[STPM] Mx T

[Daniel]

Find the modulus and argument of the following complex number: (6+8i)/(3-4i)
so i got the modulus for |Z| which is 2.

My problem is how to find the argument for Z. I got the Q where the value is -1.287. But when i substitute it in PIE(22/7) - Q(-1.287), the answer would be 4.43 which is wrong. The correct answer is 1.85. Which step that i did mistake? izzit the PIE - Q formula? Thank you for advance.

[Daniel]

1) Find the modulus and argument of the two complex numbers that satisfy the equation (1+Z^2)/(1-Z^2) = i

2) In an argand diagram, O is the origin and P represets the complex number 1 + 3i .The point Q represents a + bi (a>0 , b>0) and triangle OPQ is equilateral. Find the values of a and b.

[gohweihan]

Find the modulus and argument of the following complex number: (6+8i)/(3-4i)
so i got the modulus for |Z| which is 2.

My problem is how to find the argument for Z. I got the Q where the value is -1.287. But when i substitute it in PIE(22/7) - Q(-1.287), the answer would be 4.43 which is wrong. The correct answer is 1.85. Which step that i did mistake? izzit the PIE - Q formula? Thank you for advance.

Actually, it's a little hard to explain without a diagram. After you get the result of the division of the complex numbers (which I assume you should get -0.56 + 1.92i) you must look at the signs of both the numbers (real and imaginary).

If you draw an Argand diagram (which you should), you would notice that the line is on the negative side of the x-axis. Arguments are always be the angle from the positive x-axis to the line you drawn.

Your mistake comes from using pi - (-1.287). The answer of -1.287 comes from calculating the angle from the negative x-axis to the line you drawn. Instead of -1.287, you should just use 1.287 as it is merely showing the angle size, and is not relative to anything. If you substitute that in your calculation, your end result should be 1.85.

[Daniel]

Actually i already prepare a diagram to place it here as well to make things easier. Unfortunately i am not allowed to do so. My message was block due to security reason. Sigh.... By the way, Thank you gohweihan

[Daniel]

1) Find the modulus and argument of the two complex numbers that satisfy the equation (1+Z^2)/(1-Z^2) = i

2) In an argand diagram, O is the origin and P represets the complex number 1 + 3i .The point Q represents a + bi (a>0 , b>0) and triangle OPQ is equilateral. Find the values of a and b.

Anyone who can help me to solve this questions?

[chenchow]

Actually i already prepare a diagram to place it here as well to make things easier. Unfortunately i am not allowed to do so. My message was block due to security reason. Sigh.... By the way, Thank you gohweihan

Daniel, you can save it somewhere, like in the gallery, or somewhere else in www and then link it here. It should work.

[Daniel]

Help!
1) Find the cube of x^1/3 + y^1/3 and show that if x^1/3 + y^1/3 = z^1/3 , then (x+y-z)^3 + 27xyz = 0

2)Given z = x + iy, and |z+2| = 2 |z-i|, show that 3x^2 + 3y^2 -4x -8y = 0

3)the complex number z satisfies the equation |z|=|z+2|. Show that the real part of z is -1

Thank you

[Daniel]

Help!
1) Find the cube of x^1/3 + y^1/3 and show that if x^1/3 + y^1/3 = z^1/3 , then (x+y-z)^3 + 27xyz = 0

2)Given z = x + iy, and |z+2| = 2 |z-i|, show that 3x^2 + 3y^2 -4x -8y = 0

3)the complex number z satisfies the equation |z|=|z+2|. Show that the real part of z is -1

I know the calculation is very long. To make thing easier, Just tell me roughly how to solve the questions. Thank you. :D

[infested_ysy]

Ok, here's a problem I'm facing (embarassing, to imagine that I've totaly forgotton such basic questions which i've done over and over again in Form 5.... =\)

http://server6.uploadit.org/files/superrobot-scan0001.jpg
Find the value of x.
normally, I would log left and right, but thanks to the annoying 10 multiplier.... I'm stuck! (I've been figuring out the solutions for 20 minutes already, still have come to no avail)

[kevinkhoo1986]

Ermmm... can i try it? Sorry for the untidy writing...
http://www.xpphotoalbum.com/data/10965/9666math.JPG?4129

By the way, how did you upload your photo in here? how come my access where denied when i am trying to post this message along with the image...?

[infested_ysy]

Um... I upload the file in uploadit.org then direct link it here, using the [img] tag.

[chenchow]

Help!
1) Find the cube of x^1/3 + y^1/3 and show that if x^1/3 + y^1/3 = z^1/3 , then (x+y-z)^3 + 27xyz = 0



For Q1,
Just write through, the cube of first part and I guess you can just expand it. And then since you are doing cube of the expression,
(x^1/3 + y^1/3 )^3 , you can equate that whole thing to z,
so, that would give you x^1/3 + y^1/3 = z^1/3 ,so basically you just do some algebraic manipulations for the question.

[infested_ysy]

I'm having problems again with these 2 questions... I just don't know what I'm supposed to do =\
http://server6.uploadit.org/files/superrobot-mathsT.jpg

http://server6.uploadit.org/files/superrobot-mathsT2.jpg

[gohweihan]

For question 8, first find the value for z?. Then, replace z? and z into your equation. Expand the equation, and group the real numbers and imaginary numbers separately. From these two groups, you can find the value of a and b. Then, with the value of a and b, replace z = -1 - i, and you should get 0. Upon prooving that, you have prooved that z = -1 - i is a root of the equation.

For question 17, just represent z as x + yi, and it's conjugate z* as x - yi, and expand the equation. Group the imaginary and real numbers separately, and you should be able to get your final answer set for z and z*.

[infested_ysy]

For question 8, first find the value for z?. Then, replace z? and z into your equation. Expand the equation, and group the real numbers and imaginary numbers separately. From these two groups, you can find the value of a and b. Then, with the value of a and b, replace z = -1 - i, and you should get 0. Upon prooving that, you have prooved that z = -1 - i is a root of the equation.

I've figured out question 17, thanks =D

But I'm kinda still stuck with question 8.
I found that z?=2+2i, replaced them, and got myself 2-a+b+2i+ai=0.... now what? Do I replace the "0" at the right hand side with "a+bi"? @<hidden>@<hidden>

edit: Ok, got it already, I need to do something like 2+a=0 and 2-a+b=0

[david_david]

Ok, here's a problem I'm facing (embarassing, to imagine that I've totaly forgotton such basic questions which i've done over and over again in Form 5.... =\)

http://server6.uploadit.org/files/superrobot-scan0001.jpg
Find the value of x.
normally, I would log left and right, but thanks to the annoying 10 multiplier.... I'm stuck! (I've been figuring out the solutions for 20 minutes already, still have come to no avail)

3^2x=2^4x/240

3^2x=4^2x/240

240 =(4/3)^2x

then you can log.......does this way work????

Anybody know where to get STPM Futher maths books...

[Daniel]

For question 8, first find the value for z?. Then, replace z? and z into your equation. Expand the equation, and group the real numbers and imaginary numbers separately. From these two groups, you can find the value of a and b. Then, with the value of a and b, replace z = -1 - i, and you should get 0. Upon prooving that, you have prooved that z = -1 - i is a root of the equation.

I've figured out question 17, thanks =D

But I'm kinda still stuck with question 8.
I found that z?=2+2i, replaced them, and got myself 2-a+b+2i+ai=0.... now what? Do I replace the "0" at the right hand side with "a+bi"? @<hidden>@<hidden>

edit: Ok, got it already, I need to do something like 2+a=0 and 2-a+b=0

2-a+b+2i+ai=0
(2-a+b)+(2+a)i=0
I dont understand why we can put is as 2+a=0 and 2-a+b = 0 in order to find a and b.

[gohweihan]

For question 8, first find the value for z?. Then, replace z? and z into your equation. Expand the equation, and group the real numbers and imaginary numbers separately. From these two groups, you can find the value of a and b. Then, with the value of a and b, replace z = -1 - i, and you should get 0. Upon prooving that, you have prooved that z = -1 - i is a root of the equation.

I've figured out question 17, thanks =D

But I'm kinda still stuck with question 8.
I found that z?=2+2i, replaced them, and got myself 2-a+b+2i+ai=0.... now what? Do I replace the "0" at the right hand side with "a+bi"? @<hidden>@<hidden>

edit: Ok, got it already, I need to do something like 2+a=0 and 2-a+b=0

2-a+b+2i+ai=0
(2-a+b)+(2+a)i=0
I dont understand why we can put is as 2+a=0 and 2-a+b = 0 in order to find a and b.

Because if you look at the equation, 0 is actually 0 + 0i

[infested_ysy]

Another 2 questions which I have no idea how to do it.. :oops:
http://server6.uploadit.org/files/superrobot-mathssssss.jpg

http://server6.uploadit.org/files/superrobot-mathssssssT.jpg

[Daniel]

Another 2 questions which I have no idea how to do it.. :oops:
http://server6.uploadit.org/files/superrobot-mathssssss.jpg

http://server6.uploadit.org/files/superrobot-mathssssssT.jpg

Wow....! I thought i can dealt with complex number since i can finish all the questions in Fajar Bakti book but after seeing the past year questions that infested posted, i realised there are so many questions which i do not know how to solve.... :? :(

[gohweihan]

For Question 22, here's the working. I could not find a way to explain it in pure words.

http://planet.time.net.my/Parliament/gohweihan/Q22.jpg

As for Question 23, just use method in the first part of Question 22, it's about the same.

[infested_ysy]

Another few questions I have problems with...

http://server6.uploadit.org/files/superrobot-mathsThomework.jpg
Q22. What's meant when an integer (in this case, the w) is "totaly imaginary"?

Q23. I've done the first part, but I don't understand what's meant by a "right angled" triangle

http://server6.uploadit.org/files/superrobot-mathsThomework2.jpg

This is embarassing ... :oops:

[pandaboy]

infested_ysy,

For question 22, w is "totally imaginary" means w does not have a real part. It only consists of imaginary part. w = bi

A right-angled triangle is "segitiga bersudut tegak". That means one of the angle in the triangle is 90degrees (right angle).

