IB Math Analysis & Approaches Questionbank-Topic: SL 3.7 The circular and Composite functions SL Paper 1

 

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Question

Consider \(g(x) = 3\sin 2x\) .

Write down the period of g.

[1]
a.

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


[3]
b.

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

[2]
c.
Answer/Explanation

Markscheme

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

[1 mark]

a.

      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]

b.

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]

c.

Question

Let  \(f:x \mapsto {\sin ^3}x\) .

(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.

[5]
a.

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

[2]
b.

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.

[7]
c.
Answer/Explanation

Markscheme

(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]

a.

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

[2 marks]

b.

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]

c.

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.

[2]
a(i) and (ii).

(i)     Show that \(h(8) = 90.5\).

(ii)    Find \(h(21)\) .

[4]
b(i) and (ii).

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

[3]
c.

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

[5]
d.
Answer/Explanation

Markscheme

(i) 100 (metres)     A1     N1

(ii) 50 (metres)     A1     N1

[2 marks]

a(i) and (ii).

(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]

b(i) and (ii).


     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]

c.

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]

d.

Question

The diagram below shows part of the graph of \(f(x) = a\cos (b(x – c)) – 1\) , where \(a > 0\) .


The point \({\rm{P}}\left( {\frac{\pi }{4},2} \right)\) is a maximum point and the point \({\rm{Q}}\left( {\frac{{3\pi }}{4}, – 4} \right)\) is a minimum point.

 

Find the value of a .

[2]
a.

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

(ii)    Hence, find the value of b .

[4]
b(i) and (ii).

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

[1]
c.
Answer/Explanation

Markscheme

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]

a.

(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]

b(i) and (ii).

\(c = \frac{\pi }{4}\)     A1     N1

[1 mark]

c.

Question

The following diagram shows the graph of  \(f(x) = a\cos (bx)\) , for \(0 \le x \le 4\) .


There is a minimum point at P(2, − 3) and a maximum point at Q(4, 3) .

(i)     Write down the value of a .

(ii)    Find the value of b .

[3]
a(i) and (ii).

Write down the gradient of the curve at P.

[1]
b.

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

[2]
c.
Answer/Explanation

Markscheme

(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]

a(i) and (ii).

0     A1     N1

[1 mark]

b.

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]

c.

Question

The diagram below shows part of the graph of a function \(f\) .


The graph has a maximum at A(\(1\), \(5\)) and a minimum at B(\(3\), \( -1\)) .

The function \(f\) can be written in the form \(f(x) = p\sin (qx) + r\) . Find the value of

(a)     \(p\)

(b)     \(q\)

(c)     \(r\) .

[6]
Answer/Explanation

Markscheme

(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]

 

a.

valid approach to find \(q\)     (M1)

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

\(q = \frac{\pi }{2}\)     A1     N2

[2 marks]

b.

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]

c.

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 \(k\).

[3]
a.

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

[2]
b.

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 \(q\).

[2]
c.
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] 

a.

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]

b.

\(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]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

Question

Let \(f(x) = 3\sin (\pi x)\).

Write down the amplitude of \(f\).

[1]
a.

Find the period of \(f\).

[2]
b.

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

[4]
c.
Answer/Explanation

Markscheme

amplitude is 3     A1     N1

a.

valid approach     (M1)

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

period is 2     A1     N2

b.

      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.

c.

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 \(f\).

[3]
a.

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

M16/5/MATME/SP1/ENG/TZ1/03.b

[4]
b.
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]

a.

M16/5/MATME/SP1/ENG/TZ1/03.b/M     A1A1A1A1     N4

[4 marks]

b.

Question

Let \(f(x) = 6x\sqrt {1 – {x^2}} \), for \( – 1 \leqslant x \leqslant 1\), and \(g(x) = \cos (x)\), for \(0 \leqslant x \leqslant \pi \).

Let \(h(x) = (f \circ g)(x)\).

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

[5]
a.

Hence find the range of \(h\).

[2]
b.
Answer/Explanation

Markscheme

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]

a.

