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 f(x)=-1-√x+3 for x≥-3.
Describe a sequence of transformations that transforms the graph of y=√x for x≥0 to the graph of y=-1-√x+3 for x≥-3.
State the range of f.
Find an expression for f-1(x), stating its domain.
Find the coordinates of the point(s) where the graphs of y=f(x) and y=f-1(x) intersect.
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]
range is f(x)≤-1 A1
Note: Correct alternative notations include ]-∞,-1], (-∞,-1] or y≤-1.
[1 mark]
-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+1)2 A1
so f-1(x)=(x+1)2-3 (f-1(x)=x2+2x-2) A1
domain is x≤-1 A1
Note: Correct alternative notations include ]-∞,-1] or (-∞,-1].
[5 marks]
the point of intersection lies on the line y=x
EITHER
(x+1)2-3=x M1
attempts to solve their quadratic equation M1
for example, (x+2)(x-1)=0 or x=-1±√12-4(1)(-2)2 (x=-1±32)
OR
-1-√x+3=x M1
(-1-√x+3)2=x2⇒2√x+3+x+4=x2
substitutes 2√x+3=-2(x+1) to obtain -2(x+1)+x+4=x2
attempts to solve their quadratic equation M1
for example, (x+2)(x-1)=0 or x=-1±√12-4(1)(-2)2 (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, A1
Note: Award R0A1 if is stated without a valid reason given for rejecting .
[5 marks]