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Question
A closed rectangular box has a height \(y{\text{ cm}}\) and width \(x{\text{ cm}}\). Its length is twice its width. It has a fixed outer surface area of \(300{\text{ c}}{{\text{m}}^2}\) .
Factorise \(3{x^2} + 13x – 10\).[2]
Solve the equation \(3{x^2} + 13x – 10 = 0\).[2]
Consider a function \(f(x) = 3{x^2} + 13x – 10\) .
Find the equation of the axis of symmetry on the graph of this function.[2]
Consider a function \(f(x) = 3{x^2} + 13x – 10\) .
Calculate the minimum value of this function.[2]
Show that \(4{x^2} + 6xy = 300\).[2]
Find an expression for \(y\) in terms of \(x\).[2]
Hence show that the volume \(V\) of the box is given by \(V = 100x – \frac{4}{3}{x^3}\).[2]
Find \(\frac{{{\text{d}}V}}{{{\text{d}}x}}\).[2]
(i) Hence find the value of \(x\) and of \(y\) required to make the volume of the box a maximum.
(ii) Calculate the maximum volume.[5]
Answer/Explanation
Markscheme
\((3x – 2)(x + 5)\) (A1)(A1)[2 marks]
\((3x – 2)(x + 5) = 0\)
\(x = \frac{2}{3}\) or \(x = – 5\) (A1)(ft)(A1)(ft)(G2)[2 marks]
\(x = \frac{{ – 13}}{6}{\text{ }}( – 2.17)\) (A1)(A1)(ft)(G2)
Note: (A1) is for \(x = \), (A1) for value. (ft) if value is half way between roots in (b).[2 marks]
Minimum \(y = 3{\left( {\frac{{ – 13}}{6}} \right)^2} + 13\left( {\frac{{ – 13}}{6}} \right) – 10\) (M1)
Note: (M1) for substituting their value of \(x\) from (c) into \(f(x)\) .
\( = – 24.1\) (A1)(ft)(G2)[2 marks]
\({\text{Area}} = 2(2x)x + 2xy + 2(2x)y\) (M1)(A1)
Note: (M1) for using the correct surface area formula (which can be implied if numbers in the correct place). (A1) for using correct numbers.
\(300 = 4{x^2} + 6xy\) (AG)
Note: Final line must be seen or previous (A1) mark is lost.[2 marks]
\(6xy = 300 – 4{x^2}\) (M1)
\(y = \frac{{300 – 4{x^2}}}{{6x}}\) or \(\frac{{150 – 2{x^2}}}{{3x}}\) (A1)[2 marks]
\({\text{Volume}} = x(2x)y\) (M1)
\(V = 2{x^2}\left( {\frac{{300 – 4{x^2}}}{{6x}}} \right)\) (A1)(ft)
\( = 100x – \frac{4}{3}{x^3}\) (AG)
Note: Final line must be seen or previous (A1) mark is lost.[2 marks]
\(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 100 – \frac{{12{x^2}}}{3}\) or \(100 – 4{x^2}\) (A1)(A1)
Note: (A1) for each term.[2 marks]
Unit penalty (UP) is applicable where indicated in the left hand column
(i) For maximum \(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 0\) or \(100 – 4{x^2} = 0\) (M1)
\(x = 5\) (A1)(ft)
\(y = \frac{{300 – 4{{(5)}^2}}}{{6(5)}}\) or \(\left( {\frac{{150 – 2{{(5)}^2}}}{{3(5)}}} \right)\) (M1)
\( = \frac{{20}}{3}\) (A1)(ft)
(UP) (ii) \(333\frac{1}{3}{\text{ c}}{{\text{m}}^3}{\text{ }}(333{\text{ c}}{{\text{m}}^3})\)
Note: (ft) from their (e)(i) if working for volume is seen.[5 marks]
Question
On the coordinate axes below, \({\text{D}}\) is a point on the \(y\)-axis and \({\text{E}}\) is a point on the \(x\)-axis. \({\text{O}}\) is the origin. The equation of the line \({\text{DE}}\) is \(y + \frac{1}{2}x = 4\).
Write down the coordinates of point \({\text{E}}\).[2]
\({\text{C}}\) is a point on the line \({\text{DE}}\). \({\text{B}}\) is a point on the \(x\)-axis such that \({\text{BC}}\) is parallel to the \(y\)-axis. The \(x\)-coordinate of \({\text{C}}\) is \(t\).
Show that the \(y\)-coordinate of \({\text{C}}\) is \(4 – \frac{1}{2}t\).[2]
\({\text{OBCD}}\) is a trapezium. The \(y\)-coordinate of point \({\text{D}}\) is \(4\).
