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Date May 2017 Marks available 2 Reference code 17M.2.SL.TZ1.S_10
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number S_10 Adapted from N/A

Question

Let f ( x ) = ln x and g ( x ) = 3 + ln ( x 2 ) , for x > 0 .

The graph of g can be obtained from the graph of f by two transformations:

a horizontal stretch of scale factor  q  followed by a translation of  ( h k ) .

Let h ( x ) = g ( x ) × cos ( 0.1 x ) , for 0 < x < 4 . The following diagram shows the graph of h and the line y = x .

M17/5/MATME/SP2/ENG/TZ1/10.b.c

The graph of h intersects the graph of h 1 at two points. These points have x coordinates 0.111 and 3.31 correct to three significant figures.

Write down the value of q ;

[1]
a.i.

Write down the value of h ;

[1]
a.ii.

Write down the value of k .

[1]
a.iii.

Find 0.111 3.31 ( h ( x ) x ) d x .

[2]
b.i.

Hence, find the area of the region enclosed by the graphs of h and h 1 .

[3]
b.ii.

Let d be the vertical distance from a point on the graph of h to the line y = x . There is a point P ( a ,   b ) on the graph of h where d is a maximum.

Find the coordinates of P, where 0.111 < a < 3.31 .

[7]
c.

Markscheme

q = 2     A1     N1

 

Note:     Accept q = 1 , h = 0 , and k = 3 ln ( 2 ) , 2.31 as candidate may have rewritten g ( x ) as equal to 3 + ln ( x ) ln ( 2 ) .

 

[1 mark]

a.i.

h = 0     A1     N1

 

Note:     Accept q = 1 , h = 0 , and k = 3 ln ( 2 ) , 2.31 as candidate may have rewritten g ( x ) as equal to 3 + ln ( x ) ln ( 2 ) .

 

[1 mark]

a.ii.

k = 3     A1     N1

 

Note:     Accept q = 1 , h = 0 , and k = 3 ln ( 2 ) , 2.31 as candidate may have rewritten g ( x ) as equal to 3 + ln ( x ) ln ( 2 ) .

 

[1 mark]

a.iii.

2.72409

2.72     A2     N2

[2 marks]

b.i.

recognizing area between y = x and h equals 2.72     (M1)

eg M17/5/MATME/SP2/ENG/TZ1/10.b.ii/M

recognizing graphs of h and h 1 are reflections of each other in y = x     (M1)

eg area between y = x and h equals between y = x and h 1

2 × 2.72 0.111 3.31 ( x h 1 ( x ) ) d x = 2.72

5.44819

5.45     A1     N3

[??? marks]

b.ii.

valid attempt to find d     (M1)

eg difference in y -coordinates, d = h ( x ) x

correct expression for d     (A1)

eg ( ln 1 2 x + 3 ) ( cos 0.1 x ) x

valid approach to find when d is a maximum     (M1)

eg max on sketch of d , attempt to solve d = 0

0.973679

x = 0.974     A2     N4 

substituting their x value into h ( x )     (M1)

2.26938

y = 2.27     A1     N2

[7 marks]

c.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
a.iii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.

Syllabus sections

Topic 5 —Calculus » SL 5.5—Integration introduction, areas between curve and x axis
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Topic 5 —Calculus

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