Find \(\cos \theta \) .

Find \(\tan 2\theta \) .

METHOD 1

evidence of choosing \({\sin ^2}\theta  + {\cos ^2}\theta  = 1\)     (M1)

correct working     (A1)

e.g. \({\cos ^2}\theta  = \frac{9}{{13}}\) , \(\cos \theta  =  \pm \frac{3}{{\sqrt {13} }}\) , \(\cos \theta  = \sqrt {\frac{9}{{13}}} \)

\(\cos \theta  = - \frac{3}{{\sqrt {13} }}\)    A1     N2

Note: If no working shown, award N1 for \(\frac{3}{{\sqrt {13} }}\) .

METHOD 2

approach involving Pythagoras’ theorem     (M1)

e.g. \({2^2} + {x^2} = 13\) ,

finding third side equals 3     (A1)

\(\cos \theta  = - \frac{3}{{\sqrt {13} }}\)     A1     N2

Note: If no working shown, award N1 for \(\frac{3}{{\sqrt {13} }}\) .

[3 marks]

correct substitution into \(\sin 2\theta \) (seen anywhere)     (A1)

e.g. \(2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { - \frac{3}{{\sqrt {13} }}} \right)\)

correct substitution into \(\cos 2\theta \) (seen anywhere)     (A1)

e.g. \({\left( { - \frac{3}{{\sqrt {13} }}} \right)^2} - {\left( {\frac{2}{{\sqrt {13} }}} \right)^2}\) , \(2{\left( { - \frac{3}{{\sqrt {13} }}} \right)^2} - 1\) , \(1 - 2{\left( {\frac{2}{{\sqrt {13} }}} \right)^2}\)

valid attempt to find \(\tan 2\theta \)     (M1)

e.g. \(\frac{{2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { - \frac{3}{{\sqrt {13} }}} \right)}}{{{{\left( { - \frac{3}{{\sqrt {13} }}} \right)}^2} - {{\left( {\frac{2}{{\sqrt {13} }}} \right)}^2}}}\) , \(\frac{{2\left( { - \frac{2}{3}} \right)}}{{1 - {{\left( { - \frac{2}{3}} \right)}^2}}}\)

correct working     A1

e.g. \(\frac{{\frac{{(2)(2)( - 3)}}{{13}}}}{{\frac{9}{{13}} - \frac{4}{{13}}}}\) , \(\frac{{ - \frac{{12}}{{{{\left( {\sqrt {13} } \right)}^2}}}}}{{\frac{{18}}{{13}} - 1}}\) , \(\frac{{ - \frac{{12}}{{13}}}}{{\frac{5}{{13}}}}\)

\(\tan 2\theta  = - \frac{{12}}{5}\)     A1     N4

Note: If students find answers for \(\cos \theta \) which are not in the range \([ - 1{\text{, }}1]\), award full FT in (b) for correct FT working shown.

[5 marks]

While the majority of candidates knew to use the Pythagorean identity in part (a), very few remembered that the cosine of an angle in the second quadrant will have a negative value.

In part (b), many candidates incorrectly tried to calculate \(\tan 2\theta \) as \(2 \times \tan \theta \) , rather than using the double-angle identities.

Use the quotient rule to show that \(f'(x) = \frac{{ - 1}}{{{{\sin }^2}x}}\) .

Find \(f''(x)\) .

Find the value of p and of q.

Use information from the table to explain why there is a point of inflexion on the graph of f where \(x = \frac{\pi }{2}\) .

\(\frac{{\rm{d}}}{{{\rm{d}}x}}\sin x = \cos x\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}\cos x = - \sin x\) (seen anywhere)     (A1)(A1)

evidence of using the quotient rule     M1

correct substitution     A1

e.g. \(\frac{{\sin x( - \sin x) - \cos x(\cos x)}}{{{{\sin }^2}x}}\) , \(\frac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}}\)

\(f'(x) = \frac{{ - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x}}\)     A1

\(f'(x) = \frac{{ - 1}}{{{{\sin }^2}x}}\)     AG      N0

[5 marks]

METHOD 1

appropriate approach     (M1)

e.g. \(f'(x) = - {(\sin x)^{ - 2}}\)

\(f''(x) = 2({\sin ^{ - 3}}x)(\cos x)\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\)     A1A1     N3

Note: Award A1 for \(2{\sin ^{ - 3}}x\) , A1 for \(\cos x\) .

METHOD 2

derivative of \({\sin ^2}x = 2\sin x\cos x\) (seen anywhere)     A1

evidence of choosing quotient rule     (M1)

e.g. \(u = - 1\) ,  \(v = {\sin ^2}x\) , \(f'' = \frac{{{{\sin }^2}x \times 0 - ( - 1)2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\)

\(f''(x) = \frac{{2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\)     A1     N3

[3 marks]

evidence of substituting \(\frac{\pi }{2}\)     M1

e.g. \(\frac{{ - 1}}{{{{\sin }^2}\frac{\pi }{2}}}\) , \(\frac{{2\cos \frac{\pi }{2}}}{{{{\sin }^3}\frac{\pi }{2}}}\)

\(p = - 1\) ,  \(q = 0\)    A1A1     N1N1

[3 marks]

second derivative is zero, second derivative changes sign     R1R1     N2

[2 marks]

Many candidates comfortably applied the quotient rule, although some did not completely show that the Pythagorean identity achieves the numerator of the answer given. Whether changing to \( - {(\sin x)^{ - 2}}\) , or applying the quotient rule a second time, most candidates neglected the chain rule in finding the second derivative.

Whether changing to \( - {(\sin x)^{ - 2}}\) , or applying the quotient rule a second time, most candidates neglected the chain rule in finding the second derivative.

Those who knew the trigonometric ratios at \(\frac{\pi }{2}\)typically found the values of p and of q, sometimes in follow-through from an incorrect \(f''(x)\) .

Few candidates gave two reasons from the table that supported the existence of a point of inflexion. Most stated that the second derivative is zero and neglected to consider the sign change to the left and right of q. Some discussed a change of concavity, but without supporting this statement by referencing the change of sign in \(f''(x)\) , so no marks were earned.

(i)     Find the value of a.

(ii)    Show that \(b = \frac{\pi }{6}\) .

(iii)   Find the value of d.

(iv)   Write down a value for c.

The transformation P is given by a horizontal stretch of a scale factor of \(\frac{1}{2}\) , followed by a translation of \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 10}
\end{array}} \right)\) .

Let \({M'}\) be the image of M under P. Find the coordinates of \({M'}\) .

The graph of g is the image of the graph of f under P.

Find \(g(t)\) in the form \(g(t) = 7\cos B(t - c) + D\) .

The graph of g is the image of the graph of f under P.

Give a full geometric description of the transformation that maps the graph of g to the graph of f .

(i) attempt to substitute     (M1)

e.g. \(a = \frac{{29 - 15}}{2}\)

\(a = 7\) (accept \(a = - 7\) )     A1     N2

(ii) \({\text{period}} = 12\)     (A1)

\(b = \frac{{2\pi }}{{12}}\)    A1

\(b = \frac{\pi }{6}\)    AG     N0

(iii) attempt to substitute     (M1)

e.g. \(d = \frac{{29 + 15}}{2}\)

\(d = 22\)     A1     N2

(iv) \(c = 3\) (accept \(c = 9\) from \(a = - 7\) )     A1     N1

Note: Other correct values for c can be found, \(c = 3 \pm 12k\) , \(k \in \mathbb{Z}\) .

[7 marks]

stretch takes 3 to 1.5     (A1)

translation maps \((1.5{\text{, }}29)\) to \((4.5{\text{, }}19)\) (so \({M'}\) is \((4.5{\text{, }}19)\))     A1     N2

[2 marks]

\(g(t) = 7\cos \frac{\pi }{3}\left( {t - 4.5} \right) + 12\)    A1A2A1    N4

Note: Award A1 for \(\frac{\pi }{3}\) , A2 for 4.5, A1 for 12.

Other correct values for c can be found, \(c = 4.5 \pm 6k\) , \(k \in \mathbb{Z}\) .

[4 marks]

translation \(\left( {\begin{array}{*{20}{c}}
{ - 3}\\
{10}
\end{array}} \right)\)     (A1)

horizontal stretch of a scale factor of 2     (A1)

completely correct description, in correct order     A1     N3

e.g. translation \(\left( {\begin{array}{*{20}{c}}
{ - 3}\\
{10}
\end{array}} \right)\) then horizontal stretch of a scale factor of 2

[3 marks]

This question was the most difficult on the paper. Where candidates attempted this question, part (a) was answered satisfactorily.

Few answered part (b) correctly as most could not interpret the horizontal stretch.

Few answered part (b) correctly as most could not interpret the horizontal stretch. As a result, there were many who were unable to answer part (c) although follow through marks were often obtained from incorrect answers in both parts (a) and (b). The link between the answer in (b) and the value of C in part (c) was lost on all but the most attentive.

In part (d), some candidates could name the transformations required, although only a handful provided the correct order of the transformations to return the graph to its original state.

Write down the period of g.

On the diagram below, sketch the curve of g, for \(0 \le x \le 2\pi \) .

Write down the number of solutions to the equation \(g(x) = 2\) , for \(0 \le x \le 2\pi \) .

\({\text{period}} = \pi \)     A1     N1

[1 mark]

      A1A1A1     N3

Note: Award A1 for amplitude of 3, A1 for their period, A1 for a sine curve passing through \((0{\text{, }}0)\) and \((0{\text{, }}2\pi )\) .

[3 marks]

evidence of appropriate approach     (M1)

e.g. line \(y = 2\) on graph, discussion of number of solutions in the domain

4 (solutions)     A1     N2

[2 marks]

Many candidates were unable to write down the period of the function.

Many candidates were unable to write down the period of the function. However, they were often then able to go and correctly sketch the graph with the correct period.

The final part was poorly done with many candidates finding the number of zeros instead of the intersection with the line \(y = 2\) .

Show that \(f(x)\) can be expressed as \(1 + \sin 2x\) .

The graph of f is shown below for \(0 \le x \le 2\pi \) .

Let \(g(x) = 1 + \cos x\) . On the same set of axes, sketch the graph of g for \(0 \le x \le 2\pi \) .

The graph of g can be obtained from the graph of f under a horizontal stretch of scale factor p followed by a translation by the vector \(\left( {\begin{array}{*{20}{c}}
k\\
0
\end{array}} \right)\) .

Write down the value of p and a possible value of k .

attempt to expand     (M1)

e.g. \((\sin x + \cos x)(\sin x + \cos x)\) ; at least 3 terms

correct expansion     A1

e.g. \({\sin ^2}x + 2\sin x\cos x + {\cos ^2}x\)

\(f(x) = 1 + \sin 2x\)     AG     N0

[2 marks]

     A1A1     N2

Note: Award A1 for correct sinusoidal shape with period \(2\pi \) and range \([0{\text{, }}2]\), A1 for minimum in circle.

\(p = 2\) , \(k = - \frac{\pi }{2}\)     A1A1     N2

[2 marks]

Simplifying a trigonometric expression and applying identities was generally well answered in part (a), although some candidates were certainly helped by the fact that it was a "show that" question.

More candidates had difficulty with part (b) with many assuming the first graph was \(1 + \sin (x)\) and hence sketching a horizontal translation of \(\pi /2\) for the graph of g; some attempts were not even sinusoidal. While some candidates found the stretch factor p correctly or from follow-through on their own graph, very few successfully found the value and direction for the translation.

Part (c) certainly served as a discriminator between the grade 6 and 7 candidates.

Let \(f(x) = \sqrt 3 {{\rm{e}}^{2x}}\sin x + {{\rm{e}}^{2x}}\cos x\) , for \(0 \le x \le \pi \) . Solve the equation \(f(x) = 0\) .

\({{\rm{e}}^{2x}}\left( {\sqrt 3 \sin x + \cos x} \right) = 0\)     (A1)

\({{\rm{e}}^{2x}} = 0\) not possible (seen anywhere)     (A1)

simplifying

e.g. \(\sqrt 3 \sin x + \cos x = 0\) , \(\sqrt 3 \sin x =  - \cos x\) , \(\frac{{\sin x}}{{ - \cos x}} = \frac{1}{{\sqrt 3 }}\)     A1 

EITHER

\(\tan x = - \frac{1}{{\sqrt 3 }}\)     A1

\(x = \frac{{5\pi }}{6}\)     A2     N4

OR

sketch of \(30^\circ \) , \(60^\circ \) , \(90^\circ \) triangle with sides \(1\), \(2\), \(\sqrt 3 \)     A1

work leading to \(x = \frac{{5\pi }}{6}\)     A1

verifying \(\frac{{5\pi }}{6}\) satisfies equation     A1     N4

[6 marks]

Those who realized \({{\rm{e}}^{2x}}\) was a common factor usually earned the first four marks. Few could reason with the given information to solve the equation from there. There were many candidates who attempted some fruitless algebra that did not include factorization.

(i)     Write down the amplitude of \(f\).

(ii)     Find the period of \(f\).

On the following grid sketch the graph of \(f\).

(i)     3     A1     N1

(ii)     valid attempt to find the period     (M1)

eg\(\,\,\,\,\,\)\(\frac{{2\pi }}{b},{\text{ }}\frac{{2\pi }}{{\frac{\pi }{2}}}\)

period \( = 4\)     A1     N2

[3 marks]

     A1A1A1A1     N4

[4 marks]

Almost all candidates correctly stated the amplitude but then had difficulty finding the correct period. Few students faced problems in sketching the graph of the given function, even if they had found the wrong period, thus indicating a lack of understanding of the term ‘period’ in part a(ii). Most sketches were good although care should be taken to observe the given domain and to draw a neat curve.

