Question
The following graph shows the temperature in degrees Celsius of Robert’s cup of coffee, \(t\) minutes after pouring it out. The equation of the cooling graph is \(f (t) = 16 + 74 \times 2.8^{−0.2t}\) where \(f (t)\) is the temperature and \(t\) is the time in minutes after pouring the coffee out.
Robert, who lives in the UK, travels to Belgium. The exchange rate is 1.37 euros to one British Pound (GBP) with a commission of 3 GBP, which is subtracted before the exchange takes place. Robert gives the bank 120 GBP.
Find the initial temperature of the coffee.[1]
Write down the equation of the horizontal asymptote.[1]
Find the room temperature.[1]
Find the temperature of the coffee after 10 minutes.[1]
Find the temperature of Robert’s coffee after being heated in the microwave for 30 seconds after it has reached the temperature in part (d).[3]
Calculate the length of time it would take a similar cup of coffee, initially at 20°C, to be heated in the microwave to reach 100°C.[4]
Calculate correct to 2 decimal places the amount of euros he receives.[3]
He buys 1 kilogram of Belgian chocolates at 1.35 euros per 100 g.
Calculate the cost of his chocolates in GBP correct to 2 decimal places.[3]
Answer/Explanation
Markscheme
Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)
(UP) 90°C (A1)[1 mark]
y = 16 (A1)[1 mark]
Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)
(UP) 16°C (ft) from answer to part (b) (A1)(ft)[1 mark]
Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)
(UP) 25.4°C (A1)[1 mark]
Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)
for seeing 20.75 or equivalent (A1)
for multiplying their (d) by their 20.75 (M1)
(UP) 42.8°C (A1)(ft)(G2)[3 marks]
Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)
for seeing \(20 \times 2^{1.5t} = 100\) (A1)
for seeing a value of t between 1.54 and 1.56 inclusive (M1)(A1)
(UP) 1.55 minutes or 92.9 seconds (A1)(G3)[4 marks]
Financial accuracy penalty (FP) is applicable in part (ii) only.
\(120 – 3 = 117\)
(FP) \(117 \times 1.37\) (A1)
= 160.29 euros (correct answer only) (M1)
first (A1) for 117 seen, (M1) for multiplying by 1.37 (A1)(G2)[3 marks]
Financial accuracy penalty (FP) is applicable in part (ii) only.
(FP) \(\frac{{13.5}}{{1.37}}\) (A1)(M1)
9.85 GBP (answer correct to 2dp only)
first (A1) is for 13.5 seen, (M1) for dividing by 1.37 (A1)(ft)(G3)[3 marks]
Question
Consider the function \(f:x \mapsto \frac{{kx}}{{{2^x}}}\).
The cost per person, in euros, when \(x\) people are invited to a party can be determined by the function
\(C(x) = x + \frac{{100}}{x}\)
Given that \(f(1) = 2\), show that \(k = 4\).[2]
Write down the values of \(q\) and \(r\) for the following table.
[2]
As \(x\) increases from \( – 1\), the graph of \(y = f(x)\) reaches a maximum value and then decreases, behaving asymptotically.
Draw the graph of \(y = f(x)\) for \( – 1 \leqslant x \leqslant 8\). Use a scale of \({\text{1 cm}}\) to represent 1 unit on both axes. The position of the maximum, \({\text{M}}\), the \(y\)-intercept and the asymptotic behaviour should be clearly shown.[4]
Using your graphic display calculator, find the coordinates of \({\text{M}}\), the maximum point on the graph of \(y = f(x)\).[2]
Write down the equation of the horizontal asymptote to the graph of \(y = f(x)\).[2]
(i) Draw and label the line \( y = 1\) on your graph.
(ii) The equation \(f(x) = 1\) has two solutions. One of the solutions is \(x = 4\). Use your graph to find the other solution.[4]
Find \(C'(x)\).[3]
Show that the cost per person is a minimum when \(10\) people are invited to the party.[2]
Calculate the minimum cost per person.[2]
Answer/Explanation
Markscheme
\(f(1) = \frac{k}{{{2^1}}}\) (M1)
Note: (M1) for substituting \(x = 1\) into the formula.
\(\frac{k}{2} = 2\) (M1)
Note: (M1) for equating to 2.
\(k = 4\) (AG)[2 marks]
\(q = 2\), \(r = 0.125\) (A1)(A1)[2 marks]
(A4)
Notes: (A1) for scales and labels.
(A1) for accurate smooth curve passing through \((0, 0)\) drawn at least in the given domain.
(A1) for asymptotic behaviour (curve must not go up or cross the \(x\)-axis).
(A1) for indicating the position of the maximum point.[4 marks]
\({\text{M}}\) (\(1.44\), \(2.12\)) (G1)(G1)
Note: Brackets required, if missing award (G1)(G0). Accept \(x = 1.44\) and \(y = 2.12\).[2 marks]
\(y = 0\) (A1)(A1)
Note: (A1) for ‘\(y = \)’ provided the right hand side is a constant. (A1) for 0.[2 marks]
(i) See graph (A1)(A1)
Note: (A1) for correct line, (A1) for label.
