Shape and features of graph
$$ \text{General equation: } y = a \cos bx + c $$
Shape and features when a > 0 (click to show):
\begin{align*}
\text{Center line: } & y = c \\
\\
\text{Amplitude} & = a \\
\\
\text{Period} & = {360^\circ \over b} = {2\pi \text{ radians} \over b}
\end{align*}
Shape and features when a < 0 (click to show):
\begin{align*}
\text{Center line: } & y = c \\
\\
\text{Amplitude} & = \text{Positive value of } a \\
\\
\text{Period} & = {360^\circ \over b} = {2\pi \text{ radians} \over b}
\end{align*}
Questions
Deduce equation from graph
Q1. The diagram below shows the graph of $y = a \cos (bx) + c $.
Determine the values of $a$, $b$ and $c$.
(from A Maths 360 Workbook Ex 11.3)
Answer: $ a = 3, b = 2, c = 0 $
Solutions
\begin{align*}
\text{Center line: } & y = 0 \\
c & = 0 \\
\\
\text{Amplitude} & = 3 \\
a & = 3 \\
\\
\text{Period} & = 180^\circ \\
{360^\circ \over b} & = 180^\circ \\
360^\circ & = (180^\circ) b \\
{360^\circ \over 180^\circ} & = b \\
2 & = b \\
\\ \\
\therefore a = 3, b = 2, c = 0
\end{align*}
Q2. The diagram below shows the graph of $y = a \cos (bx) + c $.
Determine the values of $a$, $b$ and $c$.
(from A Maths 360 Workbook Ex 11.3)
Answer: $ a = -2, b = 4, c = 2 $
Solutions
\begin{align*}
\text{Center line: } & y = 2 \\
c & = 2 \\
\\
\text{Amplitude} & = 2 \\
a & = -2
\phantom{000000} [\text{Due to inverted shape}] \\
\\
\text{Period} & = {\pi \over 2} \\
{2\pi \over b} & = {\pi \over 2} \\
2(2\pi) & = \pi b \\
4\pi & = \pi b \\
{4\pi \over \pi} & = b \\
4 & = b \\
\\ \\
\therefore a = -2, b & = 4, c = 2
\end{align*}
Deduce equation from information
Q3. The curve $y = a \cos \left(x \over b\right) + c$, where $a$ and $b$ are positive integers, has an amplitude of $4$ units and period of $720^\circ$. The maximum value of $y$ is $3$. State the values of $a$, $b$ and $c$.
Answer: $ a = 4, b = 2, c = -1 $
Solutions
\begin{align}
\text{Center line: } & y = 3 - 4 \\
& y = -1 \\
\\
c & = -1 \\
\\
\text{Amplitude} & = 4 \\
a & = 4 \\
\\
\text{Period} & = 720^\circ \\
{360^\circ \over {1 \over b}} & = 720^\circ \\
360 b & = 720 \\
b & = {720 \over 360} \\
b & = 2 \\
\\ \\
\therefore a = 4, b & = 2, c = -1
\end{align}
Sketch question
Q4. Sketch the graph of $y = -1 - \cos \left({x \over 2}\right)$, for $ 0 \le x \le 6 \pi $.
Solutions
\begin{align*}
y & = - \cos \left({x \over 2}\right) - 1 \\
y & = - \cos \left({1 \over 2}x\right) - 1 \phantom{000000} [\text{Inverted shape}] \\
y & = - \cos (0.5 x) - 1 \\
\\
\text{Amplitude} & = 1 \\
\\
\text{Center line: } & y = -1 \\
\\
\text{Maximum value} & = -1 + 1 \\
& = 0 \\
\\
\text{Minimum value} & = -1 - 1 \\
& = -2 \\
\\
\text{Period} & = {2\pi \over 0.5} \\
& = 4 \pi \\
\\
\text{No. of cycles} & = {6\pi \over 4\pi} \\
& = 1{1 \over 2}
\end{align*}
Symmetry question
Q5. The diagram above shows part of the graph of $y = 6 - 3 \cos 2x$, passing through the points $(k, 4)$, $(l, 8)$ and $(m, 4)$, where $k$, $l$ and $m$ are constants.
Using symmetry of the graph, or otherwise, find an equation connecting
(from think! A Maths Worksheet 9B)
(i) $\pi$, $k$ and $m$,
Answer: $ m = k + \pi $
Solutions
\begin{align}
\text{Period} & = \pi \\
\\
m & = k + \pi
\end{align}
(ii) $\pi$, $k$ and $l$,
Answer: $ k + l = {\pi \over 2} $
Solutions
\begin{align}
k + l & = {\pi \over 2}
\end{align}
Deduce number of solutions to equation
Q6(i) On the same axes, sketch the graphs of $y = 2 \cos 2x$ and $y = \sin x + 1$ for $0 \le x \le 2\pi$.
(from A Maths 360 2nd edition Ex 10.2)
Solutions
\begin{align}
y & = 2 \cos 2x \\
\\
\text{Amplitude} & = 2 \\
\\
\text{Max. value} & = 2 \\
\text{Min. value} & = -2 \\
\\
\text{Period} & = {2\pi \over 2} \\
& = \pi \\
\\
\text{No. of cycles} & = {2\pi \over \pi} \\
& = 2 \\
\\ \\
y & = \sin x + 1 \\
\\
\text{Amplitude} & = 1 \\
\\
\text{Center line: } & y = 1 \\
\\
\text{Max. value} & = 1 + 1 = 2 \\
\text{Min. value} & = 1 - 1 = 0 \\
\\
\text{Period} & = {2\pi \over 1 }\\
& = 2\pi \\
\\
\text{No. of cycles} & = {2\pi \over 2\pi} \\
& = 1
\end{align}
Q6(ii) Hence, find the number of distinct values of $x$, in the interval $0 \le x \le 2\pi$, for which $\cos 2x - {1 \over 2} = {1 \over 2} \sin x $.
Answer: $ \text{4 distinct values of } x $
Solutions
\begin{align}
\cos 2x - {1 \over 2} & = {1 \over 2} \sin x \\
\cos 2x & = {1 \over 2} \sin x + {1 \over 2} \\
2 \cos 2x & = 2 \left({1 \over 2} \sin x + {1 \over 2} \right) \\
\underbrace{2\cos 2x}_\text{First graph} & = \underbrace{\sin x + 1}_\text{Second graph} \\
\\
\therefore \text{4 distinct} & \text{ values of } x
\end{align}
Past year O level questions
Year & paper |
Comments |
2024 P1 Question 10 |
Sketch graph and deduce solutions to equation |
2023 P2 Question 7a |
Deduce equation from information provided (difficult) |
2022 P1 Question 10 |
Sketch graph |
2017 P2 Question 10 |
Sketch graph and deduce number of solutions to inequality |
2016 P1 Question 3b |
Deduce equation from graph |
2014 P2 Question 9 |
Sketch graph and deduce solutions to equation |
2012 P1 Question 8 |
Deduce equation from information and sketch graph |
2010 P2 Question 11i |
Deduce equation from information and sketch graph |
2008 P2 Question 7 |
Sketch graph |
2007 P1 Question 9 |
Sketch graph and deduce solutions to equation |
2004 P1 Question 6 |
Find amplitude, period and the coordinates of maximum point and minimum point (Link - Subscription required) |
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