42}+\frac{x}{84}\right)$                 A1

[3 marks]

valid approach to find the minimum for $T$ (may be seen in (iii))                 (M1)

graph of  $T$  OR  $T\text{'}=0$  OR  graph of $T\text{'}$

$x=177.472\dots \text{km}$

$x=177 \text{km}$                 A1

[2 marks]

$T=3.89980\dots$

$T=3.90$ (hours)                 A1

Note: Only allow FT in (b)(ii) and (iii) for $0<x<215$ and a function $T$ that has a minimum in that interval.

[1 mark]

$C=200·\frac{\sqrt{{\left(215-x\right)}^2+{65}^2}}{42}+150·\frac{x}{84}$                 (A1)

valid approach to find the minimum for $C\left(x\right)$  (may be seen in (ii))                 (M1)

graph of $C$  OR  $C\text{'}=0$  OR  graph of $C\text{'}$

$x=188.706\dots \text{km}$

$x=189 \text{km}$                 A1

Note: Only allow FT from (b) if the function $T$ has a minimum in $0<x<215$.

[3 marks]

$C=670.864$

$C=\$671$                 A1

Note: Only allow FT from (c)(i) if the function $C$ has a minimum in $0<x<215$.

[1 mark]

Find $\left(f\circ g\right)\left(x\right)$.

Solve the equation $\left(f\circ g\right)\left(x\right)=x$.

Hence or otherwise, given that $g\left(2a\right)=f^{-1}\left(2a\right)$, find the value of $a$.

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

attempt to form composite (in any order)       (M1)

eg    $f\left(\ln x\right) , g\left(4-x^3\right)$

$\left(f\circ g\right)\left(x\right)=4-{\left(\ln x\right)}^3$      A1  N2

[2 marks]

valid approach using GDC      (M1)

eg      , $\left(2.85, 2.85\right)$

$2.85056$

$2.85$      A1  N2

[2 marks]

METHOD 1 – (using properties of functions)

recognizing inverse relationship       (M1)

eg     $f\left(g\left(2a\right)\right)=f\left(f^{-1}\left(2a\right)\right) \left(=2a\right)$

equating $2a$ to their $x$ from (i)       (A1)

eg     $2a=2.85056$

$1.42528$

$a=1.43$       A1  N2

METHOD 2 – (finding inverse)

interchanging $x$ and $y$ (seen anywhere)       (M1)

eg     $x=4-y^3 , f^{-1}\left(x\right)=\sqrt[3]{4-x}$

correct working       (A1)

eg     $\sqrt[3]{4-2a}=\ln \left(2a\right)$, sketch showing intersection of $f^{-1}\left(2x\right)$ and $g\left(2x\right)$

$1.42528$

$a=1.43$       A1  N2

[3 marks]

Sketch the curve for −1 < x < 3 and −2 < y < 12.

A teacher asks her students to make some observations about the curve.

Three students responded.
Nadia said “The x-intercept of the curve is between −1 and zero”.
Rick said “The curve is decreasing when x < 1 ”.
Paula said “The gradient of the curve is less than zero between x = 1 and x = 2 ”.

State the name of the student who made an incorrect observation.

Find $\frac{\text{dy}}{\text{dx}}$.

Show that the stationary points of the curve are at x = 1 and x = 2.

Given that y = 2x³ − 9x² + 12x + 2 = k has three solutions, find the possible values of k.

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

(A1)(A1)(A1)(A1)

Note: Award (A1) for correct window (condone a window which is slightly off) and axes labels. An indication of window is necessary. −1 to 3 on the x-axis and −2 to 12 on the y-axis and a graph in that window.
(A1) for correct shape (curve having cubic shape and must be smooth).
(A1) for both stationary points in the 1^st quadrant with approximate correct position,
(A1) for intercepts (negative x-intercept and positive y intercept) with approximate correct position.

[4 marks]

Rick     (A1)

Note: Award (A0) if extra names stated.

[1 mark]

6x² − 18x + 12     (A1)(A1)(A1)

Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if extra terms seen.