(ii) \(x = 0.3\) (ft) from candidate’s graph. (A2)(ft)
Notes: Accept \( \pm 0.1\) from their x. For \(0.310\) award (G1)(G0). For other answers taken from the GDC and not given correct to 3 significant figures award (G0)(AP)(G0) or (G1)(G0) if (AP) already applied.[4 marks]
\(C'(x) = 1 – \frac{{100}}{{{x^2}}}\) (A1)(A1)(A1)
Note: (A1) for 1, (A1) for \( – 100\) , (A1) for \({x^2}\) as denominator or \({{x^{ – 2}}}\) as numerator. Award a maximum of (A2) if an extra term is seen.[3 marks]
For studying signs of the derivative at either side of \(x = 10\) (M1)
For saying there is a change of sign of the derivative (M1)(AG)
OR
For putting \(x = 10\) into \(C’\) and getting zero (M1)
For clear sketch of the function or for mentioning that the function changes from decreasing to increasing at \(x = 10\) (M1)(AG)
OR
For solving \(C'(x) = 0\) and getting \(10\) (M1)
For clear sketch of the function or for mentioning that the function changes from decreasing to increasing at \(x = 10\) (M1)(AG)
Note: For a sketch with a clear indication of the minimum or for a table with values of \(x\) at either side of \(x = 10\) award (M1)(M0).[2 marks]
\(C(10) = 10 + \frac{{100}}{{10}}\) (M1)
\(C(10) = 20\) (A1)(G2)[2 marks]
Question
The temperature in \(^ \circ {\text{C}}\) of a pot of water removed from the cooker is given by \(T(m) = 20 + 70 \times {2.72^{ – 0.4m}}\), where \(m\) is the number of minutes after the pot is removed from the cooker.
Show that the temperature of the water when it is removed from the cooker is \({90^ \circ }{\text{C}}\).[2]
The following table shows values for \(m\) and \(T(m)\).
(i) Write down the value of \(s\).
(ii) Draw the graph of \(T(m)\) for \(0 \leqslant m \leqslant 10\) . Use a scale of \(1{\text{ cm}}\) to represent \(1\) minute on the horizontal axis and a scale of \(1{\text{ cm}}\) to represent \({10^ \circ }{\text{C}}\) on the vertical axis.
(iii) Use your graph to find how long it takes for the temperature to reach \({56^ \circ }{\text{C}}\). Show your method clearly.
(iv) Write down the temperature approached by the water after a long time. Justify your answer.[9]
Consider the function \(S(m) = 20m – 40\) for \(2 \leqslant m \leqslant 6\) .
The function \(S(m)\) represents the temperature of soup in a pot placed on the cooker two minutes after the water has been removed. The soup is then heated.
Draw the graph of \(S(m)\) on the same set of axes used for part (b).[2]
Consider the function \(S(m) = 20m – 40\) for \(2 \leqslant m \leqslant 6\) .
The function \(S(m)\) represents the temperature of soup in a pot placed on the cooker two minutes after the water has been removed. The soup is then heated.
Comment on the meaning of the constant \(20\) in the formula for \(S(m)\) in relation to the temperature of the soup.[1]
Consider the function \(S(m) = 20m – 40\) for \(2 \leqslant m \leqslant 6\) .
The function \(S(m)\) represents the temperature of soup in a pot placed on the cooker two minutes after the water has been removed. The soup is then heated.
(i) Use your graph to solve the equation \(S(m) = T(m)\) . Show your method clearly.
(ii) Hence describe by using inequalities the set of values of \(m\) for which \(S(m) > T(m)\).[4]
Answer/Explanation
Markscheme
\(T(0) = 20 + 70 \times {2.72^{ – 0.4 \times 0}} = 90\) (M1)(A1)(AG)
Note: (M1) for taking \(m = 0\) , (A1) for substituting \(0\) into the formula. For the A mark to be awarded \(90\) must be justified by correct method.[2 marks]
(i) 21.3 (A1)
(ii)
(A4)(ft)
Note: Scales and labels (A1). Smooth curve (A1). All points correct including the \(y\)-intercept (A2), 1 point incorrect (A1), otherwise (A0). Follow through from their value of \(s\).
(iii) \(m = 1.7{\text{ minutes}}\) (Accept \( \pm 0.2\) ) (A2)(ft)
Note: Follow through from candidate’s graph. Accept answers in minutes and seconds if consistent with graph. If answer incorrect and correct line(s) seen on graph award (M1)(A0).
(iv) \({20^ \circ }{\text{C}}\) (A1)(ft)
The curve behaves asymptotically to the line \(y = 20\) or similar. (A1)
OR
The room temperature is 20 or similar
OR
When \({\text{m}}\) is a very large number the term \(70 \times {2.72^{ – 0.4{\text{m}}}}\) tends to zero or similar.
