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Date May 2017 Marks available 2 Reference code 17M.2.sl.TZ1.10
Level SL only Paper 2 Time zone TZ1
Command term Find Question number 10 Adapted from N/A

Question

Let f(x)=lnx and g(x)=3+ln(x2), 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 bya translation of (hk).

Let h(x)=g(x)×cos(0.1x), 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 h1 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 3.310.111(h(x)x)dx.

[2]
b.i.

Hence, find the area of the region enclosed by the graphs of h and h1.

[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=3ln(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=3ln(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=3ln(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)

egM17/5/MATME/SP2/ENG/TZ1/10.b.ii/M

recognizing graphs of h and h1 are reflections of each other in y=x     (M1)

egarea between y=x and h equals between y=x and h1

2×2.723.310.111(xh1(x))dx=2.72

5.44819

5.45     A1     N3

[??? marks]

b.ii.

valid attempt to find d     (M1)

egdifference in y-coordinates, d=h(x)x

correct expression for d     (A1)

eg(ln12x+3)(cos0.1x)x

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

egmax 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 6 - Calculus » 6.3 » Local maximum and minimum points.
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