### Random Questions from 2017 Prelims #1

Last year, I shared a handful of random interesting questions from the 2016 Prelims. Students feedback that they were quite helpful and gave them good exposure. I thought I share some that I’ve seen this year. I know, its a bit early for Prelims. But ACJC just had their paper 1. 🙂

This is from ACJC 2017 Prelims Paper 1 Question 7. And it is on complex numbers.

7
(a) Given that $2z + 1 = |w|$ and $2w-z = 4+8i$, solve for $w$ and $z$.

(b) Find the exact values of $x$ and $y$, where $x, y \in \mathbb{R}$ such that $2e^{-(\frac{3+x+iy}{i})} = 1 -i$

I’ll put the solutions up if I’m free.

But for students stuck, consider checking this link here for (a) and this link here for (b). These links hopefully enlightens students.

Just FYI, you cannot $\text{ln}$ complex numbers as they are not real…

As the prelims examinations draw really close, many students were asking me to give questions to test their concepts for several topics. In class, I had the opportunity to explore several applications questions too. We saw several physics concepts mixed. We also have some conceptual questions that need students to be able to use the entire topic to solve it.

So I’ll share one here. This involves several concepts put together. I’ll put the solution up once I find the time. Concepts that will be involved, will be

1. Vector Product
2. Equations of Plane
3. Finding foot of perpendicular of point

The question in one a reflection of a plane in another plane. I think such questions will come out in a few guided steps in exams. But should a student be able to solve it independently, it shows that he has good understanding.

The plane $p$ has equation $x + y + z = 9$ and the plane $p_1$ contains the lines passing through $(0, 2, 3)$ and are parallel to $(1, -1, 0)$ and $(0, 1, 1)$ respectively. Find, in scalar product form, the equation of the plane which is the reflection of $p_1$ in $p$.

### Thinking [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ #9

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by KS. More of him can be found here

This is a standard summation question. I’m interested in the last part only.

The answer to (ii) is written there by the student. I’ll only do the solution to (iv).

### Quick Summary (Probability)

University is starting for some students who took A’levels in 2016. And, one of my ex-students told me to share/ summarise the things to know for probability at University level. Hopefully this helps. H2 Further Mathematics Students will find some of these helpful.

Random Variables

Suppose $X$ is a random variable which can takes values $x \in \chi$.

$X$ is a discrete r.v. is $\chi$ is countable.
$\Rightarrow p(x)$ is the probability of a value of $x$ and is called the probability mass function.

$X$ is a continuous r.v. is $\chi$ is uncountable.
$\Rightarrow f(x)$ is the probability density function and can be thought of as the probability of a value $x$.

Probability Mass Function

For a discrete r.v. the probability mass function (PMF) is

$p(a) = P(X=a)$, where $a \in \mathbb{R}$.

Probability Density Function

If $B = (a, b)$

$P(X \in B) = P(a \le X \le b) = \int_a^b f(x) ~dx$.

And strictly speaking,

$P(X = a) = \int_a^a f(x) ~dx = 0$.

Intuitively,

$f(a) = P(X = a)$.

Properties of Distributions

For discrete r.v.
$p(x) \ge 0 \forall x \in \chi$.
$\sum_{x \in \chi} p(x) = 1$.

For continuous r.v.
$f(x) \ge 0 \forall x \in \chi$.
$\int_{x \in \chi} f(x) ~dx = 1$.

Cumulative Distribution Function

For discrete r.v., the Cumulative Distribution Function (CDF) is
$F(a) = P(X \le a) = \sum_{x \le a} p(x)$.

For continuous r.v., the CDF is
$F(a) = P(X \le a ) = \int_{- \infty}^a f(x) ~dx$.

Expected Value

For a discrete r.v. X, the expected value is
$\mathbb{E} (X) = \sum_{x \in \chi} x p(x)$.

For a continuous r.v. X, the expected value is
$\mathbb{E} (X) = \int_{x \in \chi} x f(x) ~dx$.

If $Y = g(X), then For a discrete r.v. X,$latex \mathbb{E} (Y) = \mathbb{E} [g(X)] = \sum_{x \in \chi} g(x) p(x)\$.