Show that the area of \({\text{OBCD}}\) is \(4t – \frac{1}{4}{t^2}\).[3]
The area of \({\text{OBCD}}\) is \(9.75\) square units. Write down a quadratic equation that expresses this information.[1]
(i) Using your graphic display calculator, or otherwise, find the two solutions to the quadratic equation written in part (d).
(ii) Hence find the correct value for \(t\). Give a reason for your answer.[4]
Answer/Explanation
Markscheme
\({\text{E}}(8{\text{, }}0)\) (A1)(A1)
Notes: Brackets required but do not penalize again if mark lost in Q4 (i)(d). If missing award (A1)(A0).
Accept \(x = 8\), \(y = 0\)
Award (A1) for \(x = 8\)
\(y + \frac{1}{2}t = 4\) (M1)(M1)
Note: (M1) for the equation of the line seen. (M1) for substituting \(t\).
\(y = 4 – \frac{1}{2}t\) (AG)
Note: Final line must be seen or previous (M1) mark is lost.[2 marks]
\({\text{Area}} = \frac{1}{2} \times (4 + 4 – \frac{1}{2}t) \times t\) (M1)(A1)
Note: (M1) for substituting in correct formula, (A1) for correct substitution.
\( = \frac{1}{2} \times (8 – \frac{1}{2}t) \times t = \frac{1}{2}(8t – \frac{1}{2}{t^2})\) (A1)
\( = 4t – \frac{1}{4}{t^2}\) (AG)
Note: Final line must be seen or previous (A1) mark is lost.[3 marks]
\(4t – \frac{1}{4}{t^2} = 9.75\) or any equivalent form. (A1)[1 mark]
(i) \(t = 3\) or \(t =13\) (A1)(ft)(A1)(ft)(G2)
Note: Follow through from candidate’s equation to part (d). Award (A0)(A1)(ft) for \((3{\text{, }}0)\) and \((13{\text{, }}0)\).
(ii) \(t\) must be a value between \(0\) and \(8\) then \(t = 3\)
Note: Accept \({\text{B}}\) is between \({\text{O}}\) and \({\text{E}}\). Do not award (R0)(A1).
Question
Consider the function f (x) = x3 – 3x– 24x + 30.
Write down f (0).[1]
Find \(f'(x)\).[3]
Find the gradient of the graph of f (x) at the point where x = 1.[2]
(i) Use f ‘(x) to find the x-coordinate of M and of N.
(ii) Hence or otherwise write down the coordinates of M and of N.[5]
Sketch the graph of f (x) for \( – 5 \leqslant x \leqslant 7\) and \( – 60 \leqslant y \leqslant 60\). Mark clearly M and N on your graph.[4]
Lines L1 and L2 are parallel, and they are tangents to the graph of f (x) at points A and B respectively. L1 has equation y = 21x + 111.
(i) Find the x-coordinate of A and of B.
(ii) Find the y-coordinate of B.[6]
Answer/Explanation
Markscheme
30 (A1)[1 mark]
f ‘(x) = 3x2 – 6x – 24 (A1)(A1)(A1)
Note: Award (A1) for each term. Award at most (A1)(A1) if extra terms present.[3 marks]
f ‘(1) = –27 (M1)(A1)(ft)(G2)
Note: Award (M1) for substituting x = 1 into their derivative.[2 marks]
(i) f ‘(x) = 0
3x2 – 6x – 24 = 0 (M1)
x = 4; x = –2 (A1)(ft)(A1)(ft)
Notes: Award (M1) for either f ‘(x) = 0 or 3x2 – 6x – 24 = 0 seen. Follow through from their derivative. Do not award the two answer marks if derivative not used.
(ii) M(–2, 58) accept x = –2, y = 58 (A1)(ft)
N(4, – 50) accept x = 4, y = –50 (A1)(ft)
Note: Follow through from their answer to part (d) (i).[5 marks]
(A1) for window
(A1) for a smooth curve with the correct shape
(A1) for axes intercepts in approximately the correct positions
(A1) for M and N marked on diagram and in approximately
correct position (A4)
Note: If window is not indicated award at most (A0)(A1)(A0)(A1)(ft).[4 marks]
(i) 3x2 – 6x – 24 = 21 (M1)
3x2 – 6x – 45 = 0 (M1)
x = 5; x = –3 (A1)(ft)(A1)(ft)(G3)
Note: Follow through from their derivative.