Note: Follow through from their graph if appropriate.[9 marks]
(A1)(A1)
Notes: (A1) for correct line, (A1) for domain. If line not drawn on same set of axes award at most (A1)(A0).[2 marks]
It indicates by how much the temperature increases per minute. (A1)[1 mark]
(i) \(m = 3.8\) (Accept \( \pm 0.1\) ) (A2)(ft)
Note: Follow through from candidate’s graph. Accept answers in minutes and seconds if consistent with graph. If answer incorrect and correct line(s) seen on graph award (M1)(A0).
(ii) \(3.8 < m \leqslant 6\) (A1)(A1)(ft)
Note: (A1) for \(m > 3.8\) and (A1) for \(m \leqslant 6\). Follow through from candidate’s answer to part (e)(i). If candidate was already penalized in (c) for domain and does not state \(m \leqslant 6\) then award (A2)(ft).
Question
A manufacturer claims that fertilizer has an effect on the height of rice plants. He measures the height of fertilized and unfertilized plants. The results are given in the following table.
A chi-squared test is performed to decide if the manufacturer’s claim is justified at the 1 % level of significance.
The population of fleas on a dog after t days, is modelled by
\[N = 4 \times {(2)^{\frac{t}{4}}},{\text{ }}t \geqslant 0\]
Some values of N are shown in the table below.
Write down the null and alternative hypotheses for this test.[2]
For the number of fertilized plants with height greater than 75 cm, show that the expected value is 97.5.[3]
Write down the value of \(\chi_{calc}^2\).[2]
Write down the number of degrees of freedom.[1]
Is the manufacturer’s claim justified? Give a reason for your answer.[2]
Write down the value of p.[1]
Write down the value of q.[2]
Using the values in the table above, draw the graph of N for 0 ≤ t ≤ 20. Use 1 cm to represent 2 days on the horizontal axis and 1 cm to represent 10 fleas on the vertical axis.[6]
Use your graph to estimate the number of days for the population of fleas to reach 55.[2]
Answer/Explanation
Markscheme
H0: The height of the rice plants is independent of the use of a fertilizer. (A1)
Notes: For independent accept “not associated”, can accept “the use of a fertilizer has no effect on the height of the plants”.
Do not accept “not correlated”.
H1: The height of the rice plants is not independent (dependent) of the use of fertilizer. (A1)(ft)
Note: If H0 and H1 are reversed award (A0)(A1)(ft).[2 marks]
\(\frac{{180 \times 195}}{{360}}\) or \(\frac{{180}}{{360}} \times \frac{{195}}{{360}} \times 360\) (A1)(A1)(M1)
= 97.5 (AG)
Notes: Award (A1) for numerator, (A1) for denominator (M1) for division.
If final 97.5 is not seen award at most (A1)(A0)(M1).[3 marks]
\( \chi_{calc}^2 = 14.01 (14.0, 14)\) (G2)
OR
If worked out by hand award (M1) for correct substituted formula with correct values, (A1) for correct answer. (M1)(A1)[2 marks]
2 (A1)[1 mark]
\( \chi_{calc}^2 > \chi_{crit}^2\) (R1)
The manufacturer’s claim is justified. (or equivalent statement) (A1)
Note: Do not accept (R0)(A1).[2 marks]
\(p = 4\) (G1)[1 mark]
\(q = 4(2)^{\frac{16}{4}}\) (M1)
\(= 64\) (A1)(G2)[2 marks]
(A1)(A1)(A1) (A3)
Notes: Award (A1) for x axis with correct scale and label, (A1) for y axis with correct scale and label.
Accept x and y for labels.
If x and y axis reversed award at most (A0)(A1)(ft).
(A1) for smooth curve.
Award (A3) for all 6 points correct, (A2) for 4 or 5 points correct, (A1) for 2 or 3 points correct, (A0) otherwise.[6 marks]
15 (±0.8) (M1)(A1)(ft)(G2)
Note: Award (M1) for line drawn shown on graph, (A1)(ft) from candidate’s graph.[2 marks]
Question
Sketch the graph of y = 2x for \( – 2 \leqslant x \leqslant 3\). Indicate clearly where the curve intersects the y-axis.[3]
Write down the equation of the asymptote of the graph of y = 2x.[2]
On the same axes sketch the graph of y = 3 + 2x − x2. Indicate clearly where this curve intersects the x and y axes.[3]
Using your graphic display calculator, solve the equation 3 + 2x − x2 = 2x.[2]
Write down the maximum value of the function f (x) = 3 + 2x − x2.[1]
Use Differential Calculus to verify that your answer to (e) is correct.[5]
The curve y = px2 + qx − 4 passes through the point (2, –10).
Use the above information to write down an equation in p and q.[2]
The gradient of the curve \(y = p{x^2} + qx – 4\) at the point (2, –10) is 1.
Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[2]
The gradient of the curve \(y = p{x^2} + qx – 4\) at the point (2, –10) is 1.
Hence, find a second equation in p and q.[1]
The gradient of the curve \(y = p{x^2} + qx – 4\) at the point (2, –10) is 1.