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]

b.

MAA SL 3.7 TRIGONOMETRIC FUNCTIONS [concise]-manav

Question

[Maximum mark: 14] [without GDC]
Complete the following table

Answer/Explanation

Ans:

Question

[Maximum mark: 12] [without GDC]
Part of the graph of a trigonometric function f(x) is given below. There is a local maximum at (π/8,180) and a local minimum at (3π/8,20).

(a) Write down the values of
(i) the amplitude (ii) the central value (iii) the period [3]
(b) Express the function in the form f (x) = Asin Bx + C [3]
(c) Complete the following table by expressing f (x) in three alternative forms [6]

Answer/Explanation

Ans:
(a) amplitude = 80, central value = 100, period = π/2
(b) f(x) = 80sin 4x + 100, since \(B=\frac{2 \pi}{\text { Period }}=\frac{2 \pi}{\pi / 2}=4\)
(c) (i) \(f(x)=-80 \sin 4\left(x-\frac{\pi}{4}\right)+100\)     (\(D=\frac{\pi}{4}\) is the position of the 2nd (↓) root)
(ii) \(f(x)=80 \cos 4\left(x-\frac{\pi}{8}\right)+100\)     (\(D=\frac{\pi}{8}\) is the position of the maximum)
(iii) \(f(x)=-80 \cos 4\left(x-\frac{3 \pi}{8}\right)+100\)     (\(D=\frac{3 \pi}{8}\) is the position of the minimum)

Question

[Maximum mark: 5] [without GDC]
Sketch the graph of the function
\(f(x)=60 \sin 4 x+100, \quad 0 \leq x \leq \pi\)

Answer/Explanation

Ans:
\(f(x)=60 \sin 4 x+100, \quad 0 \leq x \leq \pi\)

Question

[Maximum mark: 5] [without GDC]
Sketch the graph of the function
\(f(x)=60 \cos 4 x+100, \quad 0 \leq x \leq \pi\)

Answer/Explanation

Ans:
\(f(x)=60 \cos 4 x+100, \quad 0 \leq x \leq \pi\)

Question

Maximum mark: 5] [without GDC]
Sketch the graph of the function
\(f(x)=-60 \sin 4 x+80, \quad 0 \leq x \leq \pi\)

Answer/Explanation

Ans:
\(f(x)=-60 \sin 4 x+80, \quad 0 \leq x \leq \pi\)

Question

[Maximum mark: 9] [without GDC]
(a) Sketch the graph of the function
\(f(x)=-60 \cos 4 x+80, \quad 0 \leq x \leq \pi\) [5]

(b) Write down the possible values of k if the equation f (x) = k has
(i) exactly two solution ……………………………………………………….
(ii) exactly three solutions ……………………………………………………….
(iii) exactly four solutions ……………………………………………………….
(iv) no solutions ………………………………………………………. [4]

Answer/Explanation

Ans:
\(f(x)=-60 \cos 4 x+80, \quad 0 \leq x \leq \pi\)

(i) k = 140 (ii) k = 20 (iii) 20 < k < 140 (iv) k < 20 or k > 140

Question

[Maximum mark: 7] [without GDC]
The function f is defined by \(f(x)=30 \sin 3 x \cos 3 x, 0 \leq x \leq \frac{\pi}{3}\)
(a) Write down an expression for f(x) in the form a sin 6x, where a is an integer. [2]
(b) Find the period of f. [2]
(c) Solve f(x) = 0 , giving your answers in terms of π. [3]

Answer/Explanation

Ans:
(a) f(x) = 15sin(6x)
(b) period = 2π/6 = π
(c) 15 sin 6x = 0, (OR sin 3x = 0 and cos 3x = 0)
6x = 0, π, 2π
x = 0,\(\frac{\pi}{6}, \frac{\pi}{3}\)

Question

[Maximum mark: 4] [with / without GDC]
Let \(f(x)=4 \sin \left(3 x+\frac{\pi}{2}\right)\).  For what values of k will the equation f(x) = k have no solutions?

