2013 A-level H2 Mathematics (9740) Paper 1 Question 9 Suggested Solutions

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

(i)
Let P(n) be the preposition $\sum_{r=1}^{n} r(2r^{2}+1) = \frac{n}{2} (n+1)(n^{2}+n+1), n \in {\mathbb{Z}}^{+}.$

When $n=1, ~ \mathrm{LHS} = 3 ~\mathrm{and}~ \mathrm{RHS}=3$

$\therefore$, P(1) is true.

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

Want to prove that P(k+1) is also true, i.e., $\sum_{r=1}^{k+1} r(2r^{2}+1) = \frac{k+1}{2} (k+2)[(k+1)^{2}+k+2)] = \frac{k+1}{2} (k+2)(k^{2}+3k+3)$

$\mathrm{LHS} = r(2r^{2}+1)$

$= r(2r^{2}+1) + (k+1)[2(k+1)^{2}+1]$

$= \frac{k}{2} (k+1)(k^{2}+k+1) + (k+1)(2k^{2}+4k +3)$

$= \frac{k+1}{2} [k(k^{2}+k+1) + 2(2k^{2}+4k +3)]$

$= \frac{k+1}{2} [k^{3} + 5k^{2}+9k +6]$

$= \frac{k+1}{2} (k+2)(k^{2}+3k+3) = \mathrm{RHS}$

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

(ii)
$\mathrm{LHS}= f(r) - f(r-1)$

$= 2r^{3}+3r^{2}+r+24 - 2(r-1)^{3}-3(r-1)^{2}-(r-1)-24$

$= 2r^{3}+3r^{2}+r-2r^{3}+6r^{2}-6r+2-3r^{2}+6r-3-r+1$

$= 6r^{2}$

$\therefore, a=6$

$\sum_{r=1}^{n} r^{2}$

$= \frac{1}{6} \sum_{r=1}^{n} 6r^{2}$

$= \frac{1}{6} \sum_{r=1}^{n} f(r) - f(r-1)$

$= \frac{1}{6} [ ~f(1) - f(0)$
$~~~~~~~~~+f(2) - f(1)$
$~~~~~~~~~+f(3) - f(2)$

$~~~~~~~~~+f(n-2) - f(n-3)$
$~~~~~~~~~+f(n-1) - f(n-2)$
$~~~~~~~~~+f(n) - f(n-1)$

$= \frac{1}{6} [f(n)-f(0)]$

$= \frac{1}{6} [2n^{3}+3n^{2}+n+24-24]$

$= \frac{1}{6} [2n^{3}+3n^{2}+n]$

$= \frac{n}{6} [2n^{2}+3n+1]$

$= \frac{n}{6} (2n+1)(n+1)$

(iii)
$\sum_{r=1}^{n} f(r)$

$= \sum_{r=1}^{n} 2r^{3}+3r^{2}+r+24$

$= \sum_{r=1}^{n} r(2r^{2}+1) + 3 \sum_{r=1}^{n} r^{2} + \sum_{r=1}^{n} 24$

$= \frac{n}{2}(n+1)(n^{2}+n+1) + \frac{n}{2}(2n+1)(n+1) +24n$

Students mostly did not struggle with this question. The instructions were clear. Some students found it difficult to relate (iii) to (i) and were unsure of how to resolve the summation

2013 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)
Since $|z|=r, ~arg(z)=\theta$

$|w|=|(1-i\sqrt{3} )(z)|=(2)(r)=2r$

$arg(w)=arg[(1-i\sqrt{3} )(z)]=arg(1-i\sqrt{3} )+arg(z)=-\frac{\pi}{3}+\theta$

(ii)

(iii)
$arg(\frac{z^{10}}{w^{2}}) = 10 arg(z) - 2 arg(w) = \pi$

$10 \theta - 2 (\theta - \frac{\pi}{3}) = \pi$

$\theta = \frac{\pi}{24}$

This is a question that tests student on their understanding of the standard manipulation of modulus and argument. It is not difficult.

