List of Great and Helpful Math Articles

List of Great and Helpful Math Articles

JC Mathematics, Studying Tips

Here is a compilation of all the Math articles/ opinions KS has done. Students should read them when they are free to improve their mathematics skills. They will come in handy! 🙂 For ease of navigation, some bigger topics sorted according to topics.

  1. Trigonometry
  2. Integrations
  3. Permutations & Combinations (Combinatorics)
  4. Vectors
  5. Complex Numbers
  6. APGP, Sequences & Series
  7. Statistics
  8. Importance of Prelims
  9. A-levels vs IB Mathematics
  10. How I encourage Students to study Mathematics in JC.
  11. Helping Students with Math Course
  12. Common Pitfalls in A’levels Math #1
  13. Common Pitfalls in A’levels Math #2
  14. Common Pitfalls in A’levels Math #3
  15. Why is anything to the power of 0 always 1?
  16. Classical Mathematical Fallacies #1
  17. Classical Mathematical Fallacies #2
  18. Classical Mathematical Fallacies #3
  19. Confusion on when to put ± sign
  20. Difference between H1 and H2 Math
  21. Partial Fractions made easy
  22. The Modulus Sign #1
  23. The Modulus Sign #2
  24. The Modulus Sign #3
  25. The Modulus Sign #4
  26. An easier approach to remembering discriminant
  27. Finding the Coefficient of Terms
  28. Help to start Maclaurin’s Questions
  29. Understanding A-level differentiation questions
  30. A different perspective to transformation
  31. Solving roots of higher order
  32. Proving a function is symmetrical about y-axis
  33. All tangents are straight lines
  34. Prime Numbers and their uses.
  35. My Favourite Pure Mathematics Topic
  36. Python
  37. Tricks to squaring numbers
  38. Why we need to be close to zero for an approximation to be good?
  39. Learning Math or Learning to do Math
  40. H3 Mathematics
  41. Why study Mathematics?
  42. Importance of Mathematics in Finance
  43. Some things that JC students should know
  44. Review of Basic Probability (Undergraduate)

Brute forcing simultaneous equations involving Complex Numbers

JC Mathematics

So I have many students coming up to me and saying, “Mr Teng, How do you know which to substitute away or when to introduce z=x+yi when doing simultaneous equations for complex numbers?”
Here is a lesser method that will give you the answers. It is definitely a clearer method that involves less pitfalls. This method is self-explanatory so I’ll let the working do the talking. Say we want to solve the following

2w + z = 12i ---(1)

w^* + 2z = -6 - i ---(2)

Let w = a + bi (clearly, w^* = a-bi) , and z=c+di

2(a+bi) + (c+di) = 12i ---(1)

a-bi + 2(c+di) = -6 - i ---(2)

From (1), we have (2a+c) + (2b+d)i = 12i

\Rightarrow 2a+c = 0 --- (3) , and 2b+d = 12 ---(4)

From (2), we have (a+2c) + (2d-b)i = -6-i

\Rightarrow a+2c = -6 --- (5) , and 2d-b = -1 ---(6)

Solving (3), (4), (5), (6), we find

a = 2, b =5, c=-4, d=2

Thus, w = 2+5i and z = -4+2i

2010 A-level H2 Mathematics (9740) Paper 1 Question 8 Suggested Solutions

JC Mathematics

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.

KS Comments:

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.

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

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

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}