Hope that helps. :wink:

[Daniel]

http://server6.uploadit.org/files/danielseliong-simcity.JPG
What wrong with these answer? I just could not find the correct answer for this type question....

[W-iNNie]

this question is kinda easy i guess....but juz couldnt get the same answer as the one given,can sumone help me????

8. the roots of an equation are the reciprocal of the roots of the equation x^2 +2ax-c^2=0.Find that equation.

source=pelangi maths S and T ,page 62

thanks in advance

[gohweihan]

Hey people, when you post, please use ALT + 0178 for a square sign (?) or ALT + 0179 (?) for a cube sign...it's easier to read that way...

[wawa]

Daniel, because you have the x below (X-2), you must consider two cases: when X is larger than 2 and when X is smaller then you.

I don't think you can just immediately say that X must be smaller than 2.

considering it that way i find that the answer should be
X<1 and 2<X>11/3

betulkah?? :D

[W-iNNie]

ahahaha....nvm...silly me...managed to find the answer edi

[Daniel]

how to solve this questions
1) |3x+1|+|x|-5=0

[yekban81]

how to solve this questions
1) |3x+1|+|x|-5=0


Rearrange the equation into:
l3x+1l = 5-lxl

and separated them into two equation:
y=l3x+1l
y=5-lxl

Draw out graph for l3x+1l and 5-lxl in the same plot. You may plot them in scale to get direct answer. Otherwise, just plot them schematically and do the solution analytically.

How to plot schematically:
y=l3x+1l

Plot the typical y=3x+1. Modulus sign implies that y>0. Hence, the plot is mirrored at the point when y=3x+1 intersects line y=0. In other words, 3x+1 is mirrored along an axis(x=-1/3) which is parallel to y-axis. Hence, you multiply it with -1 and get y=-3x-1 (this eq. is valid for x<-1/3; obtained from y=3x+1 when y=0, as the mirrored plot of 3x+1 at x=-1/3, so as to represent the modulus plot of 3x+1.

y=5-lxl
Just plot the typical y=-lxl . And then offset the plot vertically +5 point.


You will have two cases to solve; CaseI (x>0) and CaseII (x<0); determined roughly from the plots;two intersections point.


CaseI

y=3x+1
y=5-x
Ans:x=1



CaseII

y=-3x-1
y=5+x
Ans:x=-3/2

[Daniel]

thankx a lot!!

[Daniel]

if i got y=6+|x|, is that mean y=6+x and y=6+(-x) ??

[yekban81]

if i got y=6+|x|, is that mean y=6+x and y=6+(-x) ??

y=6+x for x>0
y=6-x for x<0

Just to add some variety to the previous question:

Solve l3x+1l+5-lxl > 0

[Daniel]

y=6+x for x>0
y=6-x for x<0

why i must put x>0 and x<0 ?
By the way, if x^4 = 16, x^2 should be 4 or -4 and +4 ??

[yekban81]

y=6+x is only valid for x>0
y=6-x is only valid for x<0

so as to represent this equation y=6+lxl

For x^4=16
x^2= 4 or -4
x=2,-2, or 2i

[Daniel]

Can someone shows me the step for this questions? the answer is +-(1-i) but i keep on getting the same answer which is +-(i-1)

*Express the square roots of -2i in the form +-(a+bi), where a and b are real numbers. Thankx

[david_david]

y=6+x is only valid for x>0
y=6-x is only valid for x<0

so as to represent this equation y=6+lxl

For x^4=16
x^2= 4 or -4
x=2,-2, or 2i




can x be -2i also??coz -2i times -2i is also -4

i think it makes sense rite, coz equations to the 4th power should have 4 roots( real and unreal)

so x=2,-2,2i, and -2i

sorry if tis is a mistake, never really leared complex numbers b4,

[david_david]

Can someone shows me the step for this questions? the answer is +-(1-i) but i keep on getting the same answer which is +-(i-1)

*Express the square roots of -2i in the form +-(a+bi), where a and b are real numbers. Thankx

I think +-(1-i) and +-(i-1) are the same.

coz +-(a-b) and +-(b-a) are the same.


I think that u equate and then square both sides.

-2i=a^2+2abi-b^2

then u get ab=-1
a^2-b^2=0

solve and then

a=1,-1,i,-i
b=-1,1,i,-i....in that order of sets

[david_david]

http://server6.uploadit.org/files/danielseliong-simcity.JPG
What wrong with these answer? I just could not find the correct answer for this type question....

just multiply both sides by (x-2)^2

then u got cubic ,

(3x-11)(x+1)(x-2) less then zero

draw the graph, less headache.

[yekban81]

can x be -2i also??coz -2i times -2i is also -4

i think it makes sense rite, coz equations to the 4th power should have 4 roots( real and unreal)

so x=2,-2,2i, and -2i

sorry if tis is a mistake, never really leared complex numbers b4,


Yes, you are right. For equation to the n-th power, there are total of n roots (real and unreal) as solution.

[bill]

yekban81,u live in scudai?den which is ur sch n taman tat u live in?i m oso live there...can visit n ask u some question when u free?

[david_david]

anybody know good calculus books..... esp da ATU folks coz u ppl take cal in ADFP..

sorry , might be out of the topic

[yekban81]

yekban81,u live in scudai?den which is ur sch n taman tat u live in?i m oso live there...can visit n ask u some question when u free?

Yes, I live in Taman Pulai Utama. Currently I am studying in UTM. What would you like to know?

[bill]

oic...den which subject tat u chose?engineering ?acturial science?

[Daniel]

Help! i am in trouble again...
1) The sum of 5 consecutive terms of an arithmetic progression is 30 and the sum of the squares of these terms is 220. Find the terms.

[gohweihan]

Help! i am in trouble again...
1) The sum of 5 consecutive terms of an arithmetic progression is 30 and the sum of the squares of these terms is 220. Find the terms.

I would recommend expanding the sum into the individual units so as to see this

30 = a + ad + 2ad + 3ad + 4ad
220 = a? + (ad)? + (2ad)? + (3ad)? + (4ad)?

Simplify, and you get

30 = a(1 + 10d)
220 = a?(1 + 30d?)

Here on, you can calculate the answer...

I got a = 10, d = 0.2 and a = 50/13, d = 17/25

From here, you should be able to calculate all of the 5 terms.

[Daniel]

Thank you Gohweihan. You are helping me a lot in solving mathematic's questions since i join recom. :wink:

[gohweihan]

Thank you Gohweihan. You are helping me a lot in solving mathematic's questions since i join recom. :wink:

Glad to be able to help...

[Daniel]

A geometric series has first term 1 and the common ratio r is positive. The sum of the first 5 terms of a G.P is twice the sum of the terms from the 6th to the 15th inclusive.

Show that r^5 = 1/2 ( 3^0.5 -1 )

Thank you :wink:

[gohweihan]

A geometric series has first term 1 and the common ratio r is positive. The sum of the first 5 terms of a G.P is twice the sum of the terms from the 6th to the 15th inclusive.

Show that r^5 = 1/2 ( 3^0.5 -1 )

Thank you :wink:

Use the formula to calculate the sum of the first 5 and the first 15 terms.

Since the sum of the 6th to the 15th term (inclusive) is also the sum of the first 15 terms minus the sum of the first 5 terms, you should be able to deduce the following:

∑ (first 5 terms) = 2 [ ∑ (first 12 terms) - ∑ (first 5 terms) ]

After putting in the respective formulas, you should get something like:

1 - r^5 = 2 [ r^5 - r^15 ]

At this point, it is wise to symbolize r^5 with x, so that

1 - x = 2(x - x?)

Bring everything over to the left, and you get

2x? - 3x + 1 = 0

Using the methods taught to you to find the roots of equation (replacing the x value), you should be able to get a root x = 1.

After that, by comparing

(x - 1)(ax + bx? + c) ≡ 2x? - 3x + 1

you should be able to find the values for a, b and c, thus creating the following equation

(x - 1)(2x + 2x? - 1) = 0

Use the formula

[ -b ? √(b? - 4ac) ] / 2a

where a = 2, b = 2 and c = -1. You should then

x = [ -2 ? √12 ] / 4

which, after rearranging (√12 = 2√3), is also the same as

x = [ -1 ? √3 ] / 2

But x = [ -1 - √3 ] / 2 is not valid because it would be less than 0. Remeber that x = r^5, and when r^5 < 0, r < 0.

So you are left with

x = [ -1 + √3 ] / 2

You rearrange a little, and replace x = r^5, and you would get

r^5 = 1/2 [ √3 - 1 ]

which is what you want to prove.

[Daniel]

Thank you :D

[gohweihan]

Ironically, I'm having trouble with these things in my course's maths.

[Daniel]

1)Given that k is a positive integer:
a)Find, in terms of k, an expression for S1, which is the sum of the integers from 2k to 4k inclusive.
b)Fine, in terms of k, an expression for S2, which is the sum fo the odd integers between 2k and 4k.

2)Show that the sum of the odd numbers from 1 to (2n-1) inclusive, is n^2. show that the sum of the positive odd numbers less than 1002 that are not divisible by 3 is 6 x 167^2

[yekban81]

Hi, Daniel

Here is an optional solution for Q1:

For part a:

We have 1,2,3,4,...2k,...4k where a=1, d=1, total n=4k

S=n/2[2a+(n-1)d] make sure you understand how this formula is originated.

S1=S(4k)-S(2k-1) we put (2k-1) since 2k is included in the sum

S(4k)=4k/2[2+(4k-1)]
S(2k-1)=[(2k-1)/2][2+(2k-1-1)]

Finally we get the answer: S1=3k(2k+1)

For part b:

We will utilise the answer in part a to ease the solution for this part

S2=S1-S(all even integer)

For even integer, we have : 2k, 2k+2, 2k+4,....4k-2, 4k

where a=2k, d=2, total n = [(4k-2k)/2] +1, L=4k (the last integer)

S=n/2[2a+(n-1)d]
=n/2[a+a+(n-1)d] *L=a+(n-1)d
=n/2[a+L]
=n/2[2k+4k]
Remember that we have n in terms of k.