[3 marks]

6x² − 18x + 12 = 0     (M1)

Note: Award (M1) for equating their derivative to 0. If the derivative is not explicitly equated to 0, but a subsequent solving of their correct equation is seen, award (M1).

6(x  − 1)(x − 2) = 0  (or equivalent)      (M1)

Note: Award (M1) for correct factorization. The final (M1) is awarded only if answers are clearly stated.

Award (M0)(M0) for substitution of 1 and of 2 in their derivative.

x = 1, x = 2 (AG)

[2 marks]

6 < k < 7     (A1)(A1)(ft)(A1)

Note: Award (A1) for an inequality with 6, award (A1)(ft) for an inequality with 7 from their part (c) provided it is greater than 6, (A1) for their correct strict inequalities. Accept ]6, 7[ or (6, 7).

[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]

Show that $b=0.3-a$.

Find the difference between the greatest possible expected value and the least possible expected value.

correct approach  A1

eg       $0.2+0.5+b+a=1$,  $0.7+a+b=1$

$b=0.3-a$       AG  N0

[1 mark]

correct substitution into $\text{E}\left(X\right)$        (A1)

eg       $0.2+4\times 0.5+a\times b+\left(a+b-0.5\right)\times a$,  $0.2+2+a\times b-0.2a$

valid attempt to express $\text{E}\left(X\right)$ in one variable       M1

eg       $0.2+4\times 0.5+a\times \left(0.3-a\right)+\left(-0.2\right)\times a$,  $2.2+0.1a-a^2$,

           $0.2+4\times 0.5+\left(0.3-b\right)\times b+\left(-0.2\right)\times \left(0.3-b\right)$,   $2.14+0.5b-b^2$

correct value of greatest $\text{E}\left(X\right)$        (A1)

$2.2025$  (exact)

valid attempt to find least value        (M1)

eg       graph with minimum indicated, $\text{E}\left(0\right)$  and  $\text{E}\left(0.3\right)$

          $\left(0\text{, }2.2\right)$  and  $\left(0.3\text{, }2.14\right)$ if $\text{E}\left(X\right)$ in terms of $a$

          $\left(0\text{, }2.14\right)$  and  $\left(0.3\text{, }2.2\right)$ if $\text{E}\left(X\right)$ in terms of $b$

correct value of least $\text{E}\left(X\right)$        (A1)

eg       $2.14$  (exact)

difference $=0.0625$ (exact)       A1  N2

[6 marks]

Find the value of $p$.

The following diagram shows part of the graph of $f$.

The region enclosed by the graph of $f$, the $x$-axis and the lines $x=-p$ and $x=p$ is rotated 360° about the $x$-axis. Find the volume of the solid formed.

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

valid approach     (M1)

eg$$$f(p)=4$, intersection with $y=4,\pm 2.32$

2.32143

$p=\sqrt{{\text{e}}^2-2}$ (exact), 2.32     A1     N2

[2 marks]

attempt to substitute either their limits or the function into volume formula (must involve $f^2$, accept reversed limits and absence of $\pi$ and/or $\text{d}x$, but do not accept any other errors)     (M1)

eg$$${\int }_{-2.32}^{2.32}{f}^2,\pi \int {\left(6-\ln (x^2+2)\right)}^2\text{d}x,\text{ 105.675}$

331.989

$\text{volume}=332$     A2     N3

[3 marks]

Sketch the graph of y = f (x), for −4 ≤ x ≤ 3 and −50 ≤ y ≤ 100.

Use your graphic display calculator to find the equation of the tangent to the graph of y = f (x) at the point (–2, 38.75).

Give your answer in the form y = mx + c.

Sketch the graph of the function g (x) = 10x + 40 on the same axes.

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

(A1)(A1)(A1)(A1)

Note: Award (A1) for axis labels and some indication of scale; accept y or f(x).

Use of graph paper is not required. If no scale is given, assume the given window for zero and minimum point.

Award (A1) for smooth curve with correct general shape.

Award (A1) for x-intercept closer to y-axis than to end of sketch.