2013 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)
Since we have a geometric progression, $p = 128(\frac{2}{3})^{n-1}$

$\mathrm{ln}p = \mathrm{ln}[128(\frac{2}{3})^{n-1}]$

$\mathrm{ln}p = \mathrm{ln}128 + (n-1)\mathrm{ln}\frac{2}{3}$

$\mathrm{ln}p = \mathrm{ln}(2^{7}) + (n-1)(\mathrm{ln}2 - \mathrm{ln}3)$

$\mathrm{ln}p = 7\mathrm{ln}2 + n\mathrm{ln}2 - n\mathrm{ln}3 - \mathrm{ln}2 + \mathrm{ln}3$

$\mathrm{ln}p = (6+n)\mathrm{ln}2 + (1- n)\mathrm{ln}3$

$\therefore, A=1, B=6, C=-1, D=1$

(ii)
$S_{\infty} = \frac{128}{1-2/3}=384$.
Thus, the length cannot be greater than 384cm.

(iii)
$S_{n}~>~380$

$\frac{128[1-(2/3)^{n}]}{1-2/3}~>~380$

$384[1-(2/3)^{n}]~>~380$

$(\frac{2}{3})^{n}~\textless~\frac{1}{96}$

$n \mathrm{ln}\frac{2}{3}~\textless~\mathrm{ln}\frac{1}{96}$

$n~>~\frac{\mathrm{ln}\frac{1}{96}}{\mathrm{ln}\frac{2}{3}}=11.26$

Therefore, 12 pieces must be cut off.

Students should not struggle to find out that this is a geometric progression, what follows is normal manipulation. Many mistakes were made in the last three lines of (iii). Some students were confused about how to handle the inequality after applying the natural logarithm.

2013 A-level H2 Mathematics (9740) Paper 1 Question 6 Suggested Solutions

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

(i)
Equation of plane OABC: r $= \lambda$a $+$ $\mu$b, for $\lambda, \mu \in \mathbb{R}$
Since c is on plane OABC, c $= \lambda$a $+$ $\mu$b for $\lambda, \mu$ are constants

(ii)
Using the ratio theorem formula, $\vec{ON}=\frac{4\mathbf{a}+3\mathbf{c}}{7}$

(iii)
Area ONC = Area OMC
$\frac{1}{2} |{\frac{4\mathbf{a}+3\mathbf{c}}{7} \times \mathbf{c}}| = \frac{1}{2}|{\frac{1}{2}\mathbf{b} \times \mathbf{c}}|$

$\frac{1}{7} |{4 \mathbf{a} \times \mathbf{c} + 3\mathbf{c} \times \mathbf{c}}| = \frac{1}{2}|{\mathbf{b} \times (\lambda \mathbf{a} + \mu \mathbf{b})}|$

$\frac{2}{7} |{4\mathbf{a} \times (\lambda \mathbf{a} + \mu \mathbf{b})}| = |{\mathbf{b} \times \lambda \mathbf{a} + \mathbf{b} \times \mu \mathbf{b}}|$

$\frac{2}{7} |{4\mathbf{a} \times \lambda \mathbf{a} + 4\mathbf{a} \times \mu \mathbf{b}}| = \frac{1}{2}|{\mathbf{b} \times \lambda \mathbf{a}}|$

$\frac{8\mu}{7} |{\mathbf{a} \times \mathbf{b}}| = \lambda|{\mathbf{a} \times \mathbf{b}}|$

$\frac{8\mu}{7} = \lambda$

For (i), some students can rely on the law of parallelogram to show too. For (iii), students must know all their vector product manipulations well to do this.