S2=[3k(2k+1)]-n/2[6k]

And we get the answer: S2=3k^2

if you do not utilise the answer in part a, you still can get the solution in the same way as used in part a, but it will be more messy in dealing with the unknown.

If you understand the method of solution for this two parts, you will be able to solve the Q2 as well.

[Daniel]

Thank you Yekban81. By the way, i still do not understand why we should place 1,2,3,4,...,2k,....4k ?? Why not 1,3,5,..... or 1,5,9,..... etc... ? :?:

[yekban81]

Hi, Daniel

For part a, I have two different approach of solution. The first one which I have shown uses integer set : 1,2,3,4...2k,....4k
The sum of integers from 2k to 4k inclusive means
S1=(2k) + (2k+1) + (2k+2) + ....+ (4k-2) + (4k-1) + (4k)

from this definition, we find that d=1. In addition to the integer set which I used above, a=1.

S1= S (sum from integer 1 to 4k) - S (sum from integer 1 to 2k-1)

Another alternative, the second approach, we use this integer set:

2k, 2k+1, 2k+2,....4k-1, 4k

From this set, we still have the same d=1 but a=2k
The total of n = (4k-2k) + 1

Hence, S1= (n/2)[2(2k)+(n-1)(1)]
by replacing n in terms of k
we get the answer.


For part b, you may use this odd integer set; 1,3,5,7,9,...,2k-1,2k+1,.....,4k-3,4k-1
as another approach to solve the question.

From that integer sets, a=1, d=2, L=4k-1
total n= 4k/2=2k

S1= S (sum from integer 1 to 4k-1) - S (sum from integer 1 to 2k-1)


Another alternative, we use odd integer set:
2k+1,2k+3, 2k+5,....4k-3,4k-1

where a=2k+1, d=2, L=4k-1 and total n=(4k-2k)/2
with this variable in terms of k, you can easily get the answer.

The previous solution uses the answer in part a for solution. I prefer to use this method.

I guess you may wonder how I produce the formula for the total n.
I got this formula by trial and error method + experience. Just test the formula with k=1,2,3.

Hope you understand.

[seiken]

Hi..
Im new to this forum. Currently studying form 6 also. I have a big problem understanding Chapter 3 Series. Especially the method of differences. Is it the same as shown in Federal Study Aids...in the given example. It uses the method of cancelling each other...

[chenchow]

seiken, may be you can share with us what you understand and what you don't. Many of us here would not have that particular textbooks.

[seiken]

Im not sure about this ques. I can do the 1st part only.Im now doing a revision on this chapter since its holiday...
Its like this:
Prove that for all values of m, the line y=mx+a/m is a tangent to the parabola y?=4ax. Hence find a quadratic equation whose roots are the gradients of the tangents from the point (x',y') to the parabola. Prove that, if the tangents from a point P to the parabola are inclined at an angle k to one another, P must lie on the curve with equation (x+a)? tan?k = y? - 4ax

[seiken]

And this:
Find the equation of the tangent to the elipse from the point (x1,y1)
Deduce that the line lx + my + n = 0 is a tangent to the elipse if a?l? + b?m? = n?, and find the coordinates of the point of contact if this condition is satisfied. Find the point on the elipse 4x?+9y?=1 where the tangent is parallel to the line 8x=9y

[chenchow]

Seiken, instead of directly type out the question and ask ReComers on how to solve, tell us in each question, what you understand, what you don't understand, how much you know and we would guide you, instead of directly telling you how to do. Is that ok?

[seiken]

And also this:
A variable point P lies on the curve y?=x? and is joined to a fixed point A with coordinates (2,0). Prove that the locus of the mid-point of AP is y?=2(x-1)?

[yekban81]

Yeah, I agree with Chenchow that anyone who ask for help in this forum, should clearly state out what you understand, what you dont understand, and so forth instead of just putting up question and expecting a detailed solution in return.

Seiken,

For your third question, you can solve it based on this:
Distance from A to midpoint = Distance from midpoint to the curve

Use pythagoras teorem for the left component. For the right component, If I am not mistaken, there is a formula that computes the distance from a fixed point to other point lying on a curve.

For your second question, I don't think it's clearly stated. Which ellipse does it mean? But for sure, you may use the gradient of the ellipse,m (dy/dx of the ellipse equation) and the point(x,y) to get the linear/tangent equation.

About your first question, first part is just some tricky manipulation of equation. The second part should be of no problem. For the third part, the clue is that tan k = m (gradient). Try to solve it.

[EricFu]

I have solved and posted my solution. Please tell me if my approach is wrong.

http://www.umich.edu/~ericchfu/recom/loci.jpg

[yekban81]

That's a smart approach.
To find the equation of a curve, where M(alpha, beta) lies in, is to find the equation that relates alpha and beta ;M(alpha,beta)
Alpha and beta can be written respectively in terms of y and x based on the condition that M is midpoint from A(0,2) to P(x,y).
In this approach, the equation of relating alpha and beta is derived from the relationship of y^2=x^3.

[seiken]

Hey...
Thanks..
I really appreciated it..
Actually the questions are from the pelangi book.
I think some of the questions there rather hard...especially chapter 3...
Anyway.im trying my best to complete all the questions...

[wawa]

practice makes perfect, jia you!

[Daniel]

Help!
Integrate these:-
1) 1/(1-sin x)
2) 1/(1+cos x +sin x)
3) sin 3x cos 5x

Hope someone can help me with the questions above. I have wasted a lot of time with these questions. Thankx ya.

[EricFu]

Good questions! So far I have solved 2 of them, namely Question 1 and Question 3. I am still figuring out the solution for Question 2.

http://www.umich.edu/~ericchfu/recom/integration.JPG

[seiken]

Wow...that's an amazing way to do it!! 8O
The first question I understand....but question 3???
How come sin3x.cos5x can be changed to sin8x - sin2x??
Please enlighten me.....

[youngyew]

Wow...that's an amazing way to do it!! 8O
The first question I understand....but question 3???
How come sin3x.cos5x can be changed to sin8x - sin2x??
Please enlighten me.....
This is Product-to-Sum formulae, it's in the syllabus.
http://www.sosmath.com/trig/Trig5/trig5/img10.gif

Anyway, Eric, you missed the half there. :wink:

Kindly refer to the trigonometric identities here:
http://www.sosmath.com/trig/Trig5/trig5/trig5.html

[EricFu]

Haha :D It has been a while since I last use this identity... So I am really sorry for that careless mistake. Thanks to Chang Yang who pointed it out...

By the way, Chang Yang, any idea for Question 2?

[yekban81]

If I am not mistaken, question 2 can be solved with the help of a dummy equation of t=tan(theta/2), hence by using:
sin(theta) =2sin(theta/2)cos(theta/2)
cos(theta)=2(cos(theta/2))^2 - 1
you get:
sin(theta)=2t/(1+t^2) and
cos(theta)=(1-t^2)/(1+t^2).

I remember I had learnt by heart all these trigo identity and their transformation equations..I forgot all of them already by now

[yekban81]

Just now try to solve question 2 out of my curiousity.
When you substitute the "dummy equation" into the integral equation, and simplify it, you shall get:

integral [ 1/(1+t) ] dt
hint:
dt = 2/(1+t^2) d(theta); can be derived from t=tan(theta/2)

solve it and then replace back to theta parameters. You will get:

= In [1+tan(theta/2)] + c

[Daniel]

1)If tan x= 4/3 and x is acute, find the value of tan x/2
and finally i got two answers which is 1/2 and -2. Which one is the correct answer? how to determine it then? thanks

[youngyew]

1)If tan x= 4/3 and x is acute, find the value of tan x/2
and finally i got two answers which is 1/2 and -2. Which one is the correct answer? how to determine it then? thanks
Since x is acute, so 1/2 must be the correct answer since the tangent of an acute angle is positive. (check out elementary trigonometry for the related info)

[Daniel]

Thank you youngyew
1)Transform each of the following expressions into the compund angle form suggested:-
a) cos x + 3 sin x r cos (x+y)

2)Find the maximum and minimum values of the following functions, stating in each case the values (0 to 360 ) of x at which the turning points occur:
a) 7cos x - 24sin x + 3

[yekban81]

2)Find the maximum and minimum values of the following functions, stating in each case the values (0 to 360 ) of x at which the turning points occur:
a) 7cos x - 24sin x + 3

You may utilize the fact that dy/dx = 0 for both max and min point.
and then d^2y/dx^2<0 for max point and >0 for min point

and remember to utilize tan x = sin x / cos x to solve any resulted equation of dy/dx = 0

[EricFu]

This is how I solved it:

http://www.umich.edu/~ericchfu/recom/trigo.JPG

[seiken]

This Harmonic form is not in STPM Maths T syllabus right?
I never seen it before...

[EricFu]

I think it is in the syllabus. I remembered reading it from Ong Beng Sim's Pure Math (Malay version) textbook when I was in Form 5. But perhaps the syllabus has changed... I am not sure though.

However, I do not think it is something really "new" right? What you need to do is apply the double angle formula, and you are all set!

[Daniel]

1)The equations of the sides of a triangle are x=0, 2x+y=8 and 3x-y+3=0 .Find the equations of the altitudes of the triangle Show that the altitudes meet at a point and find its coordinates.

2)A point P moves such that its distance from the point (-3,0) is k times its distance from the point (3,0). If k>1, show that the locus of P is circle of radius 6k/{(k^2) ? 1}

3)If (h,k) is a fixed point, find the locus of the point (h+rcosy, k+rsiny) when:
a) r is a constant and y varies

[seiken]

Have you guys heard of Olimpiad Matematik Kebangsaan and joined it before? This is my first time joining it as a Form6 student. Do I need to learn anything other that inside the STPM syllabus? Like those out of this world Maths formula?

