Write down an expression for \(T\) in terms of \(r\), \(l\) and \(\pi \).

The volume of the lobster trap is \(0.75{\text{ }}{{\text{m}}^{\text{3}}}\).

Write down an equation for the volume of the lobster trap in terms of \(r\), \(l\) and \(\pi \).

The volume of the lobster trap is \(0.75{\text{ }}{{\text{m}}^{\text{3}}}\).

Show that \(T = (2\pi  + 4)r + \frac{6}{{\pi {r^2}}}\).

The volume of the lobster trap is \(0.75{\text{ }}{{\text{m}}^{\text{3}}}\).

Find \(\frac{{{\text{d}}T}}{{{\text{d}}r}}\).

The lobster trap is designed so that the length of steel used in its frame is a minimum.

Show that the value of \(r\) for which \(T\) is a minimum is \(0.719 {\text{ m}}\), correct to three significant figures.

The lobster trap is designed so that the length of steel used in its frame is a minimum.

Calculate the value of \(l\) for which \(T\) is a minimum.

The lobster trap is designed so that the length of steel used in its frame is a minimum.

Calculate the minimum value of \(T\).

\(2\pi r + 4r + 4l\)     (A1)(A1)(A1)

Notes: Award (A1) for \(2\pi r\) (“\(\pi \)” must be seen), (A1) for \(4r\), (A1) for \(4l\). Accept equivalent forms. Accept \(T = 2\pi r + 4r + 4l\). Award a maximum of (A1)(A1)(A0) if extra terms are seen.

[3 marks]

\(0.75 = \frac{{\pi {r^2}l}}{2}\)     (A1)(A1)(A1)

Notes:     Award (A1) for their formula equated to \(0.75\), (A1) for \(l\) substituted into volume of cylinder formula, (A1) for volume of cylinder formula divided by \(2\).

If “\(\pi \)” not seen in part (a) accept use of \(3.14\) or greater accuracy. Award a maximum of (A1)(A1)(A0) if extra terms are seen.

[3 marks]

\(T = 2\pi r + 4r + 4\left( {\frac{{1.5}}{{\pi {r^2}}}} \right)\)     (A1)(ft)(A1)

\( = (2\pi  + 4)r + \frac{6}{{\pi {r^2}}}\)     (AG)

Notes: Award (A1)(ft) for correct rearrangement of their volume formula in part (b) seen, award (A1) for the correct substituted formula for \(T\). The final line must be seen, with no incorrect working, for this second (A1) to be awarded.

[2 marks]

\(\frac{{{\text{d}}T}}{{{\text{d}}r}} = 2\pi  + 4 - \frac{{12}}{{\pi {r^3}}}\)     (A1)(A1)(A1)

Note: Award (A1) for \(2\pi  + 4\), (A1) for \(\frac{{ - 12}}{\pi }\), (A1) for \({r^{ - 3}}\).

     Accept 10.3 (10.2832…) for \(2\pi  + 4\),  accept \(–3.82\) \(–3.81971…\) for \(\frac{{ - 12}}{\pi }\). Award a maximum of (A1)(A1)(A0) if extra terms are seen.

[3 marks]

\(2\pi  + 4 - \frac{{12}}{{\pi {r^3}}} = 0\)   OR   \(\frac{{{\text{d}}T}}{{{\text{d}}r}} = 0\)     (M1)

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

\(r = 0.718843 \ldots \)   OR   \(\sqrt[3]{{0.371452 \ldots }}\)   OR   \(\sqrt[3]{{\frac{{12}}{{\pi (2\pi  + 4)}}}}\)   OR   \(\sqrt[3]{{\frac{{3.81971}}{{10.2832 \ldots }}}}\)     (A1)

\(r = 0.719{\text{ (m)}}\)     (AG)

Note: The rounded and unrounded or formulaic answers must be seen for the final (A1) to be awarded. The use of \(3.14\) gives an unrounded answer of \(r = 0.719039 \ldots \).

[2 marks]

\(0.75 = \frac{{\pi  \times {{(0.719)}^2}l}}{2}\)     (M1)

Note: Award (M1) for substituting \(0.719\) into their volume formula. Follow through from part (b).

\(l = 0.924{\text{ (m)}}\)  \((0.923599 \ldots )\)     (A1)(ft)(G2)

[2 marks]

\(T = (2\pi  + 4) \times 0.719 + \frac{6}{{\pi {{(0.719)}^2}}}\)     (M1)

Notes: Award (M1) for substituting \(0.719\) in their expression for \(T\). Accept alternative methods, for example substitution of their \(l\) and \(0.719\) into their part (a) (for which the answer is \(11.08961024\)). Follow through from their answer to part (a).

\( = 11.1{\text{ (m)}}\)   \((11.0880 \ldots )\)     (A1)(ft)(G2)

Show that \({\text{BD}} = 93{\text{ m}}\) correct to the nearest metre.

Calculate angle \({\rm{B\hat CD}}\).

Find the area of ABCD.

Calculate Abdallah’s estimate for the area.

Find the percentage error in Abdallah’s estimate.

\({\text{B}}{{\text{D}}^2} = {40^2} + {84^2}\)     (M1)

Note:     Award (M1) for correct substitution into Pythagoras.

Accept correct substitution into cosine rule.

\({\text{BD}} = 93.0376 \ldots \)     (A1)

\( = 93\)     (AG)

Note:     Both the rounded and unrounded value must be seen for the (A1) to be awarded.

[2 marks]

\(\cos C = \frac{{{{115}^2} + {{60}^2} - {{93}^2}}}{{2 \times 115 \times 60}}{\text{ }}({93^2} = {115^2} + {60^2} - 2 \times 115 \times 60 \times \cos C)\)     (M1)(A1)

Note:     Award (M1) for substitution into cosine formula, (A1) for correct substitutions.

\( = 53.7^\circ {\text{ }}(53.6679 \ldots ^\circ )\)     (A1)(G2)

[3 marks]

\(\frac{1}{2}(40)(84) + \frac{1}{2}(115)(60)\sin (53.6679 \ldots )\)     (M1)(M1)(A1)(ft)

Note:     Award (M1) for correct substitution into right-angle triangle area. Award (M1) for substitution into area of triangle formula and (A1)(ft) for correct substitution.

\( = 4460{\text{ }}{{\text{m}}^2}{\text{ }}(4459.30 \ldots {\text{ }}{{\text{m}}^2})\)     (A1)(ft)(G3)

Notes:     Follow through from part (b).

[4 marks]

\(\frac{{(40 + 60)(84 + 115)}}{4}\)     (M1)

Note:     Award (M1) for correct substitution in the area formula used by ‘Ancient Egyptians’.

\( = 4980{\text{ }}{{\text{m}}^2}{\text{ }}(4975{\text{ }}{{\text{m}}^2})\)     (A1)(G2)

[2 marks]

\(\left| {\frac{{4975 - 4459.30 \ldots }}{{4459.30 \ldots }}} \right| \times 100\)     (M1)

Notes:     Award (M1) for correct substitution into percentage error formula.

\( = 11.6{\text{ }}(\% ){\text{ }}(11.5645 \ldots )\)     (A1)(ft)(G2)

Notes:    Follow through from parts (c) and (d)(i).

[2 marks]

The equation of the line BC is \(y = 4\).

Write down the coordinates of point C.

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.

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 \(4y + ax - {a^2} - 16 = 0\).

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.

Calculate the value of \(a\).

\((0,{\text{ }}4)\)     (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).

Write down an equation for the area of ABCDE using the above information.

Show that the equation in part (a)(i) simplifies to \(3{x^2} + 19x - 414 = 0\).

Find the length of CD.

Show that angle \({\rm{B\hat AE}} = 67.4^\circ \), correct to one decimal place.

Find the length of the perimeter of ABCDE.

Calculate the length of CF.

\(222 = \frac{1}{2}x(x + 3) + (x + 3)(x + 5)\)     (M1)(M1)(A1)

Note:     Award (M1) for correct area of triangle, (M1) for correct area of rectangle, (A1) for equating the sum to 222.

OR

\(222 = (x + 3)(2x + 5) - 2\left( {\frac{1}{4}} \right)x(x + 3)\)     (M1)(M1)(A1)

Note:     Award (M1) for area of bounding rectangle, (M1) for area of triangle, (A1) for equating the difference to 222.

[2 marks]

\(222 = \frac{1}{2}{x^2} + \frac{3}{2}x + {x^2} + 3x + 5x + 15\)     (M1)

Note:     Award (M1) for complete expansion of the brackets, leading to the final answer, with no incorrect working seen. The final answer must be seen to award (M1).

\(3{x^2} + 19x - 414 = 0\)     (AG)

[2 marks]

\(x = 9{\text{ }}\left( {{\text{and }}x =  - \frac{{46}}{3}} \right)\)     (A1)

\({\text{CD}} = 12{\text{ (cm)}}\)     (A1)(G2)

[2 marks]

\(\frac{1}{2}({\text{their }}x + 3) = 6\)     (A1)(ft)

Note:     Follow through from part (b).

\(\tan \left( {\frac{{{\rm{B\hat AE}}}}{2}} \right) = \frac{6}{9}\)     (M1)

Note:     Award (M1) for their correct substitutions in tangent ratio.

\({\rm{B\hat AE}} = 67.3801 \ldots ^\circ \)     (A1)

\( = 67.4^\circ \)     (AG)

Note:     Do not award the final (A1) unless both the correct unrounded and rounded answers are seen.

OR

\(\frac{1}{2}({\text{their }}x + 3) = 6\)     (A1)(ft)

\(\tan ({\rm{A\hat BE}}) = \frac{9}{6}\)     (M1)

Note:     Award (M1) for their correct substitutions in tangent ratio.

\({\rm{B\hat AE}} = 180^\circ  - 2({\rm{A\hat BE}})\)

\({\rm{B\hat AE}} = 67.3801 \ldots ^\circ \)     (A1)

\( = 67.4^\circ \)     (AG)

Note:     Do not award the final (A1) unless both the correct unrounded and rounded answers are seen.

[3 marks]

\(2\sqrt {{9^2} + {6^2}}  + 12 + 2(14)\)     (M1)(M1)

Note:     Award (M1) for correct substitution into Pythagoras. Award (M1) for the addition of 5 sides of the pentagon, consistent with their \(x\).

\(61.6{\text{ (cm) }}\left( {61.6333 \ldots {\text{ (cm)}}} \right)\)     (A1)(ft)(G3)

Note:     Follow through from part (b).

[3 marks]

\({\rm{F\hat BC}} = 90 + \left( {\frac{{180 - 67.4}}{2}} \right){\text{ }}( = 146.3^\circ )\)     (M1)

OR

\(180 - \frac{{67.4}}{2}\)     (M1)

\({\rm{C}}{{\text{F}}^2} = {8^2} + {14^2} - 2(8)(14)\cos (146.3^\circ )\)     (M1)(A1)(ft)

Note:     Award (M1) for substituted cosine rule formula and (A1) for correct substitutions. Follow through from part (b).

\({\text{CF}} = 21.1{\text{ (cm) }}(21.1271 \ldots )\)     (A1)(ft)(G3)

OR

\({\rm{G\hat BF}} = \frac{{67.4}}{2} = 33.7^\circ \)     (A1)

Note:     Award (A1) for angle \({\rm{G\hat BF}} = 33.7^\circ \), where G is the point such that CG is a projection/extension of CB and triangles BGF and CGF are right-angled triangles. The candidate may use another variable.

\({\text{GF}} = 8\sin 33.7^\circ  = 4.4387 \ldots \)\(\,\,\,\)AND\(\,\,\,\)\({\text{BG}} = 8\cos 33.7^\circ  = 6.6556 \ldots \)     (M1)

Note:     Award (M1) for correct substitution into trig formulas to find both GF and BG.

\({\text{C}}{{\text{F}}^2} = {(14 + 6.6556 \ldots )^2} + {(4.4387 \ldots )^2}\)     (M1)

Note:     Award (M1) for correct substitution into Pythagoras formula to find CF.

\({\text{CF}} = 21.1{\text{ (cm) }}(21.1271 \ldots )\)     (A1)(ft)(G3)

[4 marks]

Find the gradient of the line L.

Differentiate \(f (x)\) .

Find the coordinates of the point where the tangent to P is parallel to the line L.

Find the coordinates of the point where the tangent to P is perpendicular to the line L.

Find

(i) the gradient of the tangent to P at the point with coordinates (2, − 6).

(ii) the equation of the tangent to P at this point.

State the equation of the axis of symmetry of P.

Find the coordinates of the vertex of P and state the gradient of the curve at this point.

for attempt at substituted \(\frac{{ydistance}}{{xdistance}}\)     (M1)

gradient = 2     (A1)(G2)

[2 marks]

\(2x - 3\)     (A1)(A1)

(A1) for \(2x\) , (A1) for \(-3\)

[2 marks]

for their \(2x - 3 =\) their gradient and attempt to solve     (M1)

\(x = 2.5\)     (A1)(ft)

\(y = -5.25\) ((ft) from their x value)     (A1)(ft)(G2)

[3 marks]

for seeing \(\frac{{ - 1}}{{their(a)}}\)     (M1)

solving \(2x – 3 = - \frac{1}{2}\) (or their value)     (M1)

x = 1.25     (A1)(ft)(G1)

y = – 6.1875     (A1)(ft)(G1)

[4 marks]

(i) \(2 \times 2 - 3 = 1\) ((ft) from (b))     (A1)(ft)(G1)

(ii) \(y = mx + c\) or equivalent method to find \(c \Rightarrow -6 = 2 + c\)     (M1)

\(y = x - 8\)     (A1)(ft)(G2)

[3 marks]

\(x = 1.5\)     (A1)

[1 mark]

for substituting their answer to part (f) into the equation of the parabola (1.5, −6.25) accept x = 1.5, y = −6.25     (M1)(A1)(ft)(G2)

gradient is zero (accept \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\))     (A1)

[3 marks]

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Calculate the length of \({\text{AC}}\).

Calculate the area of triangle \({\text{ABC}}\).

