# IB Math Analysis & Approaches Questionbank-Topic: SL 2.2 Concept of a function, domain, range and graph SL Paper 1

### Question

Italia’s Pizza Company supplies and delivers large cheese pizzas.

The total cost to the customer, C, in GBP, is modelled by the function

C (n) = 34.50 n + 8.50 , n 2 , n ∈ Z ,

where n , is the number of large cheese pizzas ordered. This total cost includes a fixed

cost for delivery.

1. State, in the context of the question,

1. what the value of 34.50 represents;

2. what the value of 8.50 represents. 

2. Write down the minimum number of pizzas that can be ordered.  Aayush has 450 GBP.

3. Find the maximum number of large cheese pizzas that Aayush can order from Italia’s Pizza Company. 

Ans:

(a)

(i) the cost of each (large cheese) pizza / a pizza / one pizza / per pizza

(ii) the (fixed) delivery cost

(b) 2

(c) 450=34.50n+8.50 12.8 (12.7971…)12

### Question

Part of the graph of a function f is shown in the diagram below. On the same diagram sketch the graph of $$y = – f(x)$$ .

a.

Let $$g(x) = f(x + 3)$$ .

(i)     Find $$g( – 3)$$ .

(ii)    Describe fully the transformation that maps the graph of f to the graph of g.

b(i) and (ii).

### Markscheme M1A1     N2

Note: Award M1 for evidence of reflection in x-axis, A1 for correct vertex and all intercepts approximately correct.

a.

(i) $$g( – 3) = f(0)$$     (A1)

$$f(0) = – 1.5$$     A1     N2

(ii) translation (accept shift, slide, etc.) of $$\left( {\begin{array}{*{20}{c}} { – 3}\\ 0 \end{array}} \right)$$     A1A1     N2

[4 marks]

b(i) and (ii).

### Question

Let f be the function given by $$f(x) = {{\rm{e}}^{0.5x}}$$ , $$0 \le x \le 3.5$$ . The diagram shows the graph of f .  On the same diagram, sketch the graph of $${f^{ – 1}}$$ .


a.

Write down the range of $${f^{ – 1}}$$ .

b.

Find $${f^{ – 1}}(x)$$ .

c.

### Markscheme  A1A1A1     N3

Note: Award A1 for approximately correct (reflected) shape, A1 for right end point in circle, A1 for through $$(1{\text{, }}0)$$ .

a.

$$0 \le y \le 3.5$$     A1     N1

[1 mark]

b.

interchanging x and y (seen anywhere)     M1

e.g. $$x = {e^{0.5y}}$$

evidence of changing to log form     A1

e.g. $$\ln x = 0.5y$$ , $$\ln x = \ln {{\rm{e}}^{0.5y}}$$ (any base), $$\ln x = 0.5y\ln {\rm{e}}$$ (any base)

$${f^{ – 1}}(x) = 2\ln x$$     A1     N1

[3 marks]

c.

### Question

Let $$f(x) = 2{x^3} + 3$$ and $$g(x) = {{\rm{e}}^{3x}} – 2$$ .

(i)     Find $$g(0)$$ .

(ii)    Find $$(f \circ g)(0)$$ .

a.

Find $${f^{ – 1}}(x)$$ .

b.

### Markscheme

(i) $$g(0) = {{\rm{e}}^0} – 2$$     (A1)

$$= – 1$$     A1     N2

(ii) METHOD 1

e.g. $$(f \circ g)(0) = f( – 1)$$

correct substitution $$f( – 1) = 2{( – 1)^3} + 3$$     (A1)

$$f( – 1) = 1$$     A1     N3

METHOD 2

attempt to find $$(f \circ g)(x)$$     (M1)

e.g. $$(f \circ g)(x) = f({{\rm{e}}^{3x}} – 2)$$ $$= 2{({{\rm{e}}^{3x}} – 2)^3} + 3$$

correct expression for $$(f \circ g)(x)$$     (A1)

e.g. $$2{({{\rm{e}}^{3x}} – 2)^3} + 3$$

$$(f \circ g)(0) = 1$$     A1     N3

[5 marks]

a.

interchanging x and y (seen anywhere)     (M1)

e.g. $$x = 2{y^3} + 3$$

attempt to solve     (M1)

e.g. $${y^3} = \frac{{x – 3}}{2}$$

$${f^{ – 1}}(x) = \sqrt{{\frac{{x – 3}}{2}}}$$     A1     N3

[3 marks]

b.