Solve the equations to find the value of p and of q.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1)
Note: Award (A1) for correct domain, (A1) for smooth curve, (A1) for y-intercept clearly indicated.[3 marks]
y = 0 (A1)(A1)
Note: Award (A1) for y = constant, (A1) for 0.[2 marks]
Note: Award (A1) for smooth parabola,
(A1) for vertex (maximum) in correct quadrant.
(A1) for all clearly indicated intercepts x = −1, x = 3 and y = 3.
The final mark is to be applied very strictly. (A1)(A1)(A1) [3 marks]
x = −0.857 x = 1.77 (G1)(G1)
Note: Award a maximum of (G1) if x and y coordinates are both given.[2 marks]
4 (G1)
Note: Award (G0) for (1, 4).[1 mark]
\(f'(x) = 2 – 2x\) (A1)(A1)
Note: Award (A1) for each correct term.
Award at most (A1)(A0) if any extra terms seen.
\(2 – 2x = 0\) (M1)
Note: Award (M1) for equating their gradient function to zero.
\(x = 1\) (A1)(ft)
\(f (1) = 3 + 2(1) – (1)^2 = 4\) (A1)
Note: The final (A1) is for substitution of x = 1 into \(f (x)\) and subsequent correct answer. Working must be seen for final (A1) to be awarded.[5 marks]
22 × p + 2q − 4 = −10 (M1)
Note: Award (M1) for correct substitution in the equation.
4p + 2q = −6 or 2p + q = −3 (A1)
Note: Accept equivalent simplified forms.[2 marks]
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2px + q\) (A1)(A1)
Note: Award (A1) for each correct term.
Award at most (A1)(A0) if any extra terms seen.[2 marks]
4p + q = 1 (A1)(ft)[1 mark]
4p + 2q = −6
4p + q = 1 (M1)
Note: Award (M1) for sensible attempt to solve the equations.
p = 2, q = −7 (A1)(A1)(ft)(G3)[3 marks]
Question
George leaves a cup of hot coffee to cool and measures its temperature every minute. His results are shown in the table below.
Write down the decrease in the temperature of the coffee
(i) during the first minute (between t = 0 and t =1) ;
(ii) during the second minute;
(iii) during the third minute.[3]
Assuming the pattern in the answers to part (a) continues, show that \(k = 19\).[2]
Use the seven results in the table to draw a graph that shows how the temperature of the coffee changes during the first six minutes.
Use a scale of 2 cm to represent 1 minute on the horizontal axis and 1 cm to represent 10 °C on the vertical axis.[4]
The function that models the change in temperature of the coffee is y = p (2−t )+ q.
(i) Use the values t = 0 and y = 94 to form an equation in p and q.
(ii) Use the values t =1 and y = 54 to form a second equation in p and q.[2]
Solve the equations found in part (d) to find the value of p and the value of q.[2]
The graph of this function has a horizontal asymptote.
Write down the equation of this asymptote.[2]
George decides to model the change in temperature of the coffee with a linear function using correlation and linear regression.
Use the seven results in the table to write down
(i) the correlation coefficient;
(ii) the equation of the regression line y on t.[4]
Use the equation of the regression line to estimate the temperature of the coffee at t = 3.[2]
Find the percentage error in this estimate of the temperature of the coffee at t = 3.[2]
Answer/Explanation
Markscheme
(i) 40
(ii) 20
(iii) 10 (A3)
Notes: Award (A0)(A1)(ft)(A1)(ft) for −40, −20, −10.
Award (A1)(A0)(A1)(ft) for 40, 60, 70 seen.
Award (A0)(A0)(A1)(ft) for −40, −60, −70 seen.
\(24 – k = 5\) or equivalent (A1)(M1)
Note: Award (A1) for 5 seen, (M1) for difference from 24 indicated.
\(k = 19\) (AG)
Note: If 19 is not seen award at most (A1)(M0).
(A1)(A1)(A1)(A1)
Note: Award (A1) for scales and labelled axes (t or “time” and y or “temperature”).
Accept the use of x on the horizontal axis only if “time” is also seen as the label.
Award (A2) for all seven points accurately plotted, award (A1) for 5 or 6 points accurately plotted, award (A0) for 4 points or fewer accurately plotted.
Award (A1) for smooth curve that passes through all points on domain [0, 6].
If graph paper is not used or one or more scales is missing, award a maximum of (A0)(A0)(A0)(A1).
(i) \(94 = p + q\) (A1)
(ii) \(54 = 0.5p + q\) (A1)
Note: The equations need not be simplified; accept, for example \(94 = p(2^{-0}) + q\).
p = 80, q = 14 (G1)(G1)(ft)
Note: If the equations have been incorrectly simplified, follow through even if no working is shown.
y = 14 (A1)(A1)(ft)
Note: Award (A1) for y = a constant, (A1) for their 14. Follow through from part (e) only if their q lies between 0 and 15.25 inclusive.
(i) –0.878 (–0.87787…) (G2)
Note: Award (G1) if –0.877 seen only. If negative sign omitted award a maximum of (A1)(A0).
(ii) y = –11.7t + 71.6 (y = –11.6517…t + 71.6336…) (G1)(G1)
Note: Award (G1) for –11.7t, (G1) for 71.6.
