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Date May 2021 Marks available 2 Reference code 21M.2.SL.TZ1.9
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number 9 Adapted from N/A

Question

Consider the function f defined by fx=90e-0.5x for x+.

The graph of f and the line y=x intersect at point P.

The line L has a gradient of -1 and is a tangent to the graph of f at the point Q.

The shaded region A is enclosed by the graph of f and the lines y=x and L.

Find the x-coordinate of P.

[2]
a.

Find the exact coordinates of Q.

[4]
b.

Show that the equation of L is y=-x+2ln45+2.

[2]
c.

Find the x-coordinate of the point where L intersects the line y=x.

[1]
d.i.

Hence, find the area of A.

[4]
d.ii.

The line L is tangent to the graphs of both f and the inverse function f-1.

Find the shaded area enclosed by the graphs of f and f-1 and the line L.

[2]
e.

Markscheme

Attempt to find the point of intersection of the graph of f and the line y=x         (M1)

x=5.56619

=5.57          A1

 

[2 marks]

a.

f'x=-45e-0.5x          A1

attempt to set the gradient of f equal to -1         (M1)

-45e-0.5x=-1

Q has coordinates 2ln45, 2 (accept (-2ln145, 2)          A1A1

 

Note: Award A1 for each value, even if the answer is not given as a coordinate pair.

   Do not accept ln145-0.5 or ln450.5 as a final value for x. Do not accept 2.0 or 2.00 as a final value for y.

 

[4 marks]

b.

attempt to substitute coordinates of Q (in any order) into an appropriate equation         (M1)

y-2=-x-2ln45  OR  2=-2ln45+c          A1

equation of L is y=-x+2ln45+2           AG

 

[2 marks]

c.

x=ln45+1=4.81          A1

 

[1 mark]

d.i.

appropriate method to find the sum of two areas using integrals of the difference of two functions          (M1)

 

Note: Allow absence of incorrect limits.

 

4.8065.566x--x+2ln45+2dx+5.5667.61390e-0.5x--x+2ln45+2dx        (A1)(A1)

 

Note: Award A1 for one correct integral expression including correct limits and integrand.
          Award A1 for a second correct integral expression including correct limits and integrand.

 

=1.52196

=1.52        A1

 

[4 marks]

d.ii.

by symmetry 2×1.52         (M1)

=3.04         A1

 

Note: Accept any answer that rounds to 3.0 (but do not accept 3).

  

[2 marks]

e.

Examiners report

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

Syllabus sections

Topic 2—Functions » SL 2.2—Functions, notation domain, range and inverse as reflection
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