Assuming that no wood is wasted, show that the volume of wood required to make all \(2600\) blocks is \({\text{13}}\,{\text{400}}\,{\text{000 c}}{{\text{m}}^3}\), correct to three significant figures.

Write \({\text{13}}\,{\text{400}}\,{\text{000}}\) in the form \(a \times {10^k}\) where \(1 \leqslant a < 10\) and \(k \in \mathbb{Z}\).

Show that the total surface area of one block is \({\text{2190 c}}{{\text{m}}^2}\), correct to three significant figures.

The blocks are to be painted. One litre of paint will cover \(22{\text{ }}{{\text{m}}^2}\).

Calculate the number of litres required to paint all \(2600\) blocks.

\({\text{A}}{{\text{C}}^2} = {30^2} + {24^2} - 2 \times 30 \times 24 \times \cos 35^\circ \)     (M1)(A1)

Note: Award (M1) for substituted cosine rule formula,

     (A1) for correct substitutions.

\({\text{AC}} = 17.2{\text{ cm}}\)   \((17.2168…)\)     (A1)(G2)

Notes: Use of radians gives \(52.7002…\) Award (M1)(A1)(A0).

     No marks awarded in this part of the question where candidates assume that angle \({\text{ACB}} = 90^\circ \).

[3 marks]

Units are required in part (b).

Area of triangle \({\text{ABC = }}\frac{1}{2} \times 24 \times 30 \times \sin 35^\circ \)     (M1)(A1)

Notes: Award (M1) for substitution into area formula, (A1) for correct substitutions.

     Special Case: Where a candidate has assumed that angle \({\text{ACB}} = 90^\circ \) in part (a), award (M1)(A1) for a correct alternative substituted formula for the area of the triangle \(\left( {ie{\text{ }}\frac{1}{2} \times {\text{base}} \times {\text{height}}} \right)\).

\( = 206{\text{ c}}{{\text{m}}^2}\)   \((206.487 \ldots {\text{c}}{{\text{m}}^2})\)     (A1)(G2)

Notes: Use of radians gives negative answer, \(–154.145…\) Award (M1)(A1)(A0).

     Special Case: Award (A1)(ft) where the candidate has arrived at an area which is correct to the standard rounding rules from their lengths (units required).

[3 marks]

\(206.487… \times 25 \times 2600\)     (M1)

Note: Award (M1) for multiplication of their answer to part (b) by \(25\) and \(2600\).

\({\text{13}}\,{\text{421}}\,{\text{688.61}}\)     (A1)

\({\text{13}}\,{\text{400}}\,{\text{000}}\)     (AG)

Note: The final (A1) cannot be awarded unless both the unrounded and rounded answers are seen.

[2 marks]

\(1.34 \times {10^7}\)     (A2)

Notes: Award (A2) for the correct answer.

     Award (A1)(A0) for \(1.34\) and an incorrect index value.

     Award (A0)(A0) for any other combination (including answers such as \(13.4 \times {10^6}\)).

[2 marks]

\(2 \times 206.487 \ldots  + 24 \times 25 + 30 \times 25 + 17.2168 \ldots  \times 25\)     (M1)(M1)

Note: Award (M1) for multiplication of their answer to part (b) by \(2\) for area of two triangular ends, (M1) for three correct rectangle areas using \(24\), \(30\) and their \(17.2\).

\(2193.26…\)     (A1)

Note: Accept \(2192\) for use of 3 sf answers.

\(2190\)     (AG)

Note: The final (A1) cannot be awarded unless both the unrounded and rounded answers are seen.

[3 marks]

\(\frac{{2190 \times 2600}}{{22 \times 10\,000}}\)     (M1)(M1)

Notes: Award (M1) for multiplication by \(2600\) and division by \(22\), (M1) for division by \({10\,000}\).

     The use of \(22\) may be implied ie division by \(2200\) would be acceptable.

\(25.9\) litres   \((25.8818…)\)     (A1)(G2)

Note: Accept \(26\).

[3 marks]

Some candidates assumed that triangle ACB was a right angled triangle with angle \({\text{ACB}} = 90^\circ \). Such candidates earned no marks for part (a) but were able to recover most of the marks in the remainder of the question. For those candidates who correctly used the cosine rule for part (a), most achieved all 3 marks for this part and used a correct formula for the area of the triangle in part (b) to obtain at least 2 marks for this part. The final mark was not awarded, however, if no units or the incorrect units were given. Parts (c) and (e) were generally well done with many candidates showing the unrounded answer before the required answer. Part (f) proved to be quite problematic for many candidates. Whilst many were able to earn a method mark for \(\frac{{2190 \times 2600}}{{22}}\), a significant number of these candidates were unable to convert the units correctly and very few correct answers were seen. Indeed, the most popular answer seemed to be 2590 litres.

Some candidates assumed that triangle ACB was a right angled triangle with angle \({\text{ACB}} = 90^\circ \). Such candidates earned no marks for part (a) but were able to recover most of the marks in the remainder of the question. For those candidates who correctly used the cosine rule for part (a), most achieved all 3 marks for this part and used a correct formula for the area of the triangle in part (b) to obtain at least 2 marks for this part. The final mark was not awarded, however, if no units or the incorrect units were given. Parts (c) and (e) were generally well done with many candidates showing the unrounded answer before the required answer. Part (f) proved to be quite problematic for many candidates. Whilst many were able to earn a method mark for \(\frac{{2190 \times 2600}}{{22}}\), a significant number of these candidates were unable to convert the units correctly and very few correct answers were seen. Indeed, the most popular answer seemed to be 2590 litres.

Some candidates assumed that triangle ACB was a right angled triangle with angle \({\text{ACB}} = 90^\circ \). Such candidates earned no marks for part (a) but were able to recover most of the marks in the remainder of the question. For those candidates who correctly used the cosine rule for part (a), most achieved all 3 marks for this part and used a correct formula for the area of the triangle in part (b) to obtain at least 2 marks for this part. The final mark was not awarded, however, if no units or the incorrect units were given. Parts (c) and (e) were generally well done with many candidates showing the unrounded answer before the required answer. Part (f) proved to be quite problematic for many candidates. Whilst many were able to earn a method mark for \(\frac{{2190 \times 2600}}{{22}}\), a significant number of these candidates were unable to convert the units correctly and very few correct answers were seen. Indeed, the most popular answer seemed to be 2590 litres.

Some candidates assumed that triangle ACB was a right angled triangle with angle \({\text{ACB}} = 90^\circ \). Such candidates earned no marks for part (a) but were able to recover most of the marks in the remainder of the question. For those candidates who correctly used the cosine rule for part (a), most achieved all 3 marks for this part and used a correct formula for the area of the triangle in part (b) to obtain at least 2 marks for this part. The final mark was not awarded, however, if no units or the incorrect units were given. Parts (c) and (e) were generally well done with many candidates showing the unrounded answer before the required answer. Part (f) proved to be quite problematic for many candidates. Whilst many were able to earn a method mark for \(\frac{{2190 \times 2600}}{{22}}\), a significant number of these candidates were unable to convert the units correctly and very few correct answers were seen. Indeed, the most popular answer seemed to be 2590 litres.

Some candidates assumed that triangle ACB was a right angled triangle with angle \({\text{ACB}} = 90^\circ \). Such candidates earned no marks for part (a) but were able to recover most of the marks in the remainder of the question. For those candidates who correctly used the cosine rule for part (a), most achieved all 3 marks for this part and used a correct formula for the area of the triangle in part (b) to obtain at least 2 marks for this part. The final mark was not awarded, however, if no units or the incorrect units were given. Parts (c) and (e) were generally well done with many candidates showing the unrounded answer before the required answer. Part (f) proved to be quite problematic for many candidates. Whilst many were able to earn a method mark for \(\frac{{2190 \times 2600}}{{22}}\), a significant number of these candidates were unable to convert the units correctly and very few correct answers were seen. Indeed, the most popular answer seemed to be 2590 litres.

Some candidates assumed that triangle ACB was a right angled triangle with angle \({\text{ACB}} = 90^\circ \). Such candidates earned no marks for part (a) but were able to recover most of the marks in the remainder of the question. For those candidates who correctly used the cosine rule for part (a), most achieved all 3 marks for this part and used a correct formula for the area of the triangle in part (b) to obtain at least 2 marks for this part. The final mark was not awarded, however, if no units or the incorrect units were given. Parts (c) and (e) were generally well done with many candidates showing the unrounded answer before the required answer. Part (f) proved to be quite problematic for many candidates. Whilst many were able to earn a method mark for \(\frac{{2190 \times 2600}}{{22}}\), a significant number of these candidates were unable to convert the units correctly and very few correct answers were seen. Indeed, the most popular answer seemed to be 2590 litres.

(i)     Write down the value of \(y\) when \(x\) is \(2\).

(ii)    Write down the coordinates of the point where the curve intercepts the \(y\)-axis.

Sketch the curve for \( - 4 \leqslant x \leqslant 3\) and \( - 10 \leqslant y \leqslant 10\). Indicate clearly the information found in (a).

Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) .

Let \({L_1}\) be the tangent to the curve at \(x = 2\).

Let \({L_2}\) be a tangent to the curve, parallel to \({L_1}\).

(i)     Show that the gradient of \({L_1}\) is \(12\).

(ii)    Find the \(x\)-coordinate of the point at which \({L_2}\) and the curve meet.

(iii)   Sketch and label \({L_1}\) and \({L_2}\) on the diagram drawn in (b).

It is known that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} > 0\) for \(x < - 2\) and \(x > b\) where \(b\) is positive.

(i)     Using your graphic display calculator, or otherwise, find the value of \(b\).

(ii)    Describe the behaviour of the curve in the interval \( - 2 < x < b\) .

(iii)   Write down the equation of the tangent to the curve at \(x = - 2\).

(i)     \(y = 0\)     (A1)

(ii)    \((0{\text{, }}{- 2})\)     (A1)(A1)

Note: Award (A1)(A0) if brackets missing.

OR

\(x = 0{\text{, }}y = - 2\)     (A1)(A1)

Note: If coordinates reversed award (A0)(A1)(ft). Two coordinates must be given.

[3 marks]

     (A4)

Notes: (A1) for appropriate window. Some indication of scale on the \(x\)-axis must be present (for example ticks). Labels not required. (A1) for smooth curve and shape, (A1) for maximum and minimum in approximately correct position, (A1) for \(x\) and \(y\) intercepts found in (a) in approximately correct position.

[4 marks]

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 3{x^2} + 3x - 6\)     (A1)(A1)(A1)

 Note: (A1) for each correct term. Award (A1)(A1)(A0) at most if any other term is present.

[3 marks]

(i)     \(3 \times 4 + 3 \times 2 - 6 = 12\)     (M1)(A1)(AG)

Note: (M1) for using the derivative and substituting \(x = 2\) . (A1) for correct (and clear) substitution. The \(12\) must be seen.

(ii)    Gradient of \({L_2}\) is \(12\) (can be implied)     (A1)

\(3{x^2} + 3x - 6 = 12\)     (M1)

\(x = - 3\)     (A1)(G2)

Note: (M1) for equating the derivative to \(12\) or showing a sketch of the derivative together with a line at \(y = 12\) or a table of values showing the \(12\) in the derivative column.

(iii)   (A1) for \({L_1}\) correctly drawn at approx the correct point     (A1)

(A1) for \({L_2}\) correctly drawn at approx the correct point     (A1)

(A1) for 2 parallel lines     (A1)

Note: If lines are not labelled award at most (A1)(A1)(A0). Do not accept 2 horizontal or 2 vertical parallel lines.

[8 marks]

(i)     \(b = 1\)     (G2)

(ii)    The curve is decreasing.     (A1)

Note: Accept any valid description.

(iii)   \(y = 8\)     (A1)(A1)(G2)

Note: (A1) for “\(y =\) a constant”, (A1) for \(8\).

[5 marks]

Many candidates managed to gain good marks in this question as they were able to answer the first three parts of the question. Good sketches were drawn with the required information shown on them. Very few candidates did not recognise the notation \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) but they showed that they knew how to differentiate as in (d)(i) they found the derivative to show that the gradient of \({L_1}\) was \(12\). Candidates found it difficult to find the other \(x\) for which the derivative was \(12\). However, some could draw both tangents without having found this value of \(x\) . In general, tangents were not well drawn. The last part question did act as a discriminating question. However, those candidates that had the function drawn either in their GDC or on paper recognised that at \(x = - 2\) there was a maximum and so wrote down the correct equation of the tangent at that point.

Many candidates managed to gain good marks in this question as they were able to answer the first three parts of the question. Good sketches were drawn with the required information shown on them. Very few candidates did not recognise the notation \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) but they showed that they knew how to differentiate as in (d)(i) they found the derivative to show that the gradient of \({L_1}\) was \(12\). Candidates found it difficult to find the other \(x\) for which the derivative was \(12\). However, some could draw both tangents without having found this value of \(x\) . In general, tangents were not well drawn. The last part question did act as a discriminating question. However, those candidates that had the function drawn either in their GDC or on paper recognised that at \(x = - 2\) there was a maximum and so wrote down the correct equation of the tangent at that point.

Many candidates managed to gain good marks in this question as they were able to answer the first three parts of the question. Good sketches were drawn with the required information shown on them. Very few candidates did not recognise the notation \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) but they showed that they knew how to differentiate as in (d)(i) they found the derivative to show that the gradient of \({L_1}\) was \(12\). Candidates found it difficult to find the other \(x\) for which the derivative was \(12\). However, some could draw both tangents without having found this value of \(x\) . In general, tangents were not well drawn. The last part question did act as a discriminating question. However, those candidates that had the function drawn either in their GDC or on paper recognised that at \(x = - 2\) there was a maximum and so wrote down the correct equation of the tangent at that point.