If y = is omitted award at most (G0)(G1).
If the use of x in part (c) has not been penalized (the axis has been labelled “time”) then award at most (G0)(G1).
−11.6517…(3) + 71.6339… (M1)
Note: Award (M1) for correct substitution in their part (g)(ii).
= 36.7 (36.6785…) (A1)(ft)(G2)
Note: Follow through from part (g). Accept 36.5 for use of the 3sf answers from part (g).
\(\frac{{36.6785… – 24}}{{24}} \times 100\) (M1)
Note: Award (M1) for their correct substitution in percentage error formula.
= 52.8% (52.82738…) (A1)(ft)(G2)
Note: Follow through from part (h). Accept 52.1% for use of 36.5.
Accept 52.9 % for use of 36.7. If partial working (\(\times 100\) omitted) is followed by their correct answer award (M1)(A1). If partial working is followed by an incorrect answer award (M0)(A0). The percentage sign is not required.
Question
A cup of boiling water is placed in a room to cool. The temperature of the room is 20°C . This situation can be modelled by the exponential function \(T = a + b({k^{ – m}})\), where \(T\) is the temperature of the water, in °C , and \(m\) is the number of minutes for which the cup has been placed in the room. A sketch of the situation is given as follows.
Explain why \(a = 20\).[2]
Initially, at \(m = 0\), the temperature of the water is 100°C.
Find the value of \(b\).[2]
After being placed in the room for one minute, the temperature of the water is 84°C.
Show that \(k = 1.25\).[2]
After being placed in the room for one minute, the temperature of the water is 84°C.
Find the temperature of the water three minutes after it has been placed in the room.[2]
After being placed in the room for one minute, the temperature of the water is 84°C.
Find the total time needed for the water to reach a temperature of 35°C. Give your answer in minutes and seconds, correct to the nearest second.[2]
Answer/Explanation
Markscheme
the temperature of the water cannot fall below room temperature (R1)
an (informal) explanation that as \(m \to \infty ,{\text{ }}{k^{ – m}} \to 0\) R1)
OR
recognition that there is a horizontal asymptote at \(y = a\) (R1)
Note: Award (R1) for a contextual reason involving room temperature.
Award (R1) for a mathematical reason similar to one of the two alternatives.
\(100 = 20 + b({k^0})\) (M1)
Note: Award (M1) for substituting \(100\), \(20\) and \(0\).
\(b = 80\) (A1)(G2)
Note: The (A1) is awarded only if all working seen is consistent with the final answer of \(80\).
\(84 = 20 + 80{k^{ – 1}}\) (M1)
Note: Substituting \(k = 1.25\) at any stage is an invalid method and is awarded (M0)(M0). Award (M1) for correctly substituting \(84\), \(20\) and their \(80\).
\(\frac{{64}}{{80}} = {k^{ – 1}}\) (M1)
\(k = 1.25\) (AG)
Note: Award (M1) for correct rearrangement that isolates \(k\); \(k = 1.25\) must be consistent with their working and the conclusion \(k = 1.25\) must be seen.
\(T = 20 + 80({1.25^{ – 3}})\) (M1)
Note: Award (M1) for their correct substitutions into \(T\). Follow through from part (b) and \(k = 1.25\).
\(T = 61.0\;\;\;(60.96)\) (A1)(ft)(G2)
\(35 = 20 + 80({1.25^{ – m}})\) (M1)
Note: Award (M1) for their correct substitutions into \(T\). Follow through from part (b). Accept graphical solutions. Award (M1) for sketch of function.
\((m = ){\text{ }}7.50{\text{ (minutes)}}\;\;\;{\text{(7.50179}} \ldots {\text{)}}\) (A1)(ft)(G2)
7 minutes and 30 seconds (A1)
Note: Award the final (A1) for correct conversion of their \(m\) in minutes to minutes and seconds, but only if answer in minutes is explicitly shown.
Question
A function, \(f\) , is given by
\[f(x) = 4 \times {2^{ – x}} + 1.5x – 5.\]
Calculate \(f(0)\)[2]
Use your graphic display calculator to solve \(f(x) = 0.\)[2]
Sketch the graph of \(y = f(x)\) for \( – 2 \leqslant x \leqslant 6\) and \( – 4 \leqslant y \leqslant 10\) , showing the \(x\) and \(y\) intercepts. Use a scale of \(2\,{\text{cm}}\) to represent \(2\) units on both the horizontal axis, \(x\) , and the vertical axis, \(y\) .[4]
The function \(f\) is the derivative of a function \(g\) . It is known that \(g(1) = 3.\)
i) Calculate \(g'(1).\)
ii) Find the equation of the tangent to the graph of \(y = g(x)\) at \(x = 1.\) Give your answer in the form \(y = mx + c.\)[4]
Answer/Explanation
Markscheme
\(4 \times {2^{ – 0}} + 1.5 \times 0 – 5\) (M1)
Note: Award (M1) for substitution of \(0\) into the expression for \(f(x)\) .
