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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 ) = 9 x 2 x R .

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 ) = ( x 3 ) 2 + k ,   x R .

Find the x -intercepts of the graph of f .

[2]
a.

Show that the area of PQRS is 18 b 2 b 3 .

[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,  2 x 2 6 x + 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 ) = 0 9 x 2 = 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  2 b ,   2 b ( 9 x 2 ) ,   2 b × f ( b )

correct working that clearly leads to given answer       A1

eg    2 b ( 9 b 2 )

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

area =  18 b 2 b 3                 AG  N0

[2 marks]

b.

setting derivative = 0 (seen anywhere)        (M1)

eg  A = 0 [ 18 b 2 b 3 ] = 0  

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

eg    18 6 b 2 ,   2 b ( 2 b ) + ( 9 b 2 ) × 2

correct working        (A1)

eg    6 b 2 = 18 ,   b = ± 3

b = 3               A1  N3

[5 marks]

c.

valid approach      (M1)

eg  f = g 9 x 2 = ( x 3 ) 2 + k  

correct working        (A1)

eg    9 x 2 = x 2 6 x + 9 + k ,   9 x 2 x 2 + 6 x 9 k = 0

2 x 2 6 x + k = 0              AG  N0

[2 marks]

d.

METHOD 1 (discriminant)

recognizing to use discriminant (seen anywhere)        (M1)

eg  Δ,   b 2 4 a c

discriminant = 0 (seen anywhere)      M1

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

eg    ( 6 ) 2 4 ( 2 ) ( k ) ,   ( 6 ) 2 4 ( 2 ) ( k )

correct working      (A1)

eg   36 8 k = 0 ,   8 k = 36

k = 36 8 ( = 9 2 , 4.5 )                        A1 N2

 

METHOD 2 (completing the square)

valid approach to complete the square           (M1)

eg    2 ( x 2 3 x + 9 4 ) = k + 18 4 ,   x 2 3 x + 9 4 9 4 + k 2 = 0

correct working         (A1)

eg    2 ( x 3 2 ) 2 = k + 18 4 ,   ( x 3 2 ) 2 9 4 + k 2 = 0

recognizing condition for one solution       M1

eg    ( x 3 2 ) 2 = 0 ,   9 4 + k 2 = 0

correct working           (A1)

eg    k = 18 4 ,   k 2 = 9 4

k = 18 4 ( = 9 2 , 4.5 )                        A1 N2

 

METHOD 3 (using vertex)

valid approach to find vertex (seen anywhere)          M1

eg    ( 2 x 2 6 x + k ) = 0 ,   b 2 a

correct working        (A1)

eg    ( 2 x 2 6 x + k ) = 4 x 6 ,    ( 6 ) 2 ( 2 )

x = 6 4 ( = 3 2 )         (A1)

correct substitution        (A1)

eg    2 ( 3 2 ) 2 6 ( 3 2 ) + k = 0

k = 18 4 ( = 9 2 , 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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