Find an expression for the velocity of $P_1$ at time $t$.

Particle $P_2$ also moves in a straight line. The position of $P_2$ is given by $\mathit{r}=\left(\begin{array}{c}-1\\ 6\end{array}\right)+t\left(\begin{array}{c}4\\ -3\end{array}\right)$.

The speed of $P_1$ is greater than the speed of $P_2$ when $t>q$.

Find the value of $q$.

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

recognizing velocity is derivative of displacement     (M1)

eg    $v=\frac{\text{d}s}{\text{d}t} , \frac{\text{d}}{\text{d}t}\left(10-\frac{7}{4}{t}^2\right)$

velocity$=-\frac{14}{4}t \left(=-\frac{7}{2}t\right)$        A1 N2

[2 marks]

valid approach to find speed of $P_2$     (M1)

eg    $\left|\left(\begin{array}{c}4\\ -3\end{array}\right)\right| , \sqrt{4^2+{\left(-3\right)}^2}$ , velocity$=\sqrt{4^2+{\left(-3\right)}^2}$

correct speed     (A1)

eg   $5 {\text{m\hspace{0.05em}s}}^{-1}$

recognizing relationship between speed and velocity (may be seen in inequality/equation)        R1

eg   $\left|-\frac{7}{2}t\right|$ , speed = | velocity | , graph of $P_1$ speed ,  $P_1$ speed $=\frac{7}{2}t , P_2$ velocity $=-5$

correct inequality or equation that compares speed or velocity (accept any variable for $q$)      A1

eg   $\left|-\frac{7}{2}t\right|>5 , -\frac{7}{2}q<-5 , \frac{7}{2}q>5 , \frac{7}{2}q=5$

$q=\frac{10}{7}$ (seconds) (accept $t>\frac{10}{7}$ , do not accept $t=\frac{10}{7}$)       A1   N2

Note: Do not award the last two A1 marks without the R1.

[5 marks]

Determine whether the graph of $l$ against $\theta$ is increasing or decreasing at $\theta =50°$.

Sieun observes that when the angle is $40°$, the ball will travel a horizontal distance of $205.5 \text{m}$.

Find an expression for the function $l\left(\theta \right)$.

$l\text{'}\left(50\right)=-0.2\times 50+9$            (M1)

$=-1$            A1

the curve is decreasing at $\theta =50°$            A1

Note: For the final A1, follow through within this question part for their $l\text{'}\left(50\right)$ value. Award A0 for an answer of "decreasing" with no work shown.

[3 marks]

recognition of need to integrate (e.g. reverse power rule or integral symbol or integrating at least one term correctly)            (M1)

$l\left(\theta \right)=-0.1{\theta }^2+9\theta \left(+c\right)$            A1A1

$205.5=-0.1\times {\left(40\right)}^2+9\times \left(40\right)+c$        (M1)

Note: Award M1 for correct substitution of $\theta =40°$ and $l=205.5$. A constant of integration must be seen (can be implied by a correct answer).

$c=5.5$

$\left(l\left(\theta \right)=\right)-0.1{\theta }^2+9\theta +5.5$            A1

Note: Accept any variable in the working, but for the final A1, the variable $\theta$ must be used in the expression.

[5 marks]

Find $f\text{'}\left(p\right)$ in terms of $k$ and $p$.

Show that the equation of $L_1$ is $kx+p^{2}y-2pk=0$.

Find the area of triangle $\text{AOB}$ in terms of $k$.

The graph of $f$ is translated by $\left(\begin{array}{c}4\\ 3\end{array}\right)$ to give the graph of $g$.
In the following diagram:

• point $\text{Q}$ lies on the graph of $g$
• points $\text{C}$, $\text{D}$ and $\text{E}$ lie on the vertical asymptote of $g$
• points $\text{D}$ and $\text{F}$ lie on the horizontal asymptote of $g$
• point $\text{G}$ lies on the $x$-axis such that $\text{FG}$ is parallel to $\text{DC}$.

Line $L_2$ is the tangent to the graph of $g$ at $\text{Q}$, and passes through $\text{E}$ and $\text{F}$.

Given that triangle $\text{EDF}$ and rectangle $\text{CDFG}$ have equal areas, find the gradient of $L_2$ in terms of $p$.