### Question

Consider $$f(x) = \ln ({x^4} + 1)$$ .

The second derivative is given by $$f”(x) = \frac{{4{x^2}(3 – {x^4})}}{{{{({x^4} + 1)}^2}}}$$ .

The equation $$f”(x) = 0$$ has only three solutions, when $$x = 0$$ , $$\pm \sqrt{3}$$ $$( \pm 1.316 \ldots )$$ .

Find the value of $$f(0)$$ .

a.

Find the set of values of $$x$$ for which $$f$$ is increasing.

b.

(i)     Find $$f”(1)$$ .

(ii)     Hence, show that there is no point of inflexion on the graph of $$f$$ at $$x = 0$$ .

c.

There is a point of inflexion on the graph of $$f$$ at $$x = \sqrt{3}$$ $$(x = 1.316 \ldots )$$ .

Sketch the graph of $$f$$ , for $$x \ge 0$$ .

d.

## Markscheme

substitute $$0$$ into $$f$$     (M1)

eg   $$\ln (0 + 1)$$ , $$\ln 1$$

$$f(0) = 0$$     A1 N2

[2 marks]

a.

$$f'(x) = \frac{1}{{{x^4} + 1}} \times 4{x^3}$$ (seen anywhere)     A1A1

Note: Award A1 for $$\frac{1}{{{x^4} + 1}}$$ and A1 for $$4{x^3}$$ .

recognizing $$f$$ increasing where $$f'(x) > 0$$ (seen anywhere)     R1

eg   $$f'(x) > 0$$ , diagram of signs

attempt to solve $$f'(x) > 0$$     (M1)

eg   $$4{x^3} = 0$$ , $${x^3} > 0$$

$$f$$ increasing for $$x > 0$$ (accept $$x \ge 0$$ )     A1     N1

[5 marks]

b.

(i)     substituting $$x = 1$$ into $$f”$$     (A1)

eg   $$\frac{{4(3 – 1)}}{{{{(1 + 1)}^2}}}$$ , $$\frac{{4 \times 2}}{4}$$

$$f”(1) = 2$$     A1     N2

(ii)     valid interpretation of point of inflexion (seen anywhere)     R1

eg   no change of sign in $$f”(x)$$ , no change in concavity,

$$f’$$ increasing both sides of zero

attempt to find $$f”(x)$$ for $$x < 0$$     (M1)

eg   $$f”( – 1)$$ , $$\frac{{4{{( – 1)}^2}(3 – {{( – 1)}^4})}}{{{{({{( – 1)}^4} + 1)}^2}}}$$ , diagram of signs

correct working leading to positive value     A1

eg   $$f”( – 1) = 2$$ , discussing signs of numerator and denominator

there is no point of inflexion at $$x = 0$$     AG     N0

[5 marks]

c. A1A1A1     N3

Notes: Award A1 for shape concave up left of POI and concave down right of POI.

Only if this A1 is awarded, then award the following:

A1 for curve through ($$0$$, $$0$$) , A1 for increasing throughout.

Sketch need not be drawn to scale. Only essential features need to be clear.

[3 marks]

d.

### Question

The diagram below shows the graph of a function $$f$$ , for $$– 1 \le x \le 2$$ .  Write down the value of $$f(2)$$.

a.i.

Write down the value of $${f^{ – 1}}( – 1)$$ .

a.ii.

Sketch the graph of $${f^{ – 1}}$$ on the grid below.  b.

## Markscheme

$$f(2) = 3$$     A1     N1

[1 mark]

a.i.

$${f^{ – 1}}( – 1) = 0$$     A2     N2

[2 marks]

a.ii.

