H2 Math Tue 7pm

JC Mathematics

This page contains all questions and answers asked by students from this class. The most recent questions will be at the top.

MF26

Question 10 & 11
Question 10 & 11

10.
(i)

11.
(i)
Since P is on l, \vec{OP} = \begin{bmatrix} 1\\ 2\\ 3\end{bmatrix} + t \begin{bmatrix} -1\\ 1\\ 1\end{bmatrix} for some t
\vec{OP} perpendicular to l \Rightarrow \vec{OP} \bullet \begin{bmatrix} -1\\ 1\\ 1\end{bmatrix}
-(1-t)+2+t+3+t=0
t=-\frac{4}{3}
\vec{OP} = \begin{bmatrix} 1\\ 2\\ 3\end{bmatrix} -\frac{4}{3} \begin{bmatrix} -1\\ 1\\ 1\end{bmatrix} = ...

H2 Math Tue 5pm

JC Mathematics

This page contains all questions and answers asked by students from this class. The most recent questions will be at the top.

MF26


Question 1b
Question 1b

Differentiate x=e^t with respect to t.
\frac{dx}{dt} = e^t
Substitute x = e^t into given differential equation,
\Rightarrow e^t \frac{dy}{dt} = \mathrm{cos}(\mathrm{ln}e^t)\mathrm{sin}(\mathrm{ln}(e^t)^3)
e^t \frac{dy}{dt} = \mathrm{cos}t \mathrm{sin} (3t)
Since \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}
\frac{dy}{dt} = \frac{1}{e^t} \mathrm{cos}t \mathrm{sin} (3t) \times e^t
\frac{dy}{dt} = \mathrm{cos}t \mathrm{sin} (3t)
\int \frac{dy}{dt} dt = \int \mathrm{cos}t \mathrm{sin} (3t) dy
Using the product to sum formula as shown here, we have
\int \frac{dy}{dt} dt = \frac{1}{2} \int \mathrm{sin} 4t + \mathrm{sin} 2t dt
y = \frac{1}{2}(-\frac{\mathrm{cos}4t}{4} - \frac{\mathrm{cos}2t}{2}) + C
y = \frac{1}{2}(-\frac{\mathrm{cos}(4 \mathrm{ln}x)}{4} - \frac{\mathrm{cos}(2\mathrm{ln}x)}{2}) + C


Question 2(iv)
Question 2(iv)

Note: i = \begin{bmatrix}1\\ 0\\ 0\end{bmatrix} and j = \begin{bmatrix}\\ 1\\ 0\end{bmatrix}
\pi: 2x + 3y = -6 — (1)
\pi \prime: x + y + z= 5 — (2)
\pi_1: x = 0 — (3)
Using GC, \vec{OA} = \begin{bmatrix}0\\ -2\\ 7\end{bmatrix}

\pi: 2x + 3y = -6 — (1)
\pi \prime: x + y + z= 5 — (2)
\pi_1: y = 0 — (3)
Using GC, \vec{OA} = \begin{bmatrix}-3\\ 0\\ 8\end{bmatrix}


Qn 11
Qn 11

Qn11
(i)
l: r = \begin{bmatrix}-1\\ -1\\ 0\end{bmatrix} + \alpha \begin{bmatrix}1\\ 1\\ 1\end{bmatrix}, \alpha \in \mathbb{R}
(ii) Since l is the common line of intersection on \pi_1 and \pi_2, we need l to be on \pi_3 too. For that to happen,
1. l must be parallel to \pi_3, that is, direction of l is perpendicular to normal to \pi_3
\Rightarrow \begin{bmatrix}1\\ 1\\ 1\end{bmatrix} \bullet \begin{bmatrix}5\\ {\lambda}\\ 17\end{bmatrix} = 0
\lambda = -22
2. Given that l is parallel to \pi_3 (since \lambda = -22), we need l to be on \pi_3, so we need \begin{bmatrix}-1\\ -1\\ 0\end{bmatrix} to be on \pi_3
\Rightarrow \begin{bmatrix}-1\\ -1\\ 0\end{bmatrix} \bullet \begin{bmatrix}5\\ -22\\ 17\end{bmatrix} = \mu
\mu = 17
(iii)
For 3 planes to have nothing in common, then l must be parallel to \pi_3 (Note: if l is not parallel, l will cut \pi_3 at a point, which means that 3 planes will cut at a point)
\Rightarrow \lambda = -22 from (ii)
But \mu \neq 17, \mu \in \mathbb{R}


