### Meritocracy? Junior colleges merger and its implications

It’s pretty interesting that shortly after our post on meritocracy, we have news about the junior college (JC) mergers. For more information about the news, you could take a look at this weblink.

This drastic move by MOE has sparked a lot of concerns among the public and has brought up a few issues for us to consider. First, it would be the falling demographics of Singapore. The falling birth rates is cited as the main reason for the merger of schools, so that resources would not be wasted, and there would not be under-utilized staff in the system. With such falling birth rates, what would you think is going to happen to the future of the educational landscape (would teaching/tutoring as a profession still be lucrative? We know that MOE has cut back on the hiring of teachers from 3000 at its peak yearly to about 1000 right now).

Second question to think about would be the larger implications of these schools merger. Why are these schools selected? Some have argued that it is a strategic move by the government to level the playing field by merging these colleges so that academic standards would be streamlined? of course, we cannot merge schools like RI and HCI together as it would only further consolidate their super-elite status in society (besides strong school culture and powerful alumni).

Finally, school culture and history is being destroyed when merger takes place. If that’s the case, what does it say about how the nation values history? It is all about the future and progress right, the past no longer matters if it is holding us back. Pragmatism is the view of the Singapore’s state.

### Differentiation Question #1

Given that $y = \frac{8}{x^3} - \frac{6}{x^2} + \frac{5}{2x}$, find the approximate percentage change in $y$ when $x$ increases from 2 by 2%.

### Probability Question #4

A gambler bets on one of the integers from 1 to 6. Three fair dice are then rolled. If the gambler’s number appears $k$ times ($k = 1, 2, 3$), he wins $$k$. If his number fails to appear, he loses$1. Calculate the gambler’s expected winnings

### Random Sec 4 Differentiations

B6

$y = 3e^x + \frac{4}{e^x}$

$\frac{dy}{dx} = 3e^x - \frac{4}{e^x}$

$\frac{d^2y}{dx^2} = 3e^x + \frac{4}{e^x}$

let $\frac{dy}{dx} = 0$

$3e^x - \frac{4}{e^x} = 0$

$3e^{2x} = 4$

$2x = \mathrm{ln} \frac{4}{3}$

$x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$

Sub $x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$ to $\frac{d^2y}{dx^2}$

$\frac{d^2y}{dx^2} > 0$ Thus, it is a min point.

C7

$y = \mathrm{ln} \frac{5-4x}{3+2x}$

$y = \mathrm{ln} (5-4x) - \mathrm{ln} (3+2x)$

$\frac{dy}{dx} = \frac{-4}{5-4x} - \frac{2}{3+2x}$

let $\frac{dy}{dx} = 0$

$\frac{-4}{5-4x} - \frac{2}{3+2x} = 0$

$\frac{-4}{5-4x} = \frac{2}{3+2x}$

$-4(3+2x) = 2(5-4x)$

$-12 - 8x = 10 - 8x$

$-12 = 10$ (NA).

There are no stationary points for this curve.

C8

$x = \frac{1}{3}e^{y(2x+5)}$

$\mathrm{ln}(3x) = y(2x+5)$

$\frac{\mathrm{ln}(3x)}{2x+5} = y$

$y = \frac{\mathrm{ln}(3x)}{2x+5}$

$\frac{dy}{dx} = \frac{\frac{1}{x}(2x+5) - \mathrm{ln}(3x) \times 2}{(2x+5)^2}$

Let $x = e^2$

$\frac{dy}{dx} = \frac{\frac{1}{e^2}(2e^2+5) - \mathrm{ln}(3e^2) \times 2}{(2e^2+5)^2}$

Evaluate with a calculator…

### Vectors Question #4

Another interesting vectors question.

The fixed point $A$ has position vector a relative to a fixed point $O$. A variable point $P$ has position vector r relative to $O$. Find the locus of $P$ if r $\bullet$ (ra) = 0.

### Vectors Question #3

This is a question a student sent me a few days back, and I shared with my class.

Find the Cartesian equation of the locus of all points (plane) that is equidistant of the $xy$ plane and $xz$ plane.

The following should aid students to visualise.

Sidenote: I think Vectors is a very important topic for 9758 as its applications are wide. Students should do their best to understand the topic. I will share a few more applied questions next week when I have time.

### A little reminder to students doing Calculus now

When $\frac{dy}{dx} = 0$, it implies we have a stationary point.

To determine the nature of the stationary point, we can do either the first derivative test or the second derivative.

The first derivative test:

Students should write the actual values of $\alpha^-, \alpha, \alpha^+$ and $\frac{dy}{dx}$ in the table.

We use this under these two situations:
1. $\frac{d^2y}{dx^2}$ is difficult to solve for, that is, $\frac{dy}{dx}$ is tough to be differentiated
2. $\frac{d^2y}{dx^2} = 0$

The second derivative test:

Other things students should take note is concavity and drawing of the derivative graph.

### APGP Question #3

This is a question from a recent ACJC JC2 test.

A philanthropist started a donation matching programme to encourage more people to donate regularly to a particular charity.

(a) For a person who donates $a in the first month, and for each subsequent month donates$ $b^2$ more than the previous month, the philanthropist will donate $a in the first month and for each subsequent month, b times the amount he donated the previous month. (i) John is a regular donor of the charity. Find the values of a and b such that the philanthropist donates$20 in the third month, and ten times more than John in the seventh month.

(ii) Find the total amount of money donated by John and the philanthropist in one year, leaving your answer to the nearest dollar.

(b) In a revised donation matching programme, if a donor makes a monthly donation of $c, the philanthropist will donate according to the following plan: First month : No donation ($0\%$ donation rate) Second month : $10 \%$ of the total amount donated by the donor in the first two months Third month : $20 \%$ of the total amount donated by the donor in the first three months Fourth month : $30 \%$ of the total amount donated by the donor in the first four months and so on. (i) Show that the total amount of money the philanthropist will donate at the end of the $4^{th}$ month is$2c.

(ii) By expressing the total amount of money, the philanthropist will donate at the end of the $n^{th}$ month as a summation, and using the result $\sum_{r=1}^n r^2 = \frac{n}{6}(n+1)(2n+1)$, show that the total amount of money the philanthropist will donate at the end of the $n^{th}$ month is \$ $\frac{(n-1)n(n+1)}{30} c$.

### Vectors Question #2

If $c = |a| b + |b| a$, where $a$ , $b$ and $c$ are all non-zero vectors, show that $c$ bisects the angle between $a$ and $b$.

### Post-Results 2016

Let’s face it. Some of us will not get the dream results we want. Don’t give up and let fear conquer you.

For students unsure of the available courses, they can check out the following post. It contains the grade profile for local universities.

Our Team will be here if you need help/ advice. Feel free to text us.

P.S. Today, I saw an image shared by Mr Wee, which said that “You’re the architect of your own life”. So let’s not let the grades define us.