[youngyew]

Have you guys heard of Olimpiad Matematik Kebangsaan and joined it before? This is my first time joining it as a Form6 student. Do I need to learn anything other that inside the STPM syllabus? Like those out of this world Maths formula?

Basically learning STPM syllabus won't help much in solving olympiad problems. The style of olympiad problems require other set of strategy and tools which are quite dissimilar with the one we learn in our curriculum.

Before I begin, one of our recommers prince (Shien Jin) has an excellent page and introduction about OMK. It can be found at http://www.eecs.harvard.edu/~shienjin/math.html

For recent years there have been a pattern for the OMK questions, especially for Sulung level. Firstly, there have been almost one inequality question for each year, and you have to learn the relationship between Arithmetic Means, Geometric Means and Harmonic Means in order to solve this kind of question.

Secondly, there is also always at least one question for number theory. Although the question setter seem to avoid the necessity of "specific" formula or knowledge as they set the question, but it would always ease the work if you learn a bit of modulo arithmetic. For example, try this past year question: prove that the sum of two odd square numbers can never be a square number. The formal solution provided by the PERSAMA (the organiser) used the parity (odd or even nature of a number) to provide a proof, but it's too lengthy and is not elegant. Modulo arithmetic can solve the question in 4 lines.

For geometry most questions can always be solved with constructing perpendicular triangles and use some clever application of cosine's rules, heron's formula, sin rules and all the things you have learnt in SPM level. Master everything related to a circle. (the angle of a tangent, common tangent of two circles, angles inside the circle, concyclic quadrilaterals etc) Even pythagoras' theorem is essential and sometimes the most important (and overlooked) clue. However, learning additional geometry formulas can help you if you manage to empower them. To start with, learn ptolemy theorem, power of a point, ceva's theorem, melanaus theorem. These are the essential ones for advanced geometry.

For combinatorics, have a concept about pigeonhole's principle. The principle is a common sense actually, but just make sure you do understand it throughout. Besides, have your skills honed to figure out how many ways there are to choose m balls from n balls etc. Of course it would be a bit more convulated than this.

For functions, make sure you know the real meaning of "prove" and "find all functions". If you plug in a few examples and showed that they satisfy the given situation it's useless as a proof; and if you found a function that fit the given premises, you also wouldn't score for the question which require "find all function".

That's it for OMK. If you were to go further for IMO etc then you would need much more than what i have stated. To sum it up, try to think out of the box, be open to learn from mistakes, and read books which suit your level to polish your skill. I believe that people with interest and with some knack would pave their way to make a mark in the OMK. It seemed hard but it's possible to achieve something if you know the way and work the right way.

Hope this helps.

[afmb]

Got this URL fr a friend. U can find past yr Math STPM questions here. Just scroll down 2wards the end k!! Do u guys have past yr questions 4 other subjects?? Hope we can share...




http://www.geocities.com/CollegePark/Stadium/3986/MSMT-STPM/

[swgouk87]

Ok, here's a problem I'm facing (embarassing, to imagine that I've totaly forgotton such basic questions which i've done over and over again in Form 5.... =\)

http://server6.uploadit.org/files/superrobot-scan0001.jpg
Find the value of x.
normally, I would log left and right, but thanks to the annoying 10 multiplier.... I'm stuck! (I've been figuring out the solutions for 20 minutes already, still have come to no avail)


OK...i start from the beginning...
10(3^<2x+1>) = 2^<4x-3>
(3^2x)(3) = (2^4x)/80
3^2x = (2^4x)(240)
240 = (2^4x)/(3^2x)
240 = (16/9) ^x
log 240 / ( log 16-log 9 )=x
hence,

x= 9.526

U can understand??[/img][/list][/code]

[swgouk87]

Help!
1) Find the cube of x^1/3 + y^1/3 and show that if x^1/3 + y^1/3 = z^1/3 , then (x+y-z)^3 + 27xyz = 0

2)Given z = x + iy, and |z+2| = 2 |z-i|, show that 3x^2 + 3y^2 -4x -8y = 0

3)the complex number z satisfies the equation |z|=|z+2|. Show that the real part of z is -1

Thank you

FOR QUESTION 3..

Let Z= x+yi
|z| = |z+2|
|x+yi| = |x+yi+2|
|x+yi| = |(x+2)+yi|
modulus
root square(x?+y?) = root square(<x+2>?+y?)
Square it,
x?+y? = x?+4x+4+y?
4x+4 = 0
4x = -4
x = -1

Since x is real part of z, and equals to -1,
hence proven that real part of z is -1

Can U understand?

[Jia0924]

Hi, I need your help..Here is my question:

Given that 2y=a^x + a^-x , where a>1 and x>0 , prove that a^x = y + √(y^2 - 1).
Similarly, if 2z = a^3x + a^-3x, prove that z= 4y^3 - 3y.

I don't know how to prove that second one..pls help me.. :?

[PINKPIGGY]

can someone help me to solve this?~surds

(x^2 - {2x+1}^1/2)^1/2=2-x

[pandaboy]

can someone help me to solve this?~surds

(x^2 - {2x+1}^1/2)^1/2=2-x

Just square both sides....can ar?
I think you can solve it after that...just expand and square it again....then solve quadratic equation...

Haven't done proper maths for a while..hope I'm not wrong...

What's the answer btw?

[nicodemus]

Can someone help me to solve this? Indices...

Show that (2 + root 5)^4 + (2 - root 5)^4 is an integer
and find the value of the integer.

The books method is to substitute the numbers with algebra, but i don't know how to simplify

a^4 + b^4

Can someone tell me the way to solve this?

[youngyew]

Can someone help me to solve this? Indices...

Show that (2 + root 5)^4 + (2 - root 5)^4 is an integer
and find the value of the integer.

The books method is to substitute the numbers with algebra, but i don't know how to simplify

a^4 + b^4

Can someone tell me the way to solve this?
What is so special with (2 + surd 5) and (2 - surd 5)? Surely they are conjugate of each other, so you have a sum of 4 and product of -1.

So now we have a^4 and b^4, which doesn't relate to sum and product at all. But wait!! Did I mention binomial theorem? Remember (a+b)^4??

So now work it out.

(a + b)^4
= a^4 + 4a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4
= a^4 + b^4 + 2ab [2a^2 + 3ab + 2b^2]
= a^4 + b^4 + 2ab [ 2(a+b)^2 - ab ]

See?? Now we have every thing else in terms of (a+b) and ab. Then you can rest.

Although this method works, I don't see it as elegant. Anyone with a better idea of doing the same question?

[PINKPIGGY]

(x^2 - {2x+1}^1/2)^1/2=2-x

Thanx, panda boy.

The ans: x=3/2 :oops:

[PINKPIGGY]

How bout this Q:
If surd(11+6surd2)= surdp+surdq,
find the value of p and q.

Ans:p=9,q=2@<hidden> p=2,q=9

Can anybody help me to solve this? :oops:
Thanx!

[littlebigone]

I think this is how you do it:

square both sides:

11+ 6surd(2) = p + q + 2*surd(pq)

Then comparing the left hand side and the right hand side you get 2 equations:

11 = p + q
6*surd(2) = 2*surd(pq)

6*surd(2) = 2*surd(18)
so you get pq = 18
solve the two equations and you get the answers.

[littlebigone]

Although this method works, I don't see it as elegant. Anyone with a better idea of doing the same question?

(2 + root 5)^4 + (2 - root 5)^4
= (9 + 4*root 5)^2 + (9 - 4*root 5)^2

let 9 = a and 4*root 5 = b

we know our quadratic formulas very well right, so:

(a + b)^2 = a^2 + b^2 + 2ab
(a - b)^2 = a^2 + b^2 - 2ab

so the sum above becomes 2a^2 + 2b^2

2a^2 = 2*9^2 = 162
2b^2 = 2*(root 90)^2 = 180

the integer = 342

[youngyew]

Although this method works, I don't see it as elegant. Anyone with a better idea of doing the same question?

(2 + root 5)^4 + (2 - root 5)^4
= (9 + 4*root 5)^2 + (9 - 4*root 5)^2

let 9 = a and 4*root 5 = b

we know our quadratic formulas very well right, so:

(a + b)^2 = a^2 + b^2 + 2ab
(a - b)^2 = a^2 + b^2 - 2ab

so the sum above becomes 2a^2 + 2b^2

2a^2 = 2*9^2 = 162
2b^2 = 2*(root 90)^2 = 180

the integer = 342
Cool, certainly elegant... but it still involve manual multiplication of the surd, is there anyway to manipulate the algebra so that we don't have to do the tedious multiplation of surds?

[littlebigone]

what do you mean manual multiplication of surds?

[youngyew]

what do you mean manual multiplication of surds?
The part where you transform (2 + surd 5)^4 to (9 + 4 surd 5)^2

[littlebigone]

ahhhhh??!?

it's just like the (a + b)^2 thing what...
becomes a^2 + b^2 + 2ab
2^2 = 4
(root 5)^2 = 5
2*2*root 5 = 4* root 5

i really don't know how to make it easier...but for your sake i will try

[EricFu]

Show that n! + 2004 is not a perfect square.

[youngyew]

Show that n! + 2004 is not a perfect square.
Mathematics T?

Given that 7(8^p)=9(5^q) and 7(16^{p+1})=12(5^q) show that 2^p=1/12

Anyone?? help??

[yen_05]

Given that 7(8^p)=9(5^q) and 7(16^{p+1})=12(5^q) show that 2^p=1/12

Anyone?? help??

Anyone...pleze help...the guest is actually me...hehe...forget to log in....

[yen_05]

pheww...got it...after the whole night of trying....it is actually very simple....maybe my brain is a bit rusty... so it take me so long to come out with the solution...

so juz ignore this question.... :oops: :oops:

[gohweihan]

Given that 7(8^p)=9(5^q) and 7(16^{p+1})=12(5^q) show that 2^p=1/12

Anyone?? help??

