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Date May 2019 Marks available 2 Reference code 19M.2.SL.TZ1.T_4
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term State Question number T_4 Adapted from N/A

Question

Consider the function  f ( x ) = x 3 5 x 2 + 6 x 3 + 1 x x > 0

The function f ( x ) = x 3 5 x 2 + 6 x 3 + 1 x x > 0 , models the path of a river, as shown on the following map, where both axes represent distance and are measured in kilometres. On the same map, the location of a highway is defined by the function g ( x ) = 0.5 ( 3 ) x + 1 .

The origin, O(0, 0) , is the location of the centre of a town called Orangeton.

A straight footpath, P , is built to connect the centre of Orangeton to the river at the point where x = 1 2 .

Bridges are located where the highway crosses the river.

A straight road is built from the centre of Orangeton, due north, to connect the town to the highway.

Find the value of  f ( x ) when  x = 1 2 .

[2]
a.

Find the function, P ( x ) , that would define this footpath on the map.

[3]
b.i.

State the domain of P .

[2]
b.ii.

Find the coordinates of the bridges relative to the centre of Orangeton.

[4]
c.

Find the distance from the centre of Orangeton to the point at which the road meets the highway.

[2]
d.

This straight road crosses the highway and then carries on due north.

State whether the straight road will ever cross the river. Justify your answer.

[2]
e.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

f ( 1 2 ) = ( 1 2 ) 3 5 ( 1 2 ) 2 + 6 ( 1 2 ) 3 1 ( 1 2 )      (M1)

Note: Award (M1) for correct substitution into given function.

7 8 ( 0.875 )       (A1)(G2)

[2 marks]

a.

0 7 8 0 1 2     (M1)

Note: Award (M1) for correct substitution into gradient formula. Accept equivalent forms such as 7 8 = 1 2 m .

7 4   (1.75)      (A1)(ft)

P ( x ) = 7 4 x ( 1.75 x )       (A1)(ft)(G3)

Note: Follow through from part (a).

[3 marks]

b.i.

0 < x 1 2    (A1)(A1)

Note: Award (A1) for both endpoints correct, (A1) for correct mathematical notation indicating an interval with two endpoints. Accept weak inequalities. Award at most (A1)(A0) for incorrect notation such as 0 − 0.5 or a written description of the domain with correct endpoints. Award at most (A1)(A0) for 0 < y 1 2 .

[2 marks]

b.ii.

(0.360, 1.34)   ((0.359947…, 1.33669))   (A1)(A1)

(3.63, 1.01)   ((3.63066…, 1.00926…))   (A1)(A1)

Note: Award (A1)(A1) for each correct coordinate pair. Accept correct answers in the form of  x = 0.360 y = 1.34  etc. Award at most (A0)(A1)(A1)(A1)ft if one or both parentheses are omitted.

[4 marks]

c.

g ( 0 ) = 0.5 ( 3 ) 0 + 1     (M1)

1.5 (km)   (A1)(G2)

[2 marks]

d.

domain given as x > 0 (but equation of road is x = 0 )      (R1)

OR

(equation of road is x = 0 ) the function of the river is asymptotic to x = 0        (R1)

so it does not meet the river       (A1)

Note: Award the (R1) for a correct mathematical statement about the equation of the river (and the equation of the road). Justification must be based on mathematical reasoning. Do not award (R0)(A1).

[2 marks]

e.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.
[N/A]
d.
[N/A]
e.

Syllabus sections

Topic 2—Functions » SL 2.2—Functions, notation domain, range and inverse as reflection
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