IBDP Maths analysis and approaches Topic: SL 5.6 The chain rule for composite functions HL Paper 1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 \times {{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}} = ({{\text{e}}^{3x}} + 3x{{\text{e}}^{3x}})\)     M1A1

[2 marks]

let \(P(n)\) be the statement \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = n{3^{n – 1}}{{\text{e}}^{3x}} + x{3^n}{{\text{e}}^{3x}}\)

prove for \(n = 1\)     M1

\(LHS\) of \(P(1)\) is \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) which is \(1 \times {{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}}\) and \(RHS\) is \({3^0}{{\text{e}}^{3x}} + x{3^1}{{\text{e}}^{3x}}\)     R1

as \({\text{LHS}} = {\text{RHS, }}P(1)\) is true

assume \(P(k)\) is true and attempt to prove \(P(k + 1)\) is true     M1

assuming \(\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = k{3^{k – 1}}{{\text{e}}^{3x}} + x{3^k}{{\text{e}}^{3x}}\)

\(\frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}}} \right)\)     (M1)

\( = k{3^{k – 1}} \times 3{{\text{e}}^{3x}} + 1 \times {3^k}{{\text{e}}^{3x}} + x{3^k} \times 3{{\text{e}}^{3x}}\)     A1

\( = (k + 1){3^k}{{\text{e}}^{3x}} + x{3^{k + 1}}{{\text{e}}^{3x}}\;\;\;\)(as required)     A1

Note:     Can award the A marks independent of the M marks

since \(P(1)\) is true and \(P(k)\) is true \( \Rightarrow P(k + 1)\) is true

then (by \(PMI\)), \(P(n)\) is true \((\forall n \in {\mathbb{Z}^ + })\)     R1

Note: To gain last R1 at least four of the above marks must have been gained.

[7 marks]

\({{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}} = 0 \Rightarrow 1 + 3x = 0 \Rightarrow x =  – \frac{1}{3}\)     M1A1

point is \(\left( { – \frac{1}{3},{\text{ }} – \frac{1}{{3{\text{e}}}}} \right)\)     A1

EITHER

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}}\)

when \(x =  – \frac{1}{3},{\text{ }}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} > 0\) therefore the point is a minimum     M1A1

OR

nature table shows point is a minimum     M1A1

[5 marks]

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}}\)     A1

\(2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}} = 0 \Rightarrow 2 + 3x = 0 \Rightarrow x =  – \frac{2}{3}\)     M1A1

point is \(\left( { – \frac{2}{3},{\text{ }} – \frac{2}{{3{{\text{e}}^2}}}} \right)\)     A1

since the curvature does change (concave down to concave up) it is a point of inflection     R1

Note:     Allow \({3^{{\text{rd}}}}\) derivative is not zero at \( – \frac{2}{3}\)

[5 marks]

(general shape including asymptote and through origin)     A1

showing minimum and point of inflection     A1

Note:     Only indication of position of answers to (c) and (d) required, not coordinates.

[2 marks]

Total [21 marks]

Well done.

The logic of an induction proof was not known well enough. Many candidates used what they had to prove rather than differentiating what they had assumed. They did not have enough experience in doing Induction proofs.

Good, some forgot to test for min/max, some forgot to give the \(y\) value.

Again quite good, some forgot to check for change in curvature and some forgot the \(y\) value.

Some accurate sketches, some had all the information from earlier parts but could not apply it. The asymptote was often missed.