Many candidates managed to gain good marks in this question as they were able to answer the first three parts of the question. Good sketches were drawn with the required information shown on them. Very few candidates did not recognise the notation \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) but they showed that they knew how to differentiate as in (d)(i) they found the derivative to show that the gradient of \({L_1}\) was \(12\). Candidates found it difficult to find the other \(x\) for which the derivative was \(12\). However, some could draw both tangents without having found this value of \(x\) . In general, tangents were not well drawn. The last part question did act as a discriminating question. However, those candidates that had the function drawn either in their GDC or on paper recognised that at \(x = - 2\) there was a maximum and so wrote down the correct equation of the tangent at that point.

Many candidates managed to gain good marks in this question as they were able to answer the first three parts of the question. Good sketches were drawn with the required information shown on them. Very few candidates did not recognise the notation \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) but they showed that they knew how to differentiate as in (d)(i) they found the derivative to show that the gradient of \({L_1}\) was \(12\). Candidates found it difficult to find the other \(x\) for which the derivative was \(12\). However, some could draw both tangents without having found this value of \(x\) . In general, tangents were not well drawn. The last part question did act as a discriminating question. However, those candidates that had the function drawn either in their GDC or on paper recognised that at \(x = - 2\) there was a maximum and so wrote down the correct equation of the tangent at that point.

Write down the values of x where the graph of f (x) intersects the x-axis.

Write down f ′(x).

Find the value of the local maximum of y = f (x).

Let P be the point where the graph of f (x) intersects the y axis.

Write down the coordinates of P.

Let P be the point where the graph of f (x) intersects the y axis.

Find the gradient of the curve at P.

The line, L, is the tangent to the graph of f (x) at P.

Find the equation of L in the form y = mx + c.

There is a second point, Q, on the curve at which the tangent to f (x) is parallel to L.

Write down the gradient of the tangent at Q.

There is a second point, Q, on the curve at which the tangent to f (x) is parallel to L.

Calculate the x-coordinate of Q.

–1.10, 0.218, 3.13     (A1)(A1)(A1)

[3 marks]

f ′(x) = 12x² – 18x – 12     (A1)(A1)(A1)

Note: Award (A1) for each correct term and award maximum of (A1)(A1) if other terms seen.

[3 marks]

f ′(x) = 0     (M1)
x = –0.5, 2
x = –0.5     (A1)

Note: If x = –0.5 not stated, can be inferred from working below.

y = 4(–0.5)³ – 9(–0.5)² – 12(–0.5) + 3     (M1)
y = 6.25     (A1)(G3)

Note: Award (M1) for their value of x substituted into f (x).

Award (M1)(G2) if sketch shown as method. If coordinate pair given then award (M1)(A1)(M1)(A0). If coordinate pair given with no working award (G2).

[4 marks]

(0, 3)     (A1)

Note: Accept x = 0, y = 3.

[1 mark]

f ′(0) = –12     (M1)(A1)(ft)(G2)

Note: Award (M1) for substituting x = 0 into their derivative.

[2 marks]

Tangent: y = –12x + 3     (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for their gradient, (A1) for intercept = 3.

Award (A1)(A0) if y = not seen.

[2 marks]

–12     (A1)(ft)

Note: Follow through from their part (e).

[1 mark]

12x² – 18x – 12 = –12     (M1)

12x² – 18x = 0     (M1)

x = 1.5, 0

At Q x = 1.5     (A1)(ft)(G2)

Note: Award (M1)(G2) for 12x² – 18x – 12 = –12 followed by x = 1.5.

Follow through from their part (g).

[3 marks]

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

the length of BD;

the size of angle DAB;

the area of triangle ABD;

the area of quadrilateral ABCD;

the length of AP;

the length of the fence, BP.

\(({\text{BD}} = ){\text{ }}\sqrt {{{95}^2} + {{40}^2}} \)    (M1)

Note:     Award (M1) for correct substitution into Pythagoras’ theorem.

\( = 103{\text{ }}({\text{m}}){\text{ }}\left( {103.077 \ldots ,{\text{ }}25\sqrt {17} } \right)\)    (A1)(G2)

[2 marks]

\(\cos {\rm{B\hat AD}} = \frac{{{{105}^2} + {{70}^2} - {{(103.077 \ldots )}^2}}}{{2 \times 105 \times 70}}\)     (M1)(A1)(ft)

Note:     Award (M1) for substitution into cosine rule, (A1)(ft) for their correct substitutions. Follow through from part (a).

\(({\rm{B\hat AD}}) = 68.9^\circ {\text{ }}(68.8663 \ldots )\)    (A1)(ft)(G2)

Note:     If their 103 used, the answer is \(68.7995 \ldots \)

[3 marks]

\(({\text{Area of ABD}} = )\frac{1}{2} \times 105 \times 70 \times \sin (68.8663 \ldots )\)    (M1)(A1)(ft)

Notes:     Award (M1) for substitution into the trig form of the area of a triangle formula.

Award (A1)(ft) for their correct substitutions.

Follow through from part (b).

If 68.8° is used the area \( = 3426.28 \ldots {\text{ }}{{\text{m}}^2}\).

\( = 3430{\text{ }}{{\text{m}}^2}{\text{ }}(3427.82 \ldots )\)    (A1)(ft)(G2)

[3 marks]

\({\text{area of ABCD}} = \frac{1}{2} \times 40 \times 95 + 3427.82 \ldots \)    (M1)

Note:     Award (M1) for correctly substituted area of triangle formula added to their answer to part (c).

\( = 5330{\text{ }}{{\text{m}}^2}{\text{ }}(5327.83 \ldots )\)    (A1)(ft)(G2)

[2 marks]

\(\frac{1}{2} \times 105 \times {\text{AP}} \times \sin (68.8663 \ldots ) = 0.5 \times 5327.82 \ldots \)    (M1)(M1)

Notes:     Award (M1) for the correct substitution into triangle formula.

Award (M1) for equating their triangle area to half their part (d).

\(({\text{AP}} = ){\text{ }}54.4{\text{ }}({\text{m}}){\text{ }}(54.4000 \ldots )\)    (A1)(ft)(G2)

Notes:     Follow through from parts (b) and (d).

[3 marks]

\({\text{B}}{{\text{P}}^2} = {105^2} + {(54.4000 \ldots )^2} - 2 \times 105 \times (54.4000 \ldots ) \times \cos (68.8663 \ldots )\)    (M1)(A1)(ft)

Notes:     Award (M1) for substituted cosine rule formula.

Award (A1)(ft) for their correct substitutions. Accept the exact fraction \(\frac{{53}}{{147}}\) in place of \(\cos (68.8663 \ldots )\).

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

\(({\text{BP}} = ){\text{ }}99.3{\text{ }}({\text{m}}){\text{ }}(99.3252 \ldots )\)    (A1)(ft)(G2)

Notes:     If 54.4 and \(\cos (68.9)\) are used the answer is \(99.3567 \ldots \)

[3 marks]

Find the size of BĈA.

Calculate the length of AD.

The farmer wants to build a fence around ABD.

Calculate the total length of the fence.

The farmer wants to build a fence around ABD.

The farmer pays 802.50 USD for the fence. Find the cost per metre.

Calculate the area of the triangle ABD.

A layer of earth 3 cm thick is removed from ABD. Find the volume removed in cubic metres.

\(\frac{{\sin {\text{BCA}}}}{{35}} = \frac{{\sin 105^\circ }}{{80}}\)     (M1)(A1)

Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.

\({\text{B}}{\operatorname{\hat C}}{\text{A}} = 25.0^{\circ}\)     (A1)(G2)

[3 marks]

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

Length BD = 40 m     (A1)

Angle ABC = 180° − 105° − 25° = 50°     (A1)(ft)

Note: (ft) from their answer to (a).

AD² = 35² + 40² − (2 × 35 × 40 × cos 50°)     (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1)(ft) for correct substitutions.

(UP)     AD = 32.0 m     (A1)(ft)(G3)

Notes: If 80 is used for BD award at most (A0)(A1)(ft)(M1)(A1)(ft)(A1)(ft) for an answer of 63.4 m.

If the angle ABC is incorrectly calculated in this part award at most (A1)(A0)(M1)(A1)(ft)(A1)(ft).

If angle BCA is used award at most (A1)(A0)(M1)(A0)(A0).

[5 marks]

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

length of fence = 35 + 40 + 32     (M1)

(UP)     = 107 m     (A1)(ft)(G2)

Note: (M1) for adding 35 + 40 + their (b).

[2 marks]

cost per metre \( = \frac{802.50}{107}\)     (M1)

Note: Award (M1) for dividing 802.50 by their (c).

cost per metre = 7.50 USD (7.5 USD) (USD not required)     (A1)(ft)(G2)

[2 marks]

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

Area of ABD \( = \frac{1}{2} \times 35 \times 40 \times \sin 50^\circ \)     (M1)

= 536.2311102     (A1)(ft)

(UP)     = 536 m²     (A1)(ft)(G2)

Note: Award (M1) for correct substituted formula, (A1)(ft) for correct substitution, (ft) from their value of BD and their angle ABC in (b).

[3 marks]

Volume = 0.03 × 536     (A1)(M1)

= 16.08

= 16.1     (A1)(ft)(G2)

Note: Award (A1) for 0.03, (M1) for correct formula. (ft) from their (e).

If 3 is used award at most (A0)(M1)(A0).

[3 marks]

This was a simple application of non-right angled trigonometry and most candidates answered it well. Some candidates lost marks in both parts due to the incorrect setting of the calculators. Those that did not score well overall primarily used Pythagoras.

This was a simple application of non-right angled trigonometry and most candidates answered it well. Some candidates lost marks in both parts due to the incorrect setting of the calculators. Those that did not score well overall primarily used Pythagoras.

Most candidates scored full marks, many by follow through from an incorrect part (b). The main error was using the value for BC and not BD.

Most candidates scored full marks, many by follow through from an incorrect part (b). The main error was using the value for BC and not BD.

Done well; again some candidates used the right-angled formula.

This part was poorly done; many candidates unable to convert 3 cm to 0.03 m. A significant number used the wrong formula, multiplying their answer by 1/3.

Find the length of the path BD.

Show that angle DBC is 48.7°, correct to three significant figures.

Find the area of the park.

Find the length of the path CE.

Calculate the total length of the track.

\(({\text{B}}{{\text{D}}^2} = ){\text{ }}{95^2} + {120^2} - 2 \times 95 \times 120 \times \cos 70^\circ \)     (M1)(A1)

Note:     Award (M1) for substituted cosine rule, (A1) for correct substitution.

\(({\text{BD}} = ){\text{ }}125{\text{ (m) }}\left( {125.007 \ldots {\text{ (m)}}} \right)\)     (A1)(G2)

[3 marks]

\(\frac{{\sin {\text{DBC}}}}{{100}} = \frac{{\sin 110^\circ }}{{125.007 \ldots }}\)     (M1)(A1)(ft)

Note:     Award (M1) for substituted sine rule, (A1)(ft) for correct substitution.

Follow through from their answer to part (a).

\(({\text{DBC}} = ){\text{ }}48.7384 \ldots ^\circ \)     (A1)(ft)

\(({\text{DBC}} = ){\text{ }}48.7^\circ \)     (AG)

Notes:     Award the final (A1)(ft) only if both their unrounded answer and 48.7° is seen. Follow through from their answer to part (a), only if their unrounded answer rounds to 48.7°.

[3 marks]

\(\frac{1}{2} \times 125.007 \ldots  \times 100 \times \sin 21.3^\circ  + \frac{1}{2} \times 95 \times 120 \times \sin 70^\circ \)     (A1)(M1)(M1)

Note:     Award (A1) for 21.3° (21.2615…) seen, (M1) for substitution into (at least) one area of triangle formula in the form \(\frac{1}{2}ab\sin c\), (M1) for their correct substitutions and adding the two areas.

\(7630{\text{ }}{{\text{m}}^2}{\text{ }}(7626.70 \ldots {{\text{m}}^2})\)     (A1)(ft)(G3)

Notes:     Follow through from their answers to part (a). Accept \(7620{\text{ }}{{\text{m}}^2}{\text{ }}(7622.79 \ldots {{\text{m}}^2})\) from use of 48.7384…

[4 marks]

\(({\text{CE}} = ){\text{ }}100 \times \sin 21.3^\circ \)     (M1)

\(({\text{CE}} = ){\text{ }}36.3{\text{ (m) }}\left( {36.3251 \ldots {\text{ (m)}}} \right)\)     (A1)(ft)(G2)

Note:     Follow through from their angle 21.3° in part (c). Award (M0)(A0) for halving 110° and/or assuming E is the midpoint of BD in any method seen.

OR

\({\text{area of BCD}} = \frac{1}{2}{\text{BD}} \times {\text{CE}}\)     (M1)

\(({\text{CE}} = ){\text{ }}36.3{\text{ (m) }}\left( {36.3251 \ldots {\text{ (m)}}} \right)\)     (A1)(ft)(G2)

Note:     Follow through from parts (a) and (c). Award (M0)(A0) for halving 110° and/or assuming E is the midpoint of BD in any method seen.

[2 marks]

\(\sqrt {{{100}^2} - 36.3251{ \ldots ^2}}  + 100 + 36.3251 \ldots \)     (M1)(M1)

Note:     Award (M1) for correct use of Pythagoras to find DE (or correct trigonometric equation, \(100 \times \cos 21.3\), to find DE), (M1) for the sum of 100, their DE and their CE.

\(229{\text{ (m) }}\left( {229.494 \ldots {\text{ (m)}}} \right)\)     (A1)(ft)(G2)

Note:     Follow through from part (d). Use of 3 sf values gives an answer of \(230{\text{ (m) }}\left( {229.5{\text{ (m)}}} \right)\).

[3 marks]

Calculate the total length of the course.

It is estimated that the fastest boat in the race can travel at an average speed of \(1.5\;{\text{m}}\,{{\text{s}}^{ - 1}}\).

Calculate an estimate of the winning time of the race. Give your answer to the nearest minute.

It is estimated that the fastest boat in the race can travel at an average speed of \(1.5\;{\text{m}}\,{{\text{s}}^{ - 1}}\).

Find the size of angle \({\text{ACB}}\).

To comply with safety regulations, the area inside the triangular course must be kept clear of other boats, and the shortest distance from \({\text{B}}\) to \({\text{AC}}\) must be greater than \(375\) metres.