Anyone...pleze help...the guest is actually me...hehe...forget to log in....

http://photos1.blogger.com/img/174/2140/1024/MAT-T-SOL-0001.jpg

Being able to recall by heart from 2 to the power of 0 to 2 to the power of 30 really helps :D

[yen_05]

thanx,weihan for the advice......but i manage to remember up to 2 to the power of 12 only....but i will try my best to remember up to 2 to the power of 30 :D

[PINKPIGGY]

Prove
1/(1+x^{a-b}+x^{a-c})+1/(1+x^{b-c}+x^{b-a})+1/(1+x^{c-a}+x^{c-b})=1

Can somebody help me to solve this Q?

Thx!

[youngyew]

Prove
1/(1+x^{a-b}+x^{a-c})+1/(1+x^{b-c}+x^{b-a})+1/(1+x^{c-a}+x^{c-b})=1

Can somebody help me to solve this Q?

Thx!
Let x^(a-b) = m and x^(a-c) = n.

Then transform the whole expression in terms of m and n, you will soon arrive at the answer in 3 steps.

[scorpstreet]

Can someone please help me with Binomial Theoram? I'm never good at this chapter though...-_-"


By using Binomial Theorem, find (√2 +1)^5 in the form
of (a√2+ b) where a and b are integers.

Write down the value of (√2 - 1)^5 in the similar form.

[bush]

Can someone please help me with Binomial Theoram? I'm never good at this chapter though...-_-"


By using Binomial Theorem, find (√2 +1)^5 in the form
of (a√2+ b) where a and b are integers.

Write down the value of (√2 - 1)^5 in the similar form.

just use the binomial expansion formula for positive interger.....

the answer should be 41+29(root 2).........

for the second part, just replace the minus sign alternately in ur long expansion and u will get the answer..................

[scorpstreet]

ok thanks for the info Bush.

[bush]

....welcome

[jackfook]

Given that P#Q=P+Q+PQ,state whether the binary operation is associative or not.
(this is the question I took from real numbers-numbers and sets,math T)

[scorpstreet]

Another question from me. ^^"

Show that the equation (x-a)(x-b)=k^2, where a#b and k#0 has two distinct real roots for all real values of a,b and k.

(a) both roots cannot take values between a and b.
(b) both roots are positive if a,b and ab-k^2 are positive

Please can anyone show me how to do it? coz i mainly don't understand what the question wants me to do...-.-" Thanks for the help.

[Nelson]

Another question from me. ^^"

Show that the equation (x-a)(x-b)=k^2, where a#b and k#0 has two distinct real roots for all real values of a,b and k.

(a) both roots cannot take values between a and b.
(b) both roots are positive if a,b and ab-k^2 are positive

Please can anyone show me how to do it? coz i mainly don't understand what the question wants me to do...-.-" Thanks for the help.

I think the solution is like this :

Given that (x-a)(x-b)=k^2
hence x^2-(a+b)x+ab=k^2
x^2-(a+b)x+ab-k^2=0

By using b^2-4ac, where a=1, b=-(a+b), c=ab-k^2
[-(a+b)]^2-4(1)(ab-k^2)
=a^2+2ab+b^2-4ab+4k^2
=a^2-2ab-b^2+4k^2
=(a-b)^2+4k^2

Since (a+b)^2 and 4k^2 are always positive,
Hence b^2-4ac>0
Therefore when b^2-4ac>0, the equation (x-a)(x-b)=k^2 has two distinct real roots for all real values of a,b and k.

[Nelson]

Could anyone please help me :

Given that z=x+yi and w=(z+8i)/(z-6), z≠6. If w is totally imaginary, show that x^2+y^2+2x-48=0

The matter is I cannot prove it :oops: .

[bush]

Could anyone please help me :

Given that z=x+yi and w=(z+8i)/(z-6), z≠6. If w is totally imaginary, show that x^2+y^2+2x-48=0

The matter is I cannot prove it :oops: .

if w is totally imaginary, then the real part of w is 0..........

i couldn't get ur answer.......

i have 0=x^2+y^2-6x+8y..............

[yen_05]

Could anyone please help me :

Given that z=x+yi and w=(z+8i)/(z-6), z≠6. If w is totally imaginary, show that x^2+y^2+2x-48=0

The matter is I cannot prove it :oops: .

if w is totally imaginary, then the real part of w is 0..........

i couldn't get ur answer.......

i have 0=x^2+y^2-6x+8y..............

Yeah....i got the same answer as bush....because w is totally imaginery, so we can equate the real part with 0...but can't get that answer.....cannot equate the imaginery part because we do not know the coefficient for i in the w part.....

[lyt87]

Can anyone help?
How to Prove:
i)arc tan x + arc cot x = 90 degree
ii) arc tan x + arc tan y = arc tan [(x+y)/(1-xy)]
n how to draw graph for arc cot x

[bush]

Can anyone help?
How to Prove:
i)arc tan x + arc cot x = 90 degree
ii) arc tan x + arc tan y = arc tan [(x+y)/(1-xy)]
n how to draw graph for arc cot x

kimsiang proved it in education forum......

draw a cot x graph...........then find its inverse by relecting onto x=y

[Nelson]

Could anyone please help me :

Given that z=x+yi and w=(z+8i)/(z-6), z≠6. If w is totally imaginary, show that x^2+y^2+2x-48=0

The matter is I cannot prove it :oops: .

if w is totally imaginary, then the real part of w is 0..........

i couldn't get ur answer.......

i have 0=x^2+y^2-6x+8y..............

Yeah....i got the same answer as bush....because w is totally imaginery, so we can equate the real part with 0...but can't get that answer.....cannot equate the imaginery part because we do not know the coefficient for i in the w part.....

Hi bush and yen_05, thank you for your replies. The question should be correct as I extracted it from Pelangi Mathematics book. I have tried number of times but still failed to prove the equation as I got the same answer as you did.

Anyway I got another mathematical problem here :

The roots of the quadratic equation x^2+px+q=0, where q≠0, are α, β and one root of the quadratic equation x^2+p'x+q=0 is kα. Show that the other root of the equation is β/k.

By assuming that k^2≠1, write down an expression for the sum of the roots for each of the quadratic equations. Hence, find α and β in terms of p, p' and k. Deduce that k(kp-p')(kp'-p)=[(k^2-1)^2]q

For any p, p' and q, show that the sum of the four possible values of k is (pp')/q.

I have done the upper part just left the last part.

Any reply for both questions would be appreciated.

[scorpstreet]

Another question from me. ^^"

Show that the equation (x-a)(x-b)=k^2, where a#b and k#0 has two distinct real roots for all real values of a,b and k.

(a) both roots cannot take values between a and b.
(b) both roots are positive if a,b and ab-k^2 are positive

Please can anyone show me how to do it? coz i mainly don't understand what the question wants me to do...-.-" Thanks for the help.

I think the solution is like this :

Given that (x-a)(x-b)=k^2
hence x^2-(a+b)x+ab=k^2
x^2-(a+b)x+ab-k^2=0

By using b^2-4ac, where a=1, b=-(a+b), c=ab-k^2
[-(a+b)]^2-4(1)(ab-k^2)
=a^2+2ab+b^2-4ab+4k^2
=a^2-2ab-b^2+4k^2
=(a-b)^2+4k^2

Since (a+b)^2 and 4k^2 are always positive,
Hence b^2-4ac>0
Therefore when b^2-4ac>0, the equation (x-a)(x-b)=k^2 has two distinct real roots for all real values of a,b and k.

Thanks a tonne Nelson! ^^ Greatly appreciated!

[PINKPIGGY]

Polynomials:

1. If alpha and beta are the roots of the equation x^2 + 2ax+b=0, show that (alpha-beta)^2=4(a^2-b). Express the roots of the quadratic equation bx^2+2a(b+1)x+(b+1)^2=0 in terms of alpha and beta in its simplest form.

Can someone help me to solve the 2nd part of the Q? Thx!

2. Given that x=2 is a root of the equation ax^2-bx-c=0, state the relationship among a,b,and c. Therefore, show that x=2 is also a root of the ea=quation bx^2+cx-8a=0. If a and beta are the other root of the first and second equation above respectively, show that 4+(a+2)beta=0. Hence, or otherwise, show that k<=(smaller or equal to) 1/4 if a=kbeta and a and beta are real.

Yup, i cab't solve the bolded part too! Help...Thx!

3. Given that alpha+2 and beta+1 are the roots of the equation x^2-qx+r=0, while alpha and beta are the roots of the equation 2x^2-bx+c=0 where alpha, beta are real and alpha>= (larger or equal to)beta, find q and r in terms of b and c.
In the case alpha=beta, show that q^2=4r+1.

[Ans:q=1/2(b+6), r= 3b/4 + c/2 + 2-{square root of (b^2-8c)}/4]

I can't prove the r (beta) one, pls help. Thx a lot!! :oops:

[IchimaruGin]

Can u help me solve this question....i could not find the solution for it...its from the topic sequences and series

find the sigma from r=1 to 20 of (-1)^r r^2


ans:210

[Nelson]

Polynomials:

1. If alpha and beta are the roots of the equation x^2 + 2ax+b=0, show that (alpha-beta)^2=4(a^2-b). Express the roots of the quadratic equation bx^2+2a(b+1)x+(b+1)^2=0 in terms of alpha and beta in its simplest form.

Can someone help me to solve the 2nd part of the Q? Thx!


Is this the solution?