$f\text{'}\left(x\right)=-k{x}^{-2}$        (A1)

$f\text{'}\left(p\right)=-k{p}^{-2} \left(=-\frac{k}{p^2}\right)$     A1     N2

[2 marks]

attempt to use point and gradient to find equation of $L_1$        M1

eg    $y-\frac{k}{p}=-k{p}^{-2}\left(x-p\right), \frac{k}{p}=-\frac{k}{p^2}\left(p\right)+b$

correct working leading to answer        A1

eg    $p^{2}y-kp=-kx+kp, y-\frac{k}{p}=-\frac{k}{p^2}x+\frac{k}{p}, y=-\frac{k}{p^2}x+\frac{2k}{p}$

$kx+p^{2}y-2pk=0$     AG     N0

[2 marks]

METHOD 1 – area of a triangle

recognizing $x=0$ at $\text{B}$       (M1)

correct working to find $y$-coordinate of null       (A1)

eg   $p^{2}y-2pk=0$

$y$-coordinate of null at $y=\frac{2k}{p}$ (may be seen in area formula)        A1

correct substitution to find area of triangle       (A1)

eg   $\frac{1}{2}\left(2p\right)\left(\frac{2k}{p}\right), p\times \left(\frac{2k}{p}\right)$

area of triangle $\text{AOB}=2k$     A1     N3

METHOD 2 – integration

recognizing to integrate $L_1$ between $0$ and $2p$       (M1)

eg   ${\int }_0^{2p}{L}_{1 }dx , {\int }_0^{2p}-\frac{k}{p^2}x+\frac{2k}{p}$

correct integration of both terms        A1

eg   $-\frac{k{x}^2}{2p^2}+\frac{2kx}{p} , -\frac{k}{2p^2}{x}^2+\frac{2k}{p}x+c , {\left[-\frac{k}{2p^2}{x}^2+\frac{2k}{p}x\right]}_0^{2p}$

substituting limits into their integrated function and subtracting (in either order)       (M1)

eg    $-\frac{k{\left(2p\right)}^2}{2p^2}+\frac{2k\left(2p\right)}{p}-\left(0\right), -\frac{4k{p}^2}{2p^2}+\frac{4kp}{p}$

correct working       (A1)

eg    $-2k+4k$

area of triangle $\text{AOB}=2k$     A1     N3

[5 marks]

Note: In this question, the second M mark may be awarded independently of the other marks, so it is possible to award (M0)(A0)M1(A0)(A0)A0.

recognizing use of transformation      (M1)

eg   area of triangle $\text{AOB}$ = area of triangle $\text{DEF}$, $g\left(x\right)=\frac{k}{x-4}+3,$ gradient of $L_2=$ gradient of $L_1$, $\text{D}\left(4, 3\right)\text{, 2p+4, }$ one correct shift

correct working       (A1)

eg   area of triangle $\text{DEF}=2k, \text{CD}=3, \text{DF}=2p, \text{CG}=2p, \text{E}\left(4, \frac{2k}{p}+3\right), \text{F}\left(2p+4, 3\right), \text{Q}\left(p+4, \frac{k}{p}+3\right),$ 

gradient of $L_2=-\frac{k}{p^2}, g\text{'}\left(x\right)=-\frac{k}{{\left(x-4\right)}^2},$ area of rectangle $\text{CDFG}=2k$

valid approach      (M1)

eg   $\frac{\text{ED}\times \text{DF}}{2}=\text{CD}\times \text{DF}, 2p·3=2k , \text{ED}=2\text{CD} , {\int }_4^{2p+4}{L}_2 \text{d}x=4k$

correct working      (A1)

eg   $\text{ED}=6, \text{E}\left(4, 9\right), k=3p, \text{gradient}=\frac{3-\left({\displaystyle \frac{2k}{p}}+3\right)}{\left(2p+4\right)-4}, \frac{-6}{\left({\displaystyle \frac{2k}{3}}\right)}, -\frac{9}{k}$

correct expression for gradient (in terms of $p$)       (A1)

eg   $\frac{-6}{2p}, \frac{9-3}{4-\left(2p+4\right)}, -\frac{3p}{p^2}, \frac{3-\left({\displaystyle \frac{2\left(3p\right)}{p}}+3\right)}{\left(2p+4\right){\displaystyle -}{\displaystyle 4}}, -\frac{9}{3p}$

gradient of $L_2$ is $-\frac{3}{p} \left(=-3p^{-1}\right)$     A1     N3

[6 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 $8$.

