IB Math Analysis & Approaches Questionbank-Topic: SL 3.7 Transformations SL Paper 1

 

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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)     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.

[7]
a(i), (ii), (iii) and (iv).

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 \({M’}\) .

[2]
b.

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

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

[4]
c.

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 .

[3]
d.
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]

a(i), (ii), (iii) and (iv).

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]

b.

\(g(t) = 7\cos \frac{\pi }{3}\left( {t – 4.5} \right) + 12\)    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]

c.

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]

d.

Question

Let \(f(x) = {(\sin x + \cos x)^2}\) .

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

[2]
a.

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 \(0 \le x \le 2\pi \) .

 

[2]
b.

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 .

[2]
c.
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. \({\sin ^2}x + 2\sin x\cos x + {\cos ^2}x\)

\(f(x) = 1 + \sin 2x\)     AG     N0

[2 marks]

a.


     A1A1     N2

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

b.

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

[2 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.

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 \(f(x) = \cos x + \sqrt 3 \sin x\) , \(0 \le x \le 2\pi \) . 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 \(f'(x)\) .

[2]
a.

Hence

(i)     show that \(q = – 2\) ;

(ii)    verify that A is a minimum point.

[10]
b(i) and (ii).

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

[3]
c.

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

Write down the value of r and of a .

[2]
d.
Answer/Explanation

Markscheme

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

[2 marks]

a.

(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]

b(i) and (ii).

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]

c.

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

[2 marks]

d.

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