To find the roots,
we use x=[-b?√(b^2-4ac)]/2a where a=b, b=2a(b+1), c=(b+1)^2
x=[-2a(b+1)?√(2a(b+1))^2-4(b)((b+1)^2)]/2(b)
=[-2a(b+1)?√(4a^2(b+1)^2)-4(b)((b+1)^2)]/2(b)
={-2a(b+1)?√[4(a^2-b)(b+1)^2]}/2(b)----------------(1)

Given that α and β are the roots of the equation x^2+2ax+b=0
Hence,
Sum of roots, α+β=-2a-----------(2)
Product of roots, αβ=b------------(3)
Given also that (α-β)^2=4(a^2-b)-----------(4)
Substituting (2), (3) and (4) into (1),
x={(α+β)(αβ+1)?√[(α-β)^2][(αβ+1)^2]}/2αβ
=[(α+β)(αβ+1)?(α-β)(αβ+1)]/2αβ

x=[(α+β)(αβ+1)+(α-β)(αβ+1)]/2αβ or x=[(α+β)(αβ+1)-(α-β)(αβ+1)]/2αβ
x=[(αβ+1)(α+β+α-β)]/2αβ or x=[(αβ+1)(α+β-α+β)]/2αβ
x=[2α(αβ+1)]/2αβ or x=[2β(αβ+1)]/2αβ
x=α+1/β or x=β+1/α
Hence the roots are x=α+1/β and x=β+1/α

[Nelson]

Polynomials:

2. Given that x=2 is a root of the equation ax^2-bx-c=0, state the relationship among a,b,and c. Therefore, show that x=2 is also a root of the ea=quation bx^2+cx-8a=0. If a and beta are the other root of the first and second equation above respectively, show that 4+(a+2)beta=0. Hence, or otherwise, show that k<=(smaller or equal to) 1/4 if a=kbeta and a and beta are real.

Yup, i cab't solve the bolded part too! Help...Thx!



The propose solution:

Given that a and 2 are the roots of the equation ax^2-bx-c=0
Hence,
Sum of roots, a+2=b/a-----------(1)
Product of roots, 2a=-c/a-------------(2)
Given also that β and 2 are the roots of the equation bx^2+cx-8a=0
Hence,
Sum of roots, β+2=-c/b-----------(3)
Product of roots, 2β=-8a/b------------(4)
From (2), 2a=-c/a
-c=2a^2-----------(5)
From (3), β+2=-c/b
-c=b(β+2)-----------------(6)
Comparing (5) and (6),
2a^2=b(β+2)
b=(2a^2)/(β+2)---------------(7)
From (7), b=(2a^2)/(β+2)
2a^2=b(β+2)
a=[b(β+2)]/2a-------------(8)
Substituting (8) into (1)
[b(β+2)]/2a+2=b/a
[b(β+2)]/2a+4a/2a-2b/2a=0
b(β+2)+4a-2b=0
[(2a^2)/(β+2)](β+2)+4a-2[(2a^2)/(β+2)]=0
2a^2+4a+[(-4a^2)/(β+2)]=0
(2a^2)(β+2)+4a(β+2)-4a^2=0
2βa^2+4a^2+4aβ+8a-4a^2=0
2βa^2+4aβ+8a=0
2βa+4β+8=0
βa+2β+4=0
4+(a+2)β=0

When a=kβ,
4+(kβ+2)β=0
4+kβ^2+2β=0
kβ^2+2β+4=0
Given that a and β are real,
Hence b^2-4ac≥0 where a=k, b=2 and c=4
2^2-4(k)(4)≥0
4-16k≥0
-16k≥-4
k≤1/4

[Nelson]

Polynomials:

3. Given that alpha+2 and beta+1 are the roots of the equation x^2-qx+r=0, while alpha and beta are the roots of the equation 2x^2-bx+c=0 where alpha, beta are real and alpha>= (larger or equal to)beta, find q and r in terms of b and c.
In the case alpha=beta, show that q^2=4r+1.

[Ans:q=1/2(b+6), r= 3b/4 + c/2 + 2-{square root of (b^2-8c)}/4]

I can't prove the r (beta) one, pls help. Thx a lot!! :oops:

I think the solution is as follows:

Given that α+2 and β+1 are the roots of the equation x^2-qx+r=0
Hence,
Sum of roots,α+2+β+1=q
α+β+3=q----------(1)
Product of roots, (α+2)(β+1)=r
αβ+α+2β+2=r--------------(2)
Given that α and β are the roots of equation 2x^2-bx+c=0
Hence,
Sum of roots, α+β=b/2------------(3)
Product of roots, αβ=c/2------------(4)
Substituting (3) and (4) into (2),
c/2+b/2+β+2=r------------------(5)
Since α and β are the roots of equation 2x^2-bx+c=0
we use x=[-b?√(b^2-4ac)]/2a where a=2, b=-b, c=c
x={b?√[(-b)^2-4(2)(c)]}/2(2)
x=[b?√(b^2-8c)]/4
Since α≥β
α=[b+√(b^2-8c)]/4 and β=[b-√(b^2-8c)]/4
Therefore, β=[b-√(b^2-8c)]/4
Substituting β=[b-√(b^2-8c)]/4 into (5),
c/2+b/2+[b-√(b^2-8c)]/4+2=r
c/2+b/2+b/4-[√(b^2-8c)]/4+2=r
Hence r=3b/4+c/2+2-[√(b^2-8c)]/4

[PINKPIGGY]

Thanks a lot, Nelson! U r superb!! THX!!!! :o

[IchimaruGin]

Can u help me solve this question....i could not find the solution for it...its from the topic sequences and series

find the sigma from r=1 to 20 of (-1)^r r^2


ans:210

help me solve.....plssssssss

[youngyew]

Can u help me solve this question....i could not find the solution for it...its from the topic sequences and series

find the sigma from r=1 to 20 of (-1)^r r^2


ans:210

help me solve.....plssssssss
Sigma (r from 1 to 20) of (-1)^r r^2
= - 1 squared + 2 squared - 3 squared + 4 squared + ... + 20 squared
= (2 squared + 4 squared + ... + 20 squared) - (1 squared + ... + 19 squared)
= {Sigma (r from 1 to 10) of (2n)^2} - {Sigma (r from 1 to 10) of (2n-1)^2}
= {Sigma (r from 1 to 10) of 4n^2} - {Sigma (r from 1 to 10) of (4n^2 - 4n + 1)}
= 4 {Sigma (r from 1 to 10) of n} - {Sigma (r from 1 to 10) of 1}
= 4 (10)(11)/2 - 10
= 210

Hope that helps.

[DecentMerson]

Can u help me solve this question....i could not find the solution for it...its from the topic sequences and series

find the sigma from r=1 to 20 of (-1)^r r^2


ans:210

help me solve.....plssssssss
Sigma (r from 1 to 20) of (-1)^r r^2
= - 1 squared + 2 squared - 3 squared + 4 squared + ... + 20 squared
= (2 squared + 4 squared + ... + 20 squared) - (1 squared + ... + 19 squared)
= {Sigma (r from 1 to 10) of (2n)^2} - {Sigma (r from 1 to 10) of (2n-1)^2}
= {Sigma (r from 1 to 10) of 4n^2} - {Sigma (r from 1 to 10) of (4n^2 - 4n + 1)}
= 4 {Sigma (r from 1 to 10) of n} - {Sigma (r from 1 to 10) of 1}
= 4 (10)(11)/2 - 10
= 210

Hope that helps.

Can u help me solve this question....i could not find the solution for it...its from the topic sequences and series

find the sigma from r=1 to 20 of (-1)^r r^2


ans:210

help me solve.....plssssssss

wow...my math is so rusty... here it goes...
firstly : u shld know abt those formulas... where
summation of 1 to n for n is 1/2(n)(n+1)
summation of 1 to n for n^2 is 1/6(n)(n+1)(2n+1)

so... ur question...
to see the pattern, u can expand it a little...
u will get... summation from 1 to 20 of (-1)^r X r^2 =
-1^2 + 2^2 + -3^2 + -4^2 +.... = S

see the alternate sign and all...

group all the negative terms and let it be a...
let b = summation from 1 to 20 of r^2 (u can easily find this with the above-mentioned formula).... = 2870
(b = 1^2 + 2^2 + 3^2 + ....+ 20^2)

S = b - 2a
= 1^2 - 2(1^2) + 2^2 + 3^2 - (2)(3^2) + ...(hope that u can see the pattern, i'm too lazy to type the expansion...)

a can be written as summation from 1 to 10 (2r - 1)^2 = summation from 1 to 10 (4r^2 - 4r + 1) = 4(385) - 4(220) + 10 = 1330

S = b - 2a = 2870 - 2660 = 210

haha... changyang... couldn't believe that we are solving the question at the same time...

ur way is neater.... sweet

[IchimaruGin]

Can u help me solve this question....i could not find the solution for it...its from the topic sequences and series

find the sigma from r=1 to 20 of (-1)^r r^2


ans:210

help me solve.....plssssssss
Sigma (r from 1 to 20) of (-1)^r r^2
= - 1 squared + 2 squared - 3 squared + 4 squared + ... + 20 squared
= (2 squared + 4 squared + ... + 20 squared) - (1 squared + ... + 19 squared)
= {Sigma (r from 1 to 10) of (2n)^2} - {Sigma (r from 1 to 10) of (2n-1)^2}
= {Sigma (r from 1 to 10) of 4n^2} - {Sigma (r from 1 to 10) of (4n^2 - 4n + 1)}
= 4 {Sigma (r from 1 to 10) of n} - {Sigma (r from 1 to 10) of 1}
= 4 (10)(11)/2 - 10
= 210

Hope that helps.

thx very much for ur help.

[IchimaruGin]

Can u help me solve this question....i could not find the solution for it...its from the topic sequences and series

find the sigma from r=1 to 20 of (-1)^r r^2


ans:210

help me solve.....plssssssss
Sigma (r from 1 to 20) of (-1)^r r^2
= - 1 squared + 2 squared - 3 squared + 4 squared + ... + 20 squared
= (2 squared + 4 squared + ... + 20 squared) - (1 squared + ... + 19 squared)
= {Sigma (r from 1 to 10) of (2n)^2} - {Sigma (r from 1 to 10) of (2n-1)^2}
= {Sigma (r from 1 to 10) of 4n^2} - {Sigma (r from 1 to 10) of (4n^2 - 4n + 1)}
= 4 {Sigma (r from 1 to 10) of n} - {Sigma (r from 1 to 10) of 1}
= 4 (10)(11)/2 - 10
= 210

Hope that helps.

