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Date November Example questions Marks available 3 Reference code EXN.1.SL.TZ0.8
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Describe Question number 8 Adapted from N/A

Question

The following diagram shows the graph of y=-1-x+3 for x-3.

A function f is defined by fx=-1-x+3 for x-3.

Describe a sequence of transformations that transforms the graph of y=x for x0 to the graph of y=-1-x+3 for x-3.

[3]
a.

State the range of f.

[1]
b.

Find an expression for f-1x, stating its domain.

[5]
c.

Find the coordinates of the point(s) where the graphs of y=fx and y=f-1x intersect.

[5]
d.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

for example,

a reflection in the x-axis (in the line y=0)        A1

a horizontal translation (shift) 3 units to the left        A1

a vertical translation (shift) down by 1 unit        A1

 

Note: Award A1 for each correct transformation applied in a correct position in the sequence. Do not accept use of the “move” for a translation.

Note: Award A1A1A1 for a correct alternative sequence of transformations. For example,

a vertical translation (shift) down by 1 unit, followed by a horizontal translation (shift) 3 units to the left and then a reflection in the line y=-1.

 

[3 marks]

a.

range is fx-1        A1

 

Note: Correct alternative notations include ]-,-1](-,-1] or y-1.

 

[1 mark]

b.

-1-y+3=x        M1

 

Note: Award M1 for interchanging x and y (can be done at a later stage).

 

y+3=-x-1=-x+1        A1

y+3=x+12        A1

so f-1x=x+12-3 f-1x=x2+2x-2        A1

domain is x-1        A1

 

Note: Correct alternative notations include ]-,-1] or (-,-1].

 

[5 marks]

c.

the point of intersection lies on the line y=x

 

EITHER

x+12-3=x       M1   

attempts to solve their quadratic equation       M1

for example, x+2x-1=0 or x=-1±12-41-22  x=-1±32

 

OR

-1-x+3=x       M1

-1-x+32=x22x+3+x+4=x2

substitutes 2x+3=-2x+1 to obtain -2x+1+x+4=x2

attempts to solve their quadratic equation       M1

for example, x+2x-1=0 or x=-1±12-41-22  x=-1±32

 

THEN

x=-2,1        A1

as x-1, the only solution is x=-2        R1

so the coordinates of the point of intersection are -2,-2        A1

 

Note: Award R0A1 if -2,-2 is stated without a valid reason given for rejecting 1,1.

 

[5 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
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c.
[N/A]
d.

Syllabus sections

Topic 2—Functions » SL 2.11—Transformation of functions
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Topic 2—Functions

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