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) 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 .
Answer/Explanation
Markscheme
(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]
Question
Let \(f(x) = 6 + 6\sin x\) . Part of the graph of f is shown below.
The shaded region is enclosed by the curve of f , the x-axis, and the y-axis.
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.
Answer/Explanation
Markscheme
(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]
Question
The following diagram represents a large Ferris wheel, with a diameter of 100 metres.
Let P be a point on the wheel. The wheel starts with P at the lowest point, at ground level. The wheel rotates at a constant rate, in an anticlockwise (counter-clockwise) direction. One revolution takes 20 minutes.
Let \(h(t)\) metres be the height of P above ground level after t minutes. Some values of \(h(t)\) are given in the table below.
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 .
Answer/Explanation
Markscheme
(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]
Question
The following diagram shows the graph of \(f(x) = a\sin (b(x – c)) + d\) , for \(2 \le x \le 10\) .
There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .
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.
Answer/Explanation
Markscheme
(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]
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 \(f\) (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]
Question
Let \(f(x) = 3\sin (\pi x)\).
Write down the amplitude of \(f\).
Find the period of \(f\).
On the following grid, sketch the graph of \(y = f(x)\), for \(0 \le x \le 3\).
Answer/Explanation
Markscheme
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.
Question
Let \(f(x) = 3\sin \left( {\frac{\pi }{2}x} \right)\), for \(0 \leqslant x \leqslant 4\).
(i) Write down the amplitude of \(f\).
(ii) Find the period of \(f\).
On the following grid sketch the graph of \(f\).
Answer/Explanation
Markscheme
(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]
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) 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 .
Answer/Explanation
Markscheme
\(f'(x) = – \sin x + \sqrt 3 \cos x\) 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]