2015 A-level H2 Mathematics (9740) Paper 1 Question 7 Suggested Solutions

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

(i)
2 \vec{OC} = 3 \vec{CA} = 3 (\vec{OA} - \vec{OC})

5 \vec{OC} = 3 a

\vec{OC} = \frac{3}{5} a

11 \vec{OD} = 5 \vec{OB}

\vec{OD} = \frac{5}{11} b

(ii)
\vec{BC} = \frac{3}{5}a -b

l_{BC}: r = b + \lambda (\frac{3}{5}a - b)

l_{BC}: r = \frac{3}{5} \lambda a + (1-\lambda)b, \lambda \in \mathbb{R}

\vec{AD} = \frac{3}{5}a -b

l_{AD}: r = a + \mu (\frac{5}{11}b - a)

l_{AD}: r = \frac{5}{11} \mu b + (1-\mu)a, \mu \in \mathbb{R}

(iii)
At E, \frac{5}{11} \mu b + (1-\mu)a = \frac{3}{5} \lambda a + (1-\lambda)b

\frac{5}{11} \mu = 1 - \lambda \rightarrow(1)

\frac{3}{5} \lambda = 1 - \mu \rightarrow(1)

Solving with GC, \lambda = \frac{3}{4} and \mu = \frac{11}{20}

\vec{OC} = \frac{9}{20}a + \frac{1}{4} b

\vec{AE} = \frac{1}{4}b - \frac{11}{20}a

\vec{ED} = \frac{9}{44}b - \frac{9}{20}a

= \frac{9}{11}(\frac{1}{4}b - \frac{11}{20}a)

\vec{ED} = \frac{9}{11} \vec{AE}

\therefore, AE: ED = 11: 9

Back to 2015 A-level H2 Mathematics (9740) Paper 1 Suggested Solutions

KS Comments:

Students that faced difficulties with ratio (like me) might struggle a bit here. But if you guys do listen in class, you should know the above method I use, should help a lot.
(ii) should be manageable. For (iii), students need to be careful if they use a GC to solve the question. Its like statistics, solving \mu and \lambda.

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