A LEVEL MATHEMATICS

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Discovering vertices of a rectangle in an Argand diagram Posted by White Group Mathematics on August 4, 2015 at 9:40 PM comments (1)

Hi Mr Koh,

How do I do this question- In an Argand diagram, the points A, B, C and D represent the complex numbers a,  -2+5i,  c   and  3/2 - 1/2 i   respectively. Given that ABCD is a rectangle described in the clockwise sense with BC=2AB, find a and c.

I tried drawing B and D on the same diagram but they don't even have a right angle between them, so how can it form a rectangle? And I can't place points A and C on the diagram because they said ABCD is read clockwise but it is also mentioned BC=2AB....so I am lost. Haha. Thanks Mr Koh!

Student X

Based on the construct described, the rectangle's length is twice its breadth. I would recommend drawing a generic upright rectangle instead of trying to place the points specifically in Argand space. Do note that vectors are involved in the solving process. I have drafted a solution template for you: Hope this helps. Peace.

Best Regards,

Mr Koh

Discerning winning probabilities in national lottery Posted by White Group Mathematics on July 19, 2015 at 10:40 PM comments (0)

Hi,

Here in my country, when you buy lottery tickets, you try to guess the six (6) combination of numbers out of 42 numbers. Unlike in Sweepstakes, the order of numbers does not really matter.

http://www.philippinepcsolotto.com/6-42-lotto-result-summary

Some say that the probability of winning the jackpot is simply 6/42 = 14.29%. That is high!

Since I was in high school, that formula seemed to make sense for me. After all, what is the probability of guessing one of the six combinations? 1/42. That is the probability of guessing each number in the combination of six. That is the result when you add the probability for each number.

(1/42) X 6 = 6/42

1/42 + 1/42 + 1/42 + 1/42 + 1/42 + 1/42 = 6/42

But I've come across another possibility today. Without repetitions of numbers, you can create 5,245,786 combinations of six digits from 1 to 42. What is the probability that you will guess the right combination out of more than 5 million combinations? 1 out of 5,245,786. That is 1.906 X 10 raised to negative 7.

I'm confused now because both solutions seem to make sense, but this second idea makes more sense. Each one is a computation based on a different point of view.

I hope you can enlighten me on this one.

Thanks,

Student X

The first manner of calculating the required probability which you cited is flawed. If say, each of these 6 combinations can be recycled, the probability would be computed as (1/42)^6 = 1/5489031744.

However, I am assuming they cannot be recycled, as such there will be 42 possible combinations for the first number, 41 for the second, 40 for the third, so on and so forth. That said, the probability would be computed as (1/42)*(1/41)*(1/40)*(1/39)*(1/38)*(1/37)= 1/3776965920.

You may therefore question: why isn't this equivalent to 1/5245786, which is considerably larger in fact? The reason is (1/42)*(1/41)*(1/40)*(1/39)*(1/38)*(1/37) actually examines the instance when sequencing of numbers matters, as opposed to 1/( 42 C 6) = 1/5245786 which doesn't pay any regard to the ordering of numbers. Clearly, the former would involve much greater stakes and thus even lower likelihood of winning.

Hope this clarifies. Peace.

Best Regards,

Mr Koh

Tiny abstract vectors problem Posted by White Group Mathematics on July 14, 2015 at 10:25 PM comments (0)

Hi Mr Koh,

I encountered this question. It says: if  a = (a • b) b, where a and b are vectors, what are the possible angles between a and b?  Also, what is the magnitude of vector b?

Student X

a = (a • b) b  actually implies a is parallel to b.  Note that a • b is in fact a scalar quantity,  such that a=kb, where we can set a • b = k . In view of this, the only possible angles between a and b would be 0 and 180 degrees.

To discover |b|, we shall proceed to take modulus of both sides of the original equation a = (a • b) b ; this gives us |a| = |a • b| |b| = [ |a| |b| ] |b| = |a| |b|^2 . Cancelling |a| on both sides leads to |b|^2 =1, and thus |b|=1 (shown)

Hope this helps. Peace.

Best Regards,

Mr Koh

How to evaluate the integral of 1/(1+x^4) ? Posted by White Group Mathematics on June 20, 2015 at 10:00 PM comments (0)

I am having problems trying to figure out a way to solve for the integral of 1/(1+x^4) ; attempted some form of substitution but it didn't work out. Could you advise on how to get started properly?

Student X

You can consider factorization of the quartic polynomial, followed by the employment of partial fractions to further break things down. Thereafter , it shouldn't be all that hard to continue the solving process.