OR
Award (A1) for L1 drawn tangent to the graph of f on their sketch in approximately the correct position (x = –3), (A1) for a second tangent parallel to their L1, (A1) for x = –3, (A1) for x = 5 . (A1)(ft)(A1)(ft)(A1)(A1)
Note: If only x = –3 is shown without working award (G2). If both answers are shown irrespective of workingaward (G3).
(ii) f (5) = –40 (M1)(A1)(ft)(G2)
Notes: Award (M1) for attempting to find the image of their x = 5. Award (A1) only for (5, –40). Follow through from their x-coordinate of B only if it has been clearly identified in (f) (i).[6 marks]
Question
A geometric sequence has \(1024\) as its first term and \(128\) as its fourth term.
Consider the arithmetic sequence \(1{\text{, }}4{\text{, }}7{\text{, }}10{\text{, }}13{\text{, }} \ldots \)
Show that the common ratio is \(\frac{1}{2}\) .[2]
Find the value of the eleventh term.[2]
Find the sum of the first eight terms.[3]
Find the number of terms in the sequence for which the sum first exceeds \(2047.968\).[3]
Find the value of the eleventh term.[2]
The sum of the first \(n\) terms of this sequence is \(\frac{n}{2}(3n – 1)\).
(i) Find the sum of the first 100 terms in this arithmetic sequence.
(ii) The sum of the first \(n\) terms is \(477\).
(a) Show that \(3{n^2} – n – 954 = 0\) .
(b) Using your graphic display calculator or otherwise, find the number of terms, \(n\) .[6]
Answer/Explanation
Markscheme
\(1024{r^3} = 128\) (M1)
\({r^3} = \frac{1}{8}\) or \(r = \sqrt[3]{{\frac{1}{8}}}\) (M1)
\(r = \frac{1}{2}{\text{ }}(0.5)\) (AG)
Notes: Award at most (M1)(M0) if last line not seen. Award (M1)(M0) if \(128\) is found by repeated multiplication (division) of \(1024\) by \(0.5\) \((2)\).[2 marks]
\(1024 \times {0.5^{10}}\) (M1)
Notes: Award (M1) for correct substitution into correct formula. Accept an equivalent method.
1 (A1)(G2)[2 marks]
\({S_8} = \frac{{1024\left( {1 – {{\left( {\frac{1}{2}} \right)}^8}} \right)}}{{1 – \frac{1}{2}}}\) (M1)(A1)
Note: Award (M1) for substitution into the correct formula, (A1) for correct substitution.
OR
(A1) for complete and correct list of eight terms (A1)
(M1) for their eight terms added (M1)
\({S_8} = 2040\) (A1)(G2)[3 marks]
\(\frac{{1024\left( {1 – {{\left( {\frac{1}{2}} \right)}^n}} \right)}}{{1 – \frac{1}{2}}} > 2047.968\) (M1)(M1)(ft)
Notes: Award (M1) for correct substitution into the correct formula for the sum, (M1) for comparing to \(2047.968\) . Accept equation. Follow through from their expression for the sum used in part (c).
OR
If a list is used: \({S_{15}} = 2047.9375\) (M1)
\({S_{16}} = 2047.96875\) (M1)
\(n = 16\) (A1)(ft)(G2)
Note: Follow through from their expression for the sum used in part (c).[3 marks]
\({\text{common difference}} = 3\) (may be implied) (A1)
\({u_{11}} = 31\) (A1)(G2)[2 marks]
(i) \(\frac{{100}}{2}(3 \times 100 – 1)\) OR \(\frac{{100(2 + 99 \times 3)}}{2}\) (M1)
\(14 950\) (A1)(G2)
(ii) (a) \(\frac{n}{2}(3n – 1) = 477\) OR \(\frac{n}{2}(2 + 3(n – 1)) = 477\) (M1)
\(3{n^2} – n = 954\) (M1)
\(3{n^2} – n – 954 = 0\) (AG)
Notes: Award second (M1) for correct removal of denominator or brackets and no further incorrect working seen. Award at most
(M1)(M0) if last line not seen.
(b) \(18\) (G2)
Note: If both solutions to the quadratic equation are seen and the correct value is not identified as the required answer, award (G1)(G0).[6 marks]
Question
In a game, n small pumpkins are placed 1 metre apart in a straight line. Players start 3 metres before the first pumpkin.
Each player collects a single pumpkin by picking it up and bringing it back to the start. The nearest pumpkin is collected first. The player then collects the next nearest pumpkin and the game continues in this way until the signal is given for the end.
Sirma runs to get each pumpkin and brings it back to the start.
Write down the distance, \({a_1}\), in metres that she has to run in order to collect the first pumpkin.[1]
The distances she runs to collect each pumpkin form a sequence \({a_1},{\text{ }}{a_2},{\text{ }}{a_3}, \ldots \) .
