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Date May 2009 Marks available 7 Reference code 09M.1.sl.TZ1.10
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 10 Adapted from N/A

Question

Let f(x)=axx2+1f(x)=axx2+1 , 8x88x8 , aR .The graph of f is shown below.


The region between x=3 and x=7 is shaded.

Show that f(x)=f(x) .

[2]
a.

Given that f(x)=2ax(x23)(x2+1)3 , find the coordinates of all points of inflexion.

[7]
b.

It is given that f(x)dx=a2ln(x2+1)+C .

(i)     Find the area of the shaded region, giving your answer in the form plnq .

(ii)    Find the value of 842f(x1)dx .

[7]
c.

Markscheme

METHOD 1

evidence of substituting x for x     (M1)

f(x)=a(x)(x)2+1     A1

f(x)=axx2+1 (=f(x))     AG     N0

METHOD 2

y=f(x) is reflection of y=f(x) in x axis

and y=f(x) is reflection of y=f(x) in y axis     (M1)

sketch showing these are the same     A1

f(x)=axx2+1 (=f(x))     AG     N0

[2 marks]

a.

evidence of appropriate approach     (M1)

e.g. f(x)=0

to set the numerator equal to 0     (A1)

e.g. 2ax(x23)=0 ; (x23)=0

(0, 0) , (3,a34) , (3,a34) (accept x=0 , y=0 etc)      A1A1A1A1A1     N5

[7 marks]

b.

(i) correct expression     A2

e.g. [a2ln(x2+1)]73 , a2ln50a2ln10 , a2(ln50ln10)

area = a2ln5     A1A1     N2

(ii) METHOD 1

recognizing the shift that does not change the area     (M1)

e.g. 84f(x1)dx=73f(x)dx , a2ln5

recognizing that the factor of 2 doubles the area     (M1)

e.g. 842f(x1)dx=284f(x1)dx (=273f(x)dx)

842f(x1)dx=aln5 (i.e. 2× their answer to (c)(i))     A1     N3

METHOD 2

changing variable

let w=x1 , so dwdx=1

2f(w)dw=2a2ln(w2+1)+c     (M1)

substituting correct limits

e.g. [aln[(x1)2+1]]84[aln(w2+1)]73 , aln50aln10     (M1)

842f(x1)dx=aln5     A1     N3

[7 marks]

c.

Examiners report

Part (a) was achieved by some candidates, although brackets around the x were commonly neglected. Some attempted to show the relationship by substituting a specific value for x . This earned no marks as a general argument is required.

a.

Although many recognized the requirement to set the second derivative to zero in (b), a majority neglected to give their answers as ordered pairs, only writing the x-coordinates. Some did not consider the negative root.

b.

For those who found a correct expression in (c)(i), many finished by calculating ln50ln10=ln40 . Few recognized that the translation did not change the area, although some factored the 2 from the integrand, appreciating that the area is double that in (c)(i).

c.

Syllabus sections

Topic 6 - Calculus » 6.5 » Areas under curves (between the curve and the x-axis).
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