Use the trapezoidal rule with an interval width of $0.4$ to estimate the total amount of carbon dioxide emitted during these two hours.

The real amount of carbon dioxide emitted during these two hours was $72$ tonnes.

Find the percentage error of the estimate found in part (a).

attempt at using trapezoidal rule formula              (M1)

$\frac{1}{2}\left(\frac{2-0}{5}\right)\left(30+50+2\left(50+60+40+20\right)\right)$            (A1)

(total carbon =) $84$  tonnes            A1

[3 marks]

$\left|\frac{84-72}{72}\right|\times 100\%$              (M1)

Note: Award (M1) for correct substitution of final answer in part (a) into percentage error formula.

$=16.7\% \left(16.6666\dots \%\right)$            A1

[2 marks]

Although there were successful attempts at using the trapezoidal rule formula, there was quite a bit of confusion among candidates as to which values were to be substituted. It seemed that a significant number of candidates were approaching it with some confusion due to a lack of practice of using the trapezoidal rule formula. Calculation of percentage error in part (b) was generally well done by most candidates.

Although there were successful attempts at using the trapezoidal rule formula, there was quite a bit of confusion among candidates as to which values were to be substituted. It seemed that a significant number of candidates were approaching it with some confusion due to a lack of practice of using the trapezoidal rule formula. Calculation of percentage error in part (b) was generally well done by most candidates.

State the number of boys who answered questions in Portuguese.

Find the probability that the boy answered questions in Hindi.

Two girls are selected at random.

Calculate the probability that one girl answered questions in Mandarin and the other answered questions in Hindi.

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

20     (A1) (C1)

[1 mark]

null     (A1)(A1) (C2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

$\frac{7}{37}\times \frac{12}{36}+\frac{12}{37}\times \frac{7}{36}$     (A1)(M1)

Note: Award (A1) for first or second correct product seen, (M1) for adding their two products or for multiplying their product by two.

$=\frac{14}{111}\left(0.12612\dots ,12.6126\text{\% }\right)$     (A1) (C3)

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

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 the value of $P$ if no vases are sold.

Differentiate $P\left(x\right)$.

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

−120 (AUD)       (A1)   (C1)

[1 mark]

$-\frac{3}{5}{x}^2+14x$     (A1)(A1)     (C2)

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

[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]

Find the probability that both spins are yellow.

Find the probability that at least one of the spins is yellow.

Write down the probability that the second spin is yellow, given that the first spin is blue.

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

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

Note: Award (M1) for multiplying correct probabilities.

$\frac{1}{9}$ (0.111, 0.111111…, 11.1%)      (A1)   (C2)

[2 marks]

$\left(\frac{1}{2}\times \frac{1}{3}\right)+\left(\frac{1}{6}\times \frac{1}{3}\right)+\frac{1}{3}$       (M1)(M1)

Note: Award (M1) for $\left(\frac{1}{2}\times \frac{1}{3}\right)$ and $\left(\frac{1}{6}\times \frac{1}{3}\right)$ or equivalent, and (M1) for $\frac{1}{3}$ and adding only the three correct probabilities.

OR

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

Note: Award (M1) for $\frac{2}{3}$ seen and (M1) for subtracting ${\left(\frac{2}{3}\right)}^2$ from 1. This may be shown in a tree diagram with “yellow” and “not yellow” branches.

$\frac{5}{9}$ (0.556, 0.555555…, 55.6%)      (A1)(ft)   (C3)

Note: Follow through marks may be awarded if their answer to part (a) is used in a correct calculation.

[3 marks]

$\frac{1}{3}$  (0.333, 0.333333…, 33.3%)      (A1)   (C1)

[1 mark]

Find the value of $a$.

Find the volume of the solid formed when the shaded region is revolved $360°$ about the $x$-axis.