Award (A1) for correct local minimum with x-coordinate closer to y-axis than end of sketch and y-coordinate less than half way to top of sketch.

Award at most (A1)(A0)(A1)(A1) if the sketch intersects the y-axis or if the sketch curves away from the y-axis as x approaches zero.

[4 marks]

y = −9.25x + 20.3  (y = −9.25x + 20.25)      (A1)(A1)

Note: Award (A1) for −9.25x, award (A1) for +20.25, award a maximum of (A0)(A1) if answer is not an equation.

[2 marks]

correct line, y = 10x + 40, seen on sketch     (A1)(A1)

Note: Award (A1) for straight line with positive gradient, award (A1) for x-intercept and y-intercept in approximately the correct positions. Award at most (A0)(A1) if ruler not used. If the straight line is drawn on different axes to part (a), award at most (A0)(A1).

[2 marks]

Find the period of $f$.

Find the value of $b$.

Hence, find the value of $f$(6).

Find the value of $p$ and the value of $q$.

Find the value of $x$ for which the functions have the greatest difference.

correct approach      A1

eg   $\frac{\pi }{6}=\frac{2\pi }{period}$  (or equivalent)

period = 12        A1

[2 marks]

valid approach      (M1)

eg  $\frac{\text{max}+\text{min}}{2}b=\text{max}-\text{amplitude}$

$\frac{21.8+10.2}{2}$, or equivalent

$b$ = 16        A1

[2 marks]

attempt to substitute into their function     (M1)

$5.8\text{sin}\left(\frac{\pi }{6}\left(6+1\right)\right)+16$

$f$(6) = 13.1        A1

[2 marks]

valid attempt to set up a system of equations    (M1)

two correct equations        A1

$p\text{sin}\left(\frac{2\pi }{9}\left(3-3.75\right)\right)+q=2.5$,  $p\text{sin}\left(\frac{2\pi }{9}\left(6-3.75\right)\right)+q=15.1$

valid attempt to solve system   (M1)

$p$ = 8.4; $q$ = 6.7        A1A1

[5 marks]

attempt to use $\left|f(x)-g(x)\right|$ to find maximum difference  (M1)

$x$ = 1.64        A1

[2 marks]

write down the equation of the vertical asymptote.

find the equation of the horizontal asymptote.

Find $f^{-1}\left(x\right)$.

Using an algebraic approach, show that the graph of $f^{-1}$ is obtained by a reflection of the graph of $f$ in the $y$-axis followed by a reflection in the $x$-axis.

Find the value of $p$ and the value of $q$.

Hence, find the area enclosed by the graph of $f$ and the graph of $f^{-1}$.

$x=-4$          A1

[1 mark]

attempt to substitute into $y=\frac{a}{c}$  OR  table with large values of $x$  OR  sketch of $f$ showing asymptotic behaviour          (M1)

$y=4$          A1

[2 marks]

$y=\frac{4x+1}{x+4}$

attempt to interchange $x$ and $y$ (seen anywhere)        M1

$xy+4y=4x+1$   OR   $xy+4x=4y+1$         (A1)

$xy-4x=1-4y$   OR   $xy-4y=1-4x$         (A1)

$f^{-1}\left(x\right)=\frac{1-4x}{x-4}$  (accept $y=\frac{1-4x}{x-4}$)         A1

[4 marks]

reflection in $y$-axis given by $f\left(-x\right)$         (M1)

$f\left(-x\right)=\frac{-4x+1}{-x+4}$         (A1)

reflection of their $f\left(-x\right)$ in $x$-axis given by $-f\left(-x\right)$ accept "now $-f\left(x\right)$"        M1

$\left(-f\left(-x\right)=\right) -\frac{-4x+1}{-x+4}$

$=\frac{-4x+1}{x-4}$  OR  $\frac{4x-1}{-x+4}$         A1

$=\frac{1-4x}{x-4} \left(=f^{-1}\left(x\right)\right)$         AG

Note: If the candidate attempts to show the result using a particular coordinate on the graph of $f$ rather than a general coordinate on the graph of $f$, where appropriate, award marks as follows:
M0A0 for eg $(2,3)\to (-2,3)$
M0A0 for $(-2,3)\to (-2,-3)$

[4 marks]

attempt to solve $f\left(x\right)=f^{-1}\left(x\right)$ using graph or algebraically         (M1)

$p=-1$  AND  $q=1$         A1

Note: Award (M1)A0 if only one correct value seen.