(i) Find \({a_2}\).
(ii) Find \({a_3}\).[2]
Write down the common difference, \(d\), of the sequence.[1]
The final pumpkin Sirma collected was 24 metres from the start.
(i) Find the total number of pumpkins that Sirma collected.
(ii) Find the total distance that Sirma ran to collect these pumpkins.[5]
Peter also plays the game. When the signal is given for the end of the game he has run 940 metres.
Calculate the total number of pumpkins that Peter collected.[3]
Peter also plays the game. When the signal is given for the end of the game he has run 940 metres.
Calculate Peter’s distance from the start when the signal is given.[2]
Answer/Explanation
Markscheme
\(6{\text{ (m)}}\) (A1)(G1)
(i) \(8\) (A1)(ft)
(ii) \(10\) (A1)(ft)(G2)
Note: Follow through from part (a).
\(2{\text{ (m)}}\) (A1)(ft)
Note: Follow through from parts (a) and (b).
(i) \(2 \times 24 = 6 + 2(n – 1)\;\;\;\)OR\(\;\;\;24 = 3 + (n – 1)\) (M1)
Note: Award (M1) for correct substitution in arithmetic sequence formula.
\(n = 22\) (A1)(ft)(G1)
Note: Follow through from parts (a) and (c).
(ii) \(\frac{{(6 + 48)}}{2} \times 22\) (M1)(A1)(ft)
Note: Award (M1) for substitution in arithmetic series formula, (A1)(ft) for correct substitution.
\( = 594\) (A1)(ft)(G2)
Note: Follow through from parts (a) and (d)(i).
\(\frac{{\left[ {2 \times 6 + 2(n – 1)} \right] \times n}}{2} = 940\) (M1)(A1)(ft)
Notes: Award (M1) for substitution in arithmetic series formula, (A1) for their correct substituted formula equated to \(940\). Follow through from parts (a) and (c).
\({n^2} + 5n – 940 = 0\)
\(n = 28.2611 \ldots \)
\(n = 28\) (A1)(ft)(G2)
\(\frac{{\left[ {2 \times 6 + 2(28 – 1)} \right] \times 28}}{2}\) (M1)
Notes: Award (M1) for substituting their \(28\) into the arithmetic series formula.
\( = 16{\text{ (m)}}\) (A1)(ft)(G2)
Question
The following diagram shows two triangles, OBC and OBA, on a set of axes. Point C lies on the \(y\)-axis, and O is the origin.
The equation of the line BC is \(y = 4\).
Write down the coordinates of point C.[1]
The \(x\)-coordinate of point B is \(a\).
(i) Write down the coordinates of point B;
(ii) Write down the gradient of the line OB.[2]
Point A lies on the \(x\)-axis and the line AB is perpendicular to line OB.
(i) Write down the gradient of line AB.
(ii) Show that the equation of the line AB is \(4y + ax – {a^2} – 16 = 0\).[4]
The area of triangle AOB is three times the area of triangle OBC.
Find an expression, in terms of a, for
(i) the area of triangle OBC;
(ii) the x-coordinate of point A.[3]
Calculate the value of \(a\).[2]
Answer/Explanation
Markscheme
\((0,{\text{ }}4)\) (A1)
Notes: Accept \(x = 0,{\text{ }}y = 4\).
(i) \((a,{\text{ }}4)\) (A1)(ft)
Notes: Follow through from part (a).
(ii) \(\frac{4}{a}\) (A1)(ft)
Note: Follow through from part (b)(i).
(i) \( – \frac{a}{4}\) (A1)(ft)
Note: Follow through from part (b)(ii).
(ii) \(y = – \frac{a}{4}x + c\) (M1)
Note: Award (M1) for substitution of their gradient from part (c)(i) in the equation.
\(4 = – \frac{a}{4} \times a + c\)
\(c = \frac{1}{4} \times {a^2} + 4\)
\(y = – \frac{a}{4}x + \frac{1}{4}{a^2} + 4\) (A1)
OR
\(y – 4 = – \frac{a}{4}(x – a)\) (M1)
Note: Award (M1) for substitution of their gradient from part (c)(i) in the equation.
\(y = – \frac{{ax}}{4} + \frac{{{a^2}}}{4} + 4\) (A1)
\(4y = – ax + {a^2} + 16\)
\(4y + ax – {a^2} – 16 = 0\) (AG)
Note: Both the simplified and the not simplified equations must be seen for the final (A1) to be awarded.
(i) \(2a\) (A1)
(ii) \(\frac{{4x}}{2} = 3 \times 2a\) (M1)
Note: Award (M1) for correct equation.