Almost all candidates correctly stated the amplitude but then had difficulty finding the correct period. Few students faced problems in sketching the graph of the given function, even if they had found the wrong period, thus indicating a lack of understanding of the term ‘period’ in part a(ii). Most sketches were good although care should be taken to observe the given domain and to draw a neat curve.

find the value of \(\cos x;\)

find the value of \(\cos 2x.\)

valid approach     (M1)

eg\(\;\;\;\), \({\sin ^2}x + {\cos ^2}x = 1\)

correct working     (A1)

eg\(\;\;\;{4^2} - {3^2},{\text{ }}{\cos ^2}x = 1 - {\left( {\frac{3}{4}} \right)^2}\)

correct calculation     (A1)

eg\(\;\;\;\frac{{\sqrt 7 }}{4},{\text{ }}{\cos ^2}x = \frac{7}{{16}}\)

\(\cos x =  - \frac{{\sqrt 7 }}{4}\)     A1     N3

[4 marks]

correct substitution (accept missing minus with cos)     (A1)

eg\(\;\;\;1 - 2{\left( {\frac{3}{4}} \right)^2},{\text{ }}2{\left( { - \frac{{\sqrt 7 }}{4}} \right)^2} - 1,{\text{ }}{\left( {\frac{{\sqrt 7 }}{4}} \right)^2} - {\left( {\frac{3}{4}} \right)^2}\)

correct working     A1

eg\(\;\;\;2\left( {\frac{7}{{16}}} \right) - 1,{\text{ }}1 - \frac{{18}}{{16}},{\text{ }}\frac{7}{{16}} - \frac{9}{{16}}\)

\(\cos 2x =  - \frac{2}{{16}}\;\;\;\left( { =  - \frac{1}{8}} \right)\)     A1     N2

[3 marks]

Total [7 marks]

Many candidates were able to find the cosine ratio of \(\frac{{\sqrt 7 }}{4}\) but did not take into account the information about the obtuse angle and seldom selected the negative answer. Finding \(\cos 2x\) proved easier; the most common error seen was \(\cos 2x = 2\cos x\).

Many candidates were able to find the cosine ratio of \(\frac{{\sqrt 7 }}{4}\) but did not take into account the information about the obtuse angle and seldom selected the negative answer. Finding \(\cos 2x\) proved easier; the most common error seen was \(\cos 2x = 2\cos x\).

Find the coordinates of A.

Find \(\overrightarrow {{\text{AB}}} \);

Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).

Find \(\cos {\rm{B\hat AC}}\).

Hence or otherwise, find the distance between \({P_1}\) and \({P_2}\) two seconds after they leave A.

recognizing \(t = 0\) at A     (M1)

A is \((4,{\text{ }} - 1,{\text{ }}3)\)     A1     N2

[2 marks]

METHOD 1

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \\ 3 \end{array}} \right) + 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 2} \end{array}} \right),{\text{ }}(6,{\text{ }}3,{\text{ }} - 1)\)

correct approach to find \(\overrightarrow {{\text{AB}}} \)     (A1)

eg\(\,\,\,\,\,\)\({\text{AO}} + {\text{OB}},{\text{ B}} - {\text{A, }}\left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { - 1} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \\ 3 \end{array}} \right)\)

\(\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 4} \end{array}} \right)\)     A1     N2

METHOD 2

recognizing \(\overrightarrow {{\text{AB}}} \) is two times the direction vector     (M1)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AB}}}  = 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 2} \end{array}} \right)\)

\(\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 4} \end{array}} \right)\)     A1     N2

[3 marks]

correct substitution     (A1)

eg\(\,\,\,\,\,\)\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {{2^2} + {4^2} + {4^2}} ,{\text{ }}\sqrt {4 + 16 + 16} ,{\text{ }}\sqrt {36} \)

\(\left| {\overrightarrow {{\text{AB}}} } \right| = 6\)     A1     N2

[2 marks]

METHOD 1 (vector approach)

valid approach involving \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{AC}}} \)     (M1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AB}}}  \bullet \overrightarrow {{\text{AC}}} ,{\text{ }}\frac{{\overrightarrow {{\text{BA}}}  \bullet \overrightarrow {{\text{AC}}} }}{{{\text{AB}} \times {\text{AC}}}}\)

finding scalar product and \(\left| {\overrightarrow {{\text{AC}}} } \right|\)     (A1)(A1)

scalar product \(2(3) + 4(0) - 4(4){\text{ }}( =  - 10)\)

\(\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + {0^2} + {4^2}} {\text{ }}( = 5)\)

substitution of their scalar product and magnitudes into cosine formula     (M1)

eg\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC = }}\frac{{6 + 0 - 16}}{{6\sqrt {{3^2} + {4^2}} }}\)

\({\text{cos}}\,B\hat AC =  - \frac{{10}}{{30}}\left( { =  - \frac{1}{3}} \right)\)     A1     N2

METHOD 2 (triangle approach)

valid approach involving cosine rule     (M1)

eg\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC = }}\frac{{{\text{A}}{{\text{B}}^2} + {\text{A}}{{\text{C}}^2} - {\text{B}}{{\text{C}}^2}}}{{2 \times {\text{AB}} \times {\text{AC}}}}\)

finding lengths AC and BC     (A1)(A1)

\({\text{AC}} = 5,{\text{ BC}} = 9\)

substitution of their lengths into cosine formula     (M1)

eg\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC}} = \frac{{{5^2} + {6^2} - {9^2}}}{{2 \times 5 \times 6}}\)

\(\cos {\rm{B\hat AC}} =  - \frac{{20}}{{60}}{\text{ }}\left( { =  - \frac{1}{3}} \right)\)     A1     N2

[5 marks]

Note:     Award relevant marks for working seen to find BC in part (c) (if cosine rule used in part (c)).

METHOD 1 (using cosine rule)

recognizing need to find BC     (M1)

choosing cosine rule     (M1)

eg\(\,\,\,\,\,\)\({c^2} = {a^2} + {b^2} - 2ab\cos {\text{C}}\)

correct substitution into RHS     A1

eg\(\,\,\,\,\,\)\({\text{B}}{{\text{C}}^2} = {(6)^2} + {(5)^2} - 2(6)(5)\left( { - \frac{1}{3}} \right),{\text{ }}36 + 25 + 20\)

distance is 9     A1     N2

METHOD 2 (finding magnitude of \(\overrightarrow {BC} \)) 

recognizing need to find BC     (M1)

valid approach     (M1)

eg\(\,\,\,\,\,\)attempt to find \(\overrightarrow {{\rm{OB}}} \) or \(\overrightarrow {{\rm{OC}}} \), \(\overrightarrow {{\rm{OB}}}  = \left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { - 1} \end{array}} \right)\) or \(\overrightarrow {{\rm{OC}}}  = \left( {\begin{array}{*{20}{c}} 7 \\ { - 1} \\ 7 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{BA}}}  + \overrightarrow {{\rm{AC}}} \)

correct working     A1

eg\(\,\,\,\,\,\)\(\overrightarrow {{\rm{BC}}}  = \left( {\begin{array}{*{20}{c}} 1 \\ { - 4} \\ 8 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{CB}}}  = \left( {\begin{array}{*{20}{c}} { - 1} \\ 4 \\ { - 8} \end{array}} \right),{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}}  = \sqrt {81} \)

distance is 9     A1     N2

METHOD 3 (finding coordinates and using distance formula)

recognizing need to find BC     (M1)

valid approach     (M1)

eg\(\,\,\,\,\,\)attempt to find coordinates of B or C, \({\text{B}}(6,{\text{ }}3,{\text{ }} - 1)\) or \({\text{C}}(7,{\text{ }} - 1,{\text{ }}7)\)

correct substitution into distance formula     A1

eg\(\,\,\,\,\,\)\({\text{BC}} = \sqrt {{{(6 - 7)}^2} + {{\left( {3 - ( - 1)} \right)}^2} + {{( - 1 - 7)}^2}} ,{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}}  = \sqrt {81} \)

distance is 9     A1     N2

[4 marks]

Solve for \(0 \le x < 2\pi \)

(i)     \(6 + 6\sin x = 6\) ;

(ii)    \(6 + 6\sin x = 0\) .

Write down the exact value of the x-intercept of f , for \(0 \le x < 2\pi \) .

The area of the shaded region is k . Find the value of k , giving your answer in terms of \(\pi \) .

Let \(g(x) = 6 + 6\sin \left( {x - \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.

Give a full geometric description of this transformation.

Let \(g(x) = 6 + 6\sin \left( {x - \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.

Given that \(\int_p^{p + \frac{{3\pi }}{2}} {g(x){\rm{d}}x}  = k\) and \(0 \le p < 2\pi \) , write down the two values of p.

(i) \(\sin x = 0\)     A1

\(x = 0\) , \(x = \pi \)     A1A1     N2

(ii) \(\sin x = - 1\)     A1

\(x = \frac{{3\pi }}{2}\)    A1     N1

[5 marks]

\(\frac{{3\pi }}{2}\)     A1     N1

[1 mark]

evidence of using anti-differentiation     (M1)

e.g. \(\int_0^{\frac{{3\pi }}{2}} {(6 + 6\sin x){\rm{d}}x} \)

correct integral \(6x - 6\cos x\) (seen anywhere)     A1A1

correct substitution     (A1)

e.g. \(6\left( {\frac{{3\pi }}{2}} \right) - 6\cos \left( {\frac{{3\pi }}{2}} \right) - ( - 6\cos 0)\) , \(9\pi  - 0 + 6\)

\(k = 9\pi + 6\)     A1A1     N3

[6 marks]

translation of \(\left( {\begin{array}{*{20}{c}}
{\frac{\pi }{2}}\\
0
\end{array}} \right)\)     A1A1     N2

[2 marks]

recognizing that the area under g is the same as the shaded region in f     (M1)

\(p = \frac{\pi }{2}\) , \(p = 0\)     A1A1     N3

[3 marks]

Many candidates again had difficulty finding the common angles in the trigonometric equations. In part (a), some did not show sufficient working in solving the equations. Others obtained a single solution in (a)(i) and did not find another. Some candidates worked in degrees; the majority worked in radians.

While some candidates appeared to use their understanding of the graph of the original function to find the x-intercept in part (b), most used their working from part (a)(ii) sometimes with follow-through on an incorrect answer.

Most candidates recognized the need for integration in part (c) but far fewer were able to see the solution through correctly to the end. Some did not show the full substitution of the limits, having incorrectly assumed that evaluating the integral at 0 would be 0; without this working, the mark for evaluating at the limits could not be earned. Again, many candidates had trouble working with the common trigonometric values.

While there was an issue in the wording of the question with the given domains, this did not appear to bother candidates in part (d). This part was often well completed with candidates using a variety of language to describe the horizontal translation to the right by \(\frac{\pi }{2}\) .

Most candidates who attempted part (e) realized that the integral was equal to the value that they had found in part (c), but a majority tried to integrate the function g without success. Some candidates used sketches to find one or both values for p. The problem in the wording of the question did not appear to have been noticed by candidates in this part either.

The following diagram shows triangle \(ABC\).

Let \(\overrightarrow {{\text{AB}}}  \bullet \overrightarrow {{\text{AC}}}  =  - 5\sqrt 3 \) and \(\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right| = 10\). Find the area of triangle \(ABC\).

attempt to find \(\cos {\rm{C\hat AB}}\) (seen anywhere)     (M1)

eg\(\;\;\;\cos \theta  = \frac{{\overrightarrow {{\text{AB}}}  \bullet \overrightarrow {{\text{AC}}} }}{{\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|}}\)

\(\cos {\rm{C\hat AB}} = \frac{{ - 5\sqrt 3 }}{{10}}\;\;\;\left( { =  - \frac{{\sqrt 3 }}{2}} \right)\)     A1

valid attempt to find \(\sin {\rm{C\hat AB}}\)     (M1)

eg\(\;\;\;\)triangle, Pythagorean identity, \({\rm{C\hat AB}} = \frac{{5\pi }}{6},{\text{ }}150^\circ \)

\(\sin {\rm{C\hat AB}} = \frac{1}{2}\)     (A1)

correct substitution into formula for area     (A1)

eg\(\;\;\;\frac{1}{2} \times 10 \times \frac{1}{2},{\text{ }}\frac{1}{2} \times 10 \times \sin \frac{\pi }{6}\)

\({\text{area}} = \frac{{10}}{4}\;\;\;\left( { = \frac{5}{2}} \right)\)     A1     N3

[6 marks]

The large majority of candidates were able to find the correct expression for \(\cos {\rm{C\hat AB}}\), but few recognized that an angle with a negative cosine will be obtuse, rather than acute, and many stated that \({\rm{C\hat AB}} = 30^\circ \). When substituting into the triangle area formula, a common error was to substitute \(5\sqrt 3 \) rather than 10, as many did not understand the relationship between the magnitude of a vector and the length of a line segment in the triangle formula.

Some of the G2 comments from schools suggested that it might have been easier for their students if this question were split into two parts. While we do tend to provide more support on the earlier questions in the paper, questions 6 and 7 are usually presented with little or no scaffolding. On these later questions, the candidates are often required to use knowledge from different areas of the syllabus within a single question.

Solve \(\cos 2x - 3\cos x - 3 - {\cos ^2}x = {\sin ^2}x\) , for \(0 \le x \le 2\pi \) .

evidence of substituting for \(\cos 2x\)     (M1)

evidence of substituting into \({\sin ^2}x + {\cos ^2}x = 1\)     (M1)

correct equation in terms of \(\cos x\) (seen anywhere)     A1

e.g. \(2{\cos ^2}x - 1 - 3\cos x - 3 = 1\) , \(2{\cos ^2}x - 3\cos x - 5 = 0\)

evidence of appropriate approach to solve     (M1)

e.g. factorizing, quadratic formula

appropriate working     A1

e.g. \((2\cos x - 5)(\cos x + 1) = 0\) , \((2x - 5)(x + 1)\) , \(\cos x = \frac{{3 \pm \sqrt {49}}}{4}\)

correct solutions to the equation

e.g. \(\cos x = \frac{5}{2}\) , \(\cos x = - 1\) , \(x = \frac{5}{2}\) , \(x = - 1\)     (A1)

\(x = \pi \)     A1     N4

[7 marks]

This question was quite difficult for most candidates. A number of students earned some credit for manipulating the equation with identities, but many earned no further marks due to algebraic errors. Many did not substitute for \(\cos 2x\) ; others did this substitution but then did nothing further.