[2 marks]

attempt to set up an integral to find area between $f$ and $f^{-1}$         (M1)

$\underset{-1}{\overset{1}{\int }}\left(\frac{4x+1}{x+4}-\frac{1-4x}{x-4}\right)dx$         (A1)

$=0.675231\dots$

$=0.675$         A1

[3 marks]

Candidates mostly found the first part of this question accessible, with many knowing how to find the equation of both asymptotes. Common errors included transposing the asymptotes, or finding where an asymptote occurred but not giving it as an equation.

Candidates knew how to start part (b)(i), with most attempting to find the inverse function by firstly interchanging $x$ and $y$. However, many struggled with the algebra required to change the subject, and were not awarded all the marks. A common error in this part was for candidates to attempt to find an expression for $f\text{'}\left(x\right)$, rather than one for $f^{-1}\left(x\right)$. Few candidates were able to answer part (b)(ii). Many appeared not to know that a reflection in the $y$-axis is given by $f(-x)$, or that a reflection in the $x$-axis is given by $-f\left(x\right)$. Many of those that did, multiplied both the numerator and denominator by $-1$ when taking the negative of their $f(-x)$ , i.e. $-\left(\frac{-4x+1}{-x+4}\right)$ was often simplified as $\frac{4x-1}{x-4}$. However, the majority of candidates either did not attempt this question part or attempted to describe a graphical approach often involving a specific point, rather than an algebraic approach.

Those that attempted part (c), and had the correct expression for $f^{-1}\left(x\right)$, were usually able to gain all the marks. However, those that had an incorrect expression, or had found $f\text{'}\left(x\right)$, often proceeded to find an area, even when there was not an area enclosed by their two curves.

Given that the areas of the two shaded regions are equal, show that $\theta =2 \sin \theta$.

Hence determine the value of $\theta$.

attempt to find the area of either shaded region in terms of $r$ and $\theta$             (M1)

Note: Do not award M1 if they have only copied from the booklet and not applied to the shaded area.

Area of segment $=\frac{1}{2}{r}^2\theta -\frac{1}{2}{r}^2 \sin \theta$                 A1

Area of triangle $=\frac{1}{2}{r}^2 \sin \left(\mathrm{\pi }-\theta \right)$                 A1

correct equation in terms of $\theta$ only                 (A1)

$\theta -\sin \theta =\sin \left(\mathrm{\pi }-\theta \right)$

$\theta -\sin \theta =\sin \theta$                 A1

$\theta =2 \sin \theta$                 AG

Note: Award a maximum of M1A1A0A0A0 if a candidate uses degrees (i.e., $\frac{1}{2}{r}^2 \sin \left(180°-\theta \right)$), even if later work is correct.

Note: If a candidate directly states that the area of the triangle is $\frac{1}{2}{r}^2 \sin \theta$, award a maximum of M1A1A0A1A1.

[5 marks]

$\theta =1.89549\dots$

$\theta =1.90$                 A1

Note: Award A0 if there is more than one solution. Award A0 for an answer in degrees.

[1 mark]

Show that $b=\frac{\mathrm{\pi }}{6}$.

Find the value of $a$.

Find the value of $d$.

Find the smallest possible value of $c$.

Find the height of the water at $12:00$.

Determine the number of hours, over a 24-hour period, for which the tide is higher than $5$ metres.

