19N.1.AHL.TZ0.H_6

pestleMathematicsAAHLPaper 119N· ahl-1-15-proof-by-induction-contradiction-counterexamplessource ↗

Consider the function $f\left( x \right) = x\,{{\text{e}}^{2x}}$ , where $x \in \mathbb{R}$ . The ${n^{{\text{th}}}}$ derivative of $f\left( x \right)$ is denoted by ${f^{\left( n \right)}}\left( x \right)$ .

Prove, by mathematical induction, that ${f^{\left( n \right)}}\left( x \right) = \left( {{2^n}x + n{2^{n - 1}}} \right){{\text{e}}^{2x}}$ , $n \in {\mathbb{Z}^ + }$ .

Markscheme / solution

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

Note: This must be obtained from the candidate differentiating $f\left( x \right)$ .

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

(hence true for $n = 1$ )

assume true for $n = k$ : M1

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

Note: Award M1 if truth is assumed. Do not allow “let $n = k$ ”.

consider $n = k + 1$ :

${f^{\left( {k + 1} \right)}}\left( x \right) = \frac{{\text{d}}}{{{\text{d}}x}}\left( {\left( {{2^k}x + k{2^{k - 1}}} \right){{\text{e}}^{2x}}} \right)$

attempt to differentiate ${f^{\left( k \right)}}\left( x \right)$ M1

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

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

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

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

True for $n = 1$ and $n = k$ true implies true for $n = k + 1$ .

Therefore the statement is true for all $n\left( { \in {\mathbb{Z}^ + }} \right)$ R1

Note: Do not award final R1 if the two previous M1s are not awarded. Allow full marks for candidates who use the base case $n = 0$ .

[7 marks]

Examiners’ report

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