Solutions to Set B

Hopefully, you guys have started on the Set B. You will find the following solutions useful. Click on the question. Please do attempt them during this December Holidays. 🙂

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$x = \text{tan} \theta$

$\frac{dx}{d\theta} = \text{sec}^2 \theta$

$\int_{ \frac{1}{\sqrt{3}}}^1 \frac{x^2}{ (1 + x^2)^2} ~ dx$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{ \text{tan}^2 \theta}{ (1 + \text{tan}^2 \theta) ^2} (\text{sec}^2 \theta) ~d \theta$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \text{sin}^2 \theta ~d \theta$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1 - \text{cos} 2 \theta}{2} ~ d \theta$

$\bigg[ \frac{\theta}{2} - \frac{\text{sin}2 \theta}{4} \bigg]_{\frac{\pi}{6}}^{\frac{\pi}{4}}$

$= \frac{\pi}{24} - \frac{1}{4} + \frac{\sqrt{3}}{8}$

$\text{tan} (2 \theta + \frac{\pi}{4})$

$= \frac{ \text{tan}2 \theta + \text{tan} \frac{\pi}{4}}{ 1 - \text{tan}2\theta \text{tan} \frac{\pi}{4}}$

$= \frac{ \text{tan}2 \theta + 1}{1 - \text{tan} 2 \theta}$

$\approx \frac{2 \theta + 1}{1 - 2 \theta}$

$= (2 \theta + 1)(1 - 2 \theta) ^{-1}$

$= (2 \theta + 1)(1 + 2 \theta + \frac{(-1)(-2)}{2!}(-2 \theta)^2 + ... )$

$= (1 + 2 \theta)(1 + 2 \theta + 4 \theta ^2 + ...)$

$= 1 + 2 \theta + 4 \theta ^2 + 2 \theta + 4 \theta ^2 + ...$

$\approx 1 + 4 \theta + 8 \theta ^2$

(i)
By Pythagoras’ Theorem, Length of side of isosceles $= \sqrt{ \frac{(2x)^2}{2} } = \sqrt{2} x$

Area $= 3 \frac{1}{2} (\sqrt{2}x)(\sqrt{2}x) + 6x \times y$

$120 = 3x^2 + 6xy$

$\Rightarrow y = \frac{120-3x^2}{6x}$

$P = 6x + 6(\sqrt{2}x) + 2 y$

$P = 6x + 6\sqrt{2}x + 2(\frac{120-3x^2}{6x})$

$P = 5x + 6\sqrt{2}x + \frac{40}{x}$

(ii)
$\frac{dP}{dx} = 5 + 6 \sqrt{2} - \frac{40}{x^2}$

Let $\frac{dP}{dx} = 0$

$x^2 = \frac{40}{5 + 6\sqrt{2}}$

$x = 1.722265053$

$\frac{d^2P}{dx^2} = \frac{80}{x^3} = \frac{80}{{1.722265053}^3} \textgreater 0$

Thus $P$ is minimum when $x = 1.722265053$

$\Rightarrow P = 46.45045 \approx 46.5 \text{m}^2$ .

(i)
$\frac{AP}{PB} = \frac{2}{3}$

$3 \vec{AP} = 2 \vec{PB}$

$3 ( \vec{OP} - \vec{OA}) = 2 ( \vec{OB} - \vec{OP} )$

$\vec{OP} = \frac{2 \vec{OB} + 3 \vec{OA} }{5}$

$\vec{OP} = \frac{1}{5} ( \begin{pmatrix}{8}\\{8}\\{20}\end{pmatrix} + \begin{pmatrix}{-3}\\{-18}\\{-15}\end{pmatrix})$

$\vec{OP} = \begin{pmatrix}{1}\\{-2}\\{1}\end{pmatrix}$

(ii)
$\vec{AB} = \begin{pmatrix}{5}\\{10}\\{15}\end{pmatrix}$

$\vec{AB} \cdot \vec{OP}$

$= \begin{pmatrix}{5}\\{10}\\{15}\end{pmatrix} \cdot \begin{pmatrix}{1}\\{-2}\\{1}\end{pmatrix}$

$= 5 - 20 + 15$

$= 0 \Rightarrow AB \perp OP$

(iii)
$|a \cdot c|$

$= \frac{\big| \begin{pmatrix}{-1}\\{-6}\\{-5}\end{pmatrix} \cdot \begin{pmatrix}{1}\\{-2}\\{1}\end{pmatrix} \big|}{\big| \begin{pmatrix}{1}\\{-2}\\{1}\end{pmatrix} \big|}$

$= \frac{| -1 + 12 - 5|}{\sqrt{6}}$

$= \sqrt{6}$

$|a \cdot c|$ is the length of projection of $\vec{OA}$ onto $\vec{OC}$ .