Answer/Explanation

Ans:
From sketch of graph \(y=4 \sin \left(3 x+\frac{\pi}{2}\right)\) or by observing \(|\sin \theta| \leq 1\).
\(k>4, k<-4\)

Question

[Maximum mark: 6] [without GDC]
Consider g(x) = 3sin 2x.
(a) Write down the period of g . [1]
(b) On the diagram below, sketch the curve of g , for \(0 \leq x \leq 2 \pi\) [3]

(c) Write down the number of solutions to the equation \(g(x)=2, \text { for } 0 \leq x \leq 2 \pi\).  [2]

Answer/Explanation

Ans:
(a) period = π
(b)
(c) 4 (solutions)

Question

[Maximum mark: 4] [without GDC]
The graph of a function of the form y = pcosqx is given in the diagram below.

(a) Write down the value of p. [2]
(b) Calculate the value of q. [2]

Answer/Explanation

Ans:
(a) p = 30
(b) \(\text { Period }=\frac{2 \pi}{q}=\frac{\pi}{2} \Rightarrow q=4\)

Question

[Maximum mark: 6] [with / without GDC]
Let f(x) = sin 2x and g(x) = sin(0.5x).
(a) Write down
(i) the minimum value of the function f
(ii) the period of the function g. [3]
(b) Consider the equation f(x) = g(x) Find the number of solutions, for \(0 \leq x \leq \frac{3 \pi}{2}\). [3]

Answer/Explanation

Ans:
(a) (i) –1
(ii) 4π (accept 720°)
(b) 

Question

[Maximum mark: 4] [without GDC]
Part of the graph of y = p + qcosx is shown below. The graph passes through the points (0, 3) and (π, –1).

Find the value of (i) p ; (ii) q.

Answer/Explanation

Ans:
3 = p + q cos 0 ⇒ 3 = p + q
–1 = p + q cosπ  ⇒–1 = p – q
(i) p = 1      (ii) q = 2

Question

[Maximum mark: 6] [without GDC]
Let \(f(x)=\frac{3 x}{2}+1, g(x)=4 \cos \left(\frac{x}{3}\right)-1 . \text { Let } h(x)=(g \circ f)(x)\)
(a) Find an expression for h(x) [3]
(b) Write down the period of h. [1]
(c) Write down the range of h. [2]

Answer/Explanation

Ans:
(a) \(h(x)=4 \cos \left(\frac{\frac{3 x}{2}+1}{3}\right)-1=4 \cos \left(\frac{1}{2} x+\frac{1}{3}\right)-1\)
(b) period is 4π (12.6)
(c) range is \(-5 \leq h(x) \leq 3 \quad([-5,3])\)

Question

[Maximum mark: 7] [without GDC]
The graph of y = p + cosqx + r, for –5 ≤ x ≤ 14, is shown below.There is a minimum point at (0, –3) and a maximum point at (4, 7).
(a) Find the value of (i) p ; (ii) q ; (iii) r . [6]
(b) The equation y = k  has exactly two solutions. Write down the value of k. [1]

Answer/Explanation

Ans:
(a) (i) amplitude = \(\frac{7+3}{2}=5 \Rightarrow p=-5\)
(ii) period = 8  ⇒ q = 0.785 \(\left(=\frac{2 \pi}{8}=\frac{\pi}{4}\right)\)
(iii) \(r=\frac{7-3}{2} \Rightarrow r=2\)
(b) k = –3 (accept y = –3)

Question

[Maximum mark: 4] [without GDC]
The depth, y metres, of sea water in a bay t hours after midnight may be represented by the function
\(y=a+b \cos \left(\frac{2 \pi}{k} t\right)\)  where a , b and k are constants.
The water is at a maximum depth of 14.3 m at midnight and noon, and is at a minimum
depth of 10.3 m at 06:00 and at 18:00.
Write down the value of (ι) a ; (ιι) b ; (ιιι) k.