EITHER

attempt to draw $$y = x$$ on grid     (M1)

OR

attempt to reverse x and y coordinates     (M1)

eg   writing or plotting at least two of the points

$$( – 2, – 1)$$ , $$( – 1,0)$$ , $$(0,1)$$ , $$(3,2)$$

THEN

correct graph     A2     N3  [3 marks]

b.

### Question

Consider the functions $$f(x)$$ , $$g(x)$$ and $$h(x)$$ . The following table gives some values associated with these functions.  The following diagram shows parts of the graphs of $$h$$ and $$h”$$ .  There is a point of inflexion on the graph of $$h$$ at P, when $$x = 3$$ .

Given that $$h(x) = f(x) \times g(x)$$ ,

Write down the value of $$g(3)$$ , of $$f'(3)$$ , and of $$h”(2)$$ .

a.

Explain why P is a point of inflexion.

b.

find the $$y$$-coordinate of P.

c.

find the equation of the normal to the graph of $$h$$ at P.

d.

## Markscheme

$$g(3) = – 18$$ , $$f'(3) = 1$$ , $$h”(2) = – 6$$     A1A1A1     N3

[3 marks]

a.

$$h”(3) = 0$$     (A1)

valid reasoning     R1

eg   $${h”}$$ changes sign at $$x = 3$$ , change in concavity of $$h$$ at $$x = 3$$

so P is a point of inflexion     AG     N0

[2 marks]

b.

writing $$h(3)$$ as a product of $$f(3)$$ and $$g(3)$$     A1

eg   $$f(3) \times g(3)$$ , $$3 \times ( – 18)$$

$$h(3) = – 54$$     A1 N1

[2 marks]

c.

recognizing need to find derivative of $$h$$     (R1)

eg   $${h’}$$ , $$h'(3)$$

attempt to use the product rule (do not accept $$h’ = f’ \times g’$$ )     (M1)

eg   $$h’ = fg’ + gf’$$ ,  $$h'(3) = f(3) \times g'(3) + g(3) \times f'(3)$$

correct substitution     (A1)

eg   $$h'(3) = 3( – 3) + ( – 18) \times 1$$

$$h'(3) = – 27$$    A1

attempt to find the gradient of the normal     (M1)

eg   $$– \frac{1}{m}$$ , $$– \frac{1}{{27}}x$$

attempt to substitute their coordinates and their normal gradient into the equation of a line     (M1)

eg   $$– 54 = \frac{1}{{27}}(3) + b$$ , $$0 = \frac{1}{{27}}(3) + b$$ , $$y + 54 = 27(x – 3)$$ , $$y – 54 = \frac{1}{{27}}(x + 3)$$

correct equation in any form     A1     N4

eg   $$y + 54 = \frac{1}{{27}}(x – 3)$$ , $$y = \frac{1}{{27}}x – 54\frac{1}{9}$$

[7 marks]

d.

### Question

The following diagram shows the graph of $$y = f(x)$$, for $$– 4 \le x \le 5$$.  Write down the value of $$f( – 3)$$.

a(i).

Write down the value of  $${f^{ – 1}}(1)$$.

a(ii).

Find the domain of $${f^{ – 1}}$$.

b.

On the grid above, sketch the graph of $${f^{ – 1}}$$.

c.

### Markscheme

$$f( – 3) = – 1$$     A1     N1

[1 mark]

a(i).

$${f^{ – 1}}(1) = 0$$   (accept $$y = 0$$)     A1     N1

[1 mark]

a(ii).

domain of $${f^{ – 1}}$$ is range of $$f$$     (R1)

eg     $${\text{R}}f = {\text{D}}{f^{ – 1}}$$

eg     $$– 3 \leqslant x \leqslant 3,{\text{ }}x \in [ – 3,{\text{ }}3]{\text{ (accept }} – 3 < x < 3,{\text{ }} – 3 \leqslant y \leqslant 3)$$

[2 marks]

b.  A1A1     N2

Note: Graph must be approximately correct reflection in $$y = x$$.

Only if the shape is approximately correct, award the following:

A1 for x-intercept at $$1$$, and A1 for endpoints within circles.

[2 marks]

c.