$\big| \vec{OP} \big| = \sqrt{6}$

Area

$= \frac{1}{2} \big| \begin{pmatrix}{-1}\\{-6}\\{-5}\end{pmatrix} \times \begin{pmatrix}{4}\\{4}\\{10}\end{pmatrix} \big|$

$= \frac{1}{2} \big| \begin{pmatrix}{-40}\\{-10}\\{20}\end{pmatrix} \big|$

$= 5 \sqrt{21}$

(i)
Since $R_g = ( - \infty, 2) \subset ( - \infty, 3 ] = D_f$ , fg exists.

$fg(x) = (2-e^x)^2 - 2(2-e^x) - 24$

$fg(x) = e^{2x} - 2e^x - 24, x \in \mathbb{R}$

(ii)

Since the line $y = -24$ cuts $f(x)$ at more than one point, $f(x)$ is not a one-one function, thus the inverse does not exist.

(iii)

$\Rightarrow a = 1$

Let $f(x) = y$

$y = (x-1)^2 - 25$

$x = 1 \pm \sqrt{25 + y}$

Since $x \le 1, ~ x = 1 - \sqrt{25 + y}$

$f^{-1}(x) = 1 - \sqrt{25 + x}$

$D_{f^{-1}} = R_f = [-25, \infty)$

(a)
(i)
$15 + (50-1)(0.4) = 34.6 \text{km}$

(ii)
Let $n$ be number of sessions he needs to run
$\frac{n}{2} [2(15) + (n-1)(0.4)] = 560$

Using GC, $n = 29$

(b)
(i)
$30(0.95)^{20-1} = 11.32060 \approx 11.3$ m

(ii)
$S_{\infty} = \frac{30}{1-0.95} = 600$

Thus, he can be cycle more than 600km in total.

(i)

Area

$= \int_4^5 y ~dx$

$= \int_{0.5}^1 (4t- \frac{1}{t}) (4 - \frac{1}{t^2}) ~dt$

$= \int_{0.5}^1 (16t - \frac{8}{t} + \frac{1}{t^3}) ~dt$

$= \big[ 8t^2 - 8 \text{ln}t - \frac{1}{2t^2} \big]_{0.5}^1$

$= (8 - 0 - 0.5) - (2 + 8 \text{ln}2 - 2)$

$= 7.5 - 8 \text{ln}2$

(ii)
$x = 4t + \frac{1}{t} \Rightarrow x^2 = (4t + \frac{1}{t})^2$

$y = 4t - \frac{1}{t} \Rightarrow y^2 = (4t - \frac{1}{t})^2$

$\therefore x^2 - y^2 = 16$

Volume

$= \frac{1}{3} \pi (3)^2 \times 5 - \pi \int_4^5 x^2 - 16 ~dx$

$= 15 \pi - \pi \big[ \frac{x^3}{3} - 16x \big]_4^5$

$= \frac{32 \pi}{3}$

(i)
$y = \frac{2x+9}{x+2} = 2 + \frac{5}{x+2}$

$\Rightarrow A = 2, B = 5$

First, scale the curve 5 units parallel to the y = axis.
Then, translate the curve 2 units in the positive y – direction.

(ii)

(iii)
Using GC,

$( 2.84, 3.03)$ and $-2.95, -3.26)$

(a)
$\frac{d^2 y}{dx^2} = 9x (16 - 9x^2 )^{-1.5}$

$\frac{dy}{dx} = - \frac{1}{2} \int (-18x) (16 - 9x^2)^{-1.5} ~ dx$

$\frac{dy}{dx} = - \frac{1}{2} \big[ \frac{(16 - 9x^2)^{-0.5}}{-0.5} \big] + C$

$y = \int \frac{1}{\sqrt{16 - 9x^2}} + C ~ dx$

$y = \frac{1}{3} \text{sin}^{-1} \big[ \frac{3x}{4} \big] + Cx + D$

(b)
(i)
$\frac{dm}{dt} = \frac{1}{10} \sqrt{36 - 0.5m}$

$\int \frac{1}{\sqrt{36 - 0.5m}} ~dm = \int 0.1 ~dt$

$\big[ \frac{1}{-0.5} \big] \big[\frac{\sqrt{36 - 0.5m}}{0.5} \big] + C= 0.1 t$

$-40 \sqrt{36 - 0.5m} + C= t$

When $t = 0, m = 0, C=240$

$t = - 40 \sqrt{36 - 0.5m} + 240$

(ii)
$\frac{11}{10} \times 20 = 22$

When $m = 22, t = -40 \sqrt{36 - 11} + 240 \Rightarrow t = 40 \text{hours}$

Q10

(i)
$\text{Required~angle~} = \text{cos}^{-1} \bigg| \frac{\begin{pmatrix}{-1}\\{2}\\{1}\end{pmatrix} \cdot \begin{pmatrix}{1}\\{0}\\{0}\end{pmatrix} }{\sqrt{6} \sqrt{1} } \bigg| = 65.9 ^{\circ}$

