How to improve our H2 Mathematics?

How to improve our H2 Mathematics?

JC Mathematics

Many students of mine have asked me this question numerous times. Some lamented that they work really hard, finishing off the revision package timely. However, they still cannot improve from a C/B to an A or some are still fishing for a pass.

I feel that many students have a misunderstanding about H2 Mathematics. The H2 Mathematics (9758) Syllabus is different for the previous (9740) by a lot, primarily because its assessment objective is different. On top of that, the scheme of exam has been changed to include application questions which will test students’ abilities to wrap their heads around a given situation and solve them. Basically, students have to do of the “core values” of Mathematics, that is, problem solving.

Understanding the syllabus is one thing, next is how we should be studying it. The syllabus content is a lot lesser, so its less wide but much much deeper. Students cannot finish all the drills and practices, like in O’levels. Today, they are expected to think, so how one learns each topic is very important. One thing I pick up from NIE Mathematics is definitely the idea of examples. Every little concepts and ideas we learn, can we think of an example to relate and reinforce our understanding with?

Next, is practices. When we practice, time ourselves to finish. Do not look at the answers as we are just marginalising ourselves the opportunity to learn. If we are stuck, look at one line of the solutions and see if it will help us. Personally, I never give students solutions and even in class, I do not use solutions and rather think together with the students. It is important to train up our thinking abilities when doing H2 Mathematics.

Now students, you have about two months left to A’levels, I highly advise you to really look at how you learn the topics. A’levels will be unpredictable and questions do not repeat. To me, the TYS is more like practice papers then drills cos they will not appear again in the actual exam.

2018 A-level H2 Mathematics (9758) Paper 1 Suggested Solutions

JC Mathematics

Post will be updated again on 9th November 2018.

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

Numerical Answers (click the questions for workings/explanation)

Question 1:
Question 2:
Question 3:
Question 4:
Question 5:
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Question 8:
Question 9:
Question 10:

Relevant materials

MF26

KS Comments

2018 A-level H2 Mathematics (9758) Paper 2 Suggested Solutions

JC Mathematics, Mathematics

Post will be updated again on 14th November 2018.

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

Numerical Answers (click the questions for workings/explanation)

Question 1:
Question 2:
Question 3:
Question 4:
Question 5:
Question 6:
Question 7:
Question 8:
Question 9:
Question 10:

Relevant materials

MF26

KS Comments

Modal value & Expected value

Modal value & Expected value

JC Mathematics

Let us look at the difference between modal value and expected value. We shall start by saying they are different, albeit close.

Modal value refers to the mode, that is, the value that has the highest probability (chance) of occurring.

Expected value refers to the value, we expect to have, on average.

Before we start, I’ll do a fast recap on Binomial Distribution, X \sim \text{B}(n, p) by flashing the formulae that we can find on MF26.

\text{P}(X = x) = ^n C_x (p)^x (1-p)^{n-x}

\mathbb{E}(X) = np

\text{Var}(X) = np(1-p)

The expected value is simply given by \mathbb{E}(X).

Now to find the modal value, we have to go through a slightly nasty and long working. You may click and find out.

We have that \frac{\text{P}(X = r + 1)}{\text{P}(X = r)} = \frac{(n-r)}{(r+1)}  \frac{p}{1-p}. This is what we call the recurrence formula. We consider this to give us the ratio between successive probabilities. And to illustrate how this works, nothing beats an example question.

Consider candies are packed in packets of 20. On average the proportion of candies that are blue-colored is p. It is know that the most common number of blue-colored candies in a packet is 6. Use this information to find exactly the range of values that p can take.

First, most common number is the same as saying the modal/ highest frequency.

This means that \text{P}(X=6) is the highest/ largest probability… Let us turn our attention to the recurrence formula now. If \text{P}(X=6) is the largest, then it means that \text{P}(X=6) \textgreater \text{P}(X=7) and also \text{P}(X=6) \textgreater \text{P}(X=5).

Lets start by looking at the first one… \text{P}(X=6) \textgreater \text{P}(X=7)

\text{P}(X=6) \textgreater \text{P}(X=7)

1 > \frac{\text{P}(X=7)}{\text{P}(X=6)}

\frac{\text{P}(X=7)}{\text{P}(X=6)} \textless 1

But hold on! This looks like the recurrence formula. (ok, in exams, its either you use the recurrence formula or derive on the spot. Both works!)

Now I’ll advice you try the second one (before clicking on answer) on your own, that is, \text{P}(X=6) > \text{P}(X=5).

Now, if the question simply says that the expected number of blue-colored candies in a packet of 20 is 6. Then

\mathbb{E}(X) = 6

(20)p = 6

p = \frac{3}{10}

We observe that this value actually falls in the range of p we found.

Differential Equations (Applications)

Differential Equations (Applications)

JC Mathematics, Mathematics

When Mr. Teng retired on 1 January 2018, he put a sum of $10,000 into a senior citizen fund that has a constant rate of return of 5% at the end of every month. Starting in February 2018, he withdraws $500 at the start of each month for groceries. Denote the amount of money that Mr. Teng has at the time n years by \$x.

(i) The differential equation relating x and n can be written in the form of \frac{dx}{dn}= kx+c. State the values of k and c.

(ii) Solve the differential equation and find the amount of money that Mr. Teng has after 15 months.

(iii) In which month will Mr. Teng no longer be able to withdraw the full $400?

DRV questions with a twist

DRV questions with a twist

JC Mathematics, Mathematics

I want to share a question that is really old (older than me, actually). It is from the June 1972 paper. As most students know, Maclaurin’s series was tested in last year A’levels Paper 1 as a sum to infinity. And this DRV did the same thing. Here is the question.