Few candidates were able to get a correct equation in terms of \(\cos x\) and many who did get the equation didn't know what to do with it. Candidates who correctly solved the resulting quadratic usually found the one correct value of \(x\), earning full marks. 

(i) Write down the range of the function f .

(ii) Consider \(f(x) = 1\) , \(0 \le x \le 2\pi \) . Write down the number of solutions to this equation. Justify your answer.

Find \(f'(x)\) , giving your answer in the form \(a{\sin ^p}x{\cos ^q}x\) where \(a{\text{, }}p{\text{, }}q \in \mathbb{Z}\) .

Let \(g(x) = \sqrt 3 \sin x{(\cos x)^{\frac{1}{2}}}\) for \(0 \le x \le \frac{\pi }{2}\) . Find the volume generated when the curve of g is revolved through \(2\pi \) about the x-axis.

(i) range of f is \([ - 1{\text{, }}1]\) , \(( - 1 \le f(x) \le 1)\)     A2     N2

(ii) \({\sin ^3}x \Rightarrow 1 \Rightarrow \sin x = 1\)     A1

justification for one solution on \([0{\text{, }}2\pi ]\)    R1

e.g. \(x = \frac{\pi }{2}\) , unit circle, sketch of \(\sin x\)

1 solution (seen anywhere)     A1     N1

[5 marks]

\(f'(x) = 3{\sin ^2}x\cos x\)     A2     N2

[2 marks]

using \(V = \int_a^b {\pi {y^2}{\rm{d}}x} \)     (M1)

\(V = \int_0^{\frac{\pi }{2}} {\pi (\sqrt 3 } \sin x{\cos ^{\frac{1}{2}}}x{)^2}{\rm{d}}x\)     (A1)

\( = \pi \int_0^{\frac{\pi }{2}} {3{{\sin }^2}x\cos x{\rm{d}}x} \)     A1

\(V = \pi \left[ {{{\sin }^3}x} \right]_0^{\frac{\pi }{2}}\) \(\left( { = \pi \left( {{{\sin }^3}\left( {\frac{\pi }{2}} \right) - {{\sin }^3}0} \right)} \right)\)     A2

evidence of using \(\sin \frac{\pi }{2} = 1\) and \(\sin 0 = 0\)     (A1)

e.g. \(\pi \left( {1 - 0} \right)\)

\(V = \pi \)     A1     N1

[7 marks]

This question was not done well by most candidates. No more than one-third of them could correctly give the range of \(f(x) = {\sin ^3}x\) and few could provide adequate justification for there being exactly one solution to \(f(x) = 1\) in the interval \([0{\text{, }}2\pi ]\) .

This question was not done well by most candidates.

This question was not done well by most candidates. No more than one-third of them could correctly give the range of \(f(x) = {\sin ^3}x\) and few could provide adequate justification for there being exactly one solution to \(f(x) = 1\) in the interval \([0{\text{, }}2\pi ]\) . Finding the derivative of this function also presented major problems, thus making part (c) of the question much more difficult. In spite of the formula for volume of revolution being given in the Information Booklet, fewer than half of the candidates could correctly put the necessary function and limits into \(\pi \int_a^b {{y^2}{\rm{d}}x} \) and fewer still could square \(\sqrt 3 \sin x{\cos ^{\frac{1}{2}}}x\) correctly. From those who did square correctly, the correct antiderivative was not often recognized. All manner of antiderivatives were suggested instead.

(a)     \(p\)

(b)     \(q\)

(c)     \(r\) .

\(p\)

\(q\)

\(r\) .

(a)     valid approach to find \(p\)     (M1)

eg   amplitude \( = \frac{{{\rm{max}} - {\rm{min}}}}{2}\) , \(p = 6\)

\(p = 3\)     A1     N2

[2 marks]

(b)     valid approach to find \(q\)     (M1)

eg   period = 4 , \(q = \frac{{2\pi }}{{{\rm{period}}}}\) 

\(q = \frac{\pi }{2}\)     A1     N2

[2 marks]

(c)     valid approach to find \(r\)     (M1)

eg axis = \(\frac{{{\rm{max}} + {\rm{min}}}}{2}\) , sketch of horizontal axis, \(f(0)\)

\(r = 2\)     A1     N2

[2 marks]

Total [6 marks]

valid approach to find \(p\)     (M1)

eg   amplitude \( = \frac{{{\rm{max}} - {\rm{min}}}}{2}\) , \(p = 6\)

\(p = 3\)     A1     N2

[2 marks]

valid approach to find \(q\)     (M1)

eg   period = 4 , \(q = \frac{{2\pi }}{{{\rm{period}}}}\) 

\(q = \frac{\pi }{2}\)     A1     N2

[2 marks]

valid approach to find \(r\)     (M1)

eg axis = \(\frac{{{\rm{max}} + {\rm{min}}}}{2}\) , sketch of horizontal axis, \(f(0)\)

\(r = 2\)     A1     N2

[2 marks]

Total [6 marks]

Find the length of \({\text{arc ACB}}\).

Find the perimeter of the shaded region.

correct substitution     (A1)

eg\(\;\;\;10(1.2)\)

\(ACB\) is \(12{\text{ (cm)}}\)     A1     N2

[2 marks]

valid approach to find major arc     (M1)

eg\(\;\;\;\)circumference \( - {\text{AB}}\), major angle \({\text{AOB}} \times {\text{radius}}\)

correct working for arc length     (A1)

eg\(\;\;\;2\pi (10) - 12,{\text{ }}10(2 \times 3.142 - 1.2),{\text{ }}2\pi (10) - 12 + 20\)

perimeter is \(20\pi  + 8\;\;\;( = 70.8){\text{ (cm)}}\)     A1     N2

[3 marks]

Total [5 marks]

Most candidates were able to find the minor arc length. Similarly most candidates successfully found the major arc length in part b) but did not go on to add the two radii. Quite a few candidates worked with decimal approximations, rather than in terms of \(\pi \).

Most candidates were able to find the minor arc length. Similarly most candidates successfully found the major arc length in part b) but did not go on to add the two radii. Quite a few candidates worked with decimal approximations, rather than in terms of \(\pi \).

Show that \(\cos \theta  = \frac{3}{4}\).

Given that \(\tan \theta  > 0\), find \(\tan \theta \).

Let \(y = \frac{1}{{\cos x}}\), for \(0 < x < \frac{\pi }{2}\). The graph of \(y\)between \(x = \theta \) and \(x = \frac{\pi }{4}\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.

evidence of summing to 1     (M1)

eg\(\,\,\,\,\,\)\(\sum {p = 1} \)

correct equation     A1

eg\(\,\,\,\,\,\)\(\cos \theta  + 2\cos 2\theta  = 1\)

correct equation in \(\cos \theta \)     A1

eg\(\,\,\,\,\,\)\(\cos \theta  + 2(2{\cos ^2}\theta  - 1) = 1,{\text{ }}4{\cos ^2}\theta  + \cos \theta  - 3 = 0\)

evidence of valid approach to solve quadratic     (M1)

eg\(\,\,\,\,\,\)factorizing equation set equal to \(0,{\text{ }}\frac{{ - 1 \pm \sqrt {1 - 4 \times 4 \times ( - 3)} }}{8}\)

correct working, clearly leading to required answer     A1

eg\(\,\,\,\,\,\)\((4\cos \theta  - 3)(\cos \theta  + 1),{\text{ }}\frac{{ - 1 \pm 7}}{8}\)

correct reason for rejecting \(\cos \theta  \ne  - 1\)     R1

eg\(\,\,\,\,\,\)\(\cos \theta \) is a probability (value must lie between 0 and 1), \(\cos \theta  > 0\)

Note:     Award R0 for \(\cos \theta  \ne  - 1\) without a reason.

\(\cos \theta  = \frac{3}{4}\)    AG  N0

valid approach     (M1)

eg\(\,\,\,\,\,\)sketch of right triangle with sides 3 and 4, \({\sin ^2}x + {\cos ^2}x = 1\)

correct working     

(A1)

eg\(\,\,\,\,\,\)missing side \( = \sqrt 7 ,{\text{ }}\frac{{\frac{{\sqrt 7 }}{4}}}{{\frac{3}{4}}}\)

\(\tan \theta  = \frac{{\sqrt 7 }}{3}\)     A1     N2

[3 marks]

attempt to substitute either limits or the function into formula involving \({f^2}\)     (M1)

eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{f^2},{\text{ }}\int {{{\left( {\frac{1}{{\cos x}}} \right)}^2}} } \)

correct substitution of both limits and function     (A1)

eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{{\left( {\frac{1}{{\cos x}}} \right)}^2}{\text{d}}x} \)

correct integration     (A1)

eg\(\,\,\,\,\,\)\(\tan x\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(\tan \frac{\pi }{4} - \tan \theta \)

Note:     Award M0 if they substitute into original or differentiated function.

\(\tan \frac{\pi }{4} = 1\)    (A1)

eg\(\,\,\,\,\,\)\(1 - \tan \theta \)

\(V = \pi  - \frac{{\pi \sqrt 7 }}{3}\)     A1     N3

[6 marks]

Show that \(4 - \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin \theta + 3\) .

Hence, solve the equation \(4 - \cos 2\theta + 5\sin \theta = 0\) for \(0 \le \theta \le 2\pi \) .

attempt to substitute \(1 - 2{\sin ^2}\theta \) for \(\cos 2\theta \)     (M1)

correct substitution     A1

e.g. \(4 - (1 - 2{\sin ^2}\theta ) + 5\sin\theta \)

\(4 - \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin\theta + 3\)     AG     N0

[2 marks]

evidence of appropriate approach to solve     (M1)

e.g. factorizing, quadratic formula

correct working     A1

e.g. \((2\sin \theta + 3)(\sin \theta + 1)\) , \((2x + 3)(x + 1) = 0\) , \(\sin x = \frac{{ - 5 \pm \sqrt 1 }}{4}\)

correct solution \(\sin \theta = - 1\) (do not penalise for including \(\sin \theta = - \frac{3}{2}\)     (A1)

\(\theta = \frac{{3\pi }}{2}\)     A2     N3

[5 marks]

In part (a), most candidates successfully substituted using the double-angle formula for cosine. There were quite a few candidates who worked backward, starting with the required answer and manipulating the equation in various ways. As this was a "show that" question, working backward from the given answer is not a valid method.

In part (b), many candidates seemed to realize what was required by the word “hence”, though some had trouble factoring the quadratic-type equation. A few candidates were also successful using the quadratic formula. Some candidates got the wrong solution to the equation \(\sin \theta = - 1\) , and there were a few who did not realize that the equation \(\sin \theta = - \frac{3}{2}\) has no solution.

Solve \({\log _2}(2\sin x) + {\log _2}(\cos x) =  - 1\), for \(2\pi  < x < \frac{{5\pi }}{2}\).

correct application of \(\log a + \log b = \log ab\)     (A1)

eg\(\,\,\,\,\,\)\({\log _2}(2\sin x\cos x),{\text{ }}\log 2 + \log (\sin x) + \log (\cos x)\)

correct equation without logs     A1

eg\(\,\,\,\,\,\)\(2\sin x\cos x = {2^{ - 1}},{\text{ }}\sin x\cos x = \frac{1}{4},{\text{ }}\sin 2x = \frac{1}{2}\)

recognizing double-angle identity (seen anywhere)     A1

eg\(\,\,\,\,\,\)\(\log (\sin 2x),{\text{ }}2\sin x\cos x = \sin 2x,{\text{ }}\sin 2x = \frac{1}{2}\)

evaluating \({\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}{\text{ }}(30^\circ )\)     (A1)

correct working     A1

eg\(\,\,\,\,\,\)\(x = \frac{\pi }{{12}} + 2\pi ,{\text{ }}2x = \frac{{25\pi }}{6},{\text{ }}\frac{{29\pi }}{6},{\text{ }}750^\circ ,{\text{ }}870^\circ ,{\text{ }}x = \frac{\pi }{{12}}\)and \(x = \frac{{5\pi }}{{12}}\), one correct final answer

\(x = \frac{{25\pi }}{{12}},{\text{ }}\frac{{29\pi }}{{12}}\) (do not accept additional values)     A2     N0

[7 marks]

Write down the value of \(\tan \theta \) .

Find the value of

(i)     \(\sin 2\theta \) ;

(ii)    \(\cos 2\theta \) .

\(\tan \theta = \frac{3}{4}\) (do not accept \(\frac{3}{4}x\) )     A1     N1

[1 mark]

(i) \(\sin \theta = \frac{3}{5}\) , \(\cos \theta = \frac{4}{5}\)     (A1)(A1)

correct substitution     A1

e.g. \(\sin 2\theta = 2\left( {\frac{3}{5}} \right)\left( {\frac{4}{5}} \right)\)

\(\sin 2\theta = \frac{{24}}{{25}}\)     A1     N3

(ii) correct substitution     A1

e.g. \(\cos 2\theta = 1 - 2{\left( {\frac{3}{5}} \right)^2}\) ,  \({\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2}\)

\(\cos 2\theta = \frac{7}{{25}}\)     A1     N1

[6 marks]

Many candidates drew a diagram to correctly find \(\tan \theta \) , although few recognized that a line through the origin can be expressed as \(y = x\tan \theta \) , with gradient \(\tan \theta \) , which is explicit in the syllabus.