Can u help me solve this question....i could not find the solution for it...its from the topic sequences and series

find the sigma from r=1 to 20 of (-1)^r r^2


ans:210

help me solve.....plssssssss

wow...my math is so rusty... here it goes...
firstly : u shld know abt those formulas... where
summation of 1 to n for n is 1/2(n)(n+1)
summation of 1 to n for n^2 is 1/6(n)(n+1)(2n+1)

so... ur question...
to see the pattern, u can expand it a little...
u will get... summation from 1 to 20 of (-1)^r X r^2 =
-1^2 + 2^2 + -3^2 + -4^2 +.... = S

see the alternate sign and all...

group all the negative terms and let it be a...
let b = summation from 1 to 20 of r^2 (u can easily find this with the above-mentioned formula).... = 2870
(b = 1^2 + 2^2 + 3^2 + ....+ 20^2)

S = b - 2a
= 1^2 - 2(1^2) + 2^2 + 3^2 - (2)(3^2) + ...(hope that u can see the pattern, i'm too lazy to type the expansion...)

a can be written as summation from 1 to 10 (2r - 1)^2 = summation from 1 to 10 (4r^2 - 4r + 1) = 4(385) - 4(220) + 10 = 1330

S = b - 2a = 2870 - 2660 = 210

haha... changyang... couldn't believe that we are solving the question at the same time...

ur way is neater.... sweet

thank u..... now i understand

Could anyone please help me :

Given that z=x+yi and w=(z+8i)/(z-6), z≠6. If w is totally imaginary, show that x^2+y^2+2x-48=0

The matter is I cannot prove it :oops: .

if w is totally imaginary, then the real part of w is 0..........

i couldn't get ur answer.......

i have 0=x^2+y^2-6x+8y..............

Yeah....i got the same answer as bush....because w is totally imaginery, so we can equate the real part with 0...but can't get that answer.....cannot equate the imaginery part because we do not know the coefficient for i in the w part.....

So anyone knows how to solve it? I am facing the same problem now. Is this question wrong? I have got the same answer with you all , too. I know It's quite late to ask about it ..sigh...all because my teacher's just finished chapter 1.But please help me if your teacher has taught you how to solve it.Thanks

[doodle]

Could anyone please help me :
Given that z=x+yi and w=(z+8i)/(z-6), z≠6. If w is totally imaginary, show that x^2+y^2+2x-48=0

The matter is I cannot prove it :oops: .

The question is given wrongly. The answer is correct if w=(z+8)/(z-6).

The roots of the quadratic equation x^2+px+q=0, where q≠0, are α, β and one root of the quadratic equation x^2+p'x+q=0 is kα. Show that the other root of the equation is β/k.

By assuming that k^2≠1, write down an expression for the sum of the roots for each of the quadratic equations. Hence, find α and β in terms of p, p' and k. Deduce that k(kp-p')(kp'-p)=[(k^2-1)^2]q

For any p, p' and q, show that the sum of the four possible values of k is (pp')/q.

I have done the upper part just left the last part.

Any reply for both questions would be appreciated.

The solution is given in a pdf document (Q1.pdf) downloadable at 050703.pdf (http://www.geocities.com/thebigbagoftricks/recom/050703.pdf).

[PINKPIGGY]

Polynomials:

2. Given that x=2 is a root of the equation ax^2-bx-c=0, state the relationship among a,b,and c. Therefore, show that x=2 is also a root of the ea=quation bx^2+cx-8a=0. If a and beta are the other root of the first and second equation above respectively, show that 4+(a+2)beta=0. Hence, or otherwise, show that k<=(smaller or equal to) 1/4 if a=kbeta and a and beta are real.

Yup, i cab't solve the bolded part too! Help...Thx!



The propose solution:

Given that a and 2 are the roots of the equation ax^2-bx-c=0
Hence,
Sum of roots, a+2=b/a-----------(1)
Product of roots, 2a=-c/a-------------(2)
Given also that β and 2 are the roots of the equation bx^2+cx-8a=0
Hence,
Sum of roots, β+2=-c/b-----------(3)
Product of roots, 2β=-8a/b------------(4)


There's a shorter solution to share with u all, ~continued fr above~,

substitute (1) into (4),
2beta=-8/(a+2)
2a(beta)+4beta=-8
(a+2) beta=-4
4+(a+2) beta=0 :)

[PINKPIGGY]

Polynomials:
1. Given that 2y=a^x+a^-x, where a>1, x>0, prove that a^x=y+surd(y^2 -1). If, also given that 2z=a^3x+a^-3x, prove that z=4y^3-3y.

Can someone help me on the bolded part?

2. Prove that the equation 2^2x + 64(2^-x) = 32 has another real root besides the root x=2, hence, find the root and correct yr ans to 3 s.f (ans:1.31)

Help me 2 solve this Q. THX!!

[Nelson]

Could anyone please help me :
Given that z=x+yi and w=(z+8i)/(z-6), z≠6. If w is totally imaginary, show that x^2+y^2+2x-48=0

The matter is I cannot prove it :oops: .

The question is given wrongly. The answer is correct if w=(z+8)/(z-6).

The roots of the quadratic equation x^2+px+q=0, where q≠0, are α, β and one root of the quadratic equation x^2+p'x+q=0 is kα. Show that the other root of the equation is β/k.

By assuming that k^2≠1, write down an expression for the sum of the roots for each of the quadratic equations. Hence, find α and β in terms of p, p' and k. Deduce that k(kp-p')(kp'-p)=[(k^2-1)^2]q

For any p, p' and q, show that the sum of the four possible values of k is (pp')/q.

I have done the upper part just left the last part.

Any reply for both questions would be appreciated.

The solution is given in a pdf document (Q1.pdf) downloadable at http://doodle.mybesthost.com/Forums/.

Thank you so much.

[Nelson]

Polynomials:
1. Given that 2y=a^x+a^-x, where a>1, x>0, prove that a^x=y+surd(y^2 -1). If, also given that 2z=a^3x+a^-3x, prove that z=4y^3-3y.

Can someone help me on the bolded part?

2. Prove that the equation 2^2x + 64(2^-x) = 32 has another real root besides the root x=2, hence, find the root and correct yr ans to 3 s.f (ans:1.31)

Help me 2 solve this Q. THX!!

I think the solution is as follows:

For Q1,
Given that 2y=a^x+a^-x--------(1)
2z=a^3x+a^-3x
2z=(a^x+a^-x)(a^2x-1+a^-2x)
2z=(a^x+a^-x)[(a^2x+a^-2x)-1]
2z=(a^x+a^-x)[(a^x+a^-x)^2-2-1]---------(2)
Substituting (1) into (2),
2z=(2y)[(2y)^2-3]
z=4y^3-3y

For Q2,
Let y=2^x--------(1)
the equation becomes
y^2 + 64(1/y) = 32
y^3-32y+64=0
Given that x=2 is a root of the equation
Substituting x=2 into (1)
y=2^2
y=4
Since x=2 is a root of equation 2^2x + 64(2^-x) = 32,
y=4 is also a root of the equation y^3-32y+64=0
Using long division,
y^3-32y+64=(y-4)(y^2+4y-16)
to find the roots of y^2+4y-16,
we use y=[-b?√(b^2-4ac)]/2a where a=1, b=4 and c=-16
y=[-4?√(4^2-4(1)(-16))]/2(1)
y=(-4?√80)/2
y=-2?2√5
y=-2+2√5 or y=-2-2√5
y=2.47 or y=-6.47
2^x=2.47 or 2^x=-6.47
xlog2=log2.47 or NA
x=1.31
Hence the other root is 1.31

[PINKPIGGY]

Thx a lot, Nelson! :o

Here's another Q on polynomials:
(a) By taking y for x^m, show that x^m-1 is a factor of x^mn -1.
(b) Prove that x+y is a factor of x^n +y^n if n is an odd positive no.
(c) Factorise the expression (x+y)^6 - (x-y)^6 completely. (ans: 4xy(3x^2 + y^2)(x^2 + 3y^2)

Hope someone can help me to figure out this Q!! :oops:
Thanks!

[DecentMerson]

(c) Factorise the expression (x+y)^6 - (x-y)^6 completely. (ans: 4xy(3x^2 + y^2)(x^2 + 3y^2)

Hope someone can help me to figure out this Q!! :oops:
Thanks!

here u go... I think this is a general trick that's pretty useful... whenever u see (something of) even power - (another thing of) even power, u can always make it a^2 - b^2 = (a-b)(a+b)

so, let (x+y)^6 - (x-y)^6 = a^2 - b^2 =(a-b)(a+b), where a = (x+y)^3 and b = (x-y)^3

(x+y)^6 - (x-y)^6
= a^2 - b^2
= (a-b)(a+b)
= [(x+y)^3 - (x-y)^3][(x-y)^3+(x+y)^3]
(expand this and there's some cancelling going on...it might help if u can memorize the expansion of power 3.)
= (6x^2y + 2y^3)(2x^3 + 6xy^2)
= 2y(3x^2+y^2)(2x)(x^2 + 3y^2)
= 4xy(3x^2 + y^2)(x^2 + 3y^2) QED

will try other question later... remember this :
a^2 - b^2 =(a-b)(a+b)

[PINKPIGGY]

Thanks 4 yr help! DecentMerson.

[kimsiang]

(a) By taking y for x^m, show that x^m-1 is a factor of x^mn -1.
(b) Prove that x+y is a factor of x^n +y^n if n is an odd positive no.


(a)by substituting y = x^m into the (x^mn -1)/(x^m-1)
we can get (y^n-1)/(y-1)...this mean we hv to prove y-1 is a factor of y^n-1 which can be easily prove.
(b)let f(x) be a function of x.
write f(x) = x^n +y^n.here let n = 2k + 1 since it is odd number.
substitute into f(x),we can get
f(x) = x^(2k + 1) +y^(2k + 1)
let x = -y,
we can get
f(-y) = (-y)^(2k + 1) +y^(2k + 1)
= 0
hence,(x+y) is a factor of f(x).
which is the answer.