Calculate the area that must be kept clear of boats.

To comply with safety regulations, the area inside the triangular course must be kept clear of other boats, and the shortest distance from \({\text{B}}\) to \({\text{AC}}\) must be greater than \(375\) metres.

Determine, giving a reason, whether the course complies with the safety regulations.

The race is filmed from a helicopter, \({\text{H}}\), which is flying vertically above point \({\text{A}}\).

The angle of elevation of \({\text{H}}\) from \({\text{B}}\) is \(15^\circ\).

Calculate the vertical height, \({\text{AH}}\), of the helicopter above \({\text{A}}\).

The race is filmed from a helicopter, \({\text{H}}\), which is flying vertically above point \({\text{A}}\).

The angle of elevation of \({\text{H}}\) from \({\text{B}}\) is \(15^\circ\).

Calculate the maximum possible distance from the helicopter to a boat on the course.

\({\text{A}}{{\text{C}}^2} = {700^2} + {900^2} - 2 \times 700 \times 900 \times \cos 110^\circ \)     (M1)(A1)

\({\text{AC}} = 1315.65 \ldots \)     (A1)(G2)

length of course \( = 2920{\text{ (m)}}\;\;\;(2915.65 \ldots {\text{ m)}}\)     (A1)

Notes: Award (M1) for substitution into cosine rule formula, (A1) for correct substitution, (A1) for correct answer.

Award (G3) for \(2920\;\;\;(2915.65 \ldots )\) seen without working.

The final (A1) is awarded for adding \(900\) and \(700\) to their \({\text{AC}}\) irrespective of working seen.

\(\frac{{2915.65}}{{1.5}}\)     (M1)

Note: Award (M1) for their length of course divided by \(1.5\).

Follow through from part (a).

\( = 1943.76 \ldots {\text{ (seconds)}}\)     (A1)(ft)

\( = 32{\text{ (minutes)}}\)     (A1)(ft)(G2)

Notes: Award the final (A1) for correct conversion of their answer in seconds to minutes, correct to the nearest minute.

Follow through from part (a).

\(\frac{{700}}{{\sin {\text{ACB}}}} = \frac{{1315.65 \ldots }}{{\sin 110^\circ }}\)     (M1)(A1)(ft)

OR

\(\cos {\text{ACB}} = \frac{{{{900}^2} + 1315.65{ \ldots ^2} - {{700}^2}}}{{2 \times 900 \times 1315.65 \ldots }}\)     (M1)(A1)(ft)

\({\text{ACB}} = 30.0^\circ \;\;\;(29.9979 \ldots ^\circ )\)     (A1)(ft)(G2)

Notes: Award (M1) for substitution into sine rule or cosine rule formula, (A1) for their correct substitution, (A1) for correct answer.

Accept \(29.9^\circ\) for sine rule and \(29.8^\circ\) for cosine rule from use of correct three significant figure values. Follow through from their answer to (a).

\(\frac{1}{2} \times 700 \times 900 \times \sin 110^\circ \)     (M1)(A1)

Note: Accept \(\frac{1}{2} \times {\text{their AC}} \times {\text{900}} \times {\text{sin(their ACB)}}\). Follow through from parts (a) and (c).

\( = 296000{\text{ }}{{\text{m}}^2}\;\;\;(296003{\text{ }}{{\text{m}}^2})\)     (A1)(G2)

Notes: Award (M1) for substitution into area of triangle formula, (A1) for correct substitution, (A1) for correct answer.

Award (G1) if \(296000\) is seen without units or working.

\(\sin 29.9979 \ldots  = \frac{{{\text{distance}}}}{{900}}\)     (M1)

\({\text{(distance}} = ){\text{ }}450{\text{ (m)}}\;\;\;{\text{(449.971}} \ldots {\text{)}}\)     (A1)(ft)(G2)

Note: Follow through from part (c).

OR

\(\frac{1}{2} \times {\text{distance}} \times 1315.65 \ldots  = 296003\)     (M1)

\(({\text{distance}} = ){\text{ }}450{\text{ (m)}}\;\;\;{\text{(449.971}} \ldots {\text{)}}\)     (A1)(ft)(G2)

Note: Follow through from part (a) and part (d).

\(450\) is greater than \(375\), thus the course complies with the safety regulations     (R1)

Notes:  A comparison of their area from (d) and the area resulting from the use of \(375\) as the perpendicular distance is a valid approach and should be given full credit. Similarly a comparison of angle \({\text{ACB}}\) and \({\sin ^{ - 1}}\left( {\frac{{375}}{{900}}} \right)\) should be given full credit.

Award (R0) for correct answer without any working seen. Award (R1)(ft) for a justified reason consistent with their working.

Do not award (M0)(A0)(R1).

\(\tan 15^\circ  = \frac{{{\text{AH}}}}{{700}}\)     (M1)

Note: Award (M1) for correct substitution into trig formula.

\({\text{AH}} = 188{\text{ (m)}}\;\;\;(187.564 \ldots )\)     (A1)(ft)(G2)

\({\text{H}}{{\text{C}}^2} = 187.564{ \ldots ^2} + 1315.65{ \ldots ^2}\)     (M1)(A1)

Note: Award (M1) for substitution into Pythagoras, (A1) for their \(1315.65{ \ldots}\) and their \(187.564{ \ldots}\) correctly substituted in formula.

\({\text{HC}} = 1330 \ldots {\text{ (m)}}\;\;\;(1328.95 \ldots )\)     (A1)(ft)(G2)

Note: Follow through from their answer to parts (a) and (f).

Most candidates were able to recognize and use the cosine rule correctly in part (a) and then to complete part (b) – though perhaps not giving the answer to the correct level of accuracy. It is expected that candidates can use “distance = speed x time” without the formula being given. The work involving sine rule was less successful, though correct responses were given by the great majority and the area of the course was again successfully completed by most candidates. A common error throughout these parts was the use of the total length of the course. A more fundamental error was the halving of the angle and/or the base in calculations – this error has been seen in a number of sessions and perhaps needs more emphasis.

Most candidates were able to recognize and use the cosine rule correctly in part (a) and then to complete part (b) – though perhaps not giving the answer to the correct level of accuracy. It is expected that candidates can use “distance = speed x time” without the formula being given. The work involving sine rule was less successful, though correct responses were given by the great majority and the area of the course was again successfully completed by most candidates. A common error throughout these parts was the use of the total length of the course. A more fundamental error was the halving of the angle and/or the base in calculations – this error has been seen in a number of sessions and perhaps needs more emphasis.

Most candidates were able to recognize and use the cosine rule correctly in part (a) and then to complete part (b) – though perhaps not giving the answer to the correct level of accuracy. It is expected that candidates can use “distance = speed x time” without the formula being given. The work involving sine rule was less successful, though correct responses were given by the great majority and the area of the course was again successfully completed by most candidates. A common error throughout these parts was the use of the total length of the course. A more fundamental error was the halving of the angle and/or the base in calculations – this error has been seen in a number of sessions and perhaps needs more emphasis.

Most candidates were able to recognize and use the cosine rule correctly in part (a) and then to complete part (b) – though perhaps not giving the answer to the correct level of accuracy. It is expected that candidates can use “distance = speed x time” without the formula being given. The work involving sine rule was less successful, though correct responses were given by the great majority and the area of the course was again successfully completed by most candidates. A common error throughout these parts was the use of the total length of the course. A more fundamental error was the halving of the angle and/or the base in calculations – this error has been seen in a number of sessions and perhaps needs more emphasis.

In part (e), unless evidence was presented, reasoning marks did not accrue; the interpretative nature of this part was a significant discriminator in determining the quality of a response.

There were many instances of parts (f) and (g) being left blank and angle of elevation is still not well understood. Again, the interpretative nature of part (g) – even when part (f) was correct – caused difficulties

There were many instances of parts (f) and (g) being left blank and angle of elevation is still not well understood. Again, the interpretative nature of part (g) – even when part (f) was correct – caused difficulties

Write down the size of angle AQB.

Calculate the length of AQ.

Calculate the length of AC.

Show that the length of CQ is 50.37 m, correct to 4 significant figures.

Find the size of the angle AQC.

Calculate the total area of the glass needed to construct

(i) the two rectangular faces of the greenhouse;

(ii) the two triangular faces of the greenhouse.

The cost of one square metre of glass used to construct the greenhouse is 4.80 USD.

Calculate the cost of glass to make the greenhouse. Give your answer correct to the nearest 100 USD.

110°     (A1)

\(\frac{{AQ}}{{\sin 35^\circ }} = \frac{{10}}{{\sin 110^\circ }}\)     (M1)(A1)

Note: Award (M1) for substituted sine rule formula, (A1) for their correct substitutions.

OR

\(AQ = \frac{5}{{\cos 35^\circ }}\)     (A1)(M1)

Note: Award (A1) for 5 seen, (M1) for correctly substituted trigonometric ratio.

\(AQ = 6.10\) (6.10387...)     (A1)(ft)(G2)

Notes: Follow through from their answer to part (a).

\(AC^2 = 10^2 + 50^2\)     (M1)

Note: Award (M1) for correctly substituted Pythagoras formula.

\(AC = 51.0 (\sqrt{2600}, 50.9901...)\)     (A1)(G2)

\(QC^2 = (6.10387...)^2 + (50)^2\)     (M1)

Note: Award (M1) for correctly substituted Pythagoras formula.

\(QC = 50.3711...\)     (A1)

\(= 50.37\)     (AG)

Note: Both the unrounded and rounded answers must be seen to award (A1).

   If 6.10 is used then 50.3707... is the unrounded answer.

   For an incorrect follow through from part (b) award a maximum of (M1)(A0) – the given answer must be reached to award the final (A1)(AG).

\(\cos AQC = \frac{{{{(6.10387...)}^2} + {{(50.3711...)}^2} - {{(50.9901...)}^2}}}{{2(6.10387...)(50.3711...)}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted cosine rule formula, (A1)(ft) for their correct substitutions.

= 92.4°   (\({92.3753...^\circ }\))     (A1)(ft)(G2)

Notes: Follow through from their answers to parts (b), (c) and (d). Accept 92.2 if the 3 sf answers to parts (b), (c) and (d) are used.

     Accept 92.5° (\({92.4858...^\circ }\)) if the 3 sf answers to parts (b), (c) and 4 sf answers to part (d) used.

(i) \(2(50 \times 6.10387...)\)     (M1)

Note: Award (M1) for their correctly substituted rectangular area formula, the area of one rectangle is not sufficient.

= 610 m² (610.387...)     (A1)(ft)(G2)

Notes: Follow through from their answer to part (b).

    The answer is 610 m². The units are required.

(ii) Area of triangular face \( = \frac{1}{2} \times 10 \times 6.10387... \times \sin 35^\circ \)     (M1)(A1)(ft)

OR

Area of triangular face \( = \frac{1}{2} \times 6.10387... \times 6.10387... \times \sin 110^\circ \)     (M1)(A1)(ft)

\(= 17.5051...\)

Note: Award (M1) for substituted triangle area formula, (A1)(ft) for correct substitutions.

OR

(Height of triangle) \( = {(6.10387...)^2} - {5^2}\)

\(= 3.50103...\)

Area of triangular face \( = \frac{1}{2} \times 10 \times their{\text{ }} height\)

\(= 17.5051...\)

Note: Award (M1) for substituted triangle area formula, (A1)(ft) for correctly substituted area formula. If 6.1 is used, the height is 3.49428... and the area of both triangular faces 34.9 m²

Area of both triangular faces = 35.0 m² (35.0103...)     (A1)(ft)(G2)

Notes: The answer is 35.0 m². The units are required. Do not penalize if already penalized in part (f)(i). Follow through from their part (b).

(610.387... + 35.0103...) × 4.80     (M1)

= 3097.90...     (A1)(ft)

Notes: Follow through from their answers to parts (f)(i) and (f)(ii).

    Accept 3096 if the 3 sf answers to part (f) are used.

= 3100     (A1)(ft)(G2)

Notes: Follow through from their unrounded answer, irrespective of whether it is correct. Award (M1)(A2) if working is shown and 3100 seen without the unrounded answer being given.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates were able to recognize sine rule, substitute correctly and reach the required result.

Most candidates were able to recognize sine rule, substitute correctly and reach the required result. The use of Pythagoras’ theorem was also successful, the major source of error being the lack of unrounded and rounded answers in part (d).

Again, most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates were able to recognize sine rule, substitute correctly and reach the required result. Part (e) was less well answered, due in part to the triangle being in three dimensions. However, all three sides had either been asked for in previous parts or given and all that was required was a sketch of a triangle with the vertices labelled; such a diagram was never on any script and this technique should be encouraged.

Again, most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Show that the length of side ST is 1.95 m, correct to 3 significant figures.

Calculate the area of the triangle PTS.

Write down the area of the rectangle STVR.

Calculate the total surface area of the tent, including the base.

Calculate the volume of the tent.

A pole is placed from V to M, the midpoint of PS.

Find in metres,

(i) the height of the tent, TM;

(ii) the length of the pole, VM.

Calculate the angle between VM and the base of the tent.

\({\text{ST}} = \frac{{1.6}}{{\cos 35^\circ }}\)     (M1)(A1)

Note: Award (M1) for correctly substituted trig equation, (A1) for 1.6 seen.

OR

\(\frac{{{\text{ST}}}}{{\sin 35^\circ }} = \frac{{3.2}}{{\sin 110^\circ }}\)     (M1)(A1)

Note: Award (M1) for substituted sine rule equation, (A1) for correct substitutions.

ST = 1.95323...     (A1)

= 1.95 (m)     (AG)

Notes: Both unrounded and rounded answer must be seen for final (A1) to be awarded.

\(\frac{1}{2} \times 3.2 \times 1.95323... \times \sin 35^\circ \) or \(\frac{1}{2} \times 1.95323... \times 1.95323... \times \sin 110^\circ \)     (M1)(A1)

Note: Award (M1) for substituted area formula, (A1) for correct substitutions. Do not award follow through marks.