\( = – 1\) (A1)(G2)
\( – 0.538\,\,\,( – 0.537670…)\) and \(3\) (A1)(A1)
Note: Award at most (A0)(A1)(ft) if answer is given as pairs of coordinates.
(A1)(A1)(A1)(ft)(A1)(ft)
Note: Award (A1) for labels and some indication of scale in the correct given window.
Award (A1) for smooth curve with correct general shape with \(f( – 2) > f(6)\) and minimum to the right of the \(y\)-axis.
Award (A1)(ft) for correct \(y\)-intercept (consistent with their part (a)).
Award (A1)(ft) for approximately correct \(x\)-intercepts (consistent with their part (b), one zero between \( – 1\) and \(0\), the other between \(2.5\) and \(3.5\)).
i) \(g'(1) = f(1) = 4 \times {2^{ – 1}} + 1.5 – 5\) (M1)
Note: Award (M1) for substitution of \(1\) into \(f(x)\).
\( = – 1.5\) (A1)(G2)
ii) \(3 = – 1.5 \times 1 + c\) OR \((y – 3) = – 1.5\,(x – 1)\) (M1)
Note: Award (M1) for correct substitution of gradient and the point \((1,\,\,3)\) into the equation of a line. Follow through from (d)(i).
\(y = – 1.5x + 4.5\) (A1)(ft)(G2)
Question
The line \({L_1}\) has equation \(2y – x – 7 = 0\) and is shown on the diagram.
The point A has coordinates \((1,{\text{ }}4)\).
The point C has coordinates \((5,{\text{ }}12)\). M is the midpoint of AC.
The straight line, \({L_2}\), is perpendicular to AC and passes through M.
The point D is the intersection of \({L_1}\) and \({L_2}\).
The length of MD is \(\frac{{\sqrt {45} }}{2}\).
The point B is such that ABCD is a rhombus.
Show that A lies on \({L_1}\).[2]
Find the coordinates of M.[2]
Find the length of AC.[2]
Show that the equation of \({L_2}\) is \(2y + x – 19 = 0\).[5]
Find the coordinates of D.[2]
Write down the length of MD correct to five significant figures.[1]
Find the area of ABCD.[3]
Markscheme
\(2 \times 4 – 1 – 7 = 0\) (or equivalent) (R1)
Note: For (R1) accept substitution of \(x = 1\) or \(y = 4\) into the equation followed by a confirmation that \(y = 4\) or \(x = 1\).
(since the point satisfies the equation of the line,) A lies on \({L_1}\) (A1)
Note: Do not award (A1)(R0).[2 marks]
\(\frac{{1 + 5}}{2}\) OR \(\frac{{4 + 12}}{2}\) seen (M1)
Note: Award (M1) for at least one correct substitution into the midpoint formula.
\((3,{\text{ }}8)\) (A1)(G2)
Notes: Accept \(x = 3,{\text{ }}y = 8\).
Award (M1)(A0) for \(\left( {\frac{{1 + 5}}{2},{\text{ }}\frac{{4 + 12}}{2}} \right)\).
Award (G1) for each correct coordinate seen without working.[2 marks]
\(\sqrt {{{(5 – 1)}^2} + {{(12 – 4)}^2}} \) (M1)
Note: Award (M1) for a correct substitution into distance between two points formula.
\( = 8.94{\text{ }}\left( {4\sqrt 5 ,{\text{ }}\sqrt {80} ,{\text{ }}8.94427 \ldots } \right)\) (A1)(G2)[2 marks]
gradient of \({\text{AC}} = \frac{{12 – 4}}{{5 – 1}}\) (M1)
Note: Award (M1) for correct substitution into gradient formula.
\( = 2\) (A1)
Note: Award (M1)(A1) for gradient of \({\text{AC}} = 2\) with or without working
gradient of the normal \( = – \frac{1}{2}\) (M1)
Note: Award (M1) for the negative reciprocal of their gradient of AC.
\(y – 8 = – \frac{1}{2}(x – 3)\) OR \(8 = – \frac{1}{2}(3) + c\) (M1)
Note: Award (M1) for substitution of their point and gradient into straight line formula. This (M1) can only be awarded where \( – \frac{1}{2}\) (gradient) is correctly determined as the gradient of the normal to AC.
\(2y – 16 = – (x – 3)\) OR \( – 2y + 16 = x – 3\) OR \(2y = – x + 19\) (A1)
Note: Award (A1) for correctly removing fractions, but only if their equation is equivalent to the given equation.
\(2y + x – 19 = 0\) (AG)
Note: The conclusion \(2y + x – 19 = 0\) must be seen for the (A1) to be awarded.
Where the candidate has shown the gradient of the normal to \({\text{AC}} = – 0.5\), award (M1) for \(2(8) + 3 – 19 = 0\) and (A1) for (therefore) \(2y + x – 19 = 0\).