$12=\frac{2\mathrm{\pi }}{b}$  OR  $b=\frac{2\mathrm{\pi }}{12}$                A1

$b=\frac{\mathrm{\pi }}{6}$                AG

[1 mark]

$a=\frac{6.8-2.2}{2}$  OR  $a=\frac{\text{max}-\text{min}}{2}$                (M1)

$=2.3 \left(\text{m}\right)$                A1

[2 marks]

$d=\frac{6.8+2.2}{2}$  OR  $d=\frac{\text{max}+\text{min}}{2}$                (M1)

$=4.5 \left(\text{m}\right)$                A1

[2 marks]

METHOD 1

substituting $t=4.5$ and $H=6.8$ for example into their equation for $H$                (A1)

$6.8=2.3 \sin \left(\frac{\mathrm{\pi }}{6}\left(4.5-c\right)\right)+4.5$

attempt to solve their equation                (M1)

$c=1.5$                A1

METHOD 2

using horizontal translation of $\frac{12}{4}$                (M1)

$4.5-c=3$                (A1)

$c=1.5$                A1

METHOD 3

$H\text{'}\left(t\right)=\left(2.3\right)\left(\frac{\mathrm{\pi }}{6}\right)\cos \left(\frac{\mathrm{\pi }}{6}\left(t-c\right)\right)$                (A1)

attempts to solve their $H\text{'}\left(4.5\right)=0$ for $c$                (M1)

$\left(2.3\right)\left(\frac{\mathrm{\pi }}{6}\right)\cos \left(\frac{\mathrm{\pi }}{6}\left(4.5-c\right)\right)=0$

$c=1.5$                A1

[3 marks]

attempt to find $H$ when $t=12$ or $t=0$, graphically or algebraically                (M1)

$H=2.87365\dots$

$H=2.87 \left(\text{m}\right)$                A1

[2 marks]

attempt to solve $5=2.3 \sin \left(\frac{\mathrm{\pi }}{6}\left(t-1.5\right)\right)+4.5$                (M1)

times are $t=1.91852\dots$ and $t=7.08147\dots , \left(t=13.9185\dots , t=19.0814\dots \right)$                (A1)

total time is $2\times \left(7.081\dots -1.919\dots \right)$

$10.3258\dots$

$=10.3$ (hours)                A1

Note: Accept $10$.

[3 marks]

Find $q$.

Find $p$.

Write down the probability of drawing three blue marbles.

Explain why the probability of drawing three white marbles is $\frac{1}{6}$.

The bag contains a total of ten marbles of which $w$ are white. Find $w$.

Jill plays the game nine times. Find the probability that she wins exactly two prizes.

Grant plays the game until he wins two prizes. Find the probability that he wins his second prize on his eighth attempt.

correct substitution into $\text{E}(X)$ formula     (A1)

eg$$$0(p)+1(0.5)+2(0.3)+3(q)=1.2$

$q=\frac{1}{30}$, 0.0333     A1     N2

[2 marks]

evidence of summing probabilities to 1     (M1)

eg$$$p+0.5+0.3+q=1$

$p=\frac{1}{6},0.167$     A1     N2

[2 marks]

$\text{P (3 blue)}=\frac{1}{30},0.0333$     A1     N1

[1 mark]

valid reasoning     R1

eg$$$\text{P (3 white)}=\text{P(0 blue)}$

$\text{P(3 white)}=\frac{1}{6}$     AG     N0

[1 mark]

valid method     (M1)

eg$$$\text{P(3 white)}=\frac{w}{10}\times \frac{w-1}{9}\times \frac{w-2}{8},\frac{{}_w{C}_3}{{}_{10}{C}_3}$

correct equation     A1

eg$$$\frac{w}{10}\times \frac{w-1}{9}\times \frac{w-2}{8}=\frac{1}{6},\frac{{}_w{C}_3}{{}_{10}{C}_3}=0.167$

$w=6$     A1     N2

[3 marks]

valid approach     (M1)

eg$$$\text{B}(n,p),\left(\begin{array}{c}n\\ r\end{array}\right)p^r{q}^{n-r},(0.167{)}^2(0.833{)}^7,\left(\begin{array}{c}9\\ 2\end{array}\right)$

0.279081

0.279     A1     N2

[2 marks]

recognizing one prize in first seven attempts     (M1)

eg$$$\left(\begin{array}{c}7\\ 1\end{array}\right),{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^6$

correct working     (A1)

eg$$$\left(\begin{array}{c}7\\ 1\end{array}\right){\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^6,0.390714$

correct approach     (A1)

eg$$$\left(\begin{array}{c}7\\ 1\end{array}\right){\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^6\times \frac{1}{6}$

0.065119

0.0651     A1     N2

[4 marks]

Write down the $y$-intercept of the graph.

