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

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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 \( – 2 \leqslant x \leqslant 4\).

N17/5/MATME/SP1/ENG/TZ0/03

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.

N17/5/MATME/SP1/ENG/TZ0/03.c/M     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.

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