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
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
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
Posted by White Group Mathematics on March 10, 2015 at 2:20 AM | comments (0) |
Hi,
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 ?
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
Posted by White Group Mathematics on February 7, 2015 at 3:30 AM | comments (0) |
Find all real numbers of x which satisfy the equation
（1+x)^6-2(1-x)^6=(1-x^2)^3
Please help.
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
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
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
Posted by White Group Mathematics on September 3, 2014 at 12:40 PM | 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
Posted by White Group Mathematics on June 21, 2014 at 9:05 PM | comments (0) |
Hi!
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
Thanks!
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
Posted by White Group Mathematics on June 11, 2014 at 9:40 AM | 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
59,57,43,00,52,00
Thanks!!!
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