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Date May 2019 Marks available 5 Reference code 19M.1.SL.TZ1.S_8
Level Standard Level Paper Paper 1 Time zone Time zone 1
Command term Find Question number S_8 Adapted from N/A

Question

Let f(x)=9x2xR.

The following diagram shows part of the graph of f.

Rectangle PQRS is drawn with P and Q on the x-axis and R and S on the graph of f.

Let OP = b.

Consider another function g(x)=(x3)2+k,  xR.

Find the x-intercepts of the graph of f.

[2]
a.

Show that the area of PQRS is 18b2b3.

[2]
b.

Hence find the value of b such that the area of PQRS is a maximum.

[5]
c.

Show that when the graphs of f and g intersect, 2x26x+k=0.

[2]
d.

Given that the graphs of f and g intersect only once, find the value of k.

[5]
e.

Markscheme

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

valid approach         (M1)

eg    f(x)=09x2=0 , one correct solution

x=3,  3  (accept (3, 0), (−3, 0))         A1 N2

[2 marks]

a.

valid approach         (M1)

eg    height = f(b),  base = 2(OP) or 2b,  2b(9x2),  2b×f(b)

correct working that clearly leads to given answer       A1

eg   2b(9b2)

Note: Do not accept sloppy notation eg 2b×9b2.

area = 18b2b3                AG  N0

[2 marks]

b.

setting derivative = 0 (seen anywhere)        (M1)

eg  A=0[18b2b3]=0 

correct derivative (must be in terms of b only) (seen anywhere)      A2

eg   186b2,  2b(2b)+(9b2)×2

correct working        (A1)

eg   6b2=18,  b=±3

b=3              A1  N3

[5 marks]

c.

valid approach      (M1)

eg  f=g9x2=(x3)2+k 

correct working        (A1)

eg   9x2=x26x+9+k,  9x2x2+6x9k=0

2x26x+k=0             AG  N0

[2 marks]

d.

METHOD 1 (discriminant)

recognizing to use discriminant (seen anywhere)        (M1)

eg  Δ,  b24ac

discriminant = 0 (seen anywhere)      M1

correct substitution into discriminant (do not accept only in quadratic formula)      (A1)

eg   (6)24(2)(k),  (6)24(2)(k)

correct working      (A1)

eg  368k=0,  8k=36

k=368(=92,4.5)                       A1 N2

 

METHOD 2 (completing the square)

valid approach to complete the square           (M1)

eg    2(x23x+94)=k+184,  x23x+9494+k2=0

correct working         (A1)

eg   2(x32)2=k+184,  (x32)294+k2=0

recognizing condition for one solution       M1

eg   (x32)2=0,  94+k2=0

correct working           (A1)

eg   k=184,  k2=94

k=184(=92,4.5)                       A1 N2

 

METHOD 3 (using vertex)

valid approach to find vertex (seen anywhere)          M1

eg   (2x26x+k)=0,  b2a

correct working        (A1)

eg   (2x26x+k)=4x6,   (6)2(2)

x=64(=32)        (A1)

correct substitution        (A1)

eg   2(32)26(32)+k=0

k=184(=92,4.5)                       A1 N2

 

[5 marks]

e.

Examiners report

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e.

Syllabus sections

Topic 5—Calculus » SL 5.7—Optimisation
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