2012 A-level H2 Mathematics (9740) Paper 1 Question 3 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$u_2 = u_{1+1} = \frac{3u_1 - 1}{6} = \frac{5}{6}$

$u_3 = \frac{3u_2 -1}{6} = \frac{1}{4}$

(ii)
As $n \rightarrow \infty, u_n \rightarrow l, u_{n+1} \rightarrow l$

$\Rightarrow l = \frac{3l-1}{6}$

$6l = 3l - 1$

$l = -\frac{1}{3}$

(iii)
Let P(n) be the statement “ $u_n = \frac{14}{3} (0.5)^{n} - \frac{1}{3}$ ” for $n \in {\mathbb{Z}}^{+}$

When $n = 1, \mathrm{LHS} = u_1 = 2,~ \mathrm{RHS} = \frac{14}{3}(0.5) - \frac{1}{3} = 2 = \mathrm{LHS}$ .

Thus, P(1) is true.

Assume P(k) is true for some $k \in \mathbb{Z}^{+}$ , i.e.

$u_k = \frac{14}{3} (0.5)^{k} - \frac{1}{3}$

To prove P(k+1) is true, i.e.

$u_{k+1} = \frac{14}{3} (0.5)^{k+1} - \frac{1}{3}$

$\mathrm{LHS}$

$= u_{k+1}$

$= \frac{3u_{k}-1}{6}$

$= \frac{1}{6} [14 (0.5)^{k} - 1-1]$

$= \frac{14}{6} 2 (0.5)^{k+1} - \frac{1}{3}$

$= \frac{14}{3} (0.5)^{k+1} - \frac{1}{3}$

$= \mathrm{RHS}$

Since P(1) is true, by Mathematical Induction, P(k) is true $\Rightarrow$ P(k+1) is true, P(n) is true for all $n \in \mathbb{Z}^{+}$

KS Comments:

This question is quite straight forward. Some students did not realise to use the given recurrence relation to solve the MI.

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

2012 A-level H2 Mathematics (9740) Paper 1 Question 3 Suggested Solutions

KS Comments:

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