A surprising number were unable to find the ratios for \(\sin \theta \) and \(\cos \theta \) from \(\tan \theta \) . It was not uncommon for candidates to use unreasonable values, such as \(\sin \theta = 3\) and \(\cos \theta = 4\) , or to write nonsense such as \(2\sin \frac{3}{5}\cos \frac{4}{5}\) .

Find \(\cos \theta \).

Find \(\cos 2\theta \).

evidence of valid approach     (M1)

eg\(\,\,\,\,\,\)right triangle, \({\cos ^2}\theta  = 1 - {\sin ^2}\theta \)

correct working     (A1)

eg\(\,\,\,\,\,\)missing side is 2, \(\sqrt {1 - {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^2}} \)

\(\cos \theta  = \frac{2}{3}\)     A1     N2

[3 marks]

correct substitution into formula for \(\cos 2\theta \)     (A1)

eg\(\,\,\,\,\,\)\(2 \times {\left( {\frac{2}{3}} \right)^2} - 1,{\text{ }}1 - 2{\left( {\frac{{\sqrt 5 }}{3}} \right)^2},{\text{ }}{\left( {\frac{2}{3}} \right)^2} - {\left( {\frac{{\sqrt 5 }}{3}} \right)^2}\)

\(\cos 2\theta  =  - \frac{1}{9}\)    A1     N2

[2 marks]

Show that \(f(x) = \sin x\) .

Let \(\sin x = \frac{2}{3}\) . Show that \(f(2x) = - \frac{{4\sqrt 5 }}{9}\) .

changing \(\tan x\) into \(\frac{{\sin x}}{{\cos x}}\)     A1

e.g. \({\sin ^3}x + {\cos ^3}x\frac{{\sin x}}{{\cos x}}\)

simplifying     A1

e.g \(\sin x({\sin ^2}x + {\cos ^2}x)\) , \({\sin ^3}x + \sin x - {\sin ^3}x\)

\(f(x) = \sin x\)    AG     N0

[2 marks]

recognizing \(f(2x) = \sin 2x\) , seen anywhere     (A1)

evidence of using double angle identity \(\sin (2x) = 2\sin x\cos x\) , seen anywhere     (M1)

evidence of using Pythagoras with \(\sin x = \frac{2}{3}\)     M1

e.g. sketch of right triangle, \({\sin ^2}x + {\cos ^2}x = 1\)

\(\cos x = - \frac{{\sqrt 5 }}{3}\) (accept \(\frac{{\sqrt 5 }}{3}\) )     (A1)

\(f(2x) = 2\left( {\frac{2}{3}} \right)\left( { - \frac{{\sqrt 5 }}{3}} \right)\)     A1

\(f(2x) = - \frac{{4\sqrt 5 }}{9}\)     AG     N0

[5 marks]

Not surprisingly, this question provided the greatest challenge in section A. In part (a), candidates were able to use the identity \(\tan x = \frac{{\sin x}}{{\cos x}}\) , but many could not proceed any further.

Part (b) was generally well done by those candidates who attempted it, the major error arising when the negative sign "magically" appeared in the answer. Many candidates could find the value of cosx but failed to observe that cosine is negative in the given domain.

Use the graph to write down the value of

(i)     a ;

(ii)    c ;

(iii)   d .

Show that \(b = \frac{\pi }{4}\) .

Find \(f'(x)\) .

At a point R, the gradient is \( - 2\pi \) . Find the x-coordinate of R.

(i) \(a = 8\)     A1     N1

(ii) \(c = 2\)     A1     N1

(iii) \(d = 4\)     A1     N1

[3 marks]

METHOD 1

recognizing that period \( = 8\)     (A1)

correct working     A1

e.g. \(8 = \frac{{2\pi }}{b}\) , \(b = \frac{{2\pi }}{8}\)

\(b = \frac{\pi }{4}\)     AG     N0

METHOD 2

attempt to substitute     M1

e.g. \(12 = 8\sin (b(4 - 2)) + 4\)

correct working     A1

e.g. \(\sin 2b = 1\)

\(b = \frac{\pi }{4}\)     AG     N0

[2 marks]

evidence of attempt to differentiate or choosing chain rule     (M1)

e.g. \(\cos \frac{\pi }{4}(x - 2)\) , \(\frac{\pi }{4} \times 8\)

\(f'(x) = 2\pi \cos \left( {\frac{\pi }{4}(x - 2)} \right)\) (accept \(2\pi \cos \frac{\pi }{4}(x - 2)\) )     A2     N3

[3 marks]

recognizing that gradient is \(f'(x)\)     (M1)

e.g. \(f'(x) = m\)

correct equation     A1

e.g. \( - 2\pi  = 2\pi \cos \left( {\frac{\pi }{4}(x - 2)} \right)\) , \( - 1 = \cos \left( {\frac{\pi }{4}(x - 2)} \right)\)

correct working     (A1)

e.g. \({\cos ^{ - 1}}( - 1) = \frac{\pi }{4}(x - 2)\)

using \({\cos ^{ - 1}}( - 1) = \pi \) (seen anywhere)     (A1)

e.g. \(\pi  = \frac{\pi }{4}(x - 2)\)

simplifying     (A1)

e.g. \(4 = (x - 2)\)

\(x = 6\)     A1     N4

[6 marks]

Part (a) of this question proved challenging for most candidates.

Although a good number of candidates recognized that the period was 8 in part (b), there were some who did not seem to realize that this period could be found using the given coordinates of the maximum and minimum points.

In part (c), not many candidates found the correct derivative using the chain rule.

For part (d), a good number of candidates correctly set their expression equal to \( - 2\pi \) , but errors in their previous values kept most from correctly solving the equation. Most candidates who had the correct equation were able to gain full marks here.

Find the value of a and of b .

Hence or otherwise, solve the equation \(6\sin x\cos x = \frac{3}{2}\) , for \(\frac{\pi }{4} \le x \le \frac{\pi }{2}\) .

recognizing double angle     M1

e.g. \(3 \times 2\sin x\cos x\) , \(3\sin 2x\)

\(a = 3\) , \(b = 2\)     A1A1     N3

[3 marks]

substitution \(3\sin 2x = \frac{3}{2}\)     M1

\(\sin 2x = \frac{1}{2}\)     A1

finding the angle     A1

e.g. \(\frac{\pi }{6}\) , \(2x = \frac{{5\pi }}{6}\)

\(x = \frac{{5\pi }}{{12}}\)     A1     N2

Note: Award A0 if other values are included.

[4 marks]

Let \(\sin {100^ \circ } = m\). Find an expression for \(\cos {100^ \circ }\) in terms of m.

Let \(\sin {100^ \circ } = m\) . Find an expression for \(\tan {100^ \circ }\) in terms of m.

Let \(\sin {100^ \circ } = m\). Find an expression for \(\sin {200^ \circ }\) in terms of m.

Note: All answers must be given in terms of m. If a candidate makes an error that means there is no m in their answer, do not award the final A1FT mark.

METHOD 1 

valid approach involving Pythagoras     (M1)

e.g. \({\sin ^2}x + {\cos ^2}x = 1\) , labelled diagram 

correct working (may be on diagram)     (A1)

e.g. \({m^2} + {(\cos 100)^2} = 1\) , \(\sqrt {1 - {m^2}} \)

\(\cos 100 = - \sqrt {1 - {m^2}} \)    A1     N2

[3 marks]

METHOD 2

valid approach involving tan identity     (M1)

e.g. \(\tan  = \frac{{\sin }}{{\cos }}\)

correct working     (A1)

e.g. \(\cos 100 = \frac{{\sin 100}}{{\tan 100}}\)

\(\cos 100 = \frac{m}{{\tan 100}}\)     A1     N2

[3 marks]

METHOD 1

\(\tan 100 = - \frac{m}{{\sqrt {1 - {m^2}} }}\) (accept \(\frac{m}{{ - \sqrt {1 - {m^2}} }}\))     A1     N1

[1 mark]

METHOD 2

\(\tan 100 = \frac{m}{{\cos 100}}\)     A1     N1

[1 mark]

METHOD 1

valid approach involving double angle formula     (M1)

e.g. \(\sin 2\theta = 2\sin \theta cos\theta \)

\(\sin 200 = - 2m\sqrt {1 - {m^2}} \)  (accept \(2m\left( { - \sqrt {1 - {m^2}} } \right)\))     A1     N2

Note: If candidates find \(\cos 100 = \sqrt {1 - {m^2}} \) , award full FT in parts (b) and (c), even though the values may not have appropriate signs for the angles.

[2 marks]

METHOD 2

valid approach involving double angle formula     (M1)

e.g. \(\sin 2\theta  = 2\sin \theta \cos \theta \) , \(2m \times \frac{m}{{\tan 100}}\)

\(\sin 200 = \frac{{2{m^2}}}{{\tan 100}}( = 2m\cos 100)\)     A1     N2

[2 marks]

While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write \(\sqrt {1 - {m^2}} = 1 - m\) . In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because \(\sin {100^ \circ } = m\) , then \(\sin {200^ \circ } = 2m\) . In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant.

While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write \(\sqrt {1 - {m^2}} = 1 - m\) . In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because \(\sin {100^ \circ } = m\) , then \(\sin {200^ \circ } = 2m\) . In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant.

While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write \(\sqrt {1 - {m^2}} = 1 - m\) . In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because \(\sin {100^ \circ } = m\) , then \(\sin {200^ \circ } = 2m\) . In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant.

Let \(\int_\pi ^a {\cos 2x{\text{d}}x}  = \frac{1}{2}{\text{, where }}\pi  < a < 2\pi \). Find the value of \(a\).

correct integration (ignore absence of limits and “\(+C\)”)     (A1)

eg     \(\frac{{\sin (2x)}}{2},{\text{ }}\int_\pi ^a {\cos 2x = \left[ {\frac{1}{2}\sin (2x)} \right]_\pi ^a} \)

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

eg     \(\frac{1}{2}\sin (2a) - \frac{1}{2}\sin (2\pi ),{\text{ }}\sin (2\pi ) - \sin (2a)\)

\(\sin (2\pi ) = 0\)     (A1)

setting their result from an integrated function equal to \(\frac{1}{2}\)     M1

eg     \(\frac{1}{2}\sin 2a = \frac{1}{2},{\text{ }}\sin (2a) = 1\)

recognizing \({\sin ^{ - 1}}1 = \frac{\pi }{2}\)     (A1)

eg     \(2a = \frac{\pi }{2},{\text{ }}a = \frac{\pi }{4}\)

correct value     (A1)

eg     \(\frac{\pi }{2} + 2\pi ,{\text{ }}2a = \frac{{5\pi }}{2},{\text{ }}a = \frac{\pi }{4} + \pi \)

\(a = \frac{{5\pi }}{4}\)     A1     N3

[7 marks]

Write down the height of P above ground level after

(i)     10 minutes;

(ii)    15 minutes.

(i)     Show that \(h(8) = 90.5\).

(ii)    Find \(h(21)\) .

Sketch the graph of h , for \(0 \le t \le 40\) .

Given that h can be expressed in the form \(h(t) = a\cos bt + c\) , find a , b and c .

(i) 100 (metres)     A1     N1

(ii) 50 (metres)     A1     N1

[2 marks]

(i) identifying symmetry with \(h(2) = 9.5\)     (M1)

subtraction     A1

e.g. \(100 - h(2)\) , \(100 - 9.5\)

\(h(8) = 90.5\)     AG     N0

(ii) recognizing period     (M1)

e.g. \(h(21) = h(1)\)

\(h(21) = 2.4\)     A1     N2

[4 marks]

     A1A1A1     N3

Note: Award A1 for end points (0, 0) and (40, 0) , A1 for range \(0 \le h \le 100\) , A1 for approximately correct sinusoidal shape, with two cycles.

[3 marks]

evidence of a quotient involving 20, \(2\pi \) or \({360^ \circ }\) to find b     (M1)

e.g. \(\frac{{2\pi }}{b} = 20\) , \(b = \frac{{360}}{{20}}\)

\(b = \frac{{2\pi }}{{20}}\) \(\left( { = \frac{\pi }{{10}}} \right)\) (accept \(b = 18\) if working in degrees)     A1     N2

\(a = - 50\) , \(c = 50\)     A2A1     N3

[5 marks]

Nearly all candidates answered part (a) correctly, finding the height of the wheel at \(\frac{1}{2}\) and \(\frac{3}{4}\) of a revolution.

While many candidates were successful in part (b), there were many who tried to use right-angled triangles or find a function for height, rather than recognizing the symmetry of the wheel in its different positions and using the values given in the table.

In part (c), most candidates were able to sketch a somewhat accurate representation of the height of the wheel over two full cycles. However, it seems that many candidates are not familiar with the shape of a sinusoidal wave, as many of the candidates' graphs were constructed of line segments, rather than a curve.

For part (d), candidates were less successful in finding the parameters of the cosine function. Even candidates who drew accurate sketches were not always able to relate their sketch to the function. These candidates understood the context of the problem, that the position on the wheel goes up and down, but they did not relate this to a trigonometric function. Only a small number of candidates recognized that the value of a would be negative. Candidates should be aware that while working in degrees may be acceptable, the expectation is that radians will be used in these types of questions.

Let \(f(x) = 15 - {x^2}\), for \(x \in \mathbb{R}\). The following diagram shows part of the graph of \(f\) and the rectangle OABC, where A is on the negative \(x\)-axis, B is on the graph of \(f\), and C is on the \(y\)-axis.