For a continuous r.v. X,
$\mathbb{E} (Y) = \mathbb{E} [g(X)] = \int_{x \in \chi} g(x) f(x) ~dx$.

Properties of Expectation

For random variables $X$ and $Y$ and constants $a, b, \in \mathbb{R}$, the expected value has the following properties (applicable to both discrete and continuous r.v.s)

$\mathbb{E}(aX + b) = a \mathbb{E}(X) + b$

$\mathbb{E}(X + Y) = \mathbb{E}(X) + \mathbb{E}(Y)$

Realisations of $X$, denoted by $x$, may be larger or smaller than $\mathbb{E}(X)$,

If you observed many realisations of $X$, $\mathbb{E}(X)$ is roughly an average of the values you would observe.

$\mathbb{E} (aX + b)$
$= \int_{- \infty}^{\infty} (ax+b)f(x) ~dx$
$= \int_{- \infty}^{\infty} axf(x) ~dx + \int_{- \infty}^{\infty} bf(x) ~dx$
$= a \int_{- \infty}^{\infty} xf(x) ~dx + b \int_{- \infty}^{\infty} f(x) ~dx$
$= a \mathbb{E} (X) + b$

Variance

Generally speaking, variance is defined as

$Var(X) = \mathbb{E}[(X- \mathbb{E}(X)^2] = \mathbb{E}[X^2] - \mathbb{E}[X]^2$

If $X$ is discrete:

$Var(X) = \sum_{x \in \chi} ( x - \mathbb{E}[X])^2 p(x)$

If $X$ is continuous:

$Var(X) = \int_{x \in \chi} ( x - \mathbb{E}[X])^2 f(x) ~dx$

Using the properties of expectations, we can show $Var(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2$.

$Var(X)$
$= \mathbb{E} [(X - \mathbb{E}[X])^2]$
$= \mathbb{E} [(X^2 - 2X \mathbb{E}[X]) + \mathbb{E}[X]^2]$
$= \mathbb{E}[X^2] - 2\mathbb{E}[X]\mathbb{E}[X] + \mathbb{E}[X]^2$
$= \mathbb{E}[X^2] - \mathbb{E}[X]^2$

Standard Deviation

The standard deviation is defined as

$std(X) = \sqrt{Var(X)}$

Covariance

For two random variables $X$ and $Y$, the covariance is generally defined as

$Cov(X, Y) = \mathbb{E}[(X - \mathbb{E}[X])(Y - \mathbb{E}[Y])]$

Note that $Cov(X, X) = Var(X)$

$Cov(X, Y) = \mathbb{E}[XY] - \mathbb{E}[X] \mathbb{E}[y]$

Properties of Variance

Given random variables $X$ and $Y$, and constants $a, b, c \in \mathbb{R}$,

$Var(aX \pm bY \pm b ) = a^2 Var(X) + b^2 Var(Y) + 2ab Cov(X, Y)$

This proof for the above can be done using definitions of expectations and variance.

Properties of Covariance

Given random variables $W, X, Y$ and $Z$ and constants $a, b, \in \mathbb{R}$

$Cov(X, a) = 0$

$Cov(aX, bY) = ab Cov(X, Y)$

$Cov(W+X, Y+Z) = Cov(W, Y) + Cov(W, Z) + Cov(X, Y) + Cov(X, Z)$

Correlation

Correlation is defined as

$Corr(X, Y) = \dfrac{Cov(X, Y)}{Std(X) Std(Y)}$

It is clear the $-1 \le Corr(X, Y) \le 1$.

The properties of correlations of sums of random variables follow from those of covariance and standard deviations above.

### Thinking [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ #8

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by KS. More of him can be found here

This is a interesting Complex Number Question.

The complex number $z$ satisfies $z + |z| = 2 + 8i$. What is $|z|^2$

### Thinking [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ #4

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by KS. More of him can be found here.

This is a question from 1993 Paper 1.

The positive integers, starting at 1, are grouped into sets containing $1, 2, 4, 8, \ldots$ integers, as indicated below, so that the number of integers in each set after the first is twice the number of integers in the previous set.