2013 A-level H2 Mathematics (9740) Paper 1 Question 5 Suggested Solutions

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

(i)

(ii)
Since $x=asin\theta$, we have $dx=acos\theta ~d\theta$
When $x= \frac{\sqrt{3}a}{2}, \theta = \frac{\pi}{3}$
When $x= \frac{a}{2}, \theta = \frac{\pi}{6}$

$\int_{\pi / 6}^{\pi / 3} \sqrt{1-sin^{2}\theta}~(acos\theta) ~d\theta$

$= a \int_{\pi / 6}^{\pi / 3} cos^{2} \theta ~d \theta$

$= a \int_{\pi / 6}^{\pi / 3} \frac{1+cos2\theta}{2} ~d \theta$

$= \frac{a}{2}[\theta + \frac{sin 2 \theta}{2}]\biggl|_{\pi / 6}^{\pi / 3}$

$= \frac{a \pi}{12}$

The function is a periodic curve and some students do have problem reading such functions. They should be alert of the domain that they are required to draw too. Some students were unsure how the hint that $f(x+3a)=f(x)$ should help. The parametric integration should be quite simple to handle as students just need to be cautious and introduce the double angle formula from MF15 effectively.

2013 A-level H2 Mathematics (9740) Paper 1 Question 4 Suggested Solutions

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

(i)
$(1+2i)^{3}$

$=(1+2i)^{2}(1+2i)$

$=[1+4i+(2i)^{2}](1+2i)$

$=(1+4i-4)(1+2i)$

$=(4i-3)(1+2i)$

$=4i-3+4i(2i)-3(2i)$

$=4i-3-8-6i$

$=-11-2i$

(ii)
Since $w=1+2i$ is a root,
$a(1+2i)^{3}+5(1+2i)^{2}+17(1+2i)+b=0$

$a(-11-2i)+5(-3+4i)+17(1+2i)+b=0$

$-11a-2ai-15+20i+17+34i+b=0$

$-11a+2+b-2ai+54i=0$

Comparing the real and imaginary coefficients,

$-2a+54 = 0 ~\Rightarrow ~ a=27$,

$-11(27)+2+b= 0 ~\Rightarrow ~ b=295$,

$\therefore a=27, b=295$

(iii)
Since all coefficient of the polynomial is real, by conjugate root theorem, if $1+2i$ is root, then $1-2i$ is also a root.
$(z-1-2i)(z-1+2i)=z^{2}-2z+5$

$27z^{3}+5z^{2}+17z+295 \equiv (z^{2}-2z+5)(cz+d)$,

by comparing coefficients, $c=27, d=59$

$\mathrm{Roots}=1+2i, 1-2i, -\frac{59}{27}$

Since they want to see all workings, let us not skip any steps here and do dilligently. Do check your answers with the GC though. It is good if students emphasise that the coefficients of polynomial is real here, to indicate your understanding. It is shocking that some students confused a root with a factor though.

2013 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)

(ii)
Using the GC, we find the intersection point to be $(1,2)$.

$\therefore, x \textless \frac{1}{2} ~\mathrm{or}~ x>2$.

A very straight forward questions. Students just need to be alert in drawing the graph, labelling features of the graph as required.

2013 A-level H2 Mathematics (9740) Paper 1 Question 2 Suggested Solutions

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

$y= \frac{x^{2}+x+1}{x-1}$
$y(x-1) = x^{2}+x+1$
$0 = x^{2}+x-xy+y+1$
For $x\in \mathbb{R}$, y exists only when $b^{2}-4ac \ge 0$

$(1-y)^{2}-4(1)(1+y) \ge 0$

$1-2y+y^{2}-4-4y \ge 0$

$y^{2}-6y-3 \ge 0$

$y \le 3-2\sqrt{3}~\mathrm{or}~y \ge 3+2\sqrt{3}$

When I first saw this, I was really surprised as O’levels had something really similar. The above method is the most direct, yes, 5 marks for that. Students can also go and find the turning point (and justify they are maximum or minimum) to decide the range.

2013 A-level H2 Mathematics (9740) Paper 1 Question 1 Suggested Solutions

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

(i)
$x-2z=4$
$2x-2y+z=6$
$5x-4y+3z=-9$
Using GC, $x=-\frac{38}{3}, y=-\frac{119}{6}, z=-\frac{25}{3}$
$\therefore, (-\frac{38}{3}, -\frac{119}{6}, -\frac{25}{3})$

(ii)
$x-2z=4$
$2x-2y+z=6$
$5x-4y=-9$
Using GC, there is no solutions found. Thus, planes p, q, r have no common intersection at all. They form a triangular prism since none of the planes are parallel to each other.