Find the \(x\)-coordinate of A that gives the maximum area of OABC.

attempt to find the area of OABC     (M1)

eg\(\,\,\,\,\,\)\({\text{OA}} \times {\text{OC, }}x \times f(x),{\text{ }}f(x) \times ( - x)\)

correct expression for area in one variable     (A1)

eg\(\,\,\,\,\,\)\({\text{area}} = x(15 - {x^2}),{\text{ }}15x - {x^3},{\text{ }}{x^3} - 15x\)

valid approach to find maximum area (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)\(A’(x) = 0\)

correct derivative     A1

eg\(\,\,\,\,\,\)\(15 - 3{x^2},{\text{ }}(15 - {x^2}) + x( - 2x) = 0,{\text{ }} - 15 + 3{x^2}\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(15 = 3{x^2},{\text{ }}{x^2} = 5,{\text{ }}x = \sqrt 5 \)

\(x =  - \sqrt 5 {\text{ }}\left( {{\text{accept A}}\left( { - \sqrt 5 ,{\text{ }}0} \right)} \right)\)     A2     N3

[7 marks]

Find the value of \(k\).

Find the minimum value of \(f(x)\).

Let \(g(x) = \sin x\). The graph of g is translated to the graph of \(f\) by the vector \(\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right)\).

Write down the value of \(p\) and of \(q\).

METHOD 1

attempt to substitute both coordinates (in any order) into \(f\)     (M1)

eg     \(f\left( {\frac{\pi }{4}} \right) = 6,{\text{ }}\frac{\pi }{4} = \sin \left( {6 + \frac{\pi }{4}} \right) + k\)

correct working     (A1)

eg     \(\sin \frac{\pi }{2} = 1,{\text{ }}1 + k = 6\)

\(k = 5\)     A1     N2

[3 marks]

METHOD 2

recognizing shift of \(\frac{\pi }{4}\) left means maximum at \(6\)     R1)

recognizing \(k\) is difference of maximum and amplitude     (A1)

eg     \(6 - 1\)

\(k = 5\)     A1     N2

[3 marks] 

evidence of appropriate approach     (M1)

eg     minimum value of \(\sin x\) is \( - 1,{\text{ }} - 1 + k,{\text{ }}f'(x) = 0,{\text{ }}\left( {\frac{{5\pi }}{4},{\text{ }}4} \right)\)

minimum value is \(4\)     A1     N2

[2 marks]

\(p =  - \frac{\pi }{4},{\text{ }}q = 5{\text{     }}\left( {{\text{accept \(\left( \begin{array}{c} - {\textstyle{\pi  \over 4}}\\5\end{array} \right)\)}}} \right)\)     A1A1     N2

[2 marks]

Show that \(\cos A = \frac{{12}}{{13}}\).

Find \(\cos 2A\).

METHOD 1

approach involving Pythagoras’ theorem     (M1)

eg     \({5^2} + {x^2} = {13^2}\), labelling correct sides on triangle

finding third side is 12 (may be seen on diagram)     A1

\(\cos A = \frac{{12}}{{13}}\)     AG     N0

METHOD 2

approach involving \({\sin ^2}\theta  + {\cos ^2}\theta  = 1\)     (M1)

eg     \({\left( {\frac{5}{{13}}} \right)^2} + {\cos ^2}\theta  = 1,{\text{ }}{x^2} + \frac{{25}}{{169}} = 1\)

correct working     A1

eg     \({\cos ^2}\theta  = \frac{{144}}{{169}}\)

\(\cos A = \frac{{12}}{{13}}\)     AG     N0

[2 marks]

correct substitution into \(\cos 2\theta \)     (A1)

eg     \(1 - 2{\left( {\frac{5}{{13}}} \right)^2},{\text{ }}2{\left( {\frac{{12}}{{13}}} \right)^2} - 1,{\text{ }}{\left( {\frac{{12}}{{13}}} \right)^2} - {\left( {\frac{5}{{13}}} \right)^2}\)

correct working     (A1)

eg     \(1 - \frac{{50}}{{169}},{\text{ }}\frac{{288}}{{169}} - 1,{\text{ }}\frac{{144}}{{169}} - \frac{{25}}{{169}}\)

\(\cos 2A = \frac{{119}}{{169}}\)     A1     N2

[3 marks]

Find \(f\left( {\frac{\pi }{2}} \right)\) .

Find \((g \circ f)\left( {\frac{\pi }{2}} \right)\) .

Given that \((g \circ f)(x)\) can be written as \(\cos (kx)\) , find the value of k, \(k \in \mathbb{Z}\) .

\(f\left( {\frac{\pi }{2}} \right) = \cos \pi \)     (A1)

\( = - 1\)     A1     N2

[2 marks]

\((g \circ f)\left( {\frac{\pi }{2}} \right) = g( - 1)\) \(( = 2{( - 1)^2} - 1)\)    (A1)

\(= 1\)     A1     N2

[2 marks]

\((g \circ f)(x) = 2{(\cos (2x))^2} - 1\) \(( = 2{\cos ^2}(2x) - 1)\)     A1

evidence of \(2{\cos ^2}\theta - 1 = \cos 2\theta \) (seen anywhere)     (M1)

\((g \circ f)(x) = \cos 4x\)

\(k = 4\)     A1     N2

[3 marks]

In part (a), a number of candidates were not able to evaluate \(\cos \pi \) , either leaving it or evaluating it incorrectly.

Almost all candidates evaluated the composite function in part (b) in the given order, many earning follow-through marks for incorrect answers from part (a). On both parts (a) and (b), there were candidates who correctly used double-angle formulas to come up with correct answers; while this is a valid method, it required unnecessary additional work.

Candidates were not as successful in part (c). Many tried to use double-angle formulas, but either used the formula incorrectly or used it to write the expression in terms of \(\cos x\) and went no further. There were a number of cases in which the candidates "accidentally" came up with the correct answer based on errors or lucky guesses and did not earn credit for their final answer. Only a few candidates recognized the correct method of solution.

Find \({\rm{A\hat BC}}\).

Find the exact area of the sector BDC.

METHOD 1

correct substitution into formula for area of triangle     (A1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B,{\text{ }}6\sqrt 3 \sin B,{\text{ }}\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B = 3\sqrt 3 \)

correct working     (A1)

eg\(\,\,\,\,\,\)\(6\sqrt 3 \sin B = 3\sqrt 3 ,{\text{ }}\sin B = \frac{{3\sqrt 3 }}{{\frac{1}{2}(6)2\sqrt 3 }}\)

\(\sin B = \frac{1}{2}\)    (A1)

\(\frac{\pi }{6}(30^\circ )\)    (A1)

\({\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )\)     A1     N3

METHOD 2

(using height of triangle ABC by drawing perpendicular segment from C to AD)

correct substitution into formula for area of triangle     (A1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}\left( {2\sqrt 3 } \right)(h) = 3\sqrt 3 ,{\text{ }}h\sqrt 3 \)

correct working     (A1)

eg\(\,\,\,\,\,\)\(h\sqrt 3  = 3\sqrt 3 \)

height of triangle is 3     A1

\({\rm{C\hat BD}} = \frac{\pi }{6}(30^\circ )\)    (A1)

\({\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )\)     A1     N3

[5 marks]

recognizing supplementary angle     (M1)

eg\(\,\,\,\,\,\)\({\rm{C\hat BD}} = \frac{\pi }{6},{\text{ sector}} = \frac{1}{2}(180 - {\rm{A\hat BC)(}}{{\text{6}}^2})\)

correct substitution into formula for area of sector     (A1)

eg\(\,\,\,\,\,\)\(\frac{1}{2} \times \frac{\pi }{6} \times {6^2},{\text{ }}\pi ({6^2})\left( {\frac{{30}}{{360}}} \right)\)

\({\text{area}} = 3\pi {\text{ }}({\text{c}}{{\text{m}}^2})\)     A1     N2

[3 marks]

In part (a) of this question, the large majority of candidates substituted correctly into the area formula for the triangle, though algebraic errors kept some of them from simplifying the equation to \(\sin {\rm{A\hat BC = }}\frac{1}{2}\). Unfortunately, a number of candidates who got to this point often did not know the correct angles that correspond with this sine value.

In part (b), many candidates realized that \({\rm{C\hat BD}}\) was the supplement of \({\rm{A\hat BC}}\). However, at this point many candidates substituted 30°, or their follow-through angle in degrees, into the formula for the area of a sector found in the formula booklet, not understanding that this formula only works for angles in radians.

(i)     Find \(\overrightarrow {{\text{AB}}} \).

(ii)     Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).

Show that the coordinates of C are \(( - 2,{\text{ }}1,{\text{ }}3)\).

Write down an expression in terms of \(\theta \) for

(i)     angle ADB;

(ii)     area of triangle ABD.

Given that \(\frac{{{\text{area }}\Delta {\text{ABD}}}}{{{\text{area }}\Delta {\text{ACD}}}} = 3\), show that \(\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}\).

Hence or otherwise, find the coordinates of point D.

(i)     valid approach to find \(\overrightarrow {{\text{AB}}} \)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OB}}} - \overrightarrow {{\text{OA}}} ,{\text{ }}\left( {\begin{array}{*{20}{c}} {4 - ( - 1)} \\ {1 - 0} \\ {3 - 4} \end{array}} \right)\)

\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 5 \\ 1 \\ { - 1} \end{array}} \right)\)    A1     N2

(ii)     valid approach to find \(\left| {\overrightarrow {{\text{AB}}} } \right|\)     (M1)

eg\(\,\,\,\,\,\)\(\sqrt {{{(5)}^2} + {{(1)}^2} + {{( - 1)}^2}} \)

\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {27} \)     A1     N2

[4 marks]

correct approach     A1

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ { - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 1} \\ 0 \\ 4 \end{array}} \right)\)

\(C\) has coordinates \(( - 2,{\text{ }}1,{\text{ }}3)\)     AG     N0

[1 mark]

(i)     \({\rm{A\hat DB}} = \pi  - \theta ,{\rm{ \hat D}} = 180 - \theta \)     A1     N1

(ii)     any correct expression for the area involving \(\theta \)     A1     N1

eg\(\,\,\,\,\,\)\({\text{area}} = \frac{1}{2} \times {\text{AD}} \times {\text{BD}} \times \sin (180 - \theta ),{\text{ }}\frac{1}{2}ab\sin \theta ,{\text{ }}\frac{1}{2}\left| {\overrightarrow {{\text{DA}}} } \right|\left| {\overrightarrow {{\text{DB}}} } \right|\sin (\pi  - \theta )\)

[2 marks]

METHOD 1 (using sine formula for area)

correct expression for the area of triangle ACD (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}{\text{AD}} \times {\text{DC}} \times \sin \theta \)

correct equation involving areas     A1

eg\(\,\,\,\,\,\)\(\frac{{\frac{1}{2}{\text{AD}} \times {\text{BD}} \times \sin (\pi  - \theta )}}{{\frac{1}{2}{\text{AD}} \times {\text{DC}} \times \sin \theta }} = 3\)

recognizing that \(\sin (\pi  - \theta ) = \sin \theta \) (seen anywhere)     (A1)

\(\frac{{{\text{BD}}}}{{{\text{DC}}}} = 3\) (seen anywhere)     (A1)

correct approach using ratio     A1

eg\(\,\,\,\,\,\)\(3\overrightarrow {{\text{DC}}}  + \overrightarrow {{\text{DC}}}  = \overrightarrow {{\text{BC}}} ,{\text{ }}\overrightarrow {{\text{BC}}}  = 4\overrightarrow {{\text{DC}}} \)

correct ratio \(\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}\)     AG     N0

METHOD 2 (Geometric approach)

recognising \(\Delta {\text{ABD}}\) and \(\Delta {\text{ACD}}\) have same height     (A1)

eg\(\,\,\,\,\,\)use of \(h\) for both triangles, \(\frac{{\frac{1}{2}{\text{BD}} \times h}}{{\frac{1}{2}{\text{CD}} \times h}} = 3\)

correct approach     A2

eg\(\,\,\,\,\,\)\({\text{BD}} = 3x\) and \({\text{DC}} = x,{\text{ }}\frac{{{\text{BD}}}}{{{\text{DC}}}} = 3\)

correct working     A2

eg\(\,\,\,\,\,\)\({\text{BC}} = 4x,{\text{ BD}} + {\text{DC}} = 4{\text{DC, }}\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{{3x}}{{4x}},{\text{ }}\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{{3{\text{DC}}}}{{4{\text{DC}}}}\)

\(\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}\)    AG     N0

[5 marks]

correct working (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{BD}}} = \frac{3}{4}\overrightarrow {{\text{BC}}} ,{\text{ }}\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OB}}} + \frac{3}{4}\left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right),{\text{ }}\overrightarrow {{\text{CD}}} = \frac{1}{4}\overrightarrow {{\text{CB}}} \)

valid approach (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OB}}} + \overrightarrow {{\text{BD}}} ,{\text{ }}\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right)\)

correct working to find \(x\)-coordinate     (A1)

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 3 \end{array}} \right) + \frac{3}{4}\left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right),{\text{ }}x = 4 + \frac{3}{4}( - 6),{\text{ }} - 2 + \frac{1}{4}(6)\)

D is \(\left( { - \frac{1}{2},{\text{ }}1,{\text{ }}3} \right)\)     A1     N3

[4 marks]

Find the value of a .

(i)     Show that the period of f is \(\pi \) .

(ii)    Hence, find the value of b .

Given that \(0 < c < \pi \)  , write down the value of c .

evidence of valid approach     (M1)

e.g. \(\frac{{{\text{max }}y{\text{ value}} - {\text{min }}y{\text{ value}}}}{2}\) , distance from \(y = - 1\)

\(a = 3\)     A1     N2

[2 marks]

(i) evidence of valid approach     (M1)

e.g. finding difference in x-coordinates, \(\frac{\pi }{2}\)

evidence of doubling     A1

e.g. \(2 \times \left( {\frac{\pi }{2}} \right)\)

\({\text{period}} = \pi \)      AG     N0

(ii) evidence of valid approach     (M1)

e.g. \(b = \frac{{2\pi }}{\pi }\)

\(b = 2\)     A1     N2

[4 marks]

\(c = \frac{\pi }{4}\)     A1     N1

[1 mark]

A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph.

A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had trouble showing that the period was \(\pi \) , either incorrectly adding the given \(\pi /4\) and \(3\pi /4\) or using the value of b that they found first for part (b)(ii).