= 1.79 m² (1.79253...m²)     (A1)(G2)

Notes: The answer is 1.79 m², units are required. Accept 1.78955... from using 1.95.

OR

\(\frac{1}{2} \times 3.2 \times 1.12033...\)     (A1) (M1)

Note: Award (A1) for the correct value for TM (1.12033...) OR correct expression for TM (i.e. 1.6tan35°, \(\sqrt {{{(1.95323...)}^2} - {{1.6}^2}} \)), (M1) for correctly substituted formula for triangle area.

= 1.79 m² (1.79253...m²)     (A1)(G2)

Notes: The answer is 1.79 m², units are required. Accept 1.78 m² from using 1.95.

9.18 m² (9.18022 m²)     (A1)(G1)

Notes: The answer is 9.18 m², units are required. Do not penalize if lack of units was already penalized in (b). Do not award follow through marks here. Accept 9.17 m² (9.165 m²) from using 1.95.

\(2 \times 1.79253... + 2 \times 9.18022... + 4.7 \times 3.2\)     (M1)(A1)(ft)

Note: Award (M1) for addition of three products, (A1)(ft) for three correct products.

= 37.0 m² (36.9855...m²)     (A1)(ft)(G2)

Notes: The answer is 37.0 m², units are required. Accept 36.98 m² from using 3sf answers. Follow through from their answers to (b) and (c). Do not penalize if lack of units was penalized earlier in the question.

\(1.79253... \times 4.7\)     (M1)

Note: Award (M1) for their correctly substituted volume formula.

= 8.42 m³ (8.42489...m³)     (A1)(ft)(G2)

Notes: The answer is 8.42 m³, units are required. Accept 8.41 m³ from use of 1.79. An answer of 8.35, from use of TM = 1.11, will receive follow-through marks if working is shown. Follow through from their answer to part (b). Do not penalize if lack of units was penalized earlier in the question.

(i) \({\text{TM}} = 1.6\tan {35^\circ }\)     (M1)

Notes: Award (M1) for their correct substitution in trig ratio.

OR

\({\text{TM}} = \sqrt {{{(1.95323...)}^2} - {{1.6}^2}} \)     (M1)

Note: Award (M1) for correct substitution in Pythagoras’ theorem.

OR

\(\frac{{3.2 \times {\text{TM}}}}{2} = 1.79253...\)     (M1)

Note: Award (M1) for their correct substitution in area of triangle formula.

= 1.12 (m) (1.12033...)     (A1)(ft)(G2)

Notes: Follow through from their answer to (b) if area of triangle is used. Accept 1.11 (1.11467) from use of ST = 1.95.

(ii) \({\text{VM}} = \sqrt {{{1.12033...}^2} + {{4.7}^2}} \)     (M1)

Note: Award (M1) for their correct substitution in Pythagoras’ theorem.

= 4.83 (m) (4.83168 )     (A1)(ft)(G2)

Notes: Follow through from (f)(i).

\({\sin ^{ - 1}}\left( {\frac{{1.12033...}}{{4.83168...}}} \right)\)     (M1)

OR

\({\cos^{ - 1}}\left( {\frac{{4.7}}{{4.83168...}}} \right)\)     (M1)

OR

\({\tan^{ - 1}}\left( {\frac{{1.12033...}}{{4.7}}} \right)\)     (M1)

Note: Award (M1) for correctly substituted trig equation.

OR

\({\cos ^{ - 1}}\left( {\frac{{{{4.7}^2} + {{(4.83168...)}^2} - {{(1.12033...)}^2}}}{{2 \times 4.7 \times 4.83168...}}} \right)\)     (M1)

Note: Award (M1) for correctly substituted cosine formula.

= 13.4° (13.4073...)     (A1)(ft)(G2)

Notes: Accept 13.3°. Follow through from part (f).

A playground, when viewed from above, is shaped like a quadrilateral, \({\text{ABCD}}\), where \({\text{AB}} = 21.8\,{\text{m}}\) and \({\text{CD}} = 11\,{\text{m}}\) . Three of the internal angles have been measured and angle \({\text{ABC}} = 47^\circ \) , angle \({\text{ACB}} = 63^\circ \) and angle \({\text{CAD}} = 30^\circ \) . This information is represented in the following diagram.

Calculate the distance \({\text{AC}}\).

Calculate angle \({\text{ADC}}\).

There is a tree at \({\text{C}}\), perpendicular to the ground. The angle of elevation to the top of the tree from \({\text{D}}\) is \(35^\circ \).

Calculate the height of the tree.

Chavi estimates that the height of the tree is \(6\,{\text{m}}\).

Calculate the percentage error in Chavi’s estimate.

Chavi is celebrating her birthday with her friends on the playground. Her mother brings a \(2\,\,{\text{litre}}\) bottle of orange juice to share among them. She also brings cone-shaped paper cups.

Each cup has a vertical height of \(10\,{\text{cm}}\) and the top of the cup has a diameter of \(6\,{\text{cm}}\).

Calculate the volume of one paper cup.

Calculate the maximum number of cups that can be completely filled with the \(2\,\,{\text{litre}}\) bottle of orange juice.

\(\frac{{21.8}}{{\sin 63^\circ }} = \frac{{{\text{AC}}}}{{\sin 47^\circ }}\)          (M1)(A1)

Note: Award (M1) for substitution into the sine rule formula, (A1) for correct substitution.

\(({\text{AC}} = )\,\,17.9\,({\text{m}})\,\,\,(17.8938...\,({\text{m}}))\)           (A1)(G2)

\(\frac{{11}}{{\sin 30}} = \frac{{17.8938...}}{{\sin {\text{ADC}}}}\)          (M1)(A1)(ft)

Note: Award (M1) for substitution into the sine rule formula, (A1) for correct substitution.

\(\left( {{\text{Angle ADC}} = } \right)\,\,\,54.4^\circ \,\,\,(54.4250...^\circ )\)          (A1)(ft)(G2)

Note: Accept \(54.5\,\,(54.4527...)\) or \(126\,\,(125.547...)\) from using their \(3\) sf answer.
Follow through from part (a). Accept \(125.575...\)

\(11 \times \tan 35^\circ \)  (or equivalent)          (M1)

Note: Award (M1) for correct substitution into trigonometric ratio.

\(7.70\,({\text{m}})\,\,\,(7.70228...\,({\text{m}}))\)          (A1)(G2)

\(\left| {\frac{{6 - 7.70228...}}{{7.70228...}}} \right| \times 100\,\% \)          (M1)

Note: Award (M1) for correct substitution into the percentage error formula.

OR

\(100 - \left| {\frac{{6 \times 100}}{{7.70228...}}} \right|\)          (M1)

Note: Award (M1) for the alternative method.

\(22.1\,(\% )\,\,\,(22.1009...\,(\% ))\)          (A1)(ft)(G2)

Note: Award at most (M1)(A0) for a final answer that is negative. Follow through from part (c).

\(\frac{1}{3}\pi  \times {3^2} \times 10\)          (A1)(M1)

Note: Award (A1) for \(3\) seen, (M1) for their correct substitution into volume of a cone formula.

\(94.2\,{\text{c}}{{\text{m}}^3}\,\,\,(30\pi \,{\text{c}}{{\text{m}}^3},\,\,94.2477...\,{\text{c}}{{\text{m}}^3})\)          (A1)(G3)

Note: The answer is \(94.2\,{\text{c}}{{\text{m}}^3}\), units are required. Award at most (A0)(M1)(A0) if an incorrect value for \(r\) is used.

\(\frac{{2000}}{{94.2477...}}\) OR \(\frac{2}{{0.0942477...}}\)          (M1)(M1)

Note: Award (M1) for correct conversion (litres to \({\text{c}}{{\text{m}}^3}\) or \({\text{c}}{{\text{m}}^3}\) to litres), (M1) for dividing by their part (e) (or their converted part (e)).

\(21\)          (A1)(ft)(G2)

Note: The final (A1) is not awarded if the final answer is not an integer. Follow through from part (e), but only if the answer is rounded down.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.

Write down the length of VA in metres.

Sketch the triangle VCA showing clearly the length of VC and the size of angle VCA.

Show that the height of the pyramid is 18.0 metres correct to 3 significant figures.

Calculate the length of AC in metres.

Show that the length of BC is 19.1 metres correct to 3 significant figures.

Calculate the volume of the tower.

To calculate the cost of air conditioning, engineers must estimate the weight of air in the tower. They estimate that 90 % of the volume of the tower is occupied by air and they know that 1 m³ of air weighs 1.2 kg.

Calculate the weight of air in the tower.

22.5 (m)     (A1)

[1 mark]

     (A1)

[1 mark]

h = 22.5     sin 53.1°     (M1)
= 17.99     (A1)
= 18.0     (AG)

Note: Unrounded answer must be seen for (A1) to be awarded.

Accept 18 as (AG).

[2 marks]

\({\text{AC}} = 2\sqrt {{{22.5}^2} - {{17.99...}^2}} \)     (M1)(M1)

Note: Award (M1) for multiplying by 2, (M1) for correct substitution into formula.

OR

AC = 2(22.5)cos53.1°     (M1)(M1)

Notes: Award (M1) for correct use of cosine trig ratio, (M1) for multiplying by 2.

OR

AC² = 22.5² + 22.5² – 2(22.5)(22.5) cos73.8°     (M1)(A1)

Note: Award (M1) for substituted cosine formula, (A1) for correct substitutions.

OR

\(\frac{{{\text{AC}}}}{{\sin (73.8^\circ )}} = \frac{{22.5}}{{\sin (53.1^\circ )}}\)     (M1)(A1)

Note: Award (M1) for substituted sine formula, (A1) for correct substitutions.

AC = 27.0     (A1)(G2)

[3 marks]

\({\text{BC}} = \sqrt {{{13.5}^2} + {{13.5}^2}} \)     (M1)

= 19.09     (A1)

= 19.1     (AG)

OR

x² + x² = 27²     (M1)

2x² = 27²     (A1)

BC = 19.09…     (A1)

= 19.1     (AG)

Notes: Unrounded answer must be seen for (A1) to be awarded.

[2 marks]

Volume = Pyramid + Cuboid

\( = \frac{1}{3}(18)({19.1^2}) + (108)({19.1^2})\)     (A1)(M1)(M1)

Note: Award (A1) for 108, the height of the cuboid seen. Award (M1) for correctly substituted volume of cuboid and (M1) for correctly substituted volume of pyramid.

= \(41\,588\)     (41\(\,\)553 if 2(13.5²) is used)

= \(41\,600\) m³     (A1)(ft)(G3)

[4 marks]

Weight of air = \(41\,600 \times 1.2 \times 0.9\)     (M1)(M1)

= \(44\,900{\text{ kg}}\)     (A1)(ft)(G2)

Note: Award (M1) for their part (e) × 1.2, (M1) for × 0.9.

Award at most (M1)(M1)(A0) if the volume of the cuboid is used.

[3 marks]

This question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in three dimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of the question were independent of those preceding them; many candidates did not reach these parts, though for some, these were the only parts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.

This question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in three dimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of the question were independent of those preceding them; many candidates did not reach these parts, though for some, these were the only parts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.

This question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in three dimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of the question were independent of those preceding them; many candidates did not reach these parts, though for some, these were the only parts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.

This question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in three dimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of the question were independent of those preceding them; many candidates did not reach these parts, though for some, these were the only parts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.

This question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in three dimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of the question were independent of those preceding them; many candidates did not reach these parts, though for some, these were the only parts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.

This question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in three dimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of the question were independent of those preceding them; many candidates did not reach these parts, though for some, these were the only parts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.

This question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in three dimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of the question were independent of those preceding them; many candidates did not reach these parts, though for some, these were the only parts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.

Write down the size of angle ABC.

Calculate the length of AC.

Calculate the area of the field ABC.

N is the point on AB such that CN is perpendicular to AB. M is the midpoint of CN.

Calculate the length of NM.

A goat is attached to one end of a rope of length 7 m. The other end of the rope is attached to the point M.

Decide whether the goat can reach point P, the midpoint of CB. Justify your answer.

\({\text{Angle ABC}} = {50^ \circ }\)     (A1)

[1 mark]

\(\frac{{{\text{AC}}}}{{\sin {{50}^ \circ }}} = \frac{{25}}{{\sin {{55}^ \circ }}}\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for correct substitution. Follow through from their angle ABC.

\({\text{AC}} = 23.4{\text{ m}}\)     (A1)(ft)(G2)

[3 marks]

\({\text{Area of }}\Delta {\text{ ABC}} = \frac{1}{2} \times 23.379 \ldots  \times 25 \times \sin {75^ \circ }\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for correct substitution. Follow through from their AC.

OR

\({\text{Area of triangle ABC}} = \frac{{29.479 \ldots  \times 19.151 \ldots }}{2}\)     (A1)(ft)(M1)

Note: (A1)(ft) for correct values of AB (29.479…) and CN (19.151…). Follow through from their (a) and /or (b). Award (M1) for substitution of their values of AB and CN into the correct formula.

\({\text{Area of }}\Delta {\text{ ABC}} = 282{\text{ }}{{\text{m}}^2}\)     (A1)(ft)(G2)

Note: Accept \(283{\text{ }}{{\text{m}}^2}\) if \(23.4\) is used.

[3 marks]

\({\text{NM}} = \frac{{25 \times \sin {{50}^ \circ }}}{2}\)     (M1)(M1)

Note: Award (M1) for \({25 \times \sin {{50}^ \circ }}\) or equivalent for the length of CN. (M1) for dividing their CN by \(2\).

\({\text{NM}} = 9.58{\text{ m}}\)     (A1)(ft)(G2)

Note: Follow through from their angle ABC.

Notes: Premature rounding of CN leads to the answers \(9.60\) or \(9.6\). Award at most (M1)(M1)(A0) if working seen. Do not penalize with (AP). CN may be found in (c).

Note: The working for this part of the question may be in part (b).

[3 marks]

\({\text{Angle NCB}} = {40^ \circ }\) seen     (A1)(ft)

Note: Follow through from their (a).

