### 2010 A-level H2 Mathematics (9740) Paper 1 Question 11 Suggested Solutions

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

### 2010 A-level H2 Mathematics (9740) Paper 1 Question 10 Suggested Solutions

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

(i)

Direction vector of $l \text{~is~} \begin{pmatrix}{-3}\\6\\9\end{pmatrix} = -3 \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}$

Normal vector of $p \text{~is~} \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}$

Since $l$ is parallel to the normal vector of $p, l$ is perpendicular to $p$.

(ii)
$l: r = \begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}$

$[\begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}] \bullet \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix} = 0$

$10 + \lambda + 2 4 \lambda + 9 + 9 \lambda = 0$

$\lambda = - \frac{3}{2}$

Point of intersection $= (\frac{17}{2}, 2, \frac{3}{2})$.

(iii)
$\begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix} = \begin{pmatrix}{-2}\\23\\33\end{pmatrix}$

$10 + \lambda = - 2$ — (1)

$-1 - 2 \lambda = 23$ — (2)

$-3 - 3 \lambda = 33$ — (3)

Since $\lambda = -12$ satisfies all 3 equations, A lies on l.

Since $= (\frac{17}{2}, 2, \frac{3}{2})$ is the midpoint of A & B, by ratio theorem,

$\begin{pmatrix}{\frac{17}{2}}\\2\\{\frac{3}{2}}\end{pmatrix} = \frac{\vec{OA} + \vec{OB}}{2}$

$\vec{OB} = 2 \begin{pmatrix}{\frac{17}{2}}\\2\\{\frac{3}{2}}\end{pmatrix} - \begin{pmatrix}{-2}\\23\\33\end{pmatrix} = \begin{pmatrix}19\\{-19}\\{-30}\end{pmatrix}$

$B(19, -19, -30)$

(iv)
Area

$= \frac{1}{2} |\vec{OA} \times \vec{OB}|$

$= \frac{1}{2} |\begin{pmatrix}{-2}\\23\\33\end{pmatrix} \times \begin{pmatrix}19\\{-19}\\{-30}\end{pmatrix}|$

$= \frac{1}{2} |\begin{pmatrix}{-63}\\{567}\\{-399}\end{pmatrix}|$

$= \frac{1}{2} \sqrt{63^2 +567^2 + 399^2}$

$= 348$ to the nearest whole number.

Students must give answers in coordinates, rather than as a position vector.

### 2010 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.

Total Volume $= 3x(x)y = 300$

$x^2y = 100$

$y = \frac{100}{x^2}$

Total Surface Area, $A = 3x(x) + 8x(y) + 3x(x) + 8x(ky)$

$= 6x^2 + (k+1) \frac{800}{x}$

$\frac{dA}{dx} = 12x - (k+1)\frac{800}{x^2} = 0$

$x^3 = \frac{200}{3} (k+1)$

$x = ^3 \sqrt{\frac{200}{3}(k+1)}$

$\frac{d^2A}{dx^2} = 12 + (k+1) \frac{1600}{x^3} = 12 + (k+1)\frac{1600}{\frac{200}{3}(k+1)} = 36 > 0$

Thus, $x=^3 \sqrt{\frac{200}{3}(k+1)}$ gives a minimum total surface area.

(ii)
$\frac{y}{x} = \frac{100}{x^3} = \frac{3}{2(k+1)}$

(iii)
$0 \textless k \le 1$

$1 \textless k+1 \le 2$

$\frac{1}{2} \le \frac{1}{k+1} \textless 1$

$\frac{3}{4} \le \frac{3}{2(k+1)} \textless \frac{3}{2}$

$\frac{3}{4} \le \frac{y}{x} \textless \frac{3}{2}$

(iv)
If the box has square ends, then $y=x$

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

$k = \frac{1}{2}$

### 2010 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)
$|z_1| = \sqrt{1+3} = 2$

$\mathrm{arg}z_1 = \mathrm{tan}^{-1}(\sqrt{3}) = \frac{\pi}{3}$

$\Rightarrow z_1 = 2(\mathrm{cos}\frac{\pi}{3} + i \mathrm{sin} \frac{\pi}{3})$

$|z_2| = \sqrt{1+1} = \sqrt{2}$

$\mathrm{arg}z_2 = - \pi + \mathrm{tan}^{-1}(1) = - \frac{3\pi}{4}$

$\Rightarrow z_2 = \sqrt{2}(\mathrm{cos}-\frac{3\pi}{4} + i \mathrm{sin} -\frac{3\pi}{4})$

(ii)
$\frac{z_1}{z_2} = \frac{2(\mathrm{cos}\frac{\pi}{3} + i \mathrm{sin} \frac{\pi}{3})}{\sqrt{2}(\mathrm{cos}-\frac{3\pi}{4} + i \mathrm{sin} -\frac{3\pi}{4})}$

$= \sqrt{2}[\mathrm{cos}(-\frac{11\pi}{12}) + i \mathrm{sin} (-\frac{11\pi}{12})]$

$(\frac{z_1}{z_2})^* = \sqrt{2}[\mathrm{cos}(\frac{11\pi}{12}) + i \mathrm{sin} (\frac{11\pi}{12})]$

(iii)

(iv)
From (iii)
$(x-1)^2 + \sqrt{3}^2 = 2^2$

$x = \pm 2$

Since $x > 0, x = 2$

Thus, locus meets the positive real axis when x=2.