A bag contains 6 black balls and 2 white balls. Balls are drawn at random one at a time from the bag without replacement, and a white ball is drawn for the first time at R th draw.

(i) Tabulate the probability distribution for R.

(ii) Show that E( R ) = 3, and find Var( R ).

If each ball is replaced before another is drawn, show that in this case E( R ) = 4.

Solutions to Review 1

Solutions to Review 1

JC Mathematics, Mathematics

Question 1
(i)
y = f(x) = \frac{x^2 + 14x + 50}{3(x+7)}

3y(x+7) = x^2 + 14x + 50

x^2 + (14-3y)x + 50 - 21 y = 0

\text{discriminant} \ge 0

(14-3y)^2 - 4(1)(50-21y) \ge 0

196 - 84y + 9y^2 - 200 + 84y \ge 0

9y^2 - 4 \ge 0

(3y - 2)(3y + 2) \ge 0

y \le - \frac{2}{3} \text{~or~} y \ge \frac{2}{3}

(ii)
Using long division, we find that

y = \frac{x^2 + 14x + 50}{3(x+7)} = \frac{x}{3} + \frac{7}{3} + \frac{1}{3(x+7)}

So the asymptotes are y = \frac{x}{3} + \frac{7}{3} and x = -7

Curve C of 1(ii)

Question 2
(i)
x^2 - 9y^2 + 18y = 18

x^2 - 9(y^2 - 2y) = 18

x^2 - 9[(y-1)^2 - 1^2] = 18

x^2 - 9(y-1)^2 + 9 = 18

x^2 - 9(y-1)^2 = 9

\frac{x^2}{9} - (y-1)^2 = 1

This is a hyperbola with centre (0, 1), asymptotes are y = \pm \frac{x}{3} + 1, and vertices (3, 1) and (-3, 1).

y = \frac{1}{x^2} + 1 is a graph with asymptotes x = 0 and y=1.

Use GC to plot.

(ii)
\frac{x^2}{9} - (y-1)^2 = 1—(1)

y = \frac{1}{x^2} + 1 —(2)

Subst (2) to (1),

\frac{x^2}{9} - (\frac{1}{x^2} + 1 - 1)^2 = 1

\frac{x^2}{9} - (\frac{1}{x^2})^2 = 1

x^2 - \frac{9}{x^4} = 9

x^6 - 9 = 9x^4

x^6 - 9x^4 - 9 = 0

(iii)
From graph, we observe two intersections. Thus, two roots.

Question 3
(ai)
\sum_{r=1}^n (r+1)(3r-1)

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

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

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

= 3 \frac{n}{6}(n+1)(2n+1) + 2 \frac{n}{2}(1 + n) - n

= \frac{n}{2}(n+1)(2n+1) + n(1+n) - n

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

(aii)
2 \times 4 + 3 \times 10 + 4 \times 16 + ... + 21 \times 118

= 2 [2 \times 2 + 3 \times 5 + 4 \times 8 + ... + 21 \times 59]

= 2 [(1+1) \times (3 \cdot 1 - 1) + (2+1) \times (3 \cdot 2 -1) + (3+1) \times (3 \cdot 3 -1)  + ... + (20+1) \times (3 \cdot 20 -1) ]

= 2 \sum_{r=1}^{20} (r+1)(3r-1)

= 2 [\frac{n}{2}(n+1)(2n+1) + n^2 ]

= n(n+1)(2n+1) + n^2

= n(2n^2 + 3n + 1) + n^2

= 2n^3 + 4n^2 + n

(bi)
\frac{2}{(r-1)(r+1)} = \frac{A}{r-1} - \frac{B}{r+1}

2 = A(r+1) - B(r-1)

Let r = -1

2 = - B(-2) \Rightarrow B = 1

Let r = 1

2 = A(2) \Rightarrow A = 1

\therefore \frac{2}{(r-1)(r+1)} = \frac{1}{r-1} - \frac{1}{r+1}

(bii)
\sum_{r=2}^n \frac{1}{(r-1)(r+1)}

= \frac{1}{2} \sum_{r=2}^n \frac{2}{(r-1)(r+1)}

= \frac{1}{2} \sum_{r=2}^n (\frac{1}{r-1} - \frac{1}{r+1})

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

= \frac{1}{2} (\frac{3}{2} - \frac{n+1+n}{n(n+1)})

= \frac{3}{4} - \frac{2n+1}{2n(n+1)}

(biii)
As n \to \infty, \frac{1}{n} \to 0 and \frac{1}{n+1} \to 0, the sum of series tends to \frac{3}{4}, a constant. Thus, series is convergent.

(biv)

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

Replace r by r + 2. Then we have

\sum_{r=3}^{n+1} \frac{1}{(r-1)(r+1)}

= \sum_{r=2}^{n+1} \frac{1}{(r-1)(r+1)} - \frac{1}{(2-1)(2+1)}

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

= \frac{5}{12} - \frac{2n+3}{2(n+1)(n+2)}

Scatter Diagrams

JC Mathematics

I was teaching scatter diagram to some of my students the other day. A few of them are a bit confused with correlation and causation. I gave them the typical ice cream and murder rates example, which I shared here when I discussed about the r-value.

Think of correlation like a trend, it simply can be upwards, downwards or no trend. And since we only discuss about LINEAR correlation here, strong and weak simply is with respect to how linear it is, that means how close your scatter points can be close to a line.

Since A’levels, do ask students to draw certain scatter during exams to illustrate correlation. Here is a handy guide.

Scatter Diagrams
Credits: pythagorasandthat.co.uk