Qn12
Qn12

12.
(i)
\pi_1 : r = \begin{bmatrix}4\\ 2\\ 12\end{bmatrix} + \lambda \begin{bmatrix}2\\ 2\\ 1\end{bmatrix} + \mu \begin{bmatrix}5\\ -1\\ -5\end{bmatrix}, \mu , \lambda \in \mathbb{R}
\vec{OP} = \begin{bmatrix}4\\ 2\\ 12\end{bmatrix} is on \pi_1 when \lambda = \mu = 0

H2 Math Sun 2pm

JC Mathematics

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MF26


Question 15 CJC/2013
Question 15 CJC/2013

LHS = \sum_{r=2}^n \frac{1}{(r+1)(r-1)}
= \sum_{r=2}^n \frac{1}{2} (\frac{1}{r-1} - \frac{1}{r+1})
= \frac{1}{2} \sum_{r=2}^n \frac{1}{r-1} - \frac{1}{r+1}
= \frac{1}{2} [\frac{1}{1} - \frac{1}{3}
+ \frac{1}{2} - \frac{1}{4}
+ \frac{1}{3} - \frac{1}{5}

+ \frac{1}{n-3} - \frac{1}{n-1}
+ \frac{1}{n-2} - \frac{1}{n}
+ \frac{1}{n-1} - \frac{1}{n+1}]
= \frac{1}{2} [\frac{3}{2} - \frac{1}{n} - \frac{1}{n+1}]
= \frac{3}{4} - \frac{1}{2n} - \frac{1}{2n+2}

Since the lim_{n \rightarrow \infty} \sum_{r=2}^n \frac{1}{(r+1)(r-1)}
= lim_{n \rightarrow \infty} \frac{3}{4} - \frac{1}{2n} - \frac{1}{2n+2}
= \frac{3}{4} - 0 - 0
- \frac{3}{4} is a constant, the series convergences.
The sum to infinity = \frac{3}{4}


Question 14a MJC/2013
Question 14a MJC/2013

Let T_n denote the n^{th} term of the AP.
T_n = a + (n-1)d
T_1 = a
T_3 = a +2d
T_6 = a +5d
Since they are consecutive terms of a GP,
\frac{T_3}{T_1} = \frac{T_6}{T_3} = r
\Rightarrow \frac{a+2d}{a} = \frac{a+5d}{a+2d}
(a+2d)^2 = a(a+5d)
a^2 + 4ad + 4d^2 = a^2 + 5ad
4d^2 - ad =0
d(4d - a) = 0
d = 0 (NA) \mathrm{~or~} 4d = a
\Rightarrow r = \frac{a+2d}{a}
r = \frac{6d+2d}{4d} = \frac{3}{2} > 1, thus its not convergent

S_{15} = \frac{n}{2}(2a + (n-1)d)
= \frac{15}{2}(2a + 14 (\frac{a}{4}))
= 41.25 a


Question 11 DHS/2013
Question 11 DHS/2013

Sum of first 3 terms = \frac{3}{2} (2a+(3-1)d)
6 = 3a+3d
a+d=2 —(1)
Sum of last 3 terms = \frac{3}{2}[2(a+(n-1)d) + (3-1)(-d)]; Here we consider an AP that has first term T_n = a + (n-1)d and common difference -d.
\Rightarrow 231 = \frac{3}{2}(2a + 2nd - 2d -2d)
231 = 3a + 3nd - 6d
77 = a + nd -2d —(2)
Sum of n terms = \frac{n}{2}[2a+ (n-1)d]
1106 = \frac{n}{2}(2a + nd - d) —(3)
Solve for n.


