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Date May 2009 Marks available 2 Reference code 09M.1.sl.TZ1.10
Level SL only Paper 1 Time zone TZ1
Command term Show that Question number 10 Adapted from N/A

Question

Let \(f(x) = \frac{{ax}}{{{x^2} + 1}}\) , \( - 8 \le x \le 8\) , \(a \in \mathbb{R}\) .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 \(f''(x) = \frac{{2ax({x^2} - 3)}}{{{{({x^2} + 1)}^3}}}\) , find the coordinates of all points of inflexion.

[7]
b.

It is given that \(\int {f(x){\rm{d}}x = \frac{a}{2}} \ln ({x^2} + 1) + C\) .

(i)     Find the area of the shaded region, giving your answer in the form \(p\ln q\) .

(ii)    Find the value of \(\int_4^8 {2f(x - 1){\rm{d}}x} \) .

[7]
c.

Markscheme

METHOD 1

evidence of substituting \( - x\) for \(x\)     (M1)

\(f( - x) = \frac{{a( - x)}}{{{{( - x)}^2} + 1}}\)     A1

\(f( - x) = \frac{{ - ax}}{{{x^2} + 1}}\) \(( = - f(x))\)     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

\(f( - x) = \frac{{ - ax}}{{{x^2} + 1}}\) \(( = - f(x))\)     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. \(2ax({x^2} - 3) = 0\) ; \(({x^2} - 3) = 0\)

(0, 0) , \(\left( {\sqrt 3 ,\frac{{a\sqrt 3 }}{4}} \right)\) , \(\left( { - \sqrt 3 , - \frac{{a\sqrt 3 }}{4}} \right)\) (accept \(x = 0\) , \(y = 0\) etc)      A1A1A1A1A1     N5

[7 marks]

b.

(i) correct expression     A2

e.g. \(\left[ {\frac{a}{2}\ln ({x^2} + 1)} \right]_3^7\) , \(\frac{a}{2}\ln 50 - \frac{a}{2}\ln 10\) , \(\frac{a}{2}(\ln 50 - \ln 10)\)

area = \(\frac{a}{2}\ln 5\)     A1A1     N2

(ii) METHOD 1

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

e.g. \(\int_4^8 {f(x - 1){\rm{d}}x}  = \int_3^7 {f(x){\rm{d}}x} \) , \(\frac{a}{2}\ln 5\)

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

e.g. \(\int_4^8 {2f(x - 1){\rm{d}}x = } 2\int_4^8 {f(x - 1){\rm{d}}x} \) \(\left( { = 2\int_3^7 {f(x){\rm{d}}x} } \right)\)

\(\int_4^8 {2f(x - 1){\rm{d}}x = a\ln 5} \) (i.e. \(2 \times \) their answer to (c)(i))     A1     N3

METHOD 2

changing variable

let \(w = x - 1\) , so \(\frac{{{\rm{d}}w}}{{{\rm{d}}x}} = 1\)

\(2\int {f(w){\rm{d}}w = } \frac{{2a}}{2}\ln ({w^2} + 1) + c\)     (M1)

substituting correct limits

e.g. \(\left[ {a\ln \left[ {{{(x - 1)}^2} + 1} \right]} \right]_4^8\) , \(\left[ {a\ln ({w^2} + 1)} \right]_3^7\) , \(a\ln 50 - a\ln 10\)     (M1)

\(\int_4^8 {2f(x - 1){\rm{d}}x = a\ln 5} \)     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 \(\ln 50 - \ln 10 = \ln 40\) . 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 2 - Functions and equations » 2.3 » Transformations of graphs.
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