CAIE FP1 2010 June — Question 3 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve inequality: recurrence sequence
DifficultyStandard +0.8 This is a non-trivial induction proof requiring algebraic manipulation to show x_{n+1} > 2 given x_n > 2. Students must handle the rational recurrence relation carefully, requiring factorization or rearrangement of (2x_n^2 + 4x_n - 2)/(2x_n + 3) > 2, which involves clearing denominators and manipulating quadratic expressions. While the structure is standard induction, the algebraic complexity and need to work with inequalities involving fractions elevates this above routine Further Maths questions.
Spec4.01a Mathematical induction: construct proofs

3 The sequence \(x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots\) is such that \(x _ { 1 } = 3\) and $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 2 } + 4 x _ { n } - 2 } { 2 x _ { n } + 3 }$$ for \(n = 1,2,3 , \ldots\). Prove by induction that \(x _ { n } > 2\) for all \(n\).

AnswerMarks Guidance
\(H_k: x_k > 2\) for some \(k\)B1
\(x_{k+1} - 2 = (2x_k^2 - 8)/(2x_k + 3)\)M1A1
\(H_k \Rightarrow 2x_k^2 - 8 > 0 \Rightarrow x_{k+1} > 2 \Rightarrow H_{k+1}\)A1
\(x_1 = 3 > 2 \Rightarrow H_1\) is trueB1 CWO
Completion of the induction argumentA1 [6]
Alternatively for lines 2 and 3:
AnswerMarks
\(x_{k+1} = x_k + \frac{1}{2} - \frac{3\frac{1}{2}}{(2x_k + 3)}\)M1A1
\(H_k \Rightarrow 2x_k + 3 > 7 \Rightarrow H_{k+1}\)A1
OR \(x_{k+1} = x_k + \frac{x_k - 2}{(2x_k + 3)}\)M1A1
\(x_k > 2 \Rightarrow x_{k+1} > 2\)A1
OR \(x_{k+1} - x_k = \frac{x_k - 2}{(2x_k + 3)}\)M1A1
\(x_k > 2 \Rightarrow x_{k+1} > x_k > 2\)A1
Minimum conclusion is 'Hence true for \(n \geq 1\)'.
$H_k: x_k > 2$ for some $k$ | B1 |

$x_{k+1} - 2 = (2x_k^2 - 8)/(2x_k + 3)$ | M1A1 |

$H_k \Rightarrow 2x_k^2 - 8 > 0 \Rightarrow x_{k+1} > 2 \Rightarrow H_{k+1}$ | A1 |

$x_1 = 3 > 2 \Rightarrow H_1$ is true | B1 CWO |

Completion of the induction argument | A1 | [6]

**Alternatively for lines 2 and 3:**

$x_{k+1} = x_k + \frac{1}{2} - \frac{3\frac{1}{2}}{(2x_k + 3)}$ | M1A1 |

$H_k \Rightarrow 2x_k + 3 > 7 \Rightarrow H_{k+1}$ | A1 |

**OR** $x_{k+1} = x_k + \frac{x_k - 2}{(2x_k + 3)}$ | M1A1 |

$x_k > 2 \Rightarrow x_{k+1} > 2$ | A1 |

**OR** $x_{k+1} - x_k = \frac{x_k - 2}{(2x_k + 3)}$ | M1A1 |

$x_k > 2 \Rightarrow x_{k+1} > x_k > 2$ | A1 |

Minimum conclusion is 'Hence true for $n \geq 1$'. | |

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3 The sequence $x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots$ is such that $x _ { 1 } = 3$ and

$$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 2 } + 4 x _ { n } - 2 } { 2 x _ { n } + 3 }$$

for $n = 1,2,3 , \ldots$. Prove by induction that $x _ { n } > 2$ for all $n$.

\hfill \mbox{\textit{CAIE FP1 2010 Q3 [6]}}