From triangle MCP:

\({\text{M}}{{\text{P}}^2} = {(9.5756 \ldots )^2} + {12.5^2} - 2 \times 9.5756 \ldots  \times 12.5 \times \cos ({40^ \circ })\)     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(ft)(G3)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their d). Award (G3) for correct value of MP seen without working.

OR

From right triangle MCP

\({\text{CP}} = 12.5{\text{ m}}\) seen     (A1)

\({\text{M}}{{\text{P}}^2} = {(12.5)^2} - {(9.575 \ldots )^2}\)     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(G3)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their (d). Award (G3) for correct value of MP seen without working.

OR

From right triangle MCP

\({\text{Angle MCP}} = {40^ \circ }\) seen     (A1)(ft)

\(\frac{{{\text{MP}}}}{{12.5}} = \sin ({40^ \circ })\) or equivalent     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(G3)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their (a). Award (G3) for correct value of MP seen without working.

The goat cannot reach point P as \({\text{MP}} > 7{\text{ m}}\) .     (A1)(ft)

Note: Award (A1)(ft) only if their value of MP is compared to \(7{\text{ m}}\), and conclusion is stated.

[5 marks]

Many candidates assumed incorrectly that the triangle ABC is isosceles or/and that CN is an angle bisector, and those assumptions led them to use incorrect methods. Wherever those assumptions were made first, all or most of the marks were lost in that specific part of Question 4. There were provisions in the mark scheme to follow through in subsequent parts. Most candidates at least attempted parts a), b) and c). Some candidates incorrectly used an area formula for a right triangle in c) and lost all marks. Many candidates lost a mark for premature rounding in d). Part e) proved to be especially difficult for the candidates. Here many candidates offered guesses instead of sound mathematical reasoning.

Many candidates assumed incorrectly that the triangle ABC is isosceles or/and that CN is an angle bisector, and those assumptions led them to use incorrect methods. Wherever those assumptions were made first, all or most of the marks were lost in that specific part of Question 4. There were provisions in the mark scheme to follow through in subsequent parts. Most candidates at least attempted parts a), b) and c). Some candidates incorrectly used an area formula for a right triangle in c) and lost all marks. Many candidates lost a mark for premature rounding in d). Part e) proved to be especially difficult for the candidates. Here many candidates offered guesses instead of sound mathematical reasoning.

Many candidates assumed incorrectly that the triangle ABC is isosceles or/and that CN is an angle bisector, and those assumptions led them to use incorrect methods. Wherever those assumptions were made first, all or most of the marks were lost in that specific part of Question 4. There were provisions in the mark scheme to follow through in subsequent parts. Most candidates at least attempted parts a), b) and c). Some candidates incorrectly used an area formula for a right triangle in c) and lost all marks. Many candidates lost a mark for premature rounding in d). Part e) proved to be especially difficult for the candidates. Here many candidates offered guesses instead of sound mathematical reasoning.

Many candidates assumed incorrectly that the triangle ABC is isosceles or/and that CN is an angle bisector, and those assumptions led them to use incorrect methods. Wherever those assumptions were made first, all or most of the marks were lost in that specific part of Question 4. There were provisions in the mark scheme to follow through in subsequent parts. Most candidates at least attempted parts a), b) and c). Some candidates incorrectly used an area formula for a right triangle in c) and lost all marks. Many candidates lost a mark for premature rounding in d). Part e) proved to be especially difficult for the candidates. Here many candidates offered guesses instead of sound mathematical reasoning.

Many candidates assumed incorrectly that the triangle ABC is isosceles or/and that CN is an angle bisector, and those assumptions led them to use incorrect methods. Wherever those assumptions were made first, all or most of the marks were lost in that specific part of Question 4. There were provisions in the mark scheme to follow through in subsequent parts. Most candidates at least attempted parts a), b) and c). Some candidates incorrectly used an area formula for a right triangle in c) and lost all marks. Many candidates lost a mark for premature rounding in d). Part e) proved to be especially difficult for the candidates. Here many candidates offered guesses instead of sound mathematical reasoning.

Find the gradient of AB.

L₁ passes through C and is parallel to AB.

Write down the y-intercept of L₁.

L₂ passes through A and is perpendicular to AB.

Write down the equation of L₂. Give your answer in the form ax + by + d = 0 where a, b and d \( \in \mathbb{Z}\).

Write down the coordinates of the point D, the intersection of L₁ and L₂.

There is a point R on L₁ such that ABRD is a rectangle.

Write down the coordinates of R.

The distance between A and D is \(\sqrt {45} \).

(i) Find the distance between D and R .

(ii) Find the area of the triangle BDR .

\(\frac{{9 - 1}}{{0 - ( - 4)}}\)     (M1)

= 2    (A1)(G2)

Notes: Award (M1) for correct substitution into the gradient formula.

[2 marks]

–6     (A1)

Note: Accept (0, –6) .

[1 mark]

\(y =  - \frac{1}{2}x - 1\)  (or equivalent)     (A1)(ft)(A1)

Notes: Award (A1)(ft) for gradient, (A1) for correct y-intercept. Follow through from their gradient in (a).

x + 2y + 2 = 0     (A1)(ft)

Notes: Award (A1)(ft) from their gradient and their y-intercept. Accept any multiple of this equation with integer coefficients.

OR

\(y - 1 =  - \frac{1}{2}(x + 4)\) (or equivalent)     (A1)(ft)(A1)

Note: Award (A1)(ft) for gradient, (A1) for any point on the line correctly substituted in equation.

x + 2y + 2 = 0     (A1)(ft)

Notes: Award (A1)(ft) from their equation. Accept any multiple of this equation with integer coefficients.

[3 marks]

D(2, –2) or x = 2, y = –2     (A1)

Note: Award (A0) if brackets not present.

[1 mark]

R(6, 6) or x = 6, y = 6     (A1)(A1)

Note: Award at most (A0)(A1)(ft) if brackets not present and absence of brackets has not already been penalised in part (d).

[2 marks]

(i) \({\text{DR}} = \sqrt {{8^2} + {4^2}} \)     (M1)

\({\text{DR}} = \sqrt {80} \)  (8.94)     (A1)(ft)(G2)

Note: Award (M1) for correct substitution into the distance formula. Follow through from their D and R.

(ii) \({\text{Area}} = \frac{{\sqrt {80}  \times \sqrt {45} }}{2}\)     (M1)

= 30 (30.0)     (A1)(ft)(G2)

Note: Award (M1) for correct substitution in the area of triangle formula. Follow through from their answer to part (f) (i).

[4 marks]

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.

This question was in general well answered. In part (a) the gradient of the line AB was correctly found although some candidates did not substitute well in the gradient formula and found answers as \(\frac{1}{2}\) or –2. Also some students read B as (0, 8) instead of (0, 9). In part (b) many students again did not make good use of time as they found the equation of the line instead of just extending it to find the y - intercept. The equation of L₂ in (c) was correctly found in the form y = mx + c but very few students were able to rearrange the equation in the form ax + by + d = 0 where a, b, d \( \in \mathbb{Z}\). In (d) many candidates found the coordinates of point D by solving simultaneous equations which led again to a waste of time. The last two parts of this question were well done by those students that attempted them.

Sketch the graph of the function \(f(x)\), for \( - 10 \leqslant x \leqslant 10\) . Indicating clearly the axis intercepts and any asymptotes.

Write down the equation of the vertical asymptote.

On the same diagram as part (a) sketch the graph of \(g(x) = x + 0.5\) .

Using your graphical display calculator write down the coordinates of one of the points of intersection on the graphs of \(f\) and \(g\), giving your answer correct to five decimal places.

Write down the gradient of the line \(g(x) = x + 0.5\) .

The line \(L\) passes through the point with coordinates \(( - 2{\text{, }} - 3)\) and is perpendicular to the line \(g(x)\) . Find the equation of \(L\).

     (A6)

Notes: (A1) for labels and some idea of scale.
(A1) for \(x\)-intercept seen, (A1) for \(y\)-intercept seen in roughly the correct places (coordinates not required).
(A1) for vertical asymptote seen, (A1) for horizontal asymptote seen in roughly the correct places (equations of the lines not required).
(A1) for correct general shape.

[6 marks]

\(x = - 4\)     (A1)(A1)(ft)

Note: (A1) for \(x =\), (A1)(ft) for \( - 4\).

[2 marks]

     (A1)(A1)

Note: (A1) for correct axis intercepts, (A1) for straight line

[2 marks]

\(( - 2.85078{\text{, }} - 2.35078)\) OR \((0.35078{\text{, }}0.85078)\)     (G1)(G1)(A1)(ft)

Notes: (A1) for \(x\)-coordinate, (A1) for \(y\)-coordinate, (A1)(ft) for correct accuracy. Brackets required. If brackets not used award (G1)(G0)(A1)(ft).
Accept \(x = - 2.85078\), \(y = - 2.35078\) or \(x = 0.35078\), \(y = 0.85078\).

[3 marks]

\({\text{gradient}} = 1\)     (A1)

[1 mark]

\({\text{gradient of perpendicular}} = - 1\)     (A1)(ft)

(can be implied in the next step)

\(y = mx + c\)

\( - 3 = - 1 \times - 2 + c\)     (M1)

\(c = - 5\)

\(y = - x - 5\)     (A1)(ft)(G2)

OR

\(y + 3 = - (x + 2)\)     (M1)(A1)(ft)(G2)

Note: Award (G2) for correct answer with no working at all but (A1)(G1) if the gradient is mentioned as \( - 1\) then correct answer with no further working.

[3 marks]

This was not very well done. The graph was often correct but was so small that it was difficult to check if axes intercepts were correct or not. Often the vertical asymptote looked as if it were joined to the rest of the graph. Very few of the candidates put a scale and/or labels on their axes.

Reasonably well done. Some put \(y = - 4\) while others omitted the minus sign.

Fairly well done – but once again too small to check the axes intercepts properly. Also, many candidates did not appear to have a ruler to draw the straight line.

Well done.

Most could find the gradient of the line.

Many forgot to find the gradient of the perpendicular line. Others had problems with the equation of a line in general.

The Great Pyramid of Giza in Egypt is a right pyramid with a square base. The pyramid is made of solid stone. The sides of the base are \(230\,{\text{m}}\) long. The diagram below represents this pyramid, labelled \({\text{VABCD}}\).

\({\text{V}}\) is the vertex of the pyramid. \({\text{O}}\) is the centre of the base, \({\text{ABCD}}\) . \({\text{M}}\) is the midpoint of \({\text{AB}}\). Angle \({\text{ABV}} = 58.3^\circ \) .

Show that the length of \({\text{VM}}\) is \(186\) metres, correct to three significant figures.

Calculate the height of the pyramid, \({\text{VO}}\) .

Find the volume of the pyramid.

Write down your answer to part (c) in the form \(a \times {10^k}\)  where \(1 \leqslant a < 10\) and \(k \in \mathbb{Z}\) .

Ahmad is a tour guide at the Great Pyramid of Giza. He claims that the amount of stone used to build the pyramid could build a wall \(5\) metres high and \(1\) metre wide stretching from Paris to Amsterdam, which are \(430\,{\text{km}}\) apart.

Determine whether Ahmad’s claim is correct. Give a reason.

Ahmad and his friends like to sit in the pyramid’s shadow, \({\text{ABW}}\), to cool down.
At mid-afternoon, \({\text{BW}} = 160\,{\text{m}}\)  and angle \({\text{ABW}} = 15^\circ .\)

i)     Calculate the length of \({\text{AW}}\) at mid-afternoon.

ii)    Calculate the area of the shadow, \({\text{ABW}}\), at mid-afternoon.

\(\tan \,(58.3) = \frac{{{\text{VM}}}}{{115}}\)   OR \(115 \times \tan \,(58.3^\circ )\)              (A1)(M1)

Note: Award (A1) for \(115\,\,\left( {ie\,\frac{{230}}{2}} \right)\)   seen, (M1) for correct substitution into trig formula.

\(\left( {{\text{VM}} = } \right)\,\,186.200\,({\text{m}})\)        (A1)

\(\left( {{\text{VM}} = } \right)\,\,186\,({\text{m}})\)           (AG)

Note: Both the rounded and unrounded answer must be seen for the final (A1) to be awarded.

\({\text{V}}{{\text{O}}^2} + {115^2} = {186^2}\)  OR \(\sqrt {{{186}^2} - {{115}^2}} \)       (M1)

Note: Award (M1) for correct substitution into Pythagoras formula. Accept alternative methods.

\({\text{(VO}} = )\,\,146\,({\text{m}})\,\,(146.188...)\)       (A1)(G2)

Note: Use of full calculator display for \({\text{VM}}\) gives \(146.443...\,{\text{(m)}}\).

Units are required in part (c)

\(\frac{1}{3}({230^2} \times 146.188...)\)       (M1)

Note: Award (M1) for correct substitution in volume formula. Follow through from part (b).

\( = 2\,580\,000\,{{\text{m}}^3}\,\,(2\,577\,785...\,{{\text{m}}^3})\)       (A1)(ft)(G2)

Note: The answer is \(2\,580\,000\,{{\text{m}}^3}\) , the units are required. Use of \({\text{OV}} = 146.442\) gives  \(2582271...\,{{\text{m}}^3}\)

Use of \({\text{OV}} = 146\) gives  \(2574466...\,{{\text{m}}^3}.\)

\(2.58 \times {10^6}\,({{\text{m}}^3})\)       (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for \(2.58\) and (A1)(ft) for \( \times {10^6}.\,\)

Award (A0)(A0) for answers of the type: \(2.58 \times {10^5}\,({{\text{m}}^3}).\)

Follow through from part (c).

the volume of a wall would be \(430\,000 \times 5 \times 1\)       (M1)

Note: Award (M1) for correct substitution into volume formula.

