2011 A-level H2 Mathematics (9740) Paper 1 Question 8 Suggested Solutions

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

(i)
$\int \frac{1}{100-v^2}dv = \frac{1}{20} \mathrm{ln}|\frac{10+v}{10-v}| + C$

(ii)
(a)
$\frac{dv}{dt} = 10 - 0.1 v^2$

$\int \frac{1}{100-v^2} dv = \int \frac{1}{10} dt$

$\frac{1}{20} \mathrm{ln} |\frac{10+v}{10-v}| = \frac{1}{10} t + C$

When $t = 0, v = 0 \Rightarrow c = 0$

$\therefore t = 0.5 \mathrm{ln} |\frac{10+v}{10-v}|$

When $v = 5, t = 0.5\mathrm{ln}3$

(b)
When $t = 1$

$\Rightarrow 2 = \mathrm{ln}|\frac{10+v}{10-v}|$

$\pm e^2 = \frac{10 + v}{10 - v}$ Reject $- e^2$ since $v = 0$ when $t = 0$

$e^2 (10 - v) = 10 + v$

$10 e^2 - 10 = v - e^2v$

$v = 7.62$

(c)

Graph of 8(b)

From the graph, As $t \rightarrow \infty, v \rightarrow 10$

KS Comments:

(i) can be resolved easily with the MF15 formula.
(ii), students can actually plot the entire graph of t against v into the graphing calculator and calculate the v when t = 1. They should draw the graph out then. Should they do this, (c) will be solved conveniently.

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for almost two decades. 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.

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2011 A-level H2 Mathematics (9740) Paper 1 Question 8 Suggested Solutions

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