Question 12 SRJC/2012
Question 12 SRJC/2012

(i) Volume, V = \pi r^2 h
Volume of kth later, V_k = \pi [(20)(\frac{5}{6})^{k-1}]^2(22)(\frac{4}{5})^{k-1}
V_k = 8800 \pi [\frac{25}{36} \times \frac{4}{5}]^{k-1}
V_k = 8800 \pi (\frac{5}{9})^{k-1}
(ii)
Since $latex r = \frac{5}{9} <1, S_{\infty} $ exists. Theoretical Max Volume, $latex S_{\infty} = \frac{8800 \pi}{1 - \frac{5}{9}} = 19800 \pi$. Total Volume, $latex S_n = \frac{8800 \pi (1 - (\frac{5}{9})^n)}{1 - \frac{5}{9}}$ We want $latex S_n \le 0.95 S_{\infty}$

H2 Math Sun 1130am

JC Mathematics

This page contains all questions and answers asked by students from this class. The most recent questions will be at the top.

MF26


Vectors Q7 [Homework]
(i)
\vec{OL} = \begin{bmatrix}2\\ 7\\ -1\end{bmatrix}
\vec{OM} = \begin{bmatrix}9\\ 0\\ -8\end{bmatrix}
Using ratio theorem, \vec{OP} = \frac{2\vec{OM}+5\vec{OL}}{7} = \begin{bmatrix}4\\ 5\\ -3\end{bmatrix}
Since \vec{OP} is perpendicular to \begin{bmatrix}4\\ 1\\ q\end{bmatrix}
\Rightarrow \begin{bmatrix}4\\ 5\\ -3\end{bmatrix} \bullet \begin{bmatrix}4\\ 1\\ q\end{bmatrix} = 0
q = 7

(ii)
To be a parallelogram, \vec{OM} = \vec{LN} = \vec{ON} - \vec{OL}
\vec{ON} =\begin{bmatrix}11\\ 7\\ -9\end{bmatrix}
Area = |\vec{OM} \times \vec{OL}|
= |\begin{bmatrix}56\\ -7\\ 63\end{bmatrix}|
= \sqrt{7154} = 7 \sqrt{146} units^2

(iii)
Let \vec{OQ} = \begin{bmatrix}x\\ y\\ 0\end{bmatrix}
Since |\vec{OQ}| = |\vec{OP}|
\sqrt{x^2 + y^2} = \sqrt{50} — (1)
\begin{bmatrix}x\\ y\\ 0\end{bmatrix} \bullet \begin{bmatrix}1\\ 0\\ 0\end{bmatrix} = |\begin{bmatrix}x\\ y\\ 0\end{bmatrix} | |\begin{bmatrix}1\\ 0\\ 0\end{bmatrix} | \mathrm{cos} \theta — (2)
Solving, x = \sqrt{50} \mathrm{cos} \theta = 5 \sqrt{2} \mathrm{cos} \theta
y = \sqrt{50} \mathrm{sin} \theta = 5 \sqrt{2} \mathrm{sin} \theta
\Rightarrow \vec{OQ} = \begin{bmatrix}{5 \sqrt{2} \mathrm{cos} \theta}\\ {5 \sqrt{2} \mathrm{sin} \theta}\\ 0\end{bmatrix}


Vectors Q8 [Homework]
(i)
\vec{OA} = \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix}
\vec{OC} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix}
\vec{AC} = \vec{OC} - \vec{OA} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} - \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix} = \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}
l: r = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} + \lambda \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}, \lambda \in \mathbb{R}

(ii)
Let R be the top of the vertical pillar,
l_{QR}: r = \begin{bmatrix}15\\ 6\\ 0\end{bmatrix} + \mu \begin{bmatrix}0\\ 0\\ 1\end{bmatrix}, \mu \in \mathbb{R}
Since R is collinear with A and C, R is the intersection of line AC and QR.
\begin{bmatrix}{5 + 10 \mu}\\ {2 + 4 \mu}\\ {6 + 3 \mu}\end{bmatrix} = \begin{bmatrix}15\\ 6\\ {\mu}\end{bmatrix}
\Rightarrow \lambda = 1, \mu = 9
\vec{OR} = \begin{bmatrix}15\\ 6\\ 9\end{bmatrix} , and the height is 9m.