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

recognize $f\left(x\right)=0$       (M1)

eg   $\sqrt{12-2x}=0, 2x=12$ 

$a=6$ (accept $x=6$, $\left(6, 0\right)$)    A1  N2

[2 marks] 

attempt to substitute either their limits or the function into volume formula (must involve $f{ }^2$)      (M1)

eg   ${\int }_0^{6}f{ }^{2}dx , \mathrm{\pi }\int {\left(\sqrt{12-2x}\right)}^2 , \mathrm{\pi }{\int }_0^{6}12-2x dx$ 

correct integration of each term      A1  A1

eg   $12x-x^2 , 12x-x^2+c , {\left[12x-x^2\right]}_0^6$

substituting limits into their integrated function and subtracting (in any order)      (M1)

eg    $\mathrm{\pi }\left(12\left(6\right)-{\left(6\right)}^2\right)-\mathrm{\pi }\left(0\right) , 72\mathrm{\pi }-36\mathrm{\pi } , \left(12\left(6\right)-{\left(6\right)}^2\right)-\left(0\right)$

Note: Award M0 if candidate has substituted into $f, f{ }^2$ or $f\text{'}$.

volume$=36\mathrm{\pi }$      A1  N2      

[5 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]

Find an expression for $\frac{dy}{dx}$.

Show that the normal to the curve at the point where $x=1$ is $2y-x+3=0$.

* 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.

$\frac{dy}{dx}=2x-4$      A1

[1 mark]

Gradient at $x=1$ is $-2$         M1

Gradient of normal is $\frac{1}{2}$         A1

When $x=1$  $y=1-4+2=-1$          (M1)A1

EITHER

$y+1=\frac{1}{2}\left(x-1\right)$         M1

$2y+2=x-1$  or  $y+1=\frac{1}{2}x-\frac{1}{2}$         A1

OR

$-1=\frac{1}{2}\times 1+c$         M1

$y=\frac{1}{2}x-\frac{3}{2}$         A1

THEN

$2y-x+3=0$         AG

[6 marks]

Write down $f\prime \left(x\right)$.

Write down the gradient of this tangent.

Find the value of $k$.

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

$2x+\frac{k}{x^2}$     (A1)(A1)(A1)    (C3)

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

[3 marks]

$-2.5 \left(\frac{-5}{2}\right)$     (A1)    (C1)

[1 mark]

$-2.5=2\times \left(-2\right)+\frac{k}{{\left(-2\right)}^2}$       (M1)

Note: Award (M1) for equating their gradient from part (b) to their substituted derivative from part (a).

$\left(k=\right) 6$      (A1)(ft)    (C2)

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

[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]

Find the coordinates of P.

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

Hence, find the equation of L in terms of $a$.

The graph of $f$ has a local minimum at the point Q. The line L passes through Q.

Find the value of $a$.

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

valid approach      (M1)

eg   $f\left(0\right)$,  $0^3-2{\left(0\right)}^2+a\left(0\right)+6$,  $f\left(0\right)=6$,  $\left(0,y\right)$

(0, 6)  (accept $x$ = 0 and $y$ = 6)     A1 N2

[2 marks]

$f^{\prime }=3x^2-4x+a$     A2 N2

[2 marks]

valid approach      (M1)

eg   $f^{\prime }\left(0\right)$

correct working      (A1)

eg   $3{\left(0\right)}^2-4\left(0\right)+a$,  slope = $a$,  $f^{\prime }\left(0\right)=a$

attempt to substitute gradient and coordinates into linear equation      (M1)

eg   $y-6=a\left(x-0\right)$,  $y-0=a\left(x-6\right)$,  $6=a\left(0\right)+c$,  L $=ax+6$

correct equation      A1 N3

eg  $y=ax+6$,  $y-6=ax$,  $y-6=a\left(x-0\right)$

[4 marks]

valid approach to find intersection      (M1)

eg   $f\left(x\right)=L$

correct equation      (A1)

eg   $x^3-2x^2+ax+6=ax+6$

correct working      (A1)

eg   $x^3-2x^2=0$,  $x^2(x-2)=0$

$x=2$ at Q      (A1)

valid approach to find minimum      (M1)

eg   $f^{\prime }\left(x\right)=0$

correct equation      (A1)

eg   $3x^2-4x+a=0$

substitution of their value of $x$ at Q into their $f^{\prime }\left(x\right)=0$ equation      (M1)

eg   $3{\left(2\right)}^2-4\left(2\right)+a=0$,  $12-8+a=0$

$a$ = −4     A1 N0

[8 marks]

Find $S\prime \left(x\right)$.