Answer/Explanation

Ans:
METHOD 1
The value of cosine varies between –1 and +1. Therefore:
t = 0 ⇒ a + b = 14.3
t = 6 ⇒ a – b = 10.3
⇒ a = 12.3 b = 2
Period = 12 hours ⇒ \(\frac{2 \pi(12)}{k}\) = 2π⇒ k = 12
METHOD 2
From graph: Midpoint = a = 12.3
Amplitude = b = 2
\(\text { Period }=\frac{2 \pi}{\frac{2 \pi}{k}}=12 \Rightarrow k=12\)

Question

[Maximum mark: 6] [without GDC]
Let f(x) = asinb(x-c). Part of the graph of f is given below.Given that a, b and c are positive, find the value of a, of b and of c.

Answer/Explanation

Ans:
\(a=4, b=2, c=\frac{\pi}{2}\left(\text { or } \frac{3 \pi}{2} e t c\right)\)

Question

[Maximum mark: 6] [with / without GDC]
Consider \(y=\sin \left(x+\frac{\pi}{9}\right)\)
(a) The graph of y intersects the x -axis at point A. Find the x -coordinate of A,
where \(0 \leq x \leq \pi\). [2]
(b) Solve the equation \(\sin \left(x+\frac{\pi}{9}\right)=-\frac{1}{2}, \text { for } 0 \leq x \leq 2 \pi\). [4]

Answer/Explanation

Ans:
(a) \(\text { when } y=0, x+\frac{\pi}{9}=\pi \Rightarrow x=\frac{8 \pi}{9} \text { or } 2.79\)
(b) METHOD 1
GDC graph OR GDC SolveN
x = 3.32 or x = 5.41
METHOD 2
\(\sin \left(x+\frac{\pi}{9}\right)=-\frac{1}{2} \Rightarrow x+\frac{\pi}{9}=\frac{7 \pi}{6}, x+\frac{\pi}{9}=\frac{11 \pi}{6} \Rightarrow x=\frac{7 \pi}{6}-\frac{\pi}{9}, x=\frac{11 \pi}{6}-\frac{\pi}{9}\)
\(\Rightarrow x=\frac{19 \pi}{18}, x=\frac{31 \pi}{18}\) (x = 3.32, x =  5.41)

Question

[Maximum mark: 6] [without GDC]
The diagram below shows the graph of \(f(x)=1+\tan \left(\frac{x}{2}\right), \text { for }-360^{\circ} \leq x \leq 360^{\circ}\)

(a) On the same diagram, draw the asymptotes. [2]
(b) Write down
(i) the period of the function;
(ii) the value of f(90°). [2]
(c) Solve f(x) = 0 for −360° ≤ x ≤ 360°. [2]

Answer/Explanation

Ans:
(a)

(b) (i) Period = 360° (accept 2π) (ii) f(90°) = 2
(c) 270°, -90°

Question

[Maximum mark: 5] [with GDC]
Let \(f(x)=4 \tan ^{2} x-4 \sin x,-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}\)
(a) On the grid below, sketch the graph of y = f(x). [3]

(b) S
olve the equation f(x) = 1. [2]

Answer/Explanation

Ans:
(a)

(b) x = -0.207    x = 0.772

Question

[Maximum mark: 6] [with GDC]
Let \(f(x)=6 \sin \pi x \text { and } g(x)=6 \mathrm{e}^{-x}-3 \text {, for } 0 \leq x \leq 2\). The graph of f is shown on the diagram below. There is a maximum value at B (0.5, b).

(a) Write down the value of b. [1]
(b) On the same diagram, sketch the graph of g. [3]
(c) Solve \(f(x)=g(x), 0.5 \leq x \leq 1.5\). [2]

Answer/Explanation

Ans:
(a) b = 6
(b)

(c) x = 1.05 (accept (1.05, −0.896) ) (no additional solutions)

Question

[Maximum mark: 10] [with GDC]
A formula for the depth d metres of water in a harbour at a time t hours after midnight is \(d=P+Q \cos \left(\frac{\pi}{6} t\right), \quad 0 \leq t \leq 24\)
where P and Q are positive constants. In the following graph the point (6, 8.2) is a minimum point and the point (12, 14.6) is a maximum point.