(ii)
Normal to $P_2 = \begin{pmatrix}{-1}\\{2}\\{1}\end{pmatrix} \times \begin{pmatrix}{1}\\{1}\\{-1}\end{pmatrix} = \begin{pmatrix}{-3}\\{0}\\{-3}\end{pmatrix}$

$P_2 : r \cdot \begin{pmatrix}{-3}\\{0}\\{-3}\end{pmatrix} = \begin{pmatrix}{3}\\{5}\\{-2}\end{pmatrix} \cdot \begin{pmatrix}{-3}\\{0}\\{-3}\end{pmatrix}$

$\Rightarrow P_2: x + z = 1$

(iii)
Using GC: $m: r = \begin{pmatrix}{1}\\{3}\\{0}\end{pmatrix} + \lambda \begin{pmatrix}{-1}\\{2}\\{1}\end{pmatrix}, \lambda \in \mathbb{R}$

(iv)
Let F be the foot of perpendicular from A to $P_1$ .

$\vec{OF} = \begin{pmatrix}{3}\\{5}\\{-2}\end{pmatrix} + \lambda \begin{pmatrix}{1}\\{1}\\{-1}\end{pmatrix}$ for some $\lambda$

$\bigg[ \begin{pmatrix}{3}\\{5}\\{-2}\end{pmatrix} + \lambda \begin{pmatrix}{1}\\{1}\\{-1}\end{pmatrix} \bigg] \cdot \begin{pmatrix}{1}\\{1}\\{-1}\end{pmatrix} = 4$

$10 + 3 \lambda = 4$

$\Rightarrow \lambda = - 2$

$\Rightarrow \vec{OF} = \begin{pmatrix}{1}\\{3}\\{0}\end{pmatrix}$

$\vec{OF} = \frac{\vec{OA} + \vec{OB} }{2}$

$\vec{OB} = 2 \times \begin{pmatrix}{1}\\{3}\\{0}\end{pmatrix} - \begin{pmatrix}{3}\\{5}\\{-2}\end{pmatrix}$

$\vec{OB} = \begin{pmatrix}{-1}\\{1}\\{2}\end{pmatrix} = (-1, 1, 2)$

Q11

(i)
$x^2 - 3xy + 9y^2 = 3$

Differentiating wrt to $x$ ,

$2x - 3 (y + x \frac{dy}{dx}) + 18y \frac{dy}{dx} = 0$

$\frac{dy}{dx} = \frac{3y-2x}{18y-3x}$

(ii)
Let $\frac{dy}{dx} = 0$

$\Rightarrow 3y = 2x$

$x^2 - 3xy + 9y^2 = 3$

$x^2 - x(2x) + (2x)^2 = 3$

$3x^2 = 3$

$\Rightarrow x = \pm 1$

$(1, \frac{2}{3})$ and $(-1, - \frac{2}{3})$

(iii)
At P, $\frac{dy}{dx} = \frac{\sqrt{3} - 2\sqrt{3}}{6\sqrt{3} - 3 \sqrt{3}} = - \frac{1}{3}$

Tangent: $y - \frac{1}{\sqrt{3}} = - \frac{1}{3} (x - \sqrt{3})$

$\Rightarrow y = - \frac{1}{3}x + \frac{2}{\sqrt{3}}$

Normal: $y - \frac{1}{\sqrt{3}} = 3(x - \sqrt{3})$

$\Rightarrow y = 3x - \frac{8}{3} \sqrt{3}$

(iv)
At Q, $y = 0, x = 2 \sqrt{3}$

At R, $y = 0, x = \frac{8}{9} \sqrt{3}$

Area $= \frac{1}{2} \times \frac{1}{\sqrt{3}} \times \big( 2 \sqrt{3} - \frac{8}{9} \sqrt{3} \big)$

Area $= \frac{5}{9}$

Relevant Materials: MF26

Solutions to Set B2017-11-302017-12-15http://theculture.sg/wp-content/uploads/2018/03/the-culture-logo.pngThe Culture SGhttp://theculture.sg/wp-content/uploads/2017/11/charles-deluvio-456506.jpg200px200px

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.

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