\(x = 3a\) (A1)(ft)
Note: Follow through from part (d)(i).
OR
\(0 – 4 = – \frac{a}{4}(x – a)\) (M1)
Note: Award (M1) for correct substitution of their gradient and the coordinates of their point into the equation of a line.
\(\frac{{16}}{a} = x – a\)
\(x = a + \frac{{16}}{a}\) (A1)(ft)
Note: Follow through from parts (b)(i) and (c)(i).
OR
\(4 \times 0 + ax – {a^2} – 16 = 0\) (M1)
Note: Award (M1) for correct substitution of the coordinates of \({\text{A}}(x,{\text{ }}0)\) into the equation of line AB.
\(ax – {a^2} – 16 = 0\)
\(x = a + \frac{{16}}{a}\;\;\;\)OR\(\;\;\;x = \frac{{({a^2} + 16)}}{a}\) (A1)(G1)
\(4(0) + a(3a) – {a^2} – 16 = 0\) (M1)
Note: Award (M1) for correct substitution of their \(3a\) from part (d)(ii) into the equation of line AB.
OR
\(\frac{1}{2}\left( {a + \frac{{16}}{a}} \right) \times 4 = 3\left( {\frac{{4a}}{2}} \right)\) (M1)
Note: Award (M1) for area of triangle AOB (with their substituted \(a + \frac{{16}}{a}\) and 4) equated to three times their area of triangle AOB.
\(a = 2.83\;\;\;\left( {2.82842…,{\text{ }}2\sqrt 2 ,{\text{ }}\sqrt 8 } \right)\) (A1)(ft)(G1)
Note: Follow through from parts (d)(i) and (d)(ii).
Question
The sum of the first \(n\) terms of an arithmetic sequence is given by \({S_n} = 6n + {n^2}\).
Write down the value of
(i) \({S_1}\);
(ii) \({S_2}\).[2]
The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).
Show that \({u_2} = 9\).[1]
The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).
Find the common difference of the sequence.[2]
The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).
Find \({u_{10}}\).[2]
The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).
Find the lowest value of \(n\) for which \({u_n}\) is greater than \(1000\).[3]
The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).
There is a value of \(n\) for which
\[{u_1} + {u_2} + \ldots + {u_n} = 1512.\]
Find the value of \(n\).[2]
Answer/Explanation
Markscheme
(i) \({S_1} = 7\) (A1)
(ii) \({S_2} = 16\) (A1)
\(({u_2} = ){\text{ }}16 – 7 = 9\) (M1)(AG)
Note: Award (M1) for subtracting 7 from 16. The 9 must be seen.
OR
\(16 – 7 – 7 = 2\)
\(({u_2} = ){\text{ }}7 + (2 – 1)(2) = 9\) (M1)(AG)
Note: Award (M1) for subtracting twice \(7\) from \(16\) and for correct substitution in correct arithmetic sequence formula.
The \(9\) must be seen.
Do not accept: \(9 – 7 = 2,{\text{ }}{u_2} = 7 + (2 – 1)(2) = 9\).
\({u_1} = 7\) (A1)(ft)
\(d = 2{\text{ }}( = 9 – 7)\) (A1)(ft)(G2)
Notes: Follow through from their \({S_1}\) in part (a)(i).
\(7 + 2 \times (10 – 1)\) (M1)
Note: Award (M1) for correct substitution in the correct arithmetic sequence formula. Follow through from their parts (a)(i) and (c).
\( = 25\) (A1)(ft)(G2)
Note: Award (A1)(ft) for their correct tenth term.
\(7 + 2 \times (n – 1) > 1000\) (A1)(ft)(M1)
Note: Award (A1)(ft) for their correct expression for the \({n^{{\text{th}}}}\) term, (M1) for comparing their expression to \(1000\). Accept an equation. Follow through from their parts (a)(i) and (c).
\(n = 498\) (A1)(ft)(G2)
Notes: Answer must be a natural number.
\(6n + {n^2} = 1512\;\;\;\)OR\(\;\;\;\frac{n}{2}\left( {14 + 2(n – 1)} \right) = 1512\;\;\;\)OR
\({S_n} = 1512\;\;\;\)OR\(\;\;\;7 + 9 + \ldots + {u_n} = 1512\) (M1)
Notes: Award (M1) for equating the sum of the first \(n\) terms to \(1512\). Accept a sum of at least the first 7 correct terms.
\(n = 36\) (A1)(G2)
Note: If \(n = 36\) is seen without working, award (G2). Award a maximum of (M1)(A0) if \( – 42\) is also given as a solution.