$\{ 1 \}, \{ 2, 3 \}, \{ 4, 5, 6, 7 \}, \{ 8, 9, 10, 11, 12, 13, 14, 15 \}, \ldots$

(i) Write down the expressions, in terms of $r$ for

(a) the number of integers in the $r^{th}$ set,

(b) the first integer in the $r^{th}$ set,

(c) the last integer in the $r^{th}$ set.

(ii) Given that the integer $1,000,000$ occurs in the $r^{th}$ set, find the integer value of $r$.

(iii) The sum of all the integers in the $20^{th}$ set is denoted by $S$, and the sum of all the integers in all of the first $20$ sets is denoted by $T$. Show that $S$ may be expressed as $2^{18}(3 \times 2^{19} - 1)$.

Hence, evaluate $\frac{T}{S}$, correct to 4 decimal places.

### Population problems eventually solve themselves-government meddling only makes things worse. Discuss

Government intervention solves population problems such as population decline, which will be left unresolved if left to the masses. With a preference for smaller families and a general unwillingness to start a family in today’s modern society, negative or zero population growth often ensues. These have detrimental impact on affected countries, such as a fall in tax revenues, a smaller workforce and a high dependence of an ageing population on the working population. As these socioeconomic perspectives are entrenched in the minds of young urban professionals, these population problems are incapable of eventually solving themselves. In this case, government intervention is beneficial. In developed countries like Italy and Spain, where fertility rates stand at a meagre 1.25, new generations are unable to replace past generations thus leading to population decline. The implementation of pro-natal policies could possibly help to increase the incentive for couples to procreate and boost total population numbers. Implemented measures include longer maternity and paternity leave in Switzerland, as well as cash incentives in Singapore. Another method of boosting population growth is through the relaxation of immigration policies, which allows for an influx of permanent residents.

Population problems such as the rampant spread of diseases are also combated more efficiently and effectively through government intervention. If left to solve by itself, this results in a higher death toll and increased spread of illnesses. The successful results of government intervention is exemplified through the World Health Organization and governments’ collaboration to wipe out smallpox, which was deadly enough to kill one in every four infected persons. With public health measures to increase hygiene standards and mandatory vaccinations, smallpox was eradicated worldwide in the 1800s.

Despite the effectiveness of government intervention in solving population problems, some policies and measures undoubtedly create new problems for countries. Firstly, policies to reduce overpopulation are often successful to the extent that they eventually lead to population decline. This is evident in Singapore, which, due to the overwhelming success of the “stop at two” policy, currently faces a replacement rate of 1.25. This has led to national concerns of unsustainable population growth and the possibility of a population decline in the near future. Furthermore,  the policy of migration to solve population problems has led to social segregation in some countries.

### Thinking [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ #3

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by KS. More of him can be found here.

This is a question from 1976 A’levels Paper 2. I thought it is pretty interesting to discuss the question with a little extension.

(a) In how many ways can 5 copies of a book be distributed among 10 people, if no-one gets more than one copy?

(b) In how many ways can 5 different books be distributed among 10 people if each person can get any number of books?

So now, let us modify it a bit.

(c) In how many ways can 5 copies of a book be distributed among 10 people if each person can get any number of books?

Notice that the difference between (b) and (c) is that the book distributed is not identical. So for (c), we are pretty much distributing $r$ identical balls to $n$ distinct boxes. Whereas for (b) , we are pretty much distributing $r$ distinct balls to $n$ distinct boxes.

### Common Errors for Application Question(AQ)

My previous posts have touched on the points that you would need to excel in your AQ. It is also important for you to understand what are some pitfalls so that you will not make these mistakes.

• 1. Rehash of the author’s argument/ summary of the author’s argument- did not go on to elaborate or to even offer personal insight
• 2. Lack of evaluation- paragraphs tend to be descriptive
• 3.Poor selection of examples- they could be inaccurate or they are isolated case studies that are not representative of the whole society
• 4.No examples given
• 5.Evaluation lacks depth or it is done very superficially
• 6.No link back to the requirements of the question
• 7. No introduction or conclusion (of course this is a very obvious time management issue)

Alright, there you have it, the 7 deadly sins/pitfalls for your AQ.

### 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.