Solve $S\text{'}\left(x\right)=0$.

Interpret your answer to (b)(i) in context.

$\left(S\left(x\right)=\right) x^2+128x^{-1}$             (M1)

Note: Award (M1) for expressing second term with a negative power. This may be implied by $\frac{1}{x^2}$ seen as part of their answer.

$2x-\frac{128}{x^2}$  OR  $2x-128x^{-2}$             A1A1

Note: Award A1 for $2x$ and A1 for $-\frac{128}{x^2}$. The first A1 is for $x^2$ differentiated correctly and is independent of the (M1).

[3 marks]

EITHER

any correct manipulation of $2x-\frac{128}{x^2}=0$  e.g. $2x^3-128=0$              (M1)

OR

sketch of graph of $S\text{'}\left(x\right)$ with root indicated              (M1)

OR

sketch of graph of $S\left(x\right)$ with minimum indicated              (M1)

THEN

$x=4$             A1

Note: Value must be positive. Follow through from their part (a) irrespective of working.

[2 marks]

the value of $x$ that will minimize surface area of the box               A1

Note: Accept ‘optimize’ in place of minimize.

[1 mark]

In part (a), many candidates scored at least the mark for correctly differentiating $x^2$ although differentiating $\frac{128}{x}$ proved to be more problematic, not realizing that the term could be written as $128x^{-1}$. Some who did realize it, made a mistake while differentiating the negative index. In part (b)(i), the manipulation of the equation was frequently incorrect; those that used their GDC got the correct answer with no working. Many candidates could follow the instruction but where errors were made in part (a), valid solutions for part (b) proved tricky with some negative values seen. In part (b)(ii), a significant number of candidates did not appreciate what is meant by a gradient function equal to zero. Of those who had some idea, the words minimize and maximize were seen but not always in terms of the surface area. Many incorrect answers referred to the volume. Many candidates had difficulty communicating an interpretation of their answer in context. This resulted in several negative answers found for part (b)(i) being left as is, when contextually, negative answers would not make sense.

In part (a), many candidates scored at least the mark for correctly differentiating $x^2$ although differentiating $\frac{128}{x}$ proved to be more problematic, not realizing that the term could be written as $128x^{-1}$. Some who did realize it, made a mistake while differentiating the negative index. In part (b)(i), the manipulation of the equation was frequently incorrect; those that used their GDC got the correct answer with no working. Many candidates could follow the instruction but where errors were made in part (a), valid solutions for part (b) proved tricky with some negative values seen. In part (b)(ii), a significant number of candidates did not appreciate what is meant by a gradient function equal to zero. Of those who had some idea, the words minimize and maximize were seen but not always in terms of the surface area. Many incorrect answers referred to the volume. Many candidates had difficulty communicating an interpretation of their answer in context. This resulted in several negative answers found for part (b)(i) being left as is, when contextually, negative answers would not make sense.

In part (a), many candidates scored at least the mark for correctly differentiating $x^2$ although differentiating $\frac{128}{x}$ proved to be more problematic, not realizing that the term could be written as $128x^{-1}$. Some who did realize it, made a mistake while differentiating the negative index. In part (b)(i), the manipulation of the equation was frequently incorrect; those that used their GDC got the correct answer with no working. Many candidates could follow the instruction but where errors were made in part (a), valid solutions for part (b) proved tricky with some negative values seen. In part (b)(ii), a significant number of candidates did not appreciate what is meant by a gradient function equal to zero. Of those who had some idea, the words minimize and maximize were seen but not always in terms of the surface area. Many incorrect answers referred to the volume. Many candidates had difficulty communicating an interpretation of their answer in context. This resulted in several negative answers found for part (b)(i) being left as is, when contextually, negative answers would not make sense.

Determine whether the profit is increasing or decreasing when $t=2$.

One year after the company was founded, the profit was $4$ thousand dollars.

Find an expression for $P\left(t\right)$, when $t\ge 0$.