(iii)
\vec{OD} = \begin{bmatrix}-5\\ 2\\ 6\end{bmatrix}
\vec{AD} = \vec{OD} - \vec{OA} = \begin{bmatrix}0\\ 4\\ 3\end{bmatrix}
\vec{AX} = (\vec{AD} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}
= (\begin{bmatrix}0\\ 4\\ 3\end{bmatrix} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}
= \frac{25}{125} \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}
= \begin{bmatrix}2\\ 0.8\\ 0.6\end{bmatrix}
\vec{OX} = \vec{OA} + \vec{AX} = \begin{bmatrix}-3\\ 1.2\\ 3.6\end{bmatrix}


Vectors Q9 [Homework]
(i)
\vec{AB} = \begin{bmatrix}-4\\ 5\\ 3\end{bmatrix}
\vec{AC} = \begin{bmatrix}1\\ -3\\ 6\end{bmatrix}
Normal of \pi_1, ~n_1=\begin{bmatrix}-4\\ 5\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -3\\ 6\end{bmatrix} = \begin{bmatrix}-21\\ -21\\ -7\end{bmatrix} = -7 \begin{bmatrix}3\\ 3\\ 1\end{bmatrix}
\pi_1: r \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = \begin{bmatrix}5\\ -1\\ 0\end{bmatrix} \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = 12

(ii)
Let \theta be the acute angle
\theta - \mathrm{cos}^{-1} |\frac{\begin{bmatrix}3\\ 3\\ 1\end{bmatrix} \bullet \begin{bmatrix}1\\ -1\\ 1\end{bmatrix}}{\sqrt{19}} \sqrt{3}|
\theta = 82.4 ^{\circ}

(iii)
3x + 3 y + z = 12 — (1)
x - y + z = 1 — (2)

Using GC, l: r = \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} + \lambda \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix}, \lambda \in \mathbb{R}

(iv)
Let n_3 be the normal of \pi_3
Length of projection = |\vec{AB} \times n_3|
= \frac{1}{\sqrt{26}} |\begin{bmatrix}4\\ -5\\ 3\end{bmatrix} \times \begin{bmatrix}5\\ -1\\ 0\end{bmatrix}| = 15\sqrt{\frac{3}{26}}

(v)
Required distance = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \sqrt{3} units

(vi)
Let normal of \pi_4 = n_4 = \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -1\\ 1\end{bmatrix} = \begin{bmatrix}4\\ 5\\ 1\end{bmatrix}
\pi_4: r \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6
If \pi_1, \pi_2 \mathrm{~and~} \pi_4 intersect at l,n\begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} lies on pi_4
\Rightarrow \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6
k = \frac{23}{8}

H2 Math Sun 930am

JC Mathematics

This page contains all questions and answers asked by students from this class. The most recent questions will be at the top.

MF26


Vectors Q7 [Homework]
(i)
\vec{OL} = \begin{bmatrix}2\\ 7\\ -1\end{bmatrix}
\vec{OM} = \begin{bmatrix}9\\ 0\\ -8\end{bmatrix}
Using ratio theorem, \vec{OP} = \frac{2\vec{OM}+5\vec{OL}}{7} = \begin{bmatrix}4\\ 5\\ -3\end{bmatrix}
Since \vec{OP} is perpendicular to \begin{bmatrix}4\\ 1\\ q\end{bmatrix}
\Rightarrow \begin{bmatrix}4\\ 5\\ -3\end{bmatrix} \bullet \begin{bmatrix}4\\ 1\\ q\end{bmatrix} = 0
q = 7

(ii)
To be a parallelogram, \vec{OM} = \vec{LN} = \vec{ON} - \vec{OL}
\vec{ON} =\begin{bmatrix}11\\ 7\\ -9\end{bmatrix}
Area = |\vec{OM} \times \vec{OL}|
= |\begin{bmatrix}56\\ -7\\ 63\end{bmatrix}|
= \sqrt{7154} = 7 \sqrt{146} units^2

(iii)
Let \vec{OQ} = \begin{bmatrix}x\\ y\\ 0\end{bmatrix}
Since |\vec{OQ}| = |\vec{OP}|
\sqrt{x^2 + y^2} = \sqrt{50} — (1)
\begin{bmatrix}x\\ y\\ 0\end{bmatrix} \bullet \begin{bmatrix}1\\ 0\\ 0\end{bmatrix} = |\begin{bmatrix}x\\ y\\ 0\end{bmatrix} | |\begin{bmatrix}1\\ 0\\ 0\end{bmatrix} | \mathrm{cos} \theta — (2)
Solving, x = \sqrt{50} \mathrm{cos} \theta = 5 \sqrt{2} \mathrm{cos} \theta
y = \sqrt{50} \mathrm{sin} \theta = 5 \sqrt{2} \mathrm{sin} \theta
\Rightarrow \vec{OQ} = \begin{bmatrix}{5 \sqrt{2} \mathrm{cos} \theta}\\ {5 \sqrt{2} \mathrm{sin} \theta}\\ 0\end{bmatrix}


