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Question
Let \(f(t) = a\cos b(t – c) + d\) , \(t \ge 0\) . Part of the graph of \(y = f(t)\) is given below.
When \(t = 3\) , there is a maximum value of 29, at M.
When \(t = 9\) , there is a minimum value of 15.
(i)Â Â Â Â Find the value of a.
(ii) Â Â Show that \(b = \frac{\pi }{6}\) .
(iii)Â Â Find the value of d.
(iv) Â 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 .
Answer/Explanation
Markscheme
(i) attempt to substitute    (M1)
e.g. \(a = \frac{{29 – 15}}{2}\)
\(a = 7\) (accept \(a = – 7\)Â )Â Â Â Â A1Â Â Â Â N2
(ii) \({\text{period}} = 12\)Â Â Â Â (A1)
\(b = \frac{{2\pi }}{{12}}\)Â Â Â A1
\(b = \frac{\pi }{6}\)Â Â Â AGÂ Â Â N0
(iii) attempt to substitute    (M1)
e.g. \(d = \frac{{29 + 15}}{2}\)
\(d = 22\)Â Â Â Â A1Â Â Â Â N2
(iv) \(c = 3\)Â (accept \(c = 9\)Â from \(a = – 7\)Â )Â Â Â Â A1 Â Â N1
Note: Other correct values for c can be found, \(c = 3 \pm 12k\) , \(k \in \mathbb{Z}\) .
[7 marks]
stretch takes 3 to 1.5Â Â Â Â (A1)
translation maps \((1.5{\text{, }}29)\) to \((4.5{\text{, }}19)\) (so \({M’}\) is \((4.5{\text{, }}19)\))Â Â Â Â A1Â Â Â Â N2
[2 marks]
\(g(t) = 7\cos \frac{\pi }{3}\left( {t – 4.5} \right) + 12\)Â Â Â A1A2A1 Â Â N4
Note: Award A1 for \(\frac{\pi }{3}\)Â , 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 \(\left( {\begin{array}{*{20}{c}}
{ – 3}\\
{10}
\end{array}} \right)\)Â Â Â Â (A1)
horizontal stretch of a scale factor of 2Â Â Â Â (A1)
completely correct description, in correct order    A1   N3
e.g. translation \(\left( {\begin{array}{*{20}{c}}
{ – 3}\\
{10}
\end{array}} \right)\)Â then horizontal stretch of a scale factor of 2
[3 marks]
Question
Let \(f(x) = {(\sin x + \cos x)^2}\) .
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 \(0 \le x \le 2\pi \) .
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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 .
Answer/Explanation
Markscheme
attempt to expand    (M1)
e.g. \((\sin x + \cos x)(\sin x + \cos x)\) ; at least 3 terms
correct expansion    A1
e.g. \({\sin ^2}x + 2\sin x\cos x + {\cos ^2}x\)
\(f(x) = 1 + \sin 2x\)Â Â Â Â AGÂ Â Â Â 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]
Question
Let \(f(x) = \sin \left( {x + \frac{\pi }{4}} \right) + k\). The graph of f passes through the point \(\left( {\frac{\pi }{4},{\text{ }}6} \right)\).
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 \(q\).
Answer/Explanation
Markscheme
METHOD 1
attempt to substitute both coordinates (in any order) into \(f\) Â Â (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   \(\sin \frac{\pi }{2} = 1,{\text{ }}1 + k = 6\)
\(k = 5\) Â Â A1 Â Â N2
[3 marks]
METHOD 2
recognizing shift of \(\frac{\pi }{4}\) left means maximum at \(6\) Â Â Â R1)
recognizing \(k\) is difference of maximum and amplitude   (A1)
eg   \(6 – 1\)
\(k = 5\) Â Â A1 Â Â N2
[3 marks]Â
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 \(4\) Â Â Â A1 Â Â 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]
Question
Let \(f(x) = 3\sin \left( {\frac{\pi }{2}x} \right)\), for \(0 \leqslant x \leqslant 4\).
(i) Â Â Write down the amplitude of \(f\).
(ii) Â Â Find the period of \(f\).
On the following grid sketch the graph of \(f\).
Answer/Explanation
Markscheme
(i) Â Â 3 Â Â A1 Â Â N1
(ii)   valid attempt to find the period   (M1)
eg\(\,\,\,\,\,\)\(\frac{{2\pi }}{b},{\text{ }}\frac{{2\pi }}{{\frac{\pi }{2}}}\)
period \( = 4\)Â Â Â A1 Â Â N2
[3 marks]
   A1A1A1A1   N4
[4 marks]
Question
Let \(f(x) = \cos x + \sqrt 3 \sin x\) , \(0 \le x \le 2\pi \) . The following diagram shows the graph of \(f\) .
The \(y\)-intercept is at (\(0\), \(1\)) , there is a minimum point at A (\(p\), \(q\)) and a maximum point at B.
Find \(f'(x)\) .
Hence
(i) Â Â show that \(q = – 2\) ;
(ii)Â Â Â 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 .
Answer/Explanation
Markscheme
\(f'(x) = – \sin x + \sqrt 3 \cos x\)Â Â Â Â A1A1 Â Â N2
[2 marks]
(i) at A, \(f'(x) = 0\)Â Â Â Â Â R1
correct working    A1
e.g. \(\sin x = \sqrt 3 \cos x\)
\(\tan x = \sqrt 3 \)Â Â Â Â Â A1
\(x = \frac{\pi }{3}\) , \(\frac{{4\pi }}{3}\)Â Â Â Â A1
attempt to substitute their x into \(f(x)\)Â Â Â Â Â 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 \( – 2\) Â Â A1
e.g. \( – \frac{1}{2} – \frac{3}{2}\)
\(q = – 2\)Â Â Â Â Â AGÂ Â Â Â N0
(ii) correct calculations to find \(f'(x)\) either side of \(x = \frac{{4\pi }}{3}\)Â Â Â Â 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 \(x = \frac{\pi }{3}\)Â Â Â Â R1
correctly substituting \(x = \frac{\pi }{3}\)Â into \(f(x)\)Â Â Â Â A1
e.g. \(\frac{1}{2} + \sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right)\)
max value is 2Â Â Â Â A1Â Â Â Â N1
[3 marks]
\(r = 2\) , \(a = \frac{\pi }{3}\)Â Â Â Â A1A1Â Â Â Â N2
[2 marks]