(a) Find the value of (i) Q ; (ii) P . [3]
(b) Find the first time in the 24-hour period when the depth of the water is 10 meters. [3]
(c) (i) Use symmetry of the graph to find the next time when the depth d is 10m.
(ii) Hence find the time intervals in the 24-hour period during which the water is
less than 10 metres deep. [4]

Answer/Explanation

Ans:
(a) (i) Q=\frac{1}{2}(14.6-8.2)=3.2           (ii) \(P=\frac{1}{2}(14.6+8.2)=11.4\)
(b) \(10=11.4+3.2 \cos \left(\frac{\pi}{6} t\right)\)
t = 3.8648. t = 3.86(3 s.f.)
(c) (i) By symmetry, next time is 12 – 3.86… = 8.135… t = 8.14 (3 s.f.)
(ii) From above, first interval is 3.86 < t < 8.14
This will happen again, 12 hours later, so 15.9 < t < 20.1

Question

[Maximum mark: 18] [without GDC]
The diagram shows the graph of the function f given by
\(f(x)=A \sin \left(\frac{\pi}{2} x\right)+B, \text { for } 0 \leq x \leq 5\)
where A and B are constants, and x is measured in radians.

The graph includes the points (1, 3) and (5, 3), which are maximum points of the graph.
(a) Write down the values of f(1) and f(5). [2]
(b) Show that the period of f is 4. [2]
The point (3, –1) is a minimum point of the graph.
(c) Show that A = 2, and find the value of B. [5]
(d) Solve the equation f(x) = 2 for \(0 \leq x \leq 5\). [5]
Extra question
(e) Consider the equation \(f(x)=k, \text { for } 0 \leq x \leq 5\).
Write down the possible values of k if the equation has
(i) exactly one solution (ii) exactly three solutions
(iii) exactly two solutions (iv) no solutions

Answer/Explanation

Ans:
(a)      f(1) = 3      f(5) = 3
(b) EITHER distance between successive maxima = period = 5 – 1 = 4     OR    \(\text { period }=\frac{2 \pi}{\frac{\pi}{2}}=4\)
(c) EITHER \(A \sin \left(\frac{\pi}{2}\right)+B=3 \text { and } A \sin \left(\frac{3 \pi}{2}\right)+B=-1\)
⇔ A + B = 3, – A + B = –1
⇔ A = 2, B = 1
OR
Amplitude = A = \(\frac{3-(-1)}{2}=\frac{4}{2}=2\)
Midpoint value = B = \(\frac{3+(-1)}{2}=\frac{2}{2}=1\)
(d) \(f(x)=2 \Rightarrow 2 \sin \left(\frac{\pi}{2} x\right)+1=2 \Rightarrow x=\frac{1}{3} \text { or } \frac{5}{3} \text { or } \frac{13}{3}\)
Extra question
(e) (i) k = -1    (ii) 1≤ k < 3    (iii) -1< k < 1 or k = 3    (iv) k < -1 or k > 3

Question

[Maximum mark: 13] [with GDC]
The depth y metres of water in a harbour is given by the equation
\(y=10+4 \sin \left(\frac{t}{2}\right)\)
where t is the number of hours after midnight.
(a) Calculate the depth of the water (i) when t = 2; (ii) at 21:00. [3]
The sketch below shows the depth y, of water, at time t, during one day (24 hours).
b) (i) Write down the maximum depth of water in the harbour.
(ii) Calculate the value of t when the water is first at its maximum depth during the day. [3]
The harbour gates are closed when the depth of the water is less than seven metres. An alarm rings when the gates are opened or closed.
(c) (i) How many times does the alarm sound during the day?
(ii) Find the value of t when the alarm sounds first.
(iii) Use the graph to find the length of time during the day when the harbour gates are closed. Give your answer in hours, to the nearest hour. [7]