Find $f^{\prime }(x)$.

Show that $a=8$.

Find $f(2)$.

Write down the $x$-coordinates of these two points;

Write down the intervals where the gradient of the graph of $y=f(x)$ is positive.

Write down the range of $f(x)$.

Write down the number of possible solutions to the equation $f(x)=5$.

The equation $f(x)=m$, where $m\in \mathbb{R}$, has four solutions. Find the possible values of $m$.

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

5     (A1)

Note:     Accept an answer of $(0,5)$.

[1 mark]

$\left(f^{\prime }(x)=\right)-4x^3+2ax$     (A1)(A1)

Note:     Award (A1) for $-4x^3$ and (A1) for $+2ax$. Award at most (A1)(A0) if extra terms are seen.

[2 marks]

$-4\times 2^3+2a\times 2=0$     (M1)(M1)

Note:     Award (M1) for substitution of $x=2$ into their derivative, (M1) for equating their derivative, written in terms of $a$, to 0 leading to a correct answer (note, the 8 does not need to be seen).

$a=8$     (AG)

[2 marks]

$\left(f(2)=\right)-2^4+8\times 2^2+5$     (M1)

Note:     Award (M1) for correct substitution of $x=2$ and  $a=8$ into the formula of the function.

21     (A1)(G2)

[2 marks]

$(x=)-2,(x=)\text{ 0}$     (A1)(A1)

Note:     Award (A1) for each correct solution. Award at most (A0)(A1)(ft) if answers are given as $(-2,21)$ and $(0,5)$ or $(-2,0)$ and $(0,0)$.

[2 marks]

    (A1)(ft)(A1)(ft)

Note:     Award (A1)(ft) for , follow through from part (d)(i) provided their value is negative.

Award (A1)(ft) for , follow through only from their 0 from part (d)(i); 2 must be the upper limit.

Accept interval notation.

[2 marks]

$y⩽21$     (A1)(ft)(A1)

Notes:     Award (A1)(ft) for 21 seen in an interval or an inequality, (A1) for “$y⩽$”.

Accept interval notation.

Accept or $f(x)⩽21$.

Follow through from their answer to part (c)(ii). Award at most (A1)(ft)(A0) if $x$ is seen instead of $y$. Do not award the second (A1) if a (finite) lower limit is seen.

[2 marks]

3 (solutions)     (A1)

[1 mark]

or equivalent     (A1)(ft)(A1)

Note:     Award (A1)(ft) for 5 and 21 seen in an interval or an inequality, (A1) for correct strict inequalities. Follow through from their answers to parts (a) and (c)(ii).

Accept interval notation.

[2 marks]

Show that $A_0=100$.

Show that $k=\frac{\ln 2}{5730}$.

Find, correct to the nearest $10$ years, the time taken after the plant’s death for $25\%$ of the carbon-14 to decay.

$100=A_0{\text{e}}^0$             A1

$A_0=100$             AG

[1 mark]

correct substitution of values into exponential equation             (M1)

$50=100{\text{e}}^{-5730k}$  OR  ${\text{e}}^{-5730k}=\frac{1}{2}$

EITHER

$-5730k=\ln \frac{1}{2}$             A1

$\ln \frac{1}{2}=-\ln 2$  OR  $-\ln \frac{1}{2}=\ln 2$             A1

OR

${\text{e}}^{5730k}=2$             A1

$5730k=\ln 2$             A1

THEN

$k=\frac{\ln 2}{5730}$             AG

Note: There are many different ways of showing that $k=\frac{\ln 2}{5730}$ which involve showing different steps. Award full marks for at least two correct algebraic steps seen.