Simply substituting \((3,{\text{ }}8)\) into the equation of \({L_2}\) with no other prior working, earns no marks.[5 marks]
\((6,{\text{ }}6.5)\) (A1)(A1)(G2)
Note: Award (A1) for 6, (A1) for 6.5. Award a maximum of (A1)(A0) if answers are not given as a coordinate pair. Accept \(x = 6,{\text{ }}y = 6.5\).
Award (M1)(A0) for an attempt to solve the two simultaneous equations \(2y – x – 7 = 0\) and \(2y + x – 19 = 0\) algebraically, leading to at least one incorrect or missing coordinate.[2 marks]
3.3541 (A1)
Note: Answer must be to 5 significant figures.[1 mark]
\(2 \times \frac{1}{2} \times \sqrt {80} \times \frac{{\sqrt {45} }}{2}\) (M1)(M1)
Notes: Award (M1) for correct substitution into area of triangle formula.
If their triangle is a quarter of the rhombus then award (M1) for multiplying their triangle by 4.
If their triangle is a half of the rhombus then award (M1) for multiplying their triangle by 2.
OR
\(\frac{1}{2} \times \sqrt {80} \times \sqrt {45} \) (M1)(M1)
Notes: Award (M1) for doubling MD to get the diagonal BD, (M1) for correct substitution into the area of a rhombus formula.
Award (M1)(M1) for \(\sqrt {80} \times \) their (f).
\( = 30\) (A1)(ft)(G3)
Notes: Follow through from parts (c) and (f).
\(8.94 \times 3.3541 = 29.9856 \ldots \)[3 marks]
Question
Consider the function \(f(x) = 0.3{x^3} + \frac{{10}}{x} + {2^{ – x}}\).
Consider a second function, \(g(x) = 2x – 3\).
Calculate \(f(1)\).[2]
Sketch the graph of \(y = f(x)\) for \( – 7 \leqslant x \leqslant 4\) and \( – 30 \leqslant y \leqslant 30\).[4]
Write down the equation of the vertical asymptote.[2]
Write down the coordinates of the \(x\)-intercept.[2]
Write down the possible values of \(x\) for which \(x < 0\) and \(f’(x) > 0\).[2]
Find the solution of \(f(x) = g(x)\).[2]
Answer/Explanation
Markscheme
\(0.3{(1)^3} + \frac{{10}}{1} + {2^{ – 1}}\) (M1)
Note: Award (M1) for correct substitution into function.
\( = 10.8\) (A1)(G2)[2 marks]
(A1)(A1)(A1)(A1)
Note: Award (A1) for indication of correct window and labelled axes.
Award (A1) for correct shape and position for \(x < 0\) (with the local maximum, local minimum and \(x\)-intercept in relative approximate location in \({{\text{3}}^{{\text{rd}}}}\) quadrant).
Award (A1) for correct shape and position for \(x > 0\) (with the local minimum in relative approximate location in \({{\text{1}}^{{\text{st}}}}\) quadrant).
Award (A1) for smooth curve with indication of asymptote (graph should not touch \(y\)-axis and should not curve away from the \(y\)-axis). The asymptote is only assessed in this mark.[4 marks]
\(x = 0\) (A2)
Note: Award (A1) for “\(x = {\text{(a constant)}}\)” and (A1) for “\({\text{(a constant)}} = 0\)”.
The answer must be an equation.[2 marks]
\(( – 6.18,{\text{ }}0){\text{ }}( – 6.17516 \ldots ,{\text{ }}0)\) (A1)(A1)
Note: Award (A1) for each correct coordinate. Award (A0)(A1) if parentheses are missing.[2 marks]
\( – 4.99 < x < – 2.47{\text{ }}( – 4.98688 \ldots < x < – 2.46635 \ldots )\) (A1)(A1)
Note: Award (A1) for both correct end points, (A1) for strict inequalities used with 2 endpoints.[2 marks]
\(0.3{x^3} + \frac{{10}}{x} + {2^{ – x}} = 2x – 3\) (M1)
Note: Award (M1) for equating the expressions for \(f\) and \(g\) or for the line \(y = 2x – 3\) sketched (positive gradient, negative \(y\)-intercept) on their graph from part (a).
\((x = ){\text{ }} – 1.34{\text{ }}( – 1.33650 \ldots )\) (A1)(G2)
Note: Award a maximum of (M1)(A0) or (G1) for coordinate pair seen as final answer.[2 marks]
Question
A pan, in which to cook a pizza, is in the shape of a cylinder. The pan has a diameter of 35 cm and a height of 0.5 cm.
A chef had enough pizza dough to exactly fill the pan. The dough was in the shape of a sphere.
The pizza was cooked in a hot oven. Once taken out of the oven, the pizza was placed in a dining room.
The temperature, \(P\), of the pizza, in degrees Celsius, °C, can be modelled by
\[P(t) = a{(2.06)^{ – t}} + 19,{\text{ }}t \geqslant 0\]
where \(a\) is a constant and \(t\) is the time, in minutes, since the pizza was taken out of the oven.
When the pizza was taken out of the oven its temperature was 230 °C.
The pizza can be eaten once its temperature drops to 45 °C.