Here's what I mean: Hope this helps. Peace.

Best Regards,

Mr Koh

How to generate a²-b² in this complex numbers proof? Posted by White Group Mathematics on May 17, 2015 at 11:05 PM comments (0)

I am having problems trying to achieve the expression a²-b² for part (ii) of this problem. Please help. Student X

Please find the detailed solution below, hope it helps. Peace. Best Regards,

Mr Koh

For two given complex numbers a and b, is it true that |a-b| = |a|-|b| ? Posted by White Group Mathematics on April 24, 2015 at 12:40 PM comments (0)

For two given complex numbers a and b, is it true that  |a-b| = |a|-|b| ?

Student X

Generally, it isn't true. However, in specific instances, |a-b|  can be equal to  |a|-|b|.  This happens when arg(a)= arg(b). Bearing in mind |a-b| always represents geometrically the physical distance between the two complex numbers a and b, here is a visual representation for this unique case: Another instance worth noting is when arg(b)= arg(a) - π, then |a-b| actually becomes |a|+|b|: Hope the above clarifies. Peace.

Best Regards,

Mr Koh

Mechanics M1 problem-help needed with constant acceleration Posted by White Group Mathematics on March 10, 2015 at 2:20 AM comments (0)

Hi,

A body moving in a straight line with constant acceleration takes 3 sec and further 5 sec to cover two successive distances of 1m. Find acceleration.(Hint: use distances of 1m and 2m from the start of motion)

I cannot seem to figure out how to solve this. Could you please show the calculations to get the same answer presented in the book, which is -1/30 m/s^2 ?

Thanks.

Student X

Let us use s=ut+1/2 * at^2

When the first metre is covered, substituting s=1 and t=3 gives

3u+ 9a/2 = 1 -----------------(1)

When the first two metres are covered, substituting s=2 and t= 3+5=8 gives

8u+ 32a = 2 ====> 4u+16a =1 -------(2)

Solve these two equations simultaneously, you should be able to retrieve the answer for the acceleration shortly.

Peace.

Best Regards,

Mr Koh

Need assistance with solving sextic (degree 6 polynomial) equation Posted by White Group Mathematics on February 7, 2015 at 3:30 AM comments (2)

Find all real numbers of x which satisfy the equation

（1+x)^6-2(1-x)^6=(1-x^2)^3

Student X

（1+x)^6-2(1-x)^6=(1-x^2)^3

（1+x)^6-2(1-x)^6=(1-x)^3 * (1+x)^3 -----------(1)

Let a=(1+x)^3, b=(1-x)^3, then (1) becomes   a^2 -2b^2 = ab

Migrating all terms to the LHS,

a^2 -ab -2b^2 =0

(a-2b)(a+b) =0

a=2b or a=b

Resubstitute the expressions for both a and b in terms of x, you should arrive at the required answers shortly.

Peace.

Best Regards,

Mr Koh

Modulus function graph question Posted by White Group Mathematics on October 1, 2014 at 10:05 AM comments (0)

Hello, I've never encountered this sort of question before and I don't know how to approach it. Could you explain it please?

The functions f and g are defined on the domain of all real numbers by f(x)= |x-2| and g(x)= |x|-2.

Sketch the graph of f(x) - g(x).

Student X

First, let us define each of the individual modulus functions:

|x-2| = x-2 if x≥2

= 2-x if x<2

|x|= x if x≥0

= -x if x<0

There are 3 critical regions, namely x<0, 0 ≤ x < 2 and x≥2

For the extreme left critical region, ie x<0,

f(x) - g(x) = |x-2| - |x| + 2 = (2-x) - (-x) +2 = 4

In other words, you shall draw a horizontal line y=4 all the way from x=-∞ to x=0.

Thereafter,

for the next critical region 0 ≤ x < 2,

f(x) - g(x) = |x-2| - |x| + 2 = (2-x) - (x) +2 = 4-2x

In this case, you shall draw the line with equation y=4-2x from x=0 to x=2.

I shall let you figure out the final graph you need to draw for the remaining critical region, which shouldn't be all too difficult if you can understand what I have explained thus far.

Hope it helps. Peace.

Best Regards,

Mr Koh

Finding inverse of a function Posted by White Group Mathematics on September 19, 2014 at 11:20 AM comments (0)

Hi Mr Koh!  How do I find the inverse of this function f(x)= 2x - (1/x),  for  0 <x <3?

Student  X

Please find my workings as follows: Hope this helps. Peace.

Best Regards,

Mr Koh