A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had trouble showing that the period was \(\pi \) , either incorrectly adding the given \(\pi /4\) and \(\pi /3\) or using the value of b that they found first for part (b)(ii).

Write down an expression for

(i)     \(\sin 140^\circ \) ;

(ii)    \(\cos 70^\circ \) .

Find an expression for \(\cos 140^\circ \) .

Find an expression for \(\tan 140^\circ \) .

(i) \(\sin 140^\circ = p\)     A1     N1

(ii) \(\cos 70^\circ = - q\)     A1     N1

[2 marks]

METHOD 1

evidence of using \({\sin ^2}\theta + {\cos ^2}\theta = 1\)     (M1)

e.g. diagram, \(\sqrt {1 - {p^2}} \) (seen anywhere)

\(\cos 140^\circ = \pm \sqrt {1 - {p^2}} \)     (A1)

\(\cos 140^\circ = - \sqrt {1 - {p^2}} \)     A1     N2

METHOD 2

evidence of using \(\cos 2\theta = 2{\cos ^2}\theta - 1\)     (M1)

\(\cos 140^\circ = 2{\cos ^2}70 - 1\)     (A1)

\(\cos 140^\circ = 2{( - q)^2} - 1\) \(( = 2{q^2} - 1)\)     A1     N2

[3 marks]

METHOD 1

\(\tan 140^\circ = \frac{{\sin 140^\circ }}{{\cos 140^\circ }} = - \frac{p}{{\sqrt {1 - {p^2}} }}\)     A1     N1

METHOD 2

\(\tan 140^\circ = \frac{p}{{2{q^2} - 1}}\)     A1     N1

[1 mark]

This was one of the most difficult problems for the candidates. Even the strongest candidates had a hard time with this one and only a few received any marks at all.

Many did not appear to know the relationships between trigonometric functions of supplementary angles and that the use of \({\sin ^2}x + {\cos ^2}x = 1\) results in a \( \pm \) value. The application of a double angle formula also seemed weak.

This was one of the most difficult problems for the candidates. Even the strongest candidates had a hard time with this one and only a few received any marks at all. Many did not appear to know the relationships between trigonometric functions of supplementary angles and that the use of \({\sin ^2}x + {\cos ^2}x = 1\) results in a \( \pm \) value. The application of a double angle formula also seemed weak.

Show that \({\text{AC}} = 7{\text{ cm}}\).

The shape in the following diagram is formed by adding a semicircle with diameter [AC] to the triangle.

Find the exact perimeter of this shape.

evidence of choosing the cosine rule     (M1)

eg\(\,\,\,\,\,\)\({c^2} = {a^2} + {b^2} - ab\cos C\)

correct substitution into RHS of cosine rule     (A1)

eg\(\,\,\,\,\,\)\({3^2} + {8^2} - 2 \times 3 \times 8 \times \cos \frac{\pi }{3}\)

evidence of correct value for \(\cos \frac{\pi }{3}\) (may be seen anywhere, including in cosine rule)     A1

eg\(\,\,\,\,\,\)\(\cos \frac{\pi }{3} = \frac{1}{2},{\text{ A}}{{\text{C}}^2} = 9 + 64 - \left( {48 \times \frac{1}{2}} \right),{\text{ }}9 + 64 - 24\)

correct working clearly leading to answer     A1

eg\(\,\,\,\,\,\)\({\text{A}}{{\text{C}}^2} = 49,{\text{ }}b = \sqrt {49} \)

\({\text{AC}} = 7{\text{ (cm)}}\)     AG     N0

Note:     Award no marks if the only working seen is \({\text{A}}{{\text{C}}^2} = 49\) or \({\text{AC}} = \sqrt {49} \) (or similar).

[4 marks]

correct substitution for semicircle     (A1)

eg\(\,\,\,\,\,\)\({\text{semicircle}} = \frac{1}{2}(2\pi  \times 3.5),{\text{ }}\frac{1}{2} \times \pi  \times 7,{\text{ }}3.5\pi \)

valid approach (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)\({\text{perimeter}} = {\text{AB}} + {\text{BC}} + {\text{semicircle, }}3 + 8 + \left( {\frac{1}{2} \times 2 \times \pi  \times \frac{7}{2}} \right),{\text{ }}8 + 3 + 3.5\pi \)

\(11 + \frac{7}{2}\pi {\text{ }}( = 3.5\pi  + 11){\text{ (cm)}}\)     A1     N2

[3 marks]

The following diagram shows triangle PQR.

Find PR.

METHOD 1 

evidence of choosing the sine rule     (M1)

eg\(\,\,\,\,\,\)\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}}\)

correct substitution     A1

eg\(\,\,\,\,\,\)\(\frac{x}{{\sin 30}} = \frac{{13}}{{\sin 45}},{\text{ }}\frac{{13\sin 30}}{{\sin 45}}\)

\(\sin 30 = \frac{1}{2},{\text{ }}\sin 45 = \frac{1}{{\sqrt 2 }}\)     (A1)(A1)

correct working     A1

eg\(\,\,\,\,\,\)\(\frac{1}{2} \times \frac{{13}}{{\frac{1}{{\sqrt 2 }}}},{\text{ }}\frac{1}{2} \times 13 \times \frac{2}{{\sqrt 2 }},{\text{ }}13 \times \frac{1}{2} \times \sqrt 2 \)

correct answer     A1     N3

eg\(\,\,\,\,\,\)\({\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}\)

METHOD 2 (using height of ΔPQR)

valid approach to find height of ΔPQR     (M1)

eg\(\,\,\,\,\,\)\(\sin 30 = \frac{x}{{13}},{\text{ }}\cos 60 = \frac{x}{{13}}\)

\(\sin 30 = \frac{1}{2}{\text{ or }}\cos 60 = \frac{1}{2}\)     (A1)

\({\text{height}} = 6.5\)     A1

correct working     A1

eg\(\,\,\,\,\,\)\(\sin 45 = \frac{{6.5}}{{{\text{PR}}}},{\text{ }}\sqrt {{{6.5}^2} + {{6.5}^2}} \)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\sin 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\cos 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\sqrt {\frac{{169 \times 2}}{4}} \)

correct answer     A1     N3

eg\(\,\,\,\,\,\)\({\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}\)

[6 marks]

Write down

(i)     \(f'(x)\) ;

(ii)    \(g'(x)\) .

Let \(h(x) = {{\rm{e}}^{ - 3x}}\sin \left( {x - \frac{\pi }{3}} \right)\) . Find the exact value of \(h'\left( {\frac{\pi }{3}} \right)\) .

(i) \( - 3{{\rm{e}}^{ - 3x}}\)     A1     N1

(ii) \(\cos \left( {x - \frac{\pi }{3}} \right)\)     A1     N1

[4 marks]

evidence of choosing product rule     (M1)

e.g. \(uv' + vu'\)

correct expression     A1

e.g. \( - 3{{\rm{e}}^{ - 3x}}\sin \left( {x - \frac{\pi }{3}} \right) + {{\rm{e}}^{ - 3x}}\cos \left( {x - \frac{\pi }{3}} \right)\)

complete correct substitution of \(x = \frac{\pi }{3}\)     (A1)

e.g. \( - 3{{\rm{e}}^{ - 3\frac{\pi }{3}}}\sin \left( {\frac{\pi }{3} - \frac{\pi }{3}} \right) + {{\rm{e}}^{ - 3\frac{\pi }{3}}}\cos \left( {\frac{\pi }{3} - \frac{\pi }{3}} \right)\)        

\(h'\left( {\frac{\pi }{3}} \right) = {{\rm{e}}^{ - \pi }}\)     A1     N3

[4 marks]

A good number of candidates found the correct derivative expressions in (a). Many applied the product rule, although with mixed success.

Often the substitution of \({\frac{\pi }{3}}\) was incomplete or not done at all.

Given that \(\cos A = \frac{1}{3}\) and \(0 \le A \le \frac{\pi }{2}\) , find \(\cos 2A\) .

Given that \(\sin B = \frac{2}{3}\)  and \(\frac{\pi }{2} \le B \le \pi \) , find \(\cos B\) .

evidence of choosing the formula \(\cos 2A = 2{\cos ^2}A - 1\)     (M1)

Note: If they choose another correct formula, do not award the M1 unless there is evidence of finding \({\sin ^2}A = 1 - \frac{1}{9}\)

correct substitution     A1

 e.g.\(\cos 2A = {\left( {\frac{1}{3}} \right)^2} - \frac{8}{9}\) , \(\cos 2A = 2 \times {\left( {\frac{1}{3}} \right)^2} - 1\)

 \(\cos 2A = - \frac{7}{9}\)     A1     N2

[3 marks]

METHOD 1

evidence of using \({\sin ^2}B + {\cos ^2}B = 1\)     (M1)

e.g. \({\left( {\frac{2}{3}} \right)^2} + {\cos ^2}B = 1\) , \(\sqrt {\frac{5}{9}} \) (seen anywhere),

\(\cos B = \pm \sqrt {\frac{5}{9}} \) \(\left( { = \pm \frac{{\sqrt 5 }}{3}} \right)\)     (A1)

 \(\cos B = - \sqrt {\frac{5}{9}} \) \(\left( { = - \frac{{\sqrt 5 }}{3}} \right)\)     A1     N2

METHOD 2

diagram     M1

for finding third side equals \(\sqrt 5 \)     (A1)

\(\cos B = - \frac{{\sqrt 5 }}{3}\)     A1     N2

[3 marks]

This question was very poorly done, and knowledge of basic trigonometric identities and values of trigonometric functions of obtuse angles seemed distinctly lacking. Candidates who recognized the need of an identity for finding \(\cos 2A\) given \(\cos A\) seldom chose the most appropriate of the three and even when they did often used it incorrectly with expressions such as \(2{\cos ^2}\frac{1}{9} - 1\) .

This question was very poorly done, and knowledge of basic trigonometric identities and values of trigonometric functions of obtuse angles seemed distinctly lacking. Candidates who recognized the need of an identity for finding \(\cos 2A\) given \(\cos A\) seldom chose the most appropriate of the three and even when they did often used it incorrectly with expressions such as \(2{\cos ^2}\frac{1}{9} - 1\) .

Write down the amplitude of \(f\).

Find the period of \(f\).

On the following grid, sketch the graph of \(y = f(x)\), for \(0 \le x \le 3\).

amplitude is 3     A1     N1

valid approach     (M1)

eg\(\;\;\;{\text{period}} = \frac{{2\pi }}{\pi },{\text{ }}\frac{{360}}{\pi }\)

period is 2     A1     N2

      A1

A1A1A1     N4

Note:     Award A1 for sine curve starting at (0, 0) and correct period.

Only if this A1 is awarded, award the following for points in circles:

A1 for correct x-intercepts;

A1 for correct max and min points;

A1 for correct domain.

Find an expression for r in terms of θ.

Find the possible values of r.

Show that the sum of the infinite sequence is \(\frac{{54}}{{2 + {\text{cos}}\,\left( {2\theta } \right)}}\).

Find the values of θ which give the greatest value of the sum.

valid approach     (M1)

eg   \(\frac{{{u_2}}}{{{u_1}}},\,\,\frac{{{u_1}}}{{{u_2}}}\)

\(r = \frac{{12\,{{\sin }^2}\,\theta }}{{18}}\left( { = \frac{{2\,{{\sin }^2}\,\theta }}{3}} \right)\)      A1 N2

[2 marks]

recognizing that sinθ is bounded      (M1)

eg    0 ≤ sin² θ ≤ 1, −1 ≤ sinθ ≤ 1, −1 < sinθ < 1

0 < r ≤ \(\frac{2}{3}\)      A2 N3

Note: If working shown, award M1A1 for correct values with incorrect inequality sign(s).
If no working shown, award N1 for correct values with incorrect inequality sign(s).

[3 marks]

correct substitution into formula for infinite sum       A1

eg  \(\frac{{18}}{{1 - \frac{{2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

evidence of choosing an appropriate rule for cos 2θ (seen anywhere)         (M1)

eg   cos 2θ = 1 − 2 sin² θ

correct substitution of identity/working (seen anywhere)      (A1)

eg   \(\frac{{18}}{{1 - \frac{2}{3}\left( {\frac{{1 - {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 - 2\left( {\frac{{1 - {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 - 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

correct working that clearly leads to the given answer       A1

eg  \(\frac{{18 \times 3}}{{2 + \left( {1 - 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)}},\,\,\frac{{54}}{{3 - \left( {1 - {\text{cos}}\,2\theta } \right)}}\)

\(\frac{{54}}{{2 + {\text{cos}}\left( {2\theta } \right)}}\)    AG N0

[4 marks]

METHOD 1 (using differentiation)

recognizing \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }} = 0\) (seen anywhere)       (M1)

finding any correct expression for \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }}\)       (A1)

eg  \(\frac{{0 - 54 \times \left( { - 2\,{\text{sin}}\,2\,\theta } \right)}}{{{{\left( {2 + {\text{cos}}\,2\,\theta } \right)}^2}}},\,\, - 54{\left( {2 + {\text{cos}}\,2\,\theta } \right)^{ - 2}}\,\left( { - 2\,{\text{sin}}\,2\,\theta } \right)\)

correct working       (A1)

eg  sin 2θ = 0

any correct value for sin⁻¹(0) (seen anywhere)       (A1)

eg  0, \(\pi \), … , sketch of sine curve with x-intercept(s) marked both correct values for 2θ (ignore additional values)      (A1)

2θ = \(\pi \), 3\(\pi \) (accept values in degrees)

both correct answers \(\theta  = \frac{\pi }{2},\,\frac{{3\pi }}{2}\)      A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

METHOD 2 (using denominator)

recognizing when S_∞ is greatest      (M1)

eg 2 + cos 2θ is a minimum, 1−r is smallest
correct working      (A1)

eg  minimum value of 2 + cos 2θ is 1, minimum r = \(\frac{2}{3}\)

correct working      (A1)

eg  \({\text{cos}}\,2\,\theta  =  - 1,\,\,\frac{2}{3}\,{\text{si}}{{\text{n}}^2}\,\theta  = \frac{2}{3},\,\,{\text{si}}{{\text{n}}^2}\theta  = 1\)

EITHER (using cos 2θ)

any correct value for cos⁻¹(−1) (seen anywhere)      (A1)

eg  \(\pi \), 3\(\pi \), … (accept values in degrees), sketch of cosine curve with x-intercept(s) marked

both correct values for 2θ  (ignore additional values)      (A1)

2θ = \(\pi \), 3\(\pi \) (accept values in degrees)

OR (using sinθ)

sinθ = ±1     (A1)

sin⁻¹(1) = \(\frac{\pi }{2}\) (accept values in degrees) (seen anywhere)      A1

THEN

both correct answers \(\theta  = \frac{\pi }{2},\,\frac{{3\pi }}{2}\)       A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

[6 marks]

(i)     Write down the value of a .