Calculate the volume of this pan.[3]
Find the radius of the sphere in cm, correct to one decimal place.[4]
Find the value of \(a\).[2]
Find the temperature that the pizza will be 5 minutes after it is taken out of the oven.[2]
Calculate, to the nearest second, the time since the pizza was taken out of the oven until it can be eaten.[3]
In the context of this model, state what the value of 19 represents.[1]
Answer/Explanation
Markscheme
\((V = ){\text{ }}\pi \times {{\text{(17.5)}}^2} \times 0.5\) (A1)(M1)
Notes: Award (A1) for 17.5 (or equivalent) seen.
Award (M1) for correct substitutions into volume of a cylinder formula.
\( = 481{\text{ c}}{{\text{m}}^3}{\text{ }}(481.056 \ldots {\text{ c}}{{\text{m}}^3},{\text{ }}153.125\pi {\text{ c}}{{\text{m}}^3})\) (A1)(G2)[3 marks]
\(\frac{4}{3} \times \pi \times {r^3} = 481.056 \ldots \) (M1)
Note: Award (M1) for equating their answer to part (a) to the volume of sphere.
\({r^3} = \frac{{3 \times 481.056 \ldots }}{{4\pi }}{\text{ }}( = 114.843 \ldots )\) (M1)
Note: Award (M1) for correctly rearranging so \({r^3}\) is the subject.
\(r = 4.86074 \ldots {\text{ (cm)}}\) (A1)(ft)(G2)
Note: Award (A1) for correct unrounded answer seen. Follow through from part (a).
\( = 4.9{\text{ (cm)}}\) (A1)(ft)(G3)
Note: The final (A1)(ft) is awarded for rounding their unrounded answer to one decimal place.[4 marks]
\(230 = a{(2.06)^0} + 19\) (M1)
Note: Award (M1) for correct substitution.
\(a = 211\) (A1)(G2)[2 marks]
\((P = ){\text{ }}211 \times {(2.06)^{ – 5}} + 19\) (M1)
Note: Award (M1) for correct substitution into the function, \(P(t)\). Follow through from part (c). The negative sign in the exponent is required for correct substitution.
\( = 24.7\) (°C) \((24.6878 \ldots \) (°C)) (A1)(ft)(G2)[2 marks]
\(45 = 211 \times {(2.06)^{ – t}} + 19\) (M1)
Note: Award (M1) for equating 45 to the exponential equation and for correct substitution (follow through for their \(a\) in part (c)).
\((t = ){\text{ }}2.89711 \ldots \) (A1)(ft)(G1)
\(174{\text{ (seconds) }}\left( {173.826 \ldots {\text{ (seconds)}}} \right)\) (A1)(ft)(G2)
Note: Award final (A1)(ft) for converting their \({\text{2.89711}} \ldots \) minutes into seconds.[3 marks]
the temperature of the (dining) room (A1)
OR
the lowest final temperature to which the pizza will cool (A1)[1 mark]
Question
A deep sea diver notices that the intensity of light, \(I\) , below the surface of the ocean decreases with depth, \(d\) , according to the formula
\[I = k{(1.05)^{ – d}}{\text{,}}\]where \(I\) is expressed as a percentage, \(d\) is the depth in metres below the surface and \(k\) is a constant.
The intensity of light at the surface is \(100\% \).
Calculate the value of \(k\) .[2]
Find the intensity of light at a depth \(25{\text{ m}}\) below the surface.[2]
To be able to see clearly, a diver needs the intensity of light to be at least \(65\% \).
Using your graphic display calculator, find the greatest depth below the surface at which she can see clearly.[2]
The table below gives the intensity of light (correct to the nearest integer) at different depths.
Using this information draw the graph of \(I\) against \(d\) for \(0 \leqslant d \leqslant 100\) . Use a scale of \(1{\text{ cm}}\) to represent 10 metres on the horizontal axis and 1 cm to represent \(10\% \) on the vertical axis.[4]
Some sea creatures have adapted so they can see in low intensity light and cannot tolerate too much light.
Indicate clearly on your graph the range of depths sea creatures could inhabit if they can tolerate between \(5\% \) and \(35\% \) of the light intensity at the surface.[2]
Answer/Explanation
Markscheme
\(d = 0\), \(k = 100\) (M1)(A1)(G2)
Note: Award (M1) for \(d = 0\) seen.
\(I = 100 \times {(1.05)^{ – 25}} = 29.5(\% )\) (\(29.5302 \ldots \)) (M1)(A1)(ft)(G2)
\(65 = 100 \times {(1.05)^{ – d}}\) (M1)
Note: Award (M1) for sketch with line drawn at \(y = 65\) .
\(d = 8.83{\text{ (m)}}\) (\(8.82929 \ldots \)) (A1)(ft)(G2)
(A1) for labels and scales
(A2) for all points correct, (A1) for 3 or 4 points correct
(A1) for smooth curve asymptotic to the \(x\)-axis (A4)
Lines in approx correct positions on graph (M1)
The range of values indicated (arrows or shading) \(22\)–\(60{\text{ m}}\) (A1)