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Mechanics M1 problem-help needed with constant acceleration

Posted by White Group Mathematics on March 10, 2015 at 2:20 AM Comments comments (0)



The question asks,


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 ?



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.



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 comments (0)

Find all real numbers of x which satisfy the equation


Please help.

Student X



(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.




Best Regards,



Mr Koh

Modulus function graph question

Posted by White Group Mathematics on October 1, 2014 at 10:05 AM Comments 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.




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 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

Solving trigonometric equation with high power

Posted by White Group Mathematics on September 3, 2014 at 12:40 PM Comments comments (0)

For 0 less than or equal to x which is less than 2pi, solve

What can I do about that horrible power?

Student X

This probably needs to be solved by inspection rather than brute force expansion.


Both the LHS and RHS share the same periodicity, in fact the equation can only hold if both sides yield an integer value. It might take a while for you to convince yourself of this.


The broad solution set would actually be (2k+1) pi , where k is any chosen integer value. But since the question has specified a permissible range for x, there shall only be one solution, which is x=pi.


Hope this helps. Peace.

Best Regards,

Mr Koh

Need assistance with C3 trigonometry proof question

Posted by White Group Mathematics on June 21, 2014 at 9:05 PM Comments comments (0)


Please could you help me with answering the following proof question for trigonometry in C3.


23) Prove that if P, Q and R are the angles of a triangle, then


Student X

P+Q+R=180 deg


P+Q = 180 deg -R


tan(P+Q) = tan (180 deg -R)


(tanP +tanQ)/ (1- tanP tanQ) =-tanR


(Note: tan 180 deg =0 )


Proceed with a little more housekeeping of the above, and you should arrive at the required proof shortly.


Hope this helps. Peace.

Best Regards,

Mr Koh

Need help with RSA Cryptography

Posted by White Group Mathematics on June 11, 2014 at 9:40 AM Comments comments (0)

I' m studying computer science and engineering , but I wasn't able to follow the Math lectures, and now I`m having problems understanding some Math tasks.



Within 2 days I have an exam and I need to pass it , please help me in understanding this.



The task is :


Using Rules of RSA Cryptography by having " p = 5 , q= 13 , and  e=7 to DECRYPT this message




Student X

I will get you started off then. Let's decode 59:


p*q = 5*13 =65


φ(n)= (p-1)*(q-1) = 4*12 =48


The modular multiplicative inverse of e (mod φ(n)) where e=7, φ(n)=48 gives the result of 7.


59^7 mod(65) =2488651484819 mod (65) =19


For each of the remaining encrypted messages, the corresponding decrypted message is simply given by the value of M^7 mod (65), where M is the encrypted message.


Hope this helps. Peace.

Best Regards,

Mr Koh

Sixth Term Examination Paper II

Posted by White Group Mathematics on June 6, 2014 at 7:50 AM Comments comments (0)

Hi White Group Maths,

This is Student X here, a soon to be student of mathematics in university.

In about a month, I'm going to take the STEP II paper. In preperation for this, I've been doing practice questions since February, roughly 2 hours a day.

However, despite this practice... I feel no more confident than I did 3 months ago, and am quite worried. Do you have any advice on how to further prepare for this test? Or any tuition services to to suggest?

Thanks so much for the help.


Student X

Hi there ,


                 Thanks for writing. It is encouraging to know you have commenced preparations for the STEP paper way before the actual sitting itself; it is perfectly normal to feel shaky despite months of drilling, because STEP isn't your conventional A Level exam to begin with-it demands the candidate to constantly expect the unexpected, and to be able to conjure solutions bordering on ingenuity. Above all, the compulsory possession of a rock solid Mathematical foundation so as to nimbly navigate various kinds of twists and turns.


                 Your exposure to various genres of problems by virtue of the efforts invested since February should be considerable by now, however you do need to ask yourself these three questions:


1. What percentage of the paper did you manage to complete properly in recent attempts? Can things be improved? (Time management)


2. Do you tend to linger around a problem for way too long before moving on? Have you learnt to not allow frustration overwhelm the big picture? (Cutting losses)


3. Did you make it a point to acquire the underlying moral of the story for particularly intense problems? ( enhancing solving efficiency and becoming more elegant in solution design)



Just to add, unlike the A Level papers, you may not have the luxury of excess time at the end to verify the correctness of previous scripted answers, so you must tread carefully from the onset.


                       I doubt tuition is what you really needat this very moment-it's more about readying your state of mind for the big event. An extremely nervous constitution can wreak substantial damage to your performance on that day itself, so start conditioning yourself to be cool-headed (of course easier said than done, but it has to be done).Stop worrying about the final grade you will receive or the somewhat insurmountable difficulties lying in the immediate future, try to "enjoy" the process of conquering the paper instead.


                        And do remember to have an early rest on the night before the actual STEP examination. Good luck, and god bless.



Best Regards,

Mr Koh


Evaluating an integral without actually solving it

Posted by White Group Mathematics on May 8, 2014 at 10:00 AM Comments comments (0)

Is there any way to obtain the solution for the below integral without solving it?

Student X

Certainly. The function to be integrated is odd in nature.

If we let f(x)= x* e^|x| *sec x, 

then f(-x) = (-x)* e^|-x| *sec (-x) =  - x* e^|x| *sec x = - f(x)

This implies that the graph of f(x) has rotational symmetry about the origin. (A simple example would be the sine curve) . Thus, if f(x) is integrated wrt x from -a to a for all a ∈ ℝ, then the result is simply zero.

Hope this helps. Peace.

Best Regards,

Mr Koh

Rate of change problem

Posted by White Group Mathematics on April 2, 2014 at 10:35 AM Comments comments (0)

The circle below has a radius of 6 cm and point C is moving clockwise around the circle.  Assume that point B is moving away from point A at a rate of 2 cm./sec. i.e. x is increasing at a rate of 2 cm./sec.  At this moment , assume θ is pi/3 radians.

a. What is the instantaneous rate of change of θ at this moment. i.e. how fast is θ decreasing at this moment?

b. Also, at this moment, is θ decreasing faster or slower?  By how much?

Student X

At any particular instant, let's focus on the right angled triangle ABC where AB= x units and angle CAB = θ, where both x and θ varies with time.


A relationship can be formed between x and theta, which is


x = 6 cos θ


Differentiating both sides wrt y gives


dx/dt = - 6 sin θ * dθ/dt -------(1)


Substituting dx/dt =2 cm/s, θ = π/3,


instantaneous rate of change of θ at this moment


= dθ/dt = 2 ÷ [ -6 * sqrt(3) /2 ] =-0.385 rad/s (shown)


Since x increases at a constant rate of 2cm/s, its second order derivative wrt time is zero, ie d^2 x/dt^2 = 0 cm/s^2


Differentiating (1) on both sides wrt t once again,



d^2 x/dt^2 =  -6 [ cosθ *( dθ/dt)^2 + sinθ * (d^2 θ/dt^2 )]


(Be mindful of how the product rule is employed in the RHS)



Substituting d^2 x/dt^2 = 0 cm/s^2, θ = π/3  and  dθ/dt = -0.385 rad/s,



0 =  -6 [ cos(π/3) *( -0.385)^2 + sin(π/3)  * (d^2 θ/dt^2 )]


Solving gives d^2 θ/dt^2 = -0.0856 rad/s^2


Hence, at this moment, θ is decreasing slower (because of the negative sign) by 0.0856 rad/s. (shown)



Hope this helps. Peace.

Best Regards,

Mr Koh