METHOD 1

(when $t=2$)

$\frac{dP}{dt}=-4$   OR   $\frac{dP}{dt}<0$ (equivalent in words)   OR   $3{\left(2\right)}^2-8\left(2\right)=-4$          M1

therefore $P$ is decreasing             A1

METHOD 2

sketch with $t=2$ indicated in 4th quadrant   OR   $t$-intercepts identified          M1

therefore $P$ is decreasing             A1

[2 marks]

$\left(P\left(t\right)=\right) t^3-4t^2\left(+c\right)$            A1A1

$4=1^3-4{\left(1\right)}^2+c$            (M1)

Note: Award M1 for substituting $(1, 4)$ into their equation with $+c$ seen.

$c=7$

$P\left(t\right)=t^3-4t^2+7$            A1

[4 marks]

Even some weaker candidates were able to score in this part of the question as many candidates understood the process required to determine whether the profit is increasing or decreasing.

Many candidates failed to recognize that they needed to integrate the original function. Of the few that attempted to find the value of the constant the vast majority substituted 4000 rather than 4. So, a correct final expression for $P\left(t\right)$ was rarely seen.

Use the trapezoidal rule, with the values given in the following table, to approximate the area of the shaded region.

Find the exact area of the shaded region in the painting.

Find the area of the unshaded region in the painting.

$\frac{1}{2}\left(0.6+0+2\left(1.2+1.2\right)\right)$           (A1)(M1)

 
Note: Award A1 for evidence of $h=1$, M1 for a correct substitution into trapezoidal rule (allow for an incorrect $h$ only). The zero can be omitted in the working.

$2.7 {\text{m}}^2$        A1

[3 marks]

${\int }_{-1}^2\frac{-x^3-3x^2+4x+12}{10}dx$   OR   ${\int }_{-1}^{2}f\left(x\right)dx$        (M1)

 
Note: Award M1 for using definite integration with correct limits.

$2.925 {\text{m}}^2$        A1

Note: Question requires exact answer, do not award final A1 for $2.93$.

[2 marks]

$9-2.925$        (M1)

Note: Award M1 for $9$ seen as part of a subtraction.

$=6.08 {\text{m}}^2 \left(6.075\right)$        A1

[2 marks]

There seemed a better attempt at using the trapezium rule in this session compared to the two 2021 sessions. Despite many incorrect values for $h$, candidates obtained the method mark for a correctly substituted formula (excluding $h$).

The exact answer of 2.925 was asked for in the question, yet candidates frequently rounded to three significant figures and hence lost the final mark.

Many candidates were able to correctly find the area of the unshaded region.

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

Given that $\underset{x\to +\mathrm{\infty }}{lim}(g\circ f)(x)=-3$, find the value of $b$.

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

attempt to form composite     (M1)

eg$$$g(1+{\text{e}}^{-x})$

correct function     A1     N2

eg$$$(g\circ f)(x)=2+b+2{\text{e}}^{-x},2(1+{\text{e}}^{-x})+b$

[2 marks]

evidence of $\underset{x\to \mathrm{\infty }}{lim}(2+b+2{\text{e}}^{-x})=2+b+\underset{x\to \mathrm{\infty }}{lim}(2{\text{e}}^{-x})$     (M1)

eg$$$2+b+2{\text{e}}^{-\mathrm{\infty }}$, graph with horizontal asymptote when $x\to \mathrm{\infty }$

Note:     Award M0 if candidate clearly has incorrect limit, such as $x\to 0,{\text{e}}^{\mathrm{\infty }},2{\text{e}}^0$.

evidence that ${\text{e}}^{-x}\to 0$ (seen anywhere)     (A1)

eg$$$\underset{x\to \mathrm{\infty }}{lim}({\text{e}}^{-x})=0,1+{\text{e}}^{-x}\to 1,2(1)+b=-3,{\text{e}}^{\text{large negative number}}\to 0$, graph of $y={\text{e}}^{-x}$ or

$y=2{\text{e}}^{-x}$ with asymptote $y=0$, graph of composite function with asymptote $y=-3$

correct working     (A1)

eg$$$2+b=-3$

$b=-5$     A1     N2

[4 marks]

Calculate the area of the shaded region in diagram 1.

Write down an integral for the area of the shaded region in diagram 2.

Calculate the area of this region.

Hence, determine the area enclosed between $y=6-x$ and $y=1.5x^2-2.5x+3$.