Vectors Q8 [Homework]
(i)
\vec{OA} = \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix}
\vec{OC} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix}
\vec{AC} = \vec{OC} - \vec{OA} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} - \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix} = \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}
l: r = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} + \lambda \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}, \lambda \in \mathbb{R}

(ii)
Let R be the top of the vertical pillar,
l_{QR}: r = \begin{bmatrix}15\\ 6\\ 0\end{bmatrix} + \mu \begin{bmatrix}0\\ 0\\ 1\end{bmatrix}, \mu \in \mathbb{R}
Since R is collinear with A and C, R is the intersection of line AC and QR.
\begin{bmatrix}{5 + 10 \mu}\\ {2 + 4 \mu}\\ {6 + 3 \mu}\end{bmatrix} = \begin{bmatrix}15\\ 6\\ {\mu}\end{bmatrix}
\Rightarrow \lambda = 1, \mu = 9
\vec{OR} = \begin{bmatrix}15\\ 6\\ 9\end{bmatrix} , and the height is 9m.

(iii)
\vec{OD} = \begin{bmatrix}-5\\ 2\\ 6\end{bmatrix}
\vec{AD} = \vec{OD} - \vec{OA} = \begin{bmatrix}0\\ 4\\ 3\end{bmatrix}
\vec{AX} = (\vec{AD} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}
= (\begin{bmatrix}0\\ 4\\ 3\end{bmatrix} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}
= \frac{25}{125} \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}
= \begin{bmatrix}2\\ 0.8\\ 0.6\end{bmatrix}
\vec{OX} = \vec{OA} + \vec{AX} = \begin{bmatrix}-3\\ 1.2\\ 3.6\end{bmatrix}


Vectors Q9 [Homework]
(i)
\vec{AB} = \begin{bmatrix}-4\\ 5\\ 3\end{bmatrix}
\vec{AC} = \begin{bmatrix}1\\ -3\\ 6\end{bmatrix}
Normal of \pi_1, ~n_1=\begin{bmatrix}-4\\ 5\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -3\\ 6\end{bmatrix} = \begin{bmatrix}-21\\ -21\\ -7\end{bmatrix} = -7 \begin{bmatrix}3\\ 3\\ 1\end{bmatrix}
\pi_1: r \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = \begin{bmatrix}5\\ -1\\ 0\end{bmatrix} \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = 12

(ii)
Let \theta be the acute angle
\theta - \mathrm{cos}^{-1} |\frac{\begin{bmatrix}3\\ 3\\ 1\end{bmatrix} \bullet \begin{bmatrix}1\\ -1\\ 1\end{bmatrix}}{\sqrt{19}} \sqrt{3}|
\theta = 82.4 ^{\circ}

(iii)
3x + 3 y + z = 12 — (1)
x - y + z = 1 — (2)

Using GC, l: r = \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} + \lambda \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix}, \lambda \in \mathbb{R}

(iv)
Let n_3 be the normal of \pi_3
Length of projection = |\vec{AB} \times n_3|
= \frac{1}{\sqrt{26}} |\begin{bmatrix}4\\ -5\\ 3\end{bmatrix} \times \begin{bmatrix}5\\ -1\\ 0\end{bmatrix}| = 15\sqrt{\frac{3}{26}}

(v)
Required distance = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \sqrt{3} units

(vi)
Let normal of \pi_4 = n_4 = \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -1\\ 1\end{bmatrix} = \begin{bmatrix}4\\ 5\\ 1\end{bmatrix}
\pi_4: r \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6
If \pi_1, \pi_2 \mathrm{~and~} \pi_4 intersect at l,n\begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} lies on pi_4
\Rightarrow \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6
k = \frac{23}{8}