\(2150000\,({{\text{m}}^3})\)       (A1)(G2)

which is less than the volume of the pyramid       (R1)(ft)

Ahmad is correct.       (A1)(ft)

OR

the length of the wall would be \(\frac{{{\text{their part (c)}}}}{{5 \times 1 \times 1000}}\)       (M1)

Note: Award (M1) for dividing their part (c) by \(5000.\)

\(516\,({\text{km}})\)          (A1)(ft)(G2)

which is more than the distance from Paris to Amsterdam       (R1)(ft)

Ahmad is correct.       (A1)(ft)

Note: Do not award final (A1) without an explicit comparison. Follow through from part (c) or part (d). Award (R1) for reasoning that is consistent with their working in part (e); comparing two volumes, or comparing two lengths.

Units are required in part (f)(ii).

i)     \({\text{A}}{{\text{W}}^2} = {160^2} + {230^2} - 2 \times 160 \times 230 \times \cos \,(15^\circ )\)       (M1)(A1)

Note: Award (M1) for substitution into cosine rule formula, (A1) for correct substitution.

\({\text{AW}} = 86.1\,({\text{m}})\,\,\,(86.0689...)\)       (A1)(G2)

Note: Award (M0)(A0)(A0) if \({\text{BAW}}\) or \({\text{AWB}}\)  is considered to be a right angled triangle.

ii)    \({\text{area}} = \frac{1}{2} \times 230 \times 160 \times \sin \,(15^\circ )\)         (M1)(A1)

Note: Award (M1) for substitution into area formula, (A1) for correct substitutions.

\( = 4760\,{{\text{m}}^2}\,\,\,(4762.27...\,{{\text{m}}^2})\)       (A1)(G2)

Note: The answer is \(4760\,{{\text{m}}^2}\) , the units are required.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.

Calculate the size of angle ADB.

Find the area of triangle ADB.

Calculate the size of angle BCD.

Show that the triangle ABC is not right angled.

\(\cos {\text{ADB}} = \frac{{{{12}^2} + {{20}^2} - {{28}^2}}}{{2(12)(20)}}\)     (M1)(A1)

Notes: Award (M1) for substituted cosine rule formula, (A1) for correct substitutions.

\(\angle {\text{ADB}} = 120\)     (A1)(G2)

[3 marks]

\({\text{Area}} = \frac{{(12)(20)\sin {{120}^ \circ }}}{2}\)     (M1)(A1)(ft)

Notes: Award (M1) for substituted area formula, (A1)(ft) for their correct substitutions.

\( = 104{\text{ c}}{{\text{m}}^2}\) (\(103.923 \ldots {\text{ c}}{{\text{m}}^2}\))     (A1)(ft)(G2)

Note: The final answer is \(104{\text{ c}}{{\text{m}}^2}\) , the units are required. Accept \(100{\text{ c}}{{\text{m}}^2}\) .

[3 marks]

\(\frac{{\sin {\text{BCD}}}}{{12}} = \frac{{\sin {{60}^ \circ }}}{{13}}\)     (A1)(ft)(M1)(A1)

Note: Award (A1)(ft) for their 60 seen, (M1) for substituted sine rule formula, (A1) for correct substitutions.

\({\text{BCD}} = {53.1^ \circ }\) (\(53.0736 \ldots \))     (A1)(G3)

Note: Accept \(53\), do not accept \(50\) or \(53.0\).

[4 marks]

Using triangle ABC

\(\frac{{\sin {\text{BAC}}}}{{13}} = \frac{{\sin {{53.1}^ \circ }}}{{28}}\)     (M1)(A1)(ft)

OR

Using triangle ABD

\(\frac{{\sin {\text{BAD}}}}{{12}} = \frac{{\sin {{120}^ \circ }}}{{28}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted sine rule formula (one of the above), (A1)(ft) for their correct substitutions. Follow through from (a) or (c) as appropriate.

\({\text{BAC}} = {\text{BAD}} = {21.8^ \circ }\) (\(21.7867 \ldots \))     (A1)(ft)(G2)

Notes: Accept \(22\), do not accept \(20\) or \(21.7\). Accept equivalent methods, for example cosine rule.

\({180^ \circ } - ({53.1^ \circ } + {21.8^ \circ }) \ne {90^ \circ }\), hence triangle ABC is not right angled     (R1)(AG)

OR

\(\frac{{{\text{CD}}}}{{\sin {{66.9}^ \circ }}} = \frac{{13}}{{\sin {{60}^ \circ }}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted sine rule formula, (A1)(ft) for their correct substitutions. Follow through from (a) and (c).

\({\text{CD}} = 13.8{\text{ }}(13.8075 \ldots )\)     (A1)(ft)

\({13^3} + {28^2} \ne {33.8^2}\), hence triangle ABC is not right angled.     (R1)(ft)(AG)

Note: The complete statement is required for the final (R1) to be awarded.

[4 marks]

The vast majority of candidates scored very well on this question. Those who did not attempted it using the trigonometry associated with right angled triangles. There were few problems with the use of radians and part (d), which was expected to prove challenging, was successfully overcome by more than half of the candidature. Problems arose mainly because of a lack of clarity in identifying the correct triangle.

The vast majority of candidates scored very well on this question. Those who did not attempted it using the trigonometry associated with right angled triangles. There were few problems with the use of radians and part (d), which was expected to prove challenging, was successfully overcome by more than half of the candidature. Problems arose mainly because of a lack of clarity in identifying the correct triangle.

The vast majority of candidates scored very well on this question. Those who did not attempted it using the trigonometry associated with right angled triangles. There were few problems with the use of radians and part (d), which was expected to prove challenging, was successfully overcome by more than half of the candidature. Problems arose mainly because of a lack of clarity in identifying the correct triangle.

The vast majority of candidates scored very well on this question. Those who did not attempted it using the trigonometry associated with right angled triangles. There were few problems with the use of radians and part (d), which was expected to prove challenging, was successfully overcome by more than half of the candidature. Problems arose mainly because of a lack of clarity in identifying the correct triangle.

Calculate the size of angle ACB.

Calculate the area of the building’s footprint, ABC.

Calculate the volume of the office block.

To stabilize the structure, a steel beam must be made that runs from point C to point Q.

Calculate the length of CQ.

Calculate the angle CQ makes with BC.

\(\cos {\text{ACB}} = \frac{{{{30}^2} + {{50}^2} - {{70}^2}}}{{2 \times 30 \times 50}}\)     (M1)(A1)

Note: Award (M1) for substituted cosine rule formula, (A1) for correct substitution.

\({\text{ACB}} = {120^ \circ }\)     (A1)(G2)

\({\text{Area of triangle ABC}} = \frac{{30(50)\sin {{120}^ \circ }}}{2}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted area formula, (A1)(ft) for correct substitution.

\( = 650{\text{ }}{{\text{m}}^2}\) \((649.519 \ldots {\text{ }}{{\text{m}}^2})\)     (A1)(ft)(G2)

Notes: The answer is \(650{\text{ }}{{\text{m}}^2}\) ; the units are required. Follow through from their answer in part (a).

\({\text{Volume}} = 649.519 \ldots \times 120\)     (M1)
\( = 77900{\text{ }}{{\text{m}}^3}\) (\(77942.2 \ldots {\text{ }}{{\text{m}}^3}\))     (A1)(G2)

Note: The answer is \(77900{\text{ }}{{\text{m}}^3}\) ; the units are required. Do not penalise lack of units if already penalized in part (b). Accept \(78000{\text{ }}{{\text{m}}^3}\) from use of 3sf answer \(650{\text{ }}{{\text{m}}^2}\) from part (b).

\({\text{C}}{{\text{Q}}^2} = {50^2} + {120^2}\)     (M1)
\({\text{CQ}} = 130{\text{ (m)}}\)     (A1)(G2)

Note: The units are not required.

\(\tan {\text{QCB}} = \frac{{120}}{{50}}\)     (M1)

Note: Award (M1) for correct substituted trig formula.

\({\text{QCB}} = {67.4^ \circ }\) (\(67.3801 \ldots \))     (A1)(G2)

Note: Accept equivalent methods.

Find the length of the road AB.

Find the size of the angle CAB.

Village D is halfway between A and B. A new road perpendicular to AB and passing through D is built. Let T be the point where this road cuts AC. This information is shown in the diagram below.

Write down the distance from A to D.

Show that the distance from D to T is 2.06 km correct to three significant figures.

A bus starts and ends its journey at A taking the route AD to DT to TA.

Find the total distance for this journey.

The average speed of the bus while it is moving on the road is 70 km h^–1. The bus stops for 5 minutes at each of D and T .

Estimate the time taken by the bus to complete its journey. Give your answer correct to the nearest minute.

AB² = 10² + 8² – 2 × 10 × 8 × cos150°     (M1)(A1)

AB = 17.4 km     (A1)(G2)

Note: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer.

[3 marks]

\(\frac{8}{{\sin {\text{C}}\hat {\rm A}{\text{B}}}} = \frac{{17.4}}{{\sin 150^\circ }}\)     (M1)(A1)

\({\text{C}}\hat {\rm A}{\text{B}} = 13.3^\circ \)     (A1)(ft)(G2)

Notes: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer. Follow through from their answer to part (a).

[3 marks]

AD = 8.70 km (8.7 km)     (A1)(ft)

Note: Follow through from their answer to part (a).

[1 mark]

DT = tan (13.29...°) × 8.697... = 2.0550...     (M1)(A1)

= 2.06     (AG)

Notes: Award (M1) for correct substitution in the correct formula, award (A1) for the unrounded answer seen. If 2.06 not seen award at most (M1)(AO).

[2 marks]

\(\sqrt {{{8.70}^2} + {{2.06}^2}}  + 8.70 + 2.06\)     (A1)(M1)

= 19.7 km     (A1)(ft)(G2)

Note: Award (A1) for AT, (M1) for adding the three sides of the triangle ADT, (A1)(ft) for answer. Follow through from their answer to part (c).

[3 marks]

\(\frac{{19.7}}{{70}} \times 60 + 10\)     (M1)(M1)

= 26.9     (A1)(ft)

Note: Award (M1) for time on road in minutes, (M1) for adding 10, (A1)(ft) for unrounded answer. Follow through from their answer to (e).

= 27  (nearest minute)     (A1)(ft)(G3)

Note: Award (A1)(ft) for their unrounded answer given to the nearest minute.

[4 marks]

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

Calculate the area of triangle EAD.

Calculate the total volume of the barn.

Calculate the length of MN.

Calculate the length of AE.

Show that Farmer Brown is incorrect.

Calculate the total length of metal required for one support.

(Area of EAD =) \(\frac{1}{2} \times 10 \times 7 \times {\text{sin}}15\)    (M1)(A1)

Note: Award (M1) for substitution into area of a triangle formula, (A1) for correct substitution. Award (M0)(A0)(A0) if EAD or AED is considered to be a right-angled triangle.

= 9.06 m²  (9.05866… m²)     (A1)   (G3)

[3 marks]

(10 × 5 × 16) + (9.05866… × 16)     (M1)(M1)

Note: Award (M1) for correct substitution into volume of a cuboid, (M1) for adding the correctly substituted volume of their triangular prism.

= 945 m³  (944.938… m³)     (A1)(ft)  (G3)

Note: Follow through from part (a).

[3 marks]

\(\frac{{{\text{MN}}}}{5} = \,\,\,{\text{sin}}15\)     (M1)

Note: Award (M1) for correct substitution into trigonometric equation.

(MN =) 1.29(m) (1.29409… (m))     (A1) (G2)

[2 marks]

(AE² =) 10² + 7² − 2 × 10 × 7 × cos 15     (M1)(A1)

Note: Award (M1) for substitution into cosine rule formula, and (A1) for correct substitution.

(AE =) 3.71(m)  (3.71084… (m))     (A1) (G2)

[3 marks]

ND² = 5² − (1.29409…)²     (M1)

Note: Award (M1) for correct substitution into Pythagoras theorem.

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

OR

\(\frac{{1.29409 \ldots }}{{{\text{ND}}}} = {\text{tan}}\,15^\circ \)     (M1)

Note: Award (M1) for correct substitution into tangent.

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

OR

\(\frac{{{\text{ND}}}}{5} = {\text{cos }}15^\circ \)     (M1)

Note: Award (M1) for correct substitution into cosine.

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

OR

ND² = 1.29409…² + 5² − 2 × 1.29409… × 5 × cos 75°     (M1)

Note: Award (M1) for correct substitution into cosine rule.

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

4.82962… ≠ 3.5   (ND ≠ 3.5)     (R1)(ft)

OR

4.82962… ≠ 2.17038…   (ND ≠ NE)     (R1)(ft)

(hence Farmer Brown is incorrect)

Note: Do not award (M0)(A0)(R1)(ft). Award (M0)(A0)(R0) for a correct conclusion without any working seen.

[3 marks]

(EM² =) 1.29409…² + (7 − 4.82962…)²     (M1)

Note: Award (M1) for their correct substitution into Pythagoras theorem.

OR

(EM² =) 5² + 7² − 2 × 5 × 7 × cos 15     (M1)

Note: Award (M1) for correct substitution into cosine rule formula.

(EM =) 2.53(m) (2.52689...(m))     (A1)(ft) (G2)(ft)

Note: Follow through from parts (c), (d) and (e).

(Total length =) 2.52689… + 3.71084… + 1.29409… +10 + 7     (M1)

Note: Award (M1) for adding their EM, their parts (c) and (d), and 10 and 7.

= 24.5 (m)    (24.5318… (m))     (A1)(ft) (G4)

Note: Follow through from parts (c) and (d).

[4 marks]

Express the volume of the parcel in terms of \(l\) and \(w\).

Show that \(l = \frac{{150}}{w}\).

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Show that the length of string, \(S\) cm, required to tie up the parcel can be written as

\[S = 40 + 4w + \frac{{300}}{w},{\text{ }}0 < w \leqslant 20.\]

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Draw the graph of \(S\) for \(0 < w \leqslant 20\) and \(0 < S \leqslant 500\), clearly showing the local minimum point. Use a scale of \(2\) cm to represent \(5\) units on the horizontal axis \(w\) (cm), and a scale of \(2\) cm to represent \(100\) units on the vertical axis \(S\) (cm).