### Question

Let $$f(x) = 8x + 3$$ and $$g(x) = 4x$$, for $$x \in \mathbb{R}$$.

Write down $$g(2)$$.

a.

Find $$(f \circ g)(x)$$.

b.

Find $${f^{ – 1}}(x)$$.

c.

### Markscheme

$$g(2) = 8$$    A1     N1

[1 mark]

a.

attempt to form composite (in any order)     (M1)

eg$$\,\,\,\,\,$$$$f(4x),{\text{ }}4 \times (8x + 3)$$

$$(f \circ g)(x) = 32x + 3$$     A1     N2

[2 marks]

b.

interchanging $$x$$ and $$y$$ (may be seen at any time)     (M1)

eg$$\,\,\,\,\,$$$$x = 8y + 3$$

$${f^{ – 1}}(x) = \frac{{x – 3}}{8}\,\,\,\,\,\left( {{\text{accept }}\frac{{x – 3}}{8},{\text{ }}y = \frac{{x – 3}}{8}} \right)$$     A1     N2

[2 marks]

c.

### Question

The following diagram shows the graph of a function $$f$$, with domain $$– 2 \leqslant x \leqslant 4$$. The points $$( – 2,{\text{ }}0)$$ and $$(4,{\text{ }}7)$$ lie on the graph of $$f$$.

Write down the range of $$f$$.

a.

Write down $$f(2)$$;

b.i.

Write down $${f^{ – 1}}(2)$$.

b.ii.

On the grid, sketch the graph of $${f^{ – 1}}$$.

c.

### Markscheme

correct range (do not accept $$0 \leqslant x \leqslant 7$$)     A1     N1

eg$$\,\,\,\,\,$$$$[0,{\text{ }}7],{\text{ }}0 \leqslant y \leqslant 7$$

[1 mark]

a.

$$f(2) = 3$$     A1     N1

[1 mark]

b.i.

$${f^{ – 1}}(2) = 0$$     A1     N1

[1 mark]

b.ii. A1A1A1     N3

Notes:     Award A1 for both end points within circles,

A1 for images of $$(2,{\text{ }}3)$$ and $$(0,{\text{ }}2)$$ within circles,

A1 for approximately correct reflection in $$y = x$$, concave up then concave down shape (do not accept line segments).

[3 marks]

c.

### Question

Let $$f\left( x \right) = \sqrt {x + 2}$$ for x ≥ 2 and g(x) = 3x − 7 for $$x \in \mathbb{R}$$.

Write down f (14).

a.

Find $$\left( {g \circ f} \right)$$ (14).

b.

Find g−1(x).

c.

### Markscheme

f (14) = 4     A1 N1

[1 mark]

a.

attempt to substitute     (M1)

eg   g (4), 3 × 4 − 7

5     A1 N2

[2 marks]

b.

interchanging x and y (seen anywhere)     (M1)

eg   x = 3y − 7

evidence of correct manipulation     (A1)

eg   x + 7 = 3y

$${g^{ – 1}}\left( x \right) = \frac{{x + 7}}{3}$$     A1 N3

[3 marks]

c.

### Question

Consider a function f (x) , for −2 ≤ x ≤ 2 . The following diagram shows the graph of f. Write down the value of f (0).

a.i.

Write down the value of f −1 (1).

a.ii.

Write down the range of f −1.

b.

On the grid above, sketch the graph of f −1.

c.

### Markscheme

$$f\left( 0 \right) = – \frac{1}{2}$$     A1 N1

[1 mark]

a.i.

f −1 (1) = 2     A1 N1

[1 mark]

a.ii.

−2 ≤ y ≤ 2, y∈ [−2, 2]  (accept −2 ≤ x ≤ 2)     A1 N1

[1 mark]

b. A1A1A1A1  N4

Note: Award A1 for evidence of approximately correct reflection in y = x with correct curvature.

(y = x does not need to be explicitly seen)

Only if this mark is awarded, award marks as follows:

A1 for both correct invariant points in circles,

A1 for the three other points in circles,

A1 for correct domain.

[4 marks]

c.