EITHER

attempt to substitute $3, 4$ and $7$ into area of a trapezoid formula           (M1)

$\left(A=\right)\frac{1}{2}\left(7+4\right)\left(3\right)$

OR

given line expressed as an integral           (M1)

$\left(A=\right){\int }_{-1}^2\left(6-x\right) dx$

OR

attempt to sum area of rectangle and area of triangle           (M1)

$\left(A=\right) 4\times 3+\frac{1}{2}\left(3\right)\left(3\right)$

THEN

$16.5$ (square units)            A1

[3 marks]

$\left(A=\right) {\int }_{-1}^{2}1.5x^2-2.5x+3 dx$            A1A1

Note: Award A1 for the limits $x=-1$, $x=2$ in correct location. Award A1 for an integral of the quadratic function, $dx$ must be included. Do not accept “$y$” in place of the function, given that two equations are in the question.

[2 marks]

$9.75$ (square units)            A1

[1 mark]

$16.5-9.75$            (M1)

$6.75$ (square units)            A1

[2 marks]

There were a variety of methods used or attempted – area of trapezoid, integration, area of triangle plus area of rectangle, area of large rectangle minus area of top triangle, trapezoidal rule. All these methods, except for trapezoidal rule, proved successful for candidates, with the most common being the use of integration.

This was reasonably well done except for a few notation issues such as not including dx with their integrand. Those who attempted integration manually were not successful.

Recognition that areas had to be subtracted was very evident.

Write down, in the form $y=mx+c$, the equation of $L_2$.

$y=\frac{4}{3}x-\frac{5}{2}(y=1.33\dots x-2.5)$     (A1)(ft)     (C1)

Note:     Follow through from parts (c)(i) and (a). Award (A0) if final answer is not written in the form $y=mx+c$.

[1 mark]

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]

Find $A$ in terms of $p$ and $q$.

Show that $A=\frac{192}{q}+3q+48$.

Find $\frac{dA}{dq}$.

Write down an equation Ellis could solve to find this value of $q$.

Hence, or otherwise, find this value of $q$.

$A=\frac{1}{2}\times 6\times q+\frac{1}{2}\times 8\times p+48$  OR  $A=\frac{1}{2}\left(p+6\right)\left(q+8\right)$  OR  $A=3q+4p+48$            A1

[1 mark]

valid attempt to link $p$ and $q$, using tangents, similar triangles or other method        (M1)

eg.  $\tan \theta =\frac{8}{p}$ and $\tan \theta =\frac{q}{6}$  OR  $\tan \theta =\frac{p}{8}$ and $\tan \theta =\frac{6}{q}$  OR  $\frac{8}{p}=\frac{q}{6}$

correct equation linking $p$ and $q$            A1

eg.  $pq=48$  OR  $p=\frac{48}{q}$  OR  $q=\frac{48}{p}$

substitute $p=\frac{48}{q}$ into a correct area expression        M1

eg.  $\left(A=\right)\frac{1}{2}\times 6\times q+\frac{1}{2}\times 8\times \frac{48}{q}+48$  OR  $\left(A=\right)\frac{1}{2}\left(\frac{48}{q}+6\right)\left(q+8\right)$

$A=3q+\frac{192}{q}+48$         AG

Note: The AG line must be seen with no incorrect, intermediate working, for the final M1 to be awarded.

[3 marks]

$\frac{-192}{q^2}+3$            A1A1

Note: Award A1 for $\frac{-192}{q^2}$, A1 for $3$. Award A1A0 if extra terms are seen.

[2 marks]

$\frac{-192}{q^2}+3=0$            A1

[1 mark]

$q=8 \text{cm}$            A1

[1 mark]

Write down an equation for the area, $A$, of the curved surface in terms of $r$.

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

Find the value of $r$ when the area of the curved surface is maximized.

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

$A=2\pi r\left(12-r\right)$  OR  $A=24\pi r-2\pi r^2$        (A1)(M1)  (C2)

Note: Award (A1) for  $r+h=12$  or  $h=12-r$  seen. Award (M1) for correctly substituting into curved surface area of a cylinder. Accept $A=2\pi r\left(12-r\right)$  OR  $A=24\pi r-2\pi r^2$.