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find \(\frac{{{\text{d}}S}}{{{\text{d}}w}}\).

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find the value of \(w\) for which \(S\) is a minimum.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Write down the value, \(l\), of the parcel for which the length of string is a minimum.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find the minimum length of string required to tie up the parcel.

\(20lw\)   OR   \(V = 20lw\)     (A1)

[1 mark]

\(3000 = 20lw\)     (M1)

Note: Award (M1) for equating their answer to part (a) to \(3000\).

\(l = \frac{{3000}}{{20w}}\)     (M1)

Note: Award (M1) for rearranging equation to make \(l\) subject of the formula. The above equation must be seen to award (M1).

OR

\(150 = lw\)     (M1)

Note: Award (M1) for division by \(20\) on both sides. The above equation must be seen to award (M1).

\(l = \frac{{150}}{w}\)     (AG)

[2 marks]

\(S = 2l + 4w + 2(20)\)     (M1)

Note: Award (M1) for setting up a correct expression for \(S\).

\(2\left( {\frac{{150}}{w}} \right) + 4w + 2(20)\)     (M1)

Notes: Award (M1) for correct substitution into the expression for \(S\). The above expression must be seen to award (M1).

\( = 40 + 4w + \frac{{300}}{w}\)     (AG)

[2 marks]

     (A1)(A1)(A1)(A1)

Note: Award (A1) for correct scales, window and labels on axes, (A1) for approximately correct shape, (A1) for minimum point in approximately correct position, (A1) for asymptotic behaviour at \(w = 0\).

     Axes must be drawn with a ruler and labeled \(w\) and \(S\).

     For a smooth curve (with approximately correct shape) there should be one continuous thin line, no part of which is straight and no (one-to-many) mappings of \(w\).

     The \(S\)-axis must be an asymptote. The curve must not touch the \(S\)-axis nor must the curve approach the asymptote then deviate away later.

[4 marks]

\(4 - \frac{{300}}{{{w^2}}}\)     (A1)(A1)(A1)

Notes: Award (A1) for \(4\), (A1) for \(-300\), (A1) for \(\frac{1}{{{w^2}}}\) or \({w^{ - 2}}\). If extra terms present, award at most (A1)(A1)(A0).

[3 marks]

\(4 - \frac{{300}}{{{w^2}}} = 0\)   OR   \(\frac{{300}}{{{w^2}}} = 4\)   OR   \(\frac{{{\text{d}}S}}{{{\text{d}}w}} = 0\)     (M1)

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

\(w = 8.66{\text{ }}\left( {\sqrt {75} ,{\text{ 8.66025}} \ldots } \right)\)     (A1)(ft)(G2)

Note: Follow through from their answer to part (e).

[2 marks]

\(17.3 \left( {\frac{{150}}{{\sqrt {75} }},{\text{ 17.3205}} \ldots } \right)\)     (A1)(ft)

Note: Follow through from their answer to part (f).

[1 mark]

\(40 + 4\sqrt {75}  + \frac{{300}}{{\sqrt {75} }}\)     (M1)

Note: Award (M1) for substitution of their answer to part (f) into the expression for \(S\).

\( = 110{\text{ (cm) }}\left( {40 + 40\sqrt 3 ,{\text{ 109.282}} \ldots } \right)\)     (A1)(ft)(G2)

Note: Do not accept \(109\).

     Follow through from their answers to parts (f) and (g).

[2 marks]

Find the size of angle \(ABC\).

Find the area of the triangular field.

\({\text{M}}\) is the midpoint of \({\text{AC}}\).

Find the length of \({\text{BM}}\).

A vertical mobile phone mast, \({\text{TB}}\), is built next to the field with its base at \({\text{B}}\). The angle of elevation of \({\text{T}}\) from \({\text{M}}\) is \(63.4^\circ \). \({\text{N}}\) is the midpoint of the mast.

Calculate the angle of elevation of \({\text{N}}\) from \({\text{M}}\).

\(\frac{{70}}{{\sin 78}} = \frac{{50}}{{\sin {\rm{A\hat BC}}}}\)     (M1)(A1)

Note: Award (M1) for substituted sine rule, (A1) for correct substitution.

\({\rm{A\hat BC}} = 44.3^\circ \) (\(44.3209...\))     (A1)(G3)

Note: If radians are used the answer is \(0.375918...\), award at most (M1)(A1)(A0).

[3 marks]

\({\text{area }}\Delta {\text{ABC}} = \frac{1}{2} \times 70 \times 50 \times \sin (57.6790 \ldots )\)     (A1)(M1)(A1)(ft)
 

Notes: Award (A1)(ft) for their \(57.6790…\) seen, (M1) for substituted area formula, (A1)(ft) for correct substitution.

     Follow through from part (a).

\( = 1480{\text{ }}{{\text{m}}^2}\)  \((1478.86 \ldots )\)     (A1)(ft)(G3)

Notes: The answer is \(1480{\text{ }}{{\text{m}}^2}\), units are required. \(1479.20…\) if 3 sf used.

     If radians are used the answer is \(1554.11 \ldots {{\text{m}}^2}\), award (A1)(ft)(M1)(A1)(ft)(A1)(ft)(G3).

[4 marks]

\({\text{B}}{{\text{M}}^2} = {70^2} + {25^2} - 2 \times 70 \times 25 \times \cos (57.6790 \ldots )\)     (M1)(A1)(ft)

Notes: Award (M1) for substituted cosine rule, (A1)(ft) for correct substitution. Follow through from their angle in part (b).

\({\text{BM}} = 60.4{\text{ (m)}}\)   \((60.4457 \ldots )\)     (A1)(ft)(G2)

Notes: If the 3 sf answer is used the answer is \(60.5{\text{ }}({\text{m}})\).

     If radians are used the answer is \(62.5757… {\text{ }} ({\text{m}})\), award (M1)(A1)(ft)(A1)(ft)(G2).

[3 marks]

\(\tan 63.4^\circ  = \frac{{{\text{TB}}}}{{60.4457 \ldots }}\)     (M1)

Note: Award (M1) for their correctly substituted trig equation.

\({\text{TB}} = 120.707 \ldots \)     (A1)(ft)

Notes: Follow through from part (c). If 3 sf answers are used throughout, \({\text{TB}} = 120.815 \ldots \)

If \({\text{TB}} = 120.707 \ldots \) is seen without working, award (A2).

\(\tan {\rm{N\hat MB = }}\frac{{\left( {\frac{{120.707 \ldots }}{2}} \right)}}{{60.4457 \ldots }}\)     (A1)(ft)(M1)

Notes: Award (A1)(ft) for their \({\text{TB}}\) divided by \(2\) seen, (M1) for their correctly substituted trig equation.

     Follow through from part (c) and within part (d).

\({\rm{N\hat MB = 45.0^\circ }}\)   \({\text{(44.9563}} \ldots {\text{)}}\)     (A1)(ft)(G3)

Notes:     If 3 sf are used throughout, answer is \(45^\circ \).

     If radians are used the answer is \(0.308958…\), and if full working is shown, award at most (M1)(A1)(ft)(A1)(ft)(M1)(A0).

     If no working is shown for radians answer, award (G2).

OR

\(\tan {\rm{N\hat MB}} = \frac{{{\text{NB}}}}{{{\text{BM}}}}\)     (M1)

\(\tan 63.4^\circ  = \frac{{2 \times {\text{NB}}}}{{{\text{BM}}}}\)     (A1)(M1)

Note: Award (A1) for \(2 \times {\text{NB}}\) seen.

\(\tan {\rm{N\hat MB}} = \frac{1}{2}\tan 63.4^\circ \)     (M1)

\({\rm{N\hat MB}} = 45.0^\circ \)   \((44.9563 \ldots )\)     (A1)(G3)

Notes: If radians are used the answer is \(0.308958…\), and if full working is shown, award at most (M1)(A1)(M1)(M1)(A0). If no working is shown for radians answer, award (G2).

[5 marks]

Write down a formula for \(A\), the surface area to be coated.

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

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

Show that \(A = \pi {r^2}\frac{{1\,000\,000}}{r}\).

Show that \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\).

Find \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\).

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

Find the value of this minimum area.

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

\((A = ){\text{ }}\pi {r^2} + 2\pi rh\)    (A1)(A1)

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

[2 marks]

\(500\,000\)    (A1)

Notes:     Units not required.

[1 mark]

\(500\,000 = \pi {r^2}h\)    (A1)(ft)

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

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

[1 mark]

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).

[2 marks]

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).

[2 marks]

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

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

[3 marks]

\(2\pi r - \frac{{1\,000\,000}}{{{r^2}}} = 0\)    (M1)

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

\({r^3} = \frac{{1\,000\,000}}{{2\pi }}\) OR \(r = \sqrt[3]{{\frac{{1\,000\,000}}{{2\pi }}}}\)     (M1)

Note:     Award (M1) for isolating \(r\).

OR

sketch of derivative function     (M1)

with its zero indicated     (M1)

\((r = ){\text{ }}54.2{\text{ }}({\text{cm}}){\text{ }}(54.1926 \ldots )\)    (A1)(ft)(G2)

[3 marks]

\(\pi {(54.1926 \ldots )^2} + \frac{{1\,000\,000}}{{(54.1926 \ldots )}}\)    (M1)

Note:     Award (M1) for correct substitution of their part (f) into the given equation.

\( = 27\,700{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}(27\,679.0 \ldots )\)    (A1)(ft)(G2)

[2 marks]

\(\frac{{27\,679.0 \ldots }}{{2000}}\)    (M1)

Note:     Award (M1) for dividing their part (g) by 2000.

\( = 13.8395 \ldots \)    (A1)(ft)

Notes:     Follow through from part (g).

14 (cans)     (A1)(ft)(G3)

Notes:     Final (A1) awarded for rounding up their \(13.8395 \ldots \) to the next integer.

[3 marks]

Show that A lies on \({L_1}\).

Find the coordinates of M.

Find the length of AC.

Show that the equation of \({L_2}\) is \(2y + x - 19 = 0\).

Find the coordinates of D.

Write down the length of MD correct to five significant figures.

Find the area of ABCD.

\(2 \times 4 - 1 - 7 = 0\) (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]

Calculate the volume of the cylinder in cm³. Give your answer correct to two decimal places.

The cylinder is used for storing tennis balls. Each ball has a radius of 3.25 cm.

Calculate how many balls Jenny can fit in the cylinder if it is filled to the top.

(i) Jenny fills the cylinder with the number of balls found in part (b) and puts the lid on. Calculate the volume of air inside the cylinder in the spaces between the tennis balls.

(ii) Convert your answer to (c) (i) into cubic metres.

(i) Find the value of angle \({\text{B}}\hat {\rm T}{\text{L}}\).

(ii) Use triangle BTL to calculate the sloping distance BT from the base, B to the top, T of the tower.

Calculate the vertical height TG of the top of the tower.

Leonardo now walks to point M, a distance 200 m from B on the opposite side of the tower. Calculate the distance from M to the top of the tower at T.

\(\pi \times 3.25^2 \times 39\)     (M1)(A1)

(= 1294.1398)

Answer 1294.14 (cm³)(2dp)     (A1)(ft)(G2)

(UP) not applicable in this part due to wording of question. (M1) is for substituting appropriate numbers from the problem into the correct formula, even if the units are mixed up. (A1) is for correct substitutions or correct answer with more than 2dp in cubic centimetres seen. Award (G1) for answer to > 2dp with no working and no attempt to correct to 2dp. Award (M1)(A0)(A1)(ft) for \(\pi  \times {32.5^2} \times 39{\text{ c}}{{\text{m}}^3}\) (= 129413.9824) = 129413.98

Use of \(\pi = \frac{22}{7}\) or 3.142 etc is premature rounding and is awarded at most (M1)(A1)(A0) or (M1)(A0)(A1)(ft) depending on whether the intermediate value is seen or not. For all other incorrect substitutions, award (M1)(A0) and only follow through the 2 dp correction if the intermediate answer to more decimal places is seen. Answer given as a multiple of \(\pi\) is awarded at most (M1)(A1)(A0). As usual, an unsubstituted formula followed by correct answer only receives the G marks.

[3 marks]

39/6.5 = 6     (A1)

[1 mark]

Unit penalty (UP) is applicable where indicated in the left hand column.

(UP) (i) Volume of one ball is \(\frac{4}{3} \pi \times 3.25^3 {\text{ cm}}^3\)     (M1)

\({\text{Volume of air}} = \pi  \times {3.25^2} \times 39 - 6 \times \frac{4}{3}\pi  \times {3.25^3} = 431{\text{ c}}{{\text{m}}^3}\)     (M1)(A1)(ft)(G2)

Award first (M1) for substituted volume of sphere formula or for numerical value of sphere volume seen (143.79… or 45.77… \( \times \pi\)). Award second (M1) for subtracting candidate’s sphere volume multiplied by their answer to (b). Follow through from parts (a) and (b) only, but negative or zero answer is always awarded (A0)(ft)

(UP) (ii) 0.000431m³ or  4.31×10⁻⁴ m³     (A1)(ft)

[4 marks]

Unit penalty (UP) is applicable where indicated in the left hand column.

(i) \({\text{Angle B}}\widehat {\text{T}}{\text{L}} = 180 - 80 - 26.5\) or \(180 - 90 - 26.5 - 10\)     (M1)

\(= 73.5^\circ\)     (A1)(G2)

(ii) \(\frac{{BT}}{{\sin (26.5^\circ )}} = \frac{{120}}{{\sin (73.5^\circ )}}\)     (M1)(A1)(ft)

(UP) BT = 55.8 m (3sf)     (A1)(ft)

[5 marks]

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but
negative lengths are always awarded (A0)(ft).

The answers are (all 3sf)

(ii)(a)     – 124 m (A0)(ft)

(ii)(b)     123 m (A0)

(ii)(c)     313 m (A0)

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but negative lengths are always awarded (A0)(ft)

Unit penalty (UP) is applicable where indicated in the left hand column.

TG = 55.8sin(80°) or 55.8cos(10°)     (M1)