[confused-freaker]

I need help in this maths Q:

A rectangular tank has a horizontal base with base area A m^2. Water is flowing into the tank at a constant rate p m^3s^-1 and flows out at a rate qx m^3s^-1, where x metres is the depth of the water in the tank at time t seconds and q is a constant. If the depth is 0.5m, it remains at this constant value. Show that:

dx/dt = -k(2x-1) where k is a positive constant

When t=0, the depth of water in the tank is 0.75 m and is decreasing at a rate of 0.01ms-1. Find the time at which the depth of water is 0.55m .....


i'm out of idea how to do =.=

[kimsiang]

I need help in this maths Q:

A rectangular tank has a horizontal base with base area A m^2. Water is flowing into the tank at a constant rate p m^3s^-1 and flows out at a rate qx m^3s^-1, where x metres is the depth of the water in the tank at time t seconds and q is a constant. If the depth is 0.5m, it remains at this constant value. Show that:

dx/dt = -k(2x-1) where k is a positive constant

When t=0, the depth of water in the tank is 0.75 m and is decreasing at a rate of 0.01ms-1. Find the time at which the depth of water is 0.55m .....


i'm out of idea how to do =.=

ok.....here is my solution..
since the net flow rate = p - qx
when x=0.5m,net flow rate =0
means p-q(0.5)=0
means q = 2p
and,flow rate = d(Volume)/dt =d(Ax)/dt (here,volume = Ax)
we can get d(Ax)/dt = p - qx
A (dx/dt) = p - 2px
then, A (dx/dt) = p(1 - 2x)
finally we get dx/dt = (p/A)(1-2x) = k(1-2x) = -k(2x-1) ---- (#)

At t=0 , x=0.75m and dx/dt = -0.01ms^-1,
we can substitute those into (#)
and thus -0.01 = -k(2(0.75)-1)
then , we can get k = 0.02 s^ -1
substitute into (#) again,rearrange it..we can get
dx/dt = -0.02 (2x-1)

then, integrate [dx/(2x-1)] = integrate -0.02 dt
finally we can get 0.5 ln(2x-1) = -0.02t + C..where C is a constant
substitute t=0 , x=0.75m ,
0.5 ln (1.5 - 1 ) = C
thus C = 0.5 ln 0.5
= -0.5 ln 2
Eventually,we wil get 0.5 ln(2x-1) = -0.02t - 0.5 ln 2

substitute x=0.55m,we can get 0.5 ln(2*0.55 - 1) = -0.02t - 0.5 ln 2
0.02t = - 0.5 ln(2*0.55 - 1) - 0.5 ln 2
= - 0.5 ln 0.1 - 0.5 ln 2
= - 0.5 (ln 0.1 + ln 2)
= - 0.5 [ln (0.1*2)]
t = - (0.5/0.02) ln 0.2
= (25 ln 5)s

[confused-freaker]

lolllll i just realise it before reading the solution, but ure solution seems more faster and i didnt thought about the net flow thing :P thank you so much!!! :D really pui fook

[kimsiang]

lolllll i just realise it before reading the solution, but ure solution seems more faster and i didnt thought about the net flow thing :P thank you so much!!! :D really pui fook
you are welcome :D

[confused-freaker]

hmmm thinking about the q again.....i'm just wondering is the volume Ax = volume gained? because Ax = (p-qx)t ......if p-qx = 0 when x=0.5...... so this means that the volume gain in the tank is 0 ?

[kimsiang]

hmmm thinking about the q again.....i'm just wondering is the volume Ax = volume gained? because Ax = (p-qx)t ......if p-qx = 0 when x=0.5...... so this means that the volume gain in the tank is 0 ?
er.....actually Ax is the volume of the container....and d(Ax)/dt means changes of the volume with respect to time......
so,when x=0.5m,the volume wil remain the same....
that means,from the question,the starting volume of the container is 0.75m....then it slowly decrease until it become nearly 0.5m...

[confused-freaker]

oooooh!!!!....sigh confuse myself for nothing, thanks again!!! maths sifu~

[Daniel]

Help!
Points A and C have coordinates (-1,2) and (9,7) respectively. A rectangle ABCD has AC as a diagonal. Calculate the possible coordinates of B and D if the length of AB is 10 units

[kimsiang]

Help!
Points A and C have coordinates (-1,2) and (9,7) respectively. A rectangle ABCD has AC as a diagonal. Calculate the possible coordinates of B and D if the length of AB is 10 units

Here is my solution.....but it is quite long...shd hv other shorter solution i think.......

Since AC is a diagonal , so do BD.
Hence AC = BD = 10 units (#)
since the coordinates of A,C were given....AC^2 = 125 units^2
By using the fact that AC is the hypotenus of triangles ABC and also ADC,including (#),we can form 4 equations.
First,let coordinate B = (b1,b2) , and coordinate D = (d1,d2)
then we wil have
(b1 - 9)^2 + (b2 - 7)^2 = 25 (1) BC = 5 units
(b1 +1)^2 + (b2 - 2)^2 = 100 (2) AB = 10 units
(d1 +1)^2 + (d2 - 2)^2 = 25 (3) AD = 5 units
(d1 - 9)^2 + (d2 - 7)^2 = 100 (4) DC = 10 units

solving (1) ,(2) simultanously and then (3),(4),you can get
2b1 + b2 = 20
2d1 + d2 = 5

Substitute b1 or b2,d1 or d2 back to either (1) or (2) and (3) or (4),
you will get 2 quadratic equations.....

For example,substitute d2 = 5 - 2d1 into (3),
we can get (d1 +1)^2 + ([5 - 2d1] - 2)^2 = 25
(d1 +1)^2 + (3 - 2d1)^2 = 25
.
.
.
d1^2 - 2d1 - 3 =0
Hence d1 = -1 or 3
d2 = 7 or -1
then we can have D(-1 ,7) , D(3,-1).

For b1,b2,under the same process,you shd get the answer too.
It is b1 = 5 or 9
b2 = 10 or 2
then we can have B(5,10) , B(9,2).

Now we have to find which B is corespond to which D.
since both AC and BD are diagonals,the shd have the same center point,that is (4,9/2).We can get the sets we want.

Hence,the possible coordinate sets for B and D are
B(5,10) and D(3,-1) , B(9,2) and D(-1 ,7) .

[yama]

1) sin x - sin 3x + sin 2x = 2 sin x
1 + cos x ? 2cos2 x


2) sin 3x + sin 6x = sin 9x


does anybody can help me to solve this question ?
thank you a lot!!

[yama]

there is another question

prove that the area of any rhombus in given by A = 1/2 xy where x and y are the lengths of the diagonals. [8]

thank you !!

[kimsiang]

1) sin x - sin 3x + sin 2x = 2 sin x
1 + cos x ? 2cos2 x

2) sin 3x + sin 6x = sin 9x

3)prove that the area of any rhombus is given by A = 1/2 xy where x and y are the lengths of the diagonals.

Solution)
I think there should be domain for x,for example, -180 =< x =< 180..
so,since i duno the range,i wil use -180 =< x = < 180 here...

1.when x=0 , 1 + cos(x) ? 2cos(2x)= infinity,so,x not = 0.
when x not = 0, sin(x)-sin(3x)+sin(2x)=2sin(x)[1+cos(x)?2cos(2x)]
<=> sin(x)-sin(3x)+sin(2x) =2sin(x)+2sin(x)cos(x)?4sin(x)cos(2x)
<=> -sin(3x)+sin(2x) = sin(x)+sin(2x) ?4sin(x)cos(2x)
<=> -sin(3x) = sin(x) ?4sin(x)cos(2x)
<=> 4sin(x)cos(2x) = sin(x)+sin(3x)
<=> 4sin(x)[2cos(x)^2-1] = 2sin(2x)cos(x)
<=>8sin(x)cos(x)^2-4sin(x)=4sin(x)cos(x)^2
<=> 4sin(x)cos(x)^2 - 4sin(x) = 0
<=> sin(x)[cos(x)^2 - 1] = 0
<=> sin(x) = 0 or cos(x)^2 - 1 = 0
<=> x = -180,180 or cos(x) = -1 , 1
<=> x = -180,180 or x = -180,180
so,x = -180,180

2. sin(3x) + sin(6x) = sin(9x)
Obviously, x =-180,0,180 are 3 of the solution(s).
When x not =-180,0,180,
<=>sin(3x)+2sin(3x)cos(3x)=sin(3x)cos(6x)+sin(6x)cos(3x)
<=>sin(3x)[1+2cos(3x)]=sin(3x)[2cos(3x)^2-1]+2sin(3x)cos(3x)^2
since x not =-180,0,180,sin(3x) not = 0,then we can cancel it from both sides.
<=> 1+2cos(3x)=2cos(3x)^2-1+2cos(3x)^2
<=> 0=4cos(3x)^2-2cos(3x)-2
<=> 2cos(3x)^2-cos(3x)-1=0
<=> cos(3x)= [1/4] (+/-) [sqrt( 1 + 8 )/4]
<=> cos(3x)= [1/4] (+/-) [3/4]
<=> cos(3x)= -1/2 , 1
Since -180 =< x = < 180 , -540 =< 3x = < 540
<=> 3x= -480,-240,-120,120,240,480,360,-360
<=> x= -160,-120,-80,-40,40,80,120,160
So,the solutions r x=-180,-160,-120,-80,-40,0,40,80,120,160,180.

3.
http://mathworld.wolfram.com/images/eps-gif/Rhombus_750.gif

By refering to the diagram above,
lets devide the area into 4 parts(4 right triangles)
Area of 1 right triangle = 0.5*(base*height)
= 0.5*(0.5p*0.5q) = 0.125pq
Since there r 4 right triangles, TOtal area = