The function f is defined by \(f(x) = x{{\text{e}}^{2x}}\) .

It can be shown that \({f^{(n)}}(x) = ({2^n}x + n{2^{n - 1}}){{\text{e}}^{2x}}\) for all \(n \in {\mathbb{Z}^ + }\), where \({f^{(n)}}(x)\) represents the \({n^{{\text{th}}}}\) derivative of \(f(x)\) .

(a)     By considering \({f^{(n)}}(x){\text{ for }}n = 1{\text{ and }}n = 2\) , show that there is one minimum point P on the graph of f , and find the coordinates of P.

(b)     Show that f has a point of inflexion Q at x = −1.

(c)     Determine the intervals on the domain of f where f is

(i)     concave up;

(ii)     concave down.

(d)     Sketch f , clearly showing any intercepts, asymptotes and the points P and Q.

(e)     Use mathematical induction to prove that \({f^{(n)}}(x) = ({2^n}x + n{2^{n - 1}}){{\text{e}}^{2x}}{\text{ for all }}n \in {\mathbb{Z}^ + },{\text{ where }}{f^{(n)}}{\text{ represents the }}{n^{{\text{th}}}}{\text{ derivative of }}f(x)\) .

(a)     \(f'(x) = (1 + 2x){{\text{e}}^{2x}}\)     A1

\(f'(x) = 0\)     M1

\( \Rightarrow (1 + 2x){{\text{e}}^{2x}} = 0 \Rightarrow x = - \frac{1}{2}\)     A1

\(f''(x) = ({2^2}x + 2 \times {2^{2 - 1}}){{\text{e}}^{2x}} = (4x + 4){{\text{e}}^{2x}}\)     A1

\(f''\left( { - \frac{1}{2}} \right) = \frac{2}{{\text{e}}}\)     A1

\(\frac{2}{{\text{e}}} > 0 \Rightarrow {\text{at }}x = - \frac{1}{2},{\text{ }}f(x){\text{ has a minimum.}}\)     R1

\({\text{P}}\left( { - \frac{1}{2}, - \frac{1}{{2{\text{e}}}}} \right)\)     A1

[7 marks]

 

(b)     \(f''(x) = 0 \Rightarrow 4x + 4 = 0 \Rightarrow x = - 1\)     M1A1

\({\text{Using the }}{{\text{2}}^{{\text{nd}}}}{\text{ derivative }}f''\left( { - \frac{1}{2}} \right) = \frac{2}{{\text{e}}}{\text{ and }}f''( - 2) = - \frac{4}{{{{\text{e}}^4}}},\)     M1A1

the sign change indicates a point of inflexion.     R1

[5 marks]

 

(c)     (i)     f(x) is concave up for \(x > - 1\) .     A1

 

(ii)     f(x) is concave down for \(x < - 1\) .     A1

[2 marks]

 

(d)

     A1A1A1A1

 

Note: Award A1 for P and Q, with Q above P,

A1 for asymptote at y = 0 ,

A1 for (0, 0) ,

A1 for shape.

 

[4 marks]

 

(e)     Show true for n = 1     (M1)

\(f'(x) = {{\text{e}}^{2x}} + 2x{{\text{e}}^{2x}}\)     A1

\( = {{\text{e}}^{2x}}(1 + 2x) = (2x + {2^0}){{\text{e}}^{2x}}\)

Assume true for \(n = k{\text{ , }}i.e.{\text{ }}{f^{(k)}}x = ({2^k}x + k \times {2^{k - 1}}){{\text{e}}^{2x}}{\text{, }}k \geqslant 1\)     M1A1

Consider \(n = k + 1{\text{ , }}i.e.{\text{ an attempt to find }}\frac{{\text{d}}}{{{\text{d}}x}}\left( {{f^k}(x)} \right)\) .     M1

\({F^{(k + 1)}}(x) = {2^k}{{\text{e}}^{2x}} + 2{{\text{e}}^{2x}}({2^k}x + k \times {2^{k - 1}})\)     A1

\( = \left( {{2^k} + 2({2^k}x + k \times {2^{k - 1}})} \right){{\text{e}}^{2x}}\)

\( = (2 \times {2^k}x + {2^k} + k \times 2 \times {2^{k - 1}}){{\text{e}}^{2x}}\)

\( = ({2^{k + 1}}x + {2^k} + k \times {2^k}){{\text{e}}^{2x}}\)     A1
\( = \left( {{2^{k + 1}}x + \left( {k + 1} \right){2^k}} \right){{\text{e}}^{2x}}\)     A1

P(n) is true for \(k \Rightarrow P(n)\) is true for k + 1, and since true for n = 1, result proved by mathematical induction \(\forall n \in {\mathbb{Z}^ + }\)

Note: Only award R1 if a reasonable attempt is made to prove the \({(k + 1)^{{\text{th}}}}\) step.

 

[9 marks]

Total [27 marks]

This was the most accessible question in section B for these candidates. A majority of candidates produced partially correct answers to part (a), but a significant number struggled with demonstrating that the point is a minimum, despite the hint being given in the question. Part (b) started quite successfully but many students were unable to prove it is a point of inflexion or, more commonly, did not attempt to justify it. Correct answers were often seen for part (c). Part (d) was dependent on the successful completion of the first three parts. If candidates made errors in earlier parts, this often became obvious when they came to sketch the curve. However, few candidates realised that this part was a good way of checking that the above answers were at least consistent. The quality of curve sketching was rather weak overall, with candidates not marking points appropriately and not making features such as asymptotes clear. It is not possible to tell to what extent this was an effect of candidates not having a calculator, but it should be noted that asking students to sketch curves without a calculator will continue to appear on non-calculator papers. In part (e) the basic idea of proof by induction had clearly been taught with a significant majority of students understanding this. However, many candidates did not understand that they had to differentiate again to find the result for (k + 1) .