# 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]

2. Write down the minimum number of pizzas that can be ordered. [1] Aayush has 450 GBP.

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

Answer/Explanation

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 $$y = – f(x)$$ .[2]

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.[4]

b(i) and (ii).
Answer/Explanation

### 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 $${f^{ – 1}}$$ .
[3]

a.

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

b.

Find $${f^{ – 1}}(x)$$ .[3]

c.
Answer/Explanation

### 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 $$(f \circ g)(0)$$ .[5]

a.

Find $${f^{ – 1}}(x)$$ .[3]

b.
Answer/Explanation

### Markscheme

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

$$= – 1$$     A1     N2

(ii) METHOD 1

substituting answer from (i)     (M1)

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. $${y^3} = \frac{{x – 3}}{2}$$

$${f^{ – 1}}(x) = \sqrt[3]{{\frac{{x – 3}}{2}}}$$     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 $$x = 0$$ , $$\pm \sqrt[4]{3}$$ $$( \pm 1.316 \ldots )$$ .

Find the value of $$f(0)$$ .[2]

a.

Find the set of values of $$x$$ for which $$f$$ is increasing.[5]

b.

(i)     Find $$f”(1)$$ .

(ii)     Hence, show that there is no point of inflexion on the graph of $$f$$ at $$x = 0$$ .[5]

c.

There is a point of inflexion on the graph of $$f$$ at $$x = \sqrt[4]{3}$$ $$(x = 1.316 \ldots )$$ .

Sketch the graph of $$f$$ , for $$x \ge 0$$ .[3]

d.
Answer/Explanation

## 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 $$f(2)$$.[1]

a.i.

Write down the value of $${f^{ – 1}}( – 1)$$ .[2]

a.ii.

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

[3]
b.
Answer/Explanation

## 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 $$h”(2)$$ .[3]

a.

Explain why P is a point of inflexion.[2]

b.

find the $$y$$-coordinate of P.[2]

c.

find the equation of the normal to the graph of $$h$$ at P.[7]

d.
Answer/Explanation

## 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 $$f( – 3)$$.[1]

a(i).

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

a(ii).

Find the domain of $${f^{ – 1}}$$.[2]

b.

On the grid above, sketch the graph of $${f^{ – 1}}$$.[3]

c.
Answer/Explanation

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

correct answer     A1     N2

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 $$g(2)$$.[1]

a.

Find $$(f \circ g)(x)$$.[2]

b.

Find $${f^{ – 1}}(x)$$.[2]

c.
Answer/Explanation

### 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 $$f$$.[1]

a.

Write down $$f(2)$$;[1]

b.i.

Write down $${f^{ – 1}}(2)$$.[1]

b.ii.

On the grid, sketch the graph of $${f^{ – 1}}$$.[3]

c.
Answer/Explanation

### 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).[1]

a.

Find $$\left( {g \circ f} \right)$$ (14).[2]

b.

Find g−1(x).[3]

c.
Answer/Explanation

### 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).[1]

a.i.

Write down the value of f −1 (1).[1]

a.ii.

Write down the range of f −1.[1]

b.

On the grid above, sketch the graph of f −1.[4]

c.
Answer/Explanation

### 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.