[3 marks]

if $25\%$ of the carbon-14 has decayed, $75\%$ remains ie, $75$ units remain                      (A1)

$75=100{\text{e}}^{-\frac{\ln 2}{5730}t}$

EITHER

using an appropriate graph to attempt to solve for $t$                      (M1)

OR

manipulating logs to attempt to solve for $t$                               (M1)

$\ln 0.75=-\frac{\ln 2}{5730}t$

$t=2378.164\dots$

THEN

$t=2380$ (years) (correct to the nearest 10 years)                   A1

[3 marks]

Find $f\left(1\right)$.

Solve $f\left(x\right)=0$.

The graph of $f$ has a local minimum at point $\text{A}$.

Find the coordinates of $\text{A}$.

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

attempt to substitute $x=1$       (M1)

eg       $f\left(1\right), 1^2+1+\frac{50}{1}$

$52$  (exact)       A1   N2

[2 marks]

$-4.04932$

$-4.05$       A2   N2

[2 marks]

$\left(2.76649, 28.4934\right)$

$\text{A}\left(2.77, 28.5\right)$       A1A1   N2

[2 marks]

Sketch the graph of y = f (x) for 0 < x ≤ 6 and −30 ≤ y ≤ 60.
Clearly indicate the minimum point P and the x-intercepts on your graph.

(A1)(A1)(ft)(A1)(ft)(A1)(ft)

Note: Award (A1) for correct window. Axes must be labelled.
(A1)(ft) for a smooth curve with correct shape and zeros in approximately correct positions relative to each other.
(A1)(ft) for point P indicated in approximately the correct position. Follow through from their x-coordinate in part (c). (A1)(ft) for two x-intercepts identified on the graph and curve reflecting asymptotic properties.

[4 marks]

Write down a formula for $A$, the surface area to be coated.

Express this volume in $\text{c}{\text{m}}^3$.

Write down, in terms of $r$ and $h$, an equation for the volume of this water container.

Show that $A=\pi r^2+\frac{1000000}{r}$.

Find $\frac{\text{d}A}{\text{d}r}$.

Using your answer to part (e), find the value of $r$ which minimizes $A$.

Find the value of this minimum area.

Find the least number of cans of water-resistant material that will coat the area in part (g).

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

$(A=)\pi r^2+2\pi rh$    (A1)(A1)

Note:     Award (A1) for either $\pi r^2$ OR $2\pi rh$ seen. Award (A1) for two correct terms added together.

[2 marks]

$500000$    (A1)

Notes:     Units not required.

[1 mark]

$500000=\pi r^{2}h$    (A1)(ft)

Notes:     Award (A1)(ft) for $\pi r^{2}h$ equating to their part (b).

Do not accept unless $V=\pi r^{2}h$ is explicitly defined as their part (b).

[1 mark]

$A=\pi r^2+2\pi r\left(\frac{500000}{\pi r^2}\right)$    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their $\frac{500000}{\pi r^2}$ seen.

Award (M1) for correctly substituting only $\frac{500000}{\pi r^2}$ into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to $\pi rh=\frac{500000}{r}$ and substituting for $\pi rh$ in expression for $A$.

$A=\pi r^2+\frac{1000000}{r}$    (AG)

Notes:     The conclusion, $A=\pi r^2+\frac{1000000}{r}$, must be consistent with their working seen for the (A1) to be awarded.

Accept ${10}^6$ as equivalent to $1000000$.

[2 marks]

$2\pi r-\frac{\text{1}\text{000}\text{000}}{r^2}$    (A1)(A1)(A1)

Note:     Award (A1) for $2\pi r$, (A1) for $\frac{1}{r^2}$ or $r^{-2}$, (A1) for $-\text{1}\text{000}\text{000}$.

[3 marks]

$2\pi r-\frac{1000000}{r^2}=0$    (M1)

Note:     Award (M1) for equating their part (e) to zero.

$r^3=\frac{1000000}{2\pi }$ OR $r=\sqrt[3]{\frac{1000000}{2\pi }}$     (M1)