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Date May 2014 Marks available 2 Reference code 14M.1.hl.TZ1.11
Level HL only Paper 1 Time zone TZ1
Command term Show that Question number 11 Adapted from N/A

Question

Consider the function f(x)=lnxx, x>0.

The sketch below shows the graph of y= f(x) and its tangent at a point A.


Show that f(x)=1lnxx2.

[2]
a.

Find the coordinates of B, at which the curve reaches its maximum value.

[3]
b.

Find the coordinates of C, the point of inflexion on the curve.

[5]
c.

The graph of y= f(x) crosses the x-axis at the point A.

Find the equation of the tangent to the graph of f at the point A.

[4]
d.

The graph of y= f(x) crosses the x-axis at the point A.

Find the area enclosed by the curve y=f(x), the tangent at A, and the line x=e.

[7]
e.

Markscheme

f(x)=x×1xlnxx2     M1A1

=1lnxx2     AG

[2 marks]

a.

1lnxx2=0 has solution x=e     M1A1

y=1e     A1

hence maximum at the point (e, 1e)

[3 marks]

b.

f(x)=x2(1x)2x(1lnx)x4     M1A1

=2lnx3x3

 

Note:     The M1A1 should be awarded if the correct working appears in part (b).

 

point of inflexion where f(x)=0     M1

so x=e32, y=32e32     A1A1

C has coordinates (e32, 32e32)

[5 marks]

c.

f(1)=0     A1

f(1)=1     (A1)

y=x+c     (M1)

through (1, 0)

equation is y=x1     A1

[4 marks]

d.

METHOD 1

area =e1x1lnxxdx     M1A1A1

 

Note:     Award M1 for integration of difference between line and curve, A1 for correct limits, A1 for correct expressions in either order.

 

lnxxdx=(lnx)22(+c)     (M1)A1

(x1)dx=x22x(+c)     A1

=[12x2x12(lnx)2]e1

=(12e2e12)(121)

=12e2e     A1

METHOD 2

area = area of triangle e1lnxxdx     M1A1

 

Note:     A1 is for correct integral with limits and is dependent on the M1.

 

lnxxdx=(lnx)22(+c)     (M1)A1

area of triangle =12(e1)(e1)     M1A1

12(e1)(e1)(12)=12e2e     A1

[7 marks]

e.

Examiners report

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

Syllabus sections

Topic 6 - Core: Calculus » 6.2 » The product and quotient rules.
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