The first two parts requires both students to be able to apply your knowledge of complex numbers formulas, expanding and simplifying them effectively. For (iii), the circle must cut through the origin, and student need to draw it clearly.

### 2010 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.

$\frac{d \theta}{dy} = k(20-\theta)$ where k is a positive constant.

When $\theta = 10, \frac{d\theta}{dt}=1$

$\Rightarrow k = 0.1$

$\int \frac{1}{20-\theta} d\theta = \int 0.1 dt$

$-\mathrm{ln}|20-\theta| = 0.1t + C$

Since $\theta \le 20, -\mathrm{ln}(20-\theta) = 0.1t + C$

When $t = 0, \theta = 10$

$\Rightarrow C = -\mathrm{ln}10$

$\mathrm{ln}(20-\theta) = -0.1t + ln10$

$\theta = 20 - 10 e^{-0.1t}$

When $\theta = 15, t = 6.93\mathrm{~min}$

Not a tough question, just be care when you draw the graph, consider only the positive t values. Students can omit the module too since $\theta \le 20$ always.

### 2010 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)
Using GC, $\beta = 0.347, \gamma = 1.532$

(ii)
Using GC, Area $= |\int_0.347^1.532 x^3 - 3x + 1 dx| = 0.781$

(iii)
Area $= \int_{-\sqrt{3}}^0 x^3 -3x + 1 dx - \sqrt{3} \times 1$

$= \frac{9}{4}$

(iv)
$\frac{dy}{dx}=2x^2 - 3 = 0$

$x = \pm 1$

$y = 3 \mathrm{~or~}-1$

Thus, $-1 \textless k \textless 3$

Take note to give (i) to 3 decimal places! For (iii), students can use GC to check their answers too! (iv), Students might want to consider using the GC to graph it out to help yourself find an efficient method

### 2010 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)

$y = \frac{(x-2)^3}{2} - 6$

When $x = 0, y = -10$

When $y = 0, \frac{(x-2)^3}{2} = 6$

$\Rightarrow x-2 = \sqrt[3]{12}$

$\Rightarrow x = 2 + \sqrt[3]{12}$

Coordinates are $(0,-10) \text{~and~} (2 + \sqrt[3]{12}, 0)$

(ii)

Some students wasted time to find the expression $f^{-1}$, which shows they have poor knowledge. Students should label the axial intercepts coordinates, $(0,-10) \text{~and~} (2 + \sqrt[3]{12}, 0)$.

### 2010 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)
Differentiating the given equation by $x$,

$\Rightarrow 2x - 2y \frac{dy}{dx} + 2x \frac{dy}{dx} + 2y = 0$

$(x-y)\frac{dy}{dx} = -y-x$

$\frac{dy}{dx} = \frac{x+y}{y-x}$

(ii)
Tangent parallel to x-axis $\Rightarrow \frac{dy}{dx} = 0$

Then, $y= -x$

$\Rightarrow x^2 - (-x)^2 + 2x (-x)+4=0$

$x^2 = 2$

$x = \pm \sqrt{2}$

$y = -\sqrt{2}, ~\sqrt{2}$

Therefore, coordinates are $(\sqrt{2}, -\sqrt{2})$ and $(-\sqrt{2}, \sqrt{2})$.

Quite straight forward with the implicit differentiations at (i). Students should understand what it means when tangents are parallel to x-axis. Lastly, since the question wants coordinates, students are expected to give in coordinates instead of merely the x and y values!

### 2010 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)
$U_n = S_n - S_{n-1}$

$= n(2n+c) - (n-1)[2(n-1)+c]$

$= 4n-2+c$

(ii)
$U_{n-1} = 4(n+1)-2+c$

$U_{n+1} = U_n + 4$

Formulas!!! Every simple questions involving formulas!

### 2010 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.

(i)
$e^x (1+ \mathrm{sin}2x)$

$= (1 + x + \frac{x^2}{2!} + \ldots)(1+2x-\ldots)$

$= 1 + 3x + \frac{5}{2}x^2 + \ldots$

(ii)
$(1 + \frac{4}{3}x)^n$

$= 1 + n\frac{4}{3}x + \frac{n(n-1)}{2!}(\frac{4}{3}x)^2 + \ldots$

Since the first two terms are equal,

$\frac{4}{3}n = 3 \Rightarrow n = \frac{9}{4}$

Third term $= \frac{n(n-1)}{2!}(\frac{4}{3}x)^2 = \frac{5}{2}x^2$