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

## 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.


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 $${M’}$$ .


b.

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

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


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 .


d.

## 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 $$1 + \sin 2x$$ .


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 $$0 \le x \le 2\pi$$ .


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 .


c.

## 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 $$k$$.


a.

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


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 $$q$$.


c.

## 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 $$f$$.


a.

On the following grid sketch the graph of $$f$$. b.

## 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. 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 $$f'(x)$$ .


a.

Hence

(i)     show that $$q = – 2$$ ;

(ii)    verify that A is a minimum point.


b(i) and (ii).

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


c.

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

Write down the value of r and of a .


d.

## 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.