Answer/Explanation

Ans:
(a) (i) 10 + 4sin1 = 13.4
(ii) At 2100, t = 21
10 + 4sin 10.5 = 6.48
(b) (i) 14 metres
(ii) 14 = 10 + \(4 \sin \left(\frac{t}{2}\right) \Rightarrow\) t = π (=3.14)
(c) (i) 4
     (ii) \(10+4 \sin \left(\frac{t}{2}\right)=7 \Rightarrow t=7.98\)
(iii) depth < 7 from 8 –11 = 3 hours, from 2030 – 2330 = 3 hours therefore, total = 6 hours

Question

[Maximum mark: 11] [with GDC]
The following graph shows the depth of water, y metres, at a point P, during one day. The time t is given in hours, from midnight to noon.
(a) Use the graph to write down an estimate of the value of t when
(i) the depth of water is minimum;
(ii) the depth of water is maximum;
(iii) the depth of the water is increasing most rapidly. [3]
(b) The depth of water can be modelled by the function y = Acos(B(t-1) + C.
(i) Show that A = 8.
(ii) Write down the value of C.
(iii) Find the value of B. [6]
(c) A sailor knows that he cannot sail past P when the depth of the water is less than
12 m. Calculate the values of t between which he cannot sail past P. [2]

Answer/Explanation

Ans:
(a) (i) 7 (ii) 1 (iii) 10
(b) (i) \(A=\frac{18-2}{2}=8\)
(ii)  C = 10
(iii) period = 12  \(B=\frac{\pi}{6}\) (accept 0.524 or 30)
(c) t = 3.52, t = 10.5, between 03:31 and 10:29 (accept 10:30)

Question

[Maximum mark: 13] [with GDC]
Let \(f(x)=3 \sin x+4 \cos x, \text { for }-2 \pi \leq x \leq 2 \pi \text {. }\)
(a) Sketch the graph of f. [3]
(b) Write down
(i) the amplitude; (ii) the period; (iii) the x-intercept between \(\frac{\pi}{2}\) and 0. [3]
(c) Hence write f(x) in the form psin(qx+r). [3]
(d) Write down the x -coordinates of the points where f has a maximum. [2]
(e) Write down the two values of k for which the equation f(x) = k has exactly two
solutions. [2]

Answer/Explanation

Ans:
(a)

(b) (i) 5 (ii) 2π (6.28) (iii) –0.927
(c) f(x) = 5 sin (x + 0.927) (accept p = 5, q = 1, r = 0.927)
(d) 3 s.f. values which round to –5.6, 0.64
(e) k = –5, k = 5

Question

[Maximum mark: 10] [with GDC]
A spring is suspended from the ceiling. It is pulled down and released, and then oscillates up and down. Its length, l centimetres, is modelled by the function
\(l=33+5 \cos \left((720 t)^{\circ}\right)\)
where t is time in seconds after release.
(a) Find the length of the spring after 1 second. [2]
(b) Find the minimum length of the spring. [3]
(c) Find the first time at which the length is 33 cm. [3]
(d) What is the period of the motion? [2]

Answer/Explanation


Ans:
(a) When t = 1, l = 33 + 5 cos 720 = 38
(b) \(l_{\min }\) = 33 – 5= 28
(c) 33 = 33 + 5cos720t ⇒ t = 1/8
(d) period = \(\frac{360}{720}\left(=\frac{1}{2}\right)\)

Question

[Maximum mark: 12] [with GDC]
Let \(f(x)=5 \cos \frac{\pi}{4} x \text { and } g(x)=-0.5 x^{2}+5 x-8, \text { for } 0 \leq x \leq 9\)
(a) On the same diagram, sketch the graphs of f and g. [3]
(b) Consider the graph of f. Write down
(i) the x-intercept that lies between x = 0 and x = 3;
(ii) the period;
(iii) the amplitude. [4]
(c) Consider the graph of g. Write down
(i) the two x-intercepts;
(ii) the equation of the axis of symmetry. [3]
(d) Find the x-coordinates of the points of intersection between f and g. [2]

