IB Math Analysis & Approaches Questionbank-Topic: SL 2.11 Transformations of graphs SL Paper 1

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]

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]

(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]