(ii)    Find the value of b .

Write down the gradient of the curve at P.

Write down the equation of the normal to the curve at P.

(i) \(a = 3\)     A1     N1

(ii) METHOD 1

attempt to find period     (M1)

e.g. 4 , \(b = 4\) , \(\frac{{2\pi }}{b}\)

\(b = \frac{{2\pi }}{4}\left( { = \frac{\pi }{2}} \right)\)     A1     N2

[3 marks]

METHOD 2

attempt to substitute coordinates     (M1)

e.g. \(3\cos (2b) = - 3\) , \(3\cos (4b) = 3\)

\(b = \frac{{2\pi }}{4}\left( { = \frac{\pi }{2}} \right)\)     A1     N2

[3 marks]

0     A1     N1

[1 mark]

recognizing that normal is perpendicular to tangent     (M1)

e.g. \({m_1} \times {m_2} = - 1\) , \(m = - \frac{1}{0}\) , sketch of vertical line on diagram

\(x = 2\) (do not accept 2 or \(y = 2\) )     A1     N2

[2 marks]

In part (a), many candidates were able to successfully write down the value of a as instructed by inspecting the graph and seeing the amplitude of the function is 3. Many also used a formulaic approach to reach the correct answer. When finding the value of b, there were many candidates who thought b was the period of the function, rather than \(\frac{{2\pi }}{{{\rm{period}}}}\) .

In part (b), the directions asked candidates to write down the gradient of the curve at the local minimum point P. However, many candidates spent a good deal of time finding the derivative of the function and finding the value of the derivative for the given value of x, rather than simply stating that the gradient of a curve at a minimum point is zero.

For part (c), finding the equation of the normal to the curve, many candidates tried to work with algebraic equations involving negative reciprocal gradients, rather than recognizing that the equation of the vertical line was \(x = 2\) . There were also candidates who had trouble expressing the correct equation of a line parallel to the y-axis.

Find the area of the triangle OPB, in terms of \(\theta \) .

Explain why the area of triangle OPA is the same as the area triangle OPB.

Show that \(S = 2(\pi  - 2\sin \theta )\) .

Find the value of \(\theta \) when S is a local minimum, justifying that it is a minimum.

Find a value of \(\theta \) for which S has its greatest value.

evidence of using area of a triangle     (M1)

e.g. \(A = \frac{1}{2} \times 2 \times 2 \times \sin \theta \)

\(A = 2\sin \theta \)     A1     N2

[2 marks]

METHOD 1

\({\rm{P}}\widehat {\rm{O}}{\rm{A = }}\pi  - \theta \)     (A1)

\({\text{area }}\Delta {\rm{OPA}} = \frac{1}{2}2 \times 2 \times \sin (\pi  - \theta )\) \(( = 2\sin (\pi  - \theta ))\)     A1

since \(\sin (\pi  - \theta ) = \sin \theta \)     R1

then both triangles have the same area     AG     N0

METHOD 2

triangle OPA has the same height and the same base as triangle OPB     R3

then both triangles have the same area     AG     N0

[3 marks]

area semicircle \( = \frac{1}{2} \times \pi {(2)^2}\) \(( = 2\pi )\)     A1

\({\text{area }}\Delta {\rm{APB}} = 2\sin \theta + 2\sin \theta \) \(( = 4\sin \theta )\)     A1

\(S{\text{ = area of semicircle}} - {\text{area }}\Delta {\rm{APB}}\) \(( = 2\pi - 4\sin \theta )\)     M1

\(S = 2(\pi - 2\sin \theta )\)     AG     N0

[3 marks]

METHOD 1

attempt to differentiate     (M1)

e.g. \(\frac{{{\rm{d}}S}}{{{\rm{d}}\theta }} = - 4\cos \theta \)

setting derivative equal to 0     (M1)

correct equation     A1

e.g. \( - 4\cos \theta = 0\) , \(\cos \theta = 0\) , \(4\cos \theta = 0\)

\(\theta = \frac{\pi }{2}\)     A1     N3

EITHER

evidence of using second derivative     (M1)

\(S''(\theta ) = 4\sin \theta \)   A1

\(S''\left( {\frac{\pi }{2}} \right) = 4\)     A1

it is a minimum because \(S''\left( {\frac{\pi }{2}} \right) > 0\)     R1     N0

OR

evidence of using first derivative      (M1)

for \(\theta < \frac{\pi }{2},S'(\theta ) < 0\) (may use diagram)     A1

for \(\theta > \frac{\pi }{2},S'(\theta ) > 0\) (may use diagram)    A1

it is a minimum since the derivative goes from negative to positive     R1     N0

METHOD 2

\(2\pi - 4\sin \theta \) is minimum when \(4\sin \theta \) is a maximum     R3

\(4\sin \theta \) is a maximum when \(\sin \theta = 1\)     (A2)

\(\theta = \frac{\pi }{2}\)     A3     N3

[8 marks]

S is greatest when \(4\sin \theta \) is smallest (or equivalent)     (R1)

\(\theta = 0\) (or \(\pi \) )     A1     N2

[2 marks]

Most candidates could obtain the area of triangle OPB as equal to \(2\sin \theta \) , though \(2\theta \) was given quite often as the area.

A minority recognized the equality of the sines of supplementary angles and the term complementary was frequently used instead of supplementary. Only a handful of candidates used the simple equal base and altitude argument.

Many candidates seemed to see why \(S = 2(\pi - 2\sin \theta )\) but the arguments presented for showing why this result was true were not very convincing in many cases. Explicit evidence of why the area of the semicircle was \(2\pi \) was often missing as was an explanation for \(2(2\sin \theta )\) and for subtraction.

Only a small number of candidates recognized the fact S would be minimum when sin was maximum, leading to a simple non-calculus solution. Those who chose the calculus route often had difficulty finding the derivative of S, failing in a significant number of cases to recognize that the derivative of a constant is 0, and also going through painstaking application of the product rule to find the simple derivative. When it came to justify a minimum, there was evidence in some cases of using some form of valid test, but explanation of the test being used was generally poor.

Candidates who answered part (d) correctly generally did well in part (e) as well, though answers outside the domain of \(\theta \) were frequently seen.

Write \(h(x)\) in the form \(a\sin (bx)\), where \(a,{\text{ }}b \in \mathbb{Z}\).

Hence find the range of \(h\).

attempt to form composite in any order     (M1)

eg\(\,\,\,\,\,\)\(f\left( {g(x)} \right),{\text{ }}\cos \left( {6x\sqrt {1 - {x^2}} } \right)\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(6\cos x\sqrt {1 - {{\cos }^2}x} \)

correct application of Pythagorean identity (do not accept \({\sin ^2}x + {\cos ^2}x = 1\))     (A1)

eg\(\,\,\,\,\,\)\({\sin ^2}x = 1 - {\cos ^2}x,{\text{ }}6\cos x\sin x,{\text{ }}6\cos x \left| \sin x\right|\)

valid approach (do not accept \(2\sin x\cos x = \sin 2x\))     (M1)

eg\(\,\,\,\,\,\)\(3(2\cos x\sin x)\)

\(h(x) = 3\sin 2x\)    A1     N3

[5 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)amplitude \( = 3\), sketch with max and min \(y\)-values labelled, \( - 3 < y < 3\)

correct range     A1     N2

eg\(\,\,\,\,\,\)\( - 3 \leqslant y \leqslant 3\), \([ - 3,{\text{ }}3]\) from \( - 3\) to 3

Note:     Do not award A1 for \( - 3 < y < 3\) or for “between \( - 3\) and 3”.

[2 marks]

In part (a), nearly all candidates found the correct composite function in terms of \(\cos x\), though many did not get any further than this first step in their solution to the question. While some candidates seemed to recognize the need to use trigonometric identities, most were unsuccessful in finding the correct expression in the required form.

In part (b), very few candidates were able to provide the correct range of the function.

Find the length of the arc ACB .

Find the area of the shaded region.

correct substitution in \(l = r\theta \)     (A1)

e.g. \(10 \times \frac{\pi }{3}\) , \(\frac{1}{6} \times 2\pi  \times 10\)

arc length \( = \frac{{20\pi }}{6}\) \(\left( { = \frac{{10\pi }}{3}} \right)\)     A1      N2

[2 marks]

area of large sector \( = \frac{1}{2} \times {10^2} \times \frac{\pi }{3}\) \(\left( { = \frac{{100\pi }}{6}} \right)\)     (A1)

area of small sector \( = \frac{1}{2} \times {8^2} \times \frac{\pi }{3}\) \(\left( { = \frac{{64\pi }}{6}} \right)\)     (A1)

evidence of valid approach (seen anywhere)     M1

e.g. subtracting areas of two sectors, \(\frac{1}{2} \times \frac{\pi }{3}({10^2} - {8^2})\)

area shaded \( = 6\pi \) (accept \(\frac{{36\pi }}{6}\) , etc.)     A1     N3

[4 marks]

This question was very well done by the majority of candidates. Some candidates correctly substituted the values into the formulas, but failed to do the calculations and write their answers in finished form.

Nearly all used the correct method of subtracting the sector areas in part (b), though multiplying with fractions proved challenging for some candidates.

Find

(i)     \(\overrightarrow {{\rm{PQ}}} \) ;

(ii)    \(\overrightarrow {{\rm{PR}}} \) .

Show that \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) .

(i)     Find \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}}\) .

(ii)    Hence, find the area of triangle PQR, giving your answer in the form \(a\sqrt 3 \) .

(i) evidence of approach     (M1)

e.g. \(\overrightarrow {{\rm{PQ}}}  = \overrightarrow {{\rm{PO}}}  + \overrightarrow {{\rm{OQ}}} \) , \({\rm{Q}} - {\rm{P}}\)

\(\overrightarrow {{\rm{PQ}}}  = \left( {\begin{array}{*{20}{c}}
{ - 1}\\
2\\
1
\end{array}} \right)\)     A1     N2

(ii) \(\overrightarrow {{\rm{PR}}}  = \left( {\begin{array}{*{20}{c}}
2\\
2\\
4
\end{array}} \right)\)     A1     N1

[3 marks]

METHOD 1

choosing correct vectors \(\overrightarrow {{\rm{PQ}}} \) and \(\overrightarrow {{\rm{PR}}} \)    (A1)(A1)

finding \(\overrightarrow {{\rm{PQ}}}  \bullet \overrightarrow {{\rm{PR}}} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right|\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right|\)    (A1) (A1)(A1)

\(\overrightarrow {{\rm{PQ}}}  \bullet \overrightarrow {{\rm{PR}}}  = - 2 + 4 + 4( = 6)\)

\(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt {{{( - 1)}^2} + {2^2} + {1^2}} \) \(\left( { = \sqrt 6 } \right)\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {{2^2} + {2^2} + {4^2}} \) \(\left( { = \sqrt {24} } \right)\)

substituting into formula for angle between two vectors     M1

e.g. \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{6}{{\sqrt 6  \times \sqrt {24} }}\)

simplifying to expression clearly leading to \(\frac{1}{2}\)     A1

e.g. \(\frac{6}{{\sqrt 6  \times 2\sqrt 6 }}\) , \(\frac{6}{{\sqrt {144} }}\) , \(\frac{6}{{12}}\)

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\)     AG     N0

METHOD 2

evidence of choosing cosine rule (seen anywhere)     (M1)

\(\overrightarrow {{\rm{QR}}}  = \left( {\begin{array}{*{20}{c}}
3\\
0\\
3
\end{array}} \right)\)     A1

\(\left| {\overrightarrow {{\rm{QR}}} } \right| = \sqrt {18} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt 6 \) and \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {24} \)     (A1)(A1)(A1)

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt {24} } \right)}^2} - {{\left( {\sqrt {18} } \right)}^2}}}{{2\sqrt 6  \times \sqrt {24} }}\)     A1

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{6 + 24 - 18}}{{24}}\) \(\left( { = \frac{{12}}{{24}}} \right)\)     A1

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\)      AG     N0

[7 marks]

(i) METHOD 1

evidence of appropriate approach     (M1)

e.g. using \({\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q + co}}{{\rm{s}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q}} = 1\) , diagram

substituting correctly     (A1)

e.g. \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} \)

\({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {\frac{3}{4}} \) \(\left( { = \frac{{\sqrt 3 }}{2}} \right)\)     A1     N3

METHOD 2

since \(\cos \widehat {\rm{P}} = \frac{1}{2}\) , \(\widehat {\rm{P}} = 60^\circ \)     (A1)

evidence of approach

e.g. drawing a right triangle, finding the missing side     (A1)

\(\sin \widehat {\rm{P}} = \frac{{\sqrt 3 }}{2}\)     A1     N3

(ii) evidence of appropriate approach      (M1)

e.g. attempt to substitute into \(\frac{1}{2}ab\sin C\)

correct substitution

e.g. area \( = \frac{1}{2}\sqrt 6  \times \sqrt {24}  \times \frac{{\sqrt 3 }}{2}\)     A1

area \( = 3\sqrt 3 \)     A1     N2

[6 marks]

Combining the vectors in (a) was generally well done, although some candidates reversed the subtraction, while others calculated the magnitudes.