[2 marks]

$24\pi -4\pi r$       (A1)(ft)(A1)(ft)  (C2)

Note: Award (A1)(ft) for $24\pi$ and  (A1)(ft) for $-4\pi r$ . Follow through from part (a). Award at most (A1)(ft)(A0) if additional terms are seen.

[2 marks]

$24\pi -4\pi r=0$       (M1)

Note: Award (M1) for setting their part (b) equal to zero.

6 (cm)       (A1)(ft)  (C2)

Note: Follow through from part (b).

[2 marks]

Find an expression for $P$ in terms of $x$.

The company regularly increases the number of cars it produces.

Describe how their profit changes if they increase production to over $30$ cars per month and up to $50$ cars per month. Justify your answer.

recognition of need to integrate (eg reverse power rule or integral symbol)             (M1)

$P\left(x\right)=-0.8x^2+48x \left(+c\right)$             A1A1

$260=-0.8\times {\left(15\right)}^2+48\times \left(15\right)+c$             (M1)

Note: Award M1 for correct substitution of $x=15$ and $P=260$. A constant of integration must be seen (can be implied by a correct answer).

$c=-280$

$P\left(x\right)=-0.8x^2+48x -280$          A1

[5 marks]

profit will decrease (with each new car produced)             A1

EITHER

because the profit function is decreasing / the gradient is negative / the rate of change of $P$ is negative             R1

OR

${\int }_{30}^{50}-1.6x+48 \left(dx\right)=-320$             R1

OR

evidence of finding $P\left(30\right)=440$ and $P\left(50\right)=120$             R1

Note: Award at most R1A0 if $P\left(30\right)$ or $P\left(50\right)$ or both have incorrect values.

[2 marks]

Write down the value of $f(1)$.

Find the equation of $N$. Give your answer in the form $ax+by+d=0$ where $a$, $b$, $d\in \mathbb{Z}$.

Draw the line $N$ on the diagram above.

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

3     (A1)     (C1)

Notes:     Accept $y=3$

[1 mark]

$3=0.5(1)+c$$$OR$$$y-3=0.5(x-1)$     (A1)(A1)

Note:     Award (A1) for correct gradient, (A1) for correct substitution of $\text{A}(1,3)$ in the equation of line.

$x-2y+5=0$ or any integer multiple     (A1)(ft)     (C3)

Note:     Award (A1)(ft) for their equation correctly rearranged in the indicated form.

The candidate’s answer must be an equation for this mark.

[3 marks]

     (M1)(A1)(ft)     (C2)

Note:     Award M1) for a straight line, with positive gradient, passing through $(1,3)$, (A1)(ft) for line (or extension of their line) passing approximately through 2.5 or their intercept with the $y$-axis.

[2 marks]

Write down an integral for the area of region R.

Find the area of region R.

The three points A(0, 0) , B(3, 10) and C($a$, 0) define the vertices of a triangle.

Find the value of $a$, the $x$-coordinate of C, such that the area of the triangle is equal to the area of region R.

A = ${\int }_0^2\left(6-3x\right)\left(4+x\right)\text{d}x$      A1A1

Note: Award A1 for the limits $x$ = 0, $x$  = 2. Award A1 for an integral of $f(x)$.

[2 marks]

28     A1

[1 mark]

$28=0.5\times a\times 10$    M1

$5.6\left(\frac{28}{5}\right)$      A1

[2 marks]

It was pleasing to see that, for those candidates who made a reasonable attempt at the paper, many were able to identify the correct values on the tree diagram.

Write down the gradient of the curve of $f$ at P.

Find the equation of the normal to the curve of $f$ at P.

Determine the concavity of the graph of $f$ when and justify your answer.

$-2$     A1     N1

[1 mark]

gradient of normal $=\frac{1}{2}$     (A1)

attempt to substitute their normal gradient and coordinates of P (in any order)     (M1)

eg$$$y-4=\frac{1}{2}(x-3),3=\frac{1}{2}(4)+b,b=1$

$y-3=\frac{1}{2}(x-4),y=\frac{1}{2}x+1,x-2y+2=0$     A1     N3

[3 marks]

correct answer and valid reasoning     A2     N2

answer:     eg     graph of $f$ is concave up, concavity is positive (between )

reason:     eg     slope of $f^{\prime }$ is positive, $f^{\prime }$ is increasing, $f^{″}>0$,