Answer/Explanation

Ans:
(a)
(b) (i) (2,0) (or x = 2)
(ii) period = 8
(iii) amplitude = 5
(c) (i) (2, 0), (8, 0) (or x = 2, x = 8)
(ii) x = 5 (must be an equation)
(d) intersect when x = 2 and x = 6.79

Question

[Maximum mark: 20] [with / without GDC]
A Ferris wheel with centre O and a radius of 15 metres is represented in the diagram
below. Initially seat A is at ground level. The next seat is B, where \(\mathrm{AOB}=\frac{\pi}{6}\).

(a) Find the length of the arc AB. [2]
(b) Find the area of the sector AOB. [2]
(c) The wheel turns clockwise through an angle of \(\frac{2 \pi}{3}\). Find the height of A above the ground. [3]
The height, h metres, of seat A above the ground after t minutes, can be modelled by
the function \(h(t)=-a \cos 2 t+c\)
(d) (i) Find the values of a and c.
(ii) Find the time for a complete turn of the wheel. [5]
The height, h metres, of seat C above the ground after t minutes, can be modelled by the function \(h(t)=15-15 \cos \left(2 t+\frac{\pi}{4}\right)\)
(e) (i) Find the height of seat C when \(t=\frac{\pi}{4}\).
(ii) Find the initial height of seat C.
(iii) Find the time at which seat C first reaches its highest point. [8]

Answer/Explanation

Ans:
(a) arc AB = rθ = 7.85 (m)
(b) Area of sector AOB \(A=\frac{1}{2} r^{2} \theta=58.9\left(\mathrm{~m}^{2}\right)\)
(c) METHOD 1

\(\text { angle }=\frac{\pi}{6}\left(30^{\circ}\right)\)
\(\text { height }=15+15 \sin \frac{\pi}{6}=22.5(\mathrm{~m})\)
METHOD 2

\(\text { angle }=\frac{\pi}{3}\left(60^{\circ}\right)\)
\(\text { height }=15+15 \cos \frac{\pi}{3}=22.5(\mathrm{~m})\)
(d) (i) \(h\left(\frac{\pi}{4}\right)=15-15 \cos \left(\frac{\pi}{2}+\frac{\pi}{4}\right)=25.6(\mathrm{~m})\)
(ii) \(h(0)=15-15 \cos \left(0+\frac{\pi}{4}\right)=4.39(\mathrm{~m})\)
(iii) METHOD 1
Highest point when h = 30
\(30=15-15 \cos \left(2 t+\frac{\pi}{4}\right) \Leftrightarrow t=1.18\left(\text { accept } \frac{3 \pi}{8}\right)\)
METHOD 2
Using graph t = 1.18

Question

[Maximum mark: 14] [without GDC]
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 (counterclockwise) direction. One revolution takes 20 minutes.

(a) Write down the height of P above ground level after
(i) 10 minutes;      (ii) 15 minutes. [2]
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.
\(\begin{array}{|c|c|}
\hline t & h(t) \\
\hline 0 & 0.0 \\
\hline 1 & 2.4 \\
\hline 2 & 9.5 \\
\hline 3 & 20.6 \\
\hline 4 & 34.5 \\
\hline 5 & 50.0 \\
\hline
\end{array}\)
(b) (i) Show that h(8) = 90.5. (ii) Find h(21). [4]
(c) Sketch the graph of h, for \(0 \leq t \leq 40\). [3]
(d) Given that h can be expressed in the form \(h(t)=a \cos b t+c, \text { find } a, b \text { and } c\).    [5]

Answer/Explanation

Ans:
(a) (i) 100 (metres)
(ii) 50 (metres)
(b) (i) Symmetry with h(2) = 9.5
h(8) = 100 – 9.5 = 90.5
(ii) h(21) = h(1) = 2.4
(c)
(d)  \(b=\frac{2 \pi}{20}\left(=\frac{\pi}{10}\right)\)    (accept b =18 if working in degrees)
a = –50, c = 50

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