Many candidates successfully used scalar product and magnitude calculations to complete part (b). Alternatively, some used the cosine rule, and often achieved correct results. Some assumed the triangle was a right-angled triangle and thus did not earn full marks. Although PQR is indeed right-angled, in a “show that” question this attribute must be directly established.

Many candidates attained the value for sine in (c) with little difficulty, some using the Pythagorean identity, while others knew the side relationships in a 30-60-90 triangle. Unfortunately, a good number of candidates then used the side values of \(1,2,\sqrt 3 \) to find the area of PQR , instead of the magnitudes of the vectors found in (a). Furthermore, the "hence" command was sometimes neglected as the value of sine was expected to be used in the approach.

Solve the equation \(2\cos x = \sin 2x\) , for \(0 \le x \le 3\pi \) .

METHOD 1

using double-angle identity (seen anywhere)     A1

e.g. \(\sin 2x = 2\sin x\cos x\) , \(2\cos x = 2\sin x\cos x\)

evidence of valid attempt to solve equation     (M1)

e.g. \(0 = 2\sin x\cos x - 2\cos x\) , \(2\cos x(1 - \sin x) = 0\)

\(\cos x = 0\) , \(\sin x = 1\)     A1A1

\(x = \frac{\pi }{2}\) , \(x = \frac{{3\pi }}{2}\) , \(x = \frac{{5\pi }}{2}\)     A1A1A1     N4

METHOD 2

     A1A1M1A1

Notes: Award A1 for sketch of \(\sin 2x\) , A1 for a sketch of \(2\cos x\) , M1 for at least one intersection point seen, and A1 for 3 approximately correct intersection points. Accept sketches drawn outside \(\left[ {0,3\pi } \right]\) , even those with more than 3 intersections.

\(x = \frac{\pi }{2}\) , \(x = \frac{{3\pi }}{2}\) , \(x = \frac{{5\pi }}{2}\)     A1A1A1     N4

[7 marks]

By far the most common error was to “cancel” the \(\cos x\) and find only two of the three solutions. It was disappointing how few candidates solved this by setting factors equal to zero. Some candidates wrote all three answers from \(\sin x = 1\) , which only earned two of the three final marks. On a brighter note, many candidates found the \(\frac{{5\pi }}{2}\) , which showed an appreciation for the period of the function as well as the domain restriction. A handful of candidates cleverly sketched both graphs and used the intersections to find the three solutions.

Find \(f'(x)\) .

Hence

(i)     show that \(q = - 2\) ;

(ii)    verify that A is a minimum point.

Find the maximum value of \(f(x)\) .

The function \(f(x)\) can be written in the form \(r\cos (x - a)\) .

Write down the value of r and of a .

\(f'(x) = - \sin x + \sqrt 3 \cos x\)     A1A1     N2

[2 marks]

(i) at A, \(f'(x) = 0\)     R1

correct working     A1

e.g. \(\sin x = \sqrt 3 \cos x\)

\(\tan x = \sqrt 3 \)     A1

\(x = \frac{\pi }{3}\) , \(\frac{{4\pi }}{3}\)     A1

attempt to substitute their x into \(f(x)\)     M1

e.g. \(\cos \left( {\frac{{4\pi }}{3}} \right) + \sqrt 3 \sin \left( {\frac{{4\pi }}{3}} \right)\)

correct substitution     A1

e.g. \( - \frac{1}{2} + \sqrt 3 \left( { - \frac{{\sqrt 3 }}{2}} \right)\)

correct working that clearly leads to \( - 2\)     A1

e.g. \( - \frac{1}{2} - \frac{3}{2}\)

 \(q = - 2\)     AG     N0

(ii) correct calculations to find \(f'(x)\) either side of \(x = \frac{{4\pi }}{3}\)     A1A1

e.g. \(f'(\pi ) = 0 - \sqrt 3 \) ,  \(f'(2\pi ) = 0 + \sqrt 3 \)
  

\(f'(x)\) changes sign from negative to positive     R1

so A is a minimum     AG     N0

[10 marks]

max when \(x = \frac{\pi }{3}\)     R1

correctly substituting \(x = \frac{\pi }{3}\) into \(f(x)\)     A1

e.g. \(\frac{1}{2} + \sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right)\)

max value is 2     A1     N1

[3 marks]

\(r = 2\) , \(a = \frac{\pi }{3}\)     A1A1     N2

[2 marks]

Write down an expression in terms of \(\theta \) for

(i)     \(x\) ;

(ii)    \(y\) .

Let the area of the rectangle be A.

Show that \(A = 18\sin 2\theta \) .

(i)     Find \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }}\) .

(ii)    Hence, find the exact value of \(\theta \) which maximizes the area of the rectangle.

(iii)   Use the second derivative to justify that this value of \(\theta \) does give a maximum.

(i) \(x = 3\cos \theta \)     A1     N1 

(ii) \(y = 3\sin \theta \)     A1     N1

[2 marks]

finding area     (M1)

e.g. \(A = 2x \times 2y\) , \(A = 8 \times \frac{1}{2}bh\) 

substituting     A1

e.g. \(A = 4 \times 3\sin \theta  \times 3\cos \theta \) , \(8 \times \frac{1}{2} \times 3\cos \theta  \times 3\sin \theta \)

\(A = 18(2\sin \theta \cos \theta )\)    A1

\(A = 18\sin 2\theta \)     AG     N0

[3 marks]

(i) \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 36\cos 2\theta \)     A2     N2 

(ii) for setting derivative equal to 0     (M1)

e.g. \(36\cos 2\theta  = 0\) , \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 0\)

\(2\theta  = \frac{\pi }{2}\)     (A1)

\(\theta  = \frac{\pi }{4}\)     A1     N2

(iii) valid reason (seen anywhere)     R1

e.g. at \(\frac{\pi }{4}\), \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} < 0\) ; maximum when \(f''(x) < 0\)

finding second derivative \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} =  - 72\sin 2\theta \)     A1

evidence of substituting \(\frac{\pi }{4}\)     M1

e.g. \( - 72\sin \left( {2 \times \frac{\pi }{4}} \right)\) , \( - 72\sin \left( {\frac{\pi }{2}} \right)\) , \( - 72\)

\(\theta  = \frac{\pi }{4}\) produces the maximum area     AG     N0

[8 marks]

Candidates familiar with the circular nature of sine and cosine found part (a) accessible. However, a good number of candidates left this part blank, which suggests that there was difficulty interpreting the meaning of the x and y in the diagram. 

Those with answers from (a) could begin part (b), but many worked backwards and thus earned no marks. In a "show that" question, a solution cannot begin with the answer given. The area of the rectangle could be found by using \(2x \times 2y\) , or by using the eight small triangles, but it was essential that the substitution of the double-angle formula was shown before writing the given answer.

As the area function was given in part (b), many candidates correctly found the derivative in (c) and knew to set this derivative to zero for a maximum value. Many gave answers in degrees, however, despite the given domain in radians.

Although some candidates found the second derivative function correctly, few stated that the second derivative must be negative at a maximum value. Simply calculating a negative value is not sufficient for a justification.

Let \(h(x) = \frac{{6x}}{{\cos x}}\) . Find \(h'(0)\) .

METHOD 1 (quotient)

derivative of numerator is 6     (A1)

derivative of denominator is \( - \sin x\)     (A1)

attempt to substitute into quotient rule     (M1)

correct substitution     A1

e.g. \(\frac{{(\cos x)(6) - (6x)( - \sin x)}}{{{{(\cos x)}^2}}}\)

substituting \(x = 0\)     (A1)

e.g. \(\frac{{(\cos 0)(6) - (6 \times 0)( - \sin 0)}}{{{{(\cos 0)}^2}}}\)

\(h'(0) = 6\)     A1     N2

METHOD 2 (product)

\(h(x) = 6x \times {(\cos x)^{ - 1}}\)

derivative of 6x is 6     (A1)

derivative of \({(\cos x)^{ - 1}}\) is \(( - {(\cos x)^{ - 2}}( - \sin x))\)     (A1)

attempt to substitute into product rule     (M1)

correct substitution     A1

e.g. \((6x)( - {(\cos x)^{ - 2}}( - \sin x)) + (6){(\cos x)^{ - 1}}\)

substituting \(x = 0\)    (A1)

e.g. \((6 \times 0)( - {(\cos 0)^{ - 2}}( - \sin 0)) + (6){(\cos 0)^{ - 1}}\)

\(h'(0) = 6\)     A1     N2

[6 marks]

The majority of candidates were successful in using the quotient rule, and were able to earn most of the marks for this question. However, there were a large number of candidates who substituted correctly into the quotient rule, but then went on to make mistakes in simplifying this expression. These algebraic errors kept the candidates from earning the final mark for the correct answer. A few candidates tried to use the product rule to find the derivative, but they were generally not as successful as those who used the quotient rule. It was pleasing to note that most candidates did know the correct values for the sine and cosine of zero.

The following diagram shows a circle with centre O and radius r cm.

The points A and B lie on the circumference of the circle, and \({\text{A}}\mathop {\text{O}}\limits^ \wedge  {\text{B}}\) = θ. The area of the shaded sector AOB is 12 cm² and the length of arc AB is 6 cm.

Find the value of r.

evidence of correctly substituting into circle formula (may be seen later)      A1A1
eg  \(\frac{1}{2}\theta {r^2} = 12,\,\,r\theta  = 6\)

attempt to eliminate one variable      (M1)
eg  \(r = \frac{6}{\theta },\,\,\theta  = \frac{1}{r},\,\,\frac{{\frac{1}{2}\theta {r^2}}}{{r\theta }} = \frac{{12}}{6}\)

correct elimination      (A1)
eg  \(\frac{1}{2} \times \frac{6}{r} \times {r^2} = 12,\,\,\frac{1}{2}\theta  \times {\left( {\frac{6}{\theta }} \right)^2} = 12,\,\,A = \frac{1}{2} \times {r^2} \times \frac{l}{r},\,\,\frac{{{r^2}}}{{2r}} = 2\)

correct equation     (A1)
eg  \(\frac{1}{2} \times 6r = 12,\,\,\frac{1}{2} \times \frac{{36}}{\theta } = 12,\,\,12 = \frac{1}{2} \times {r^2} \times \frac{6}{r}\)

correct working      (A1)
eg  \(3r = 12,\,\,\frac{{18}}{\theta } = 12,\,\,\frac{r}{2} = 2,\,\,24 = 6r\)

r = 4 (cm)      A1 N2

[7 marks]

Six equilateral triangles, each with side length 3 cm, are arranged to form a hexagon.
This is shown in the following diagram.

The vectors p , q and r are shown on the diagram.

Find p•(p + q + r).

METHOD 1 (using |p| |2q| cosθ)

finding p + q + r      (A1)

eg  2q, 

| p + q + r | = 2 × 3 (= 6)  (seen anywhere)     A1

correct angle between p and q (seen anywhere)      (A1)

\(\frac{\pi }{3}\)  (accept 60°)

substitution of their values     (M1)

eg  3 × 6 × cos\(\left( {\frac{\pi }{3}} \right)\)

correct value for cos\(\left( {\frac{\pi }{3}} \right)\) (seen anywhere)     (A1)

eg  \(\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}\)

p•(p + q + r) = 9     A1 N3

METHOD 2 (scalar product using distributive law)

correct expression for scalar distribution      (A1)

eg  p• p + p•q + p•r

three correct angles between the vector pairs (seen anywhere)      (A2)

eg  0° between p and p, \(\frac{\pi }{3}\) between p and q, \(\frac{{2\pi }}{3}\) between p and r

Note: Award A1 for only two correct angles.

substitution of their values      (M1)

eg  3.3.cos0 +3.3.cos\(\frac{\pi }{3}\) + 3.3.cos120

one correct value for cos0, cos\(\left( {\frac{\pi }{3}} \right)\) or cos\(\left( {\frac{2\pi }{3}} \right)\) (seen anywhere)      A1

eg  \(\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}\)

p•(p + q + r) = 9     A1 N3

METHOD 3 (scalar product using relative position vectors)

valid attempt to find one component of p or r      (M1)

eg   sin 60 = \(\frac{x}{3}\), cos 60 = \(\frac{x}{3}\), one correct value \(\frac{3}{2},\,\,\frac{{3\sqrt 3 }}{2},\,\,\frac{{ - 3\sqrt 3 }}{2}\)

one correct vector (two or three dimensions) (seen anywhere)      A1

eg  \(p = \left( \begin{gathered}
\,\,\,\frac{3}{2} \hfill \\
\frac{{3\sqrt 3 }}{2} \hfill \\
\end{gathered} \right),\,\,q = \left( \begin{gathered}
3 \hfill \\
0 \hfill \\
\end{gathered} \right),\,\,r = \left( \begin{gathered}
\,\,\,\,\frac{3}{2} \hfill \\
- \frac{{3\sqrt 3 }}{2} \hfill \\
\,\,\,\,0 \hfill \\
\end{gathered} \right)\)

three correct vectors p + q + r = 2q     (A1)

p + q + r = \(\left( \begin{gathered}
6 \hfill \\
0 \hfill \\
\end{gathered} \right)\) or \(\left( \begin{gathered}
6 \hfill \\
0 \hfill \\
0 \hfill \\
\end{gathered} \right)\) (seen anywhere, including scalar product)      (A1)

correct working       (A1)
eg  \(\left( {\frac{3}{2} \times 6} \right) + \left( {\frac{{3\sqrt 3 }}{2} \times 0} \right),\,\,9 + 0 + 0\)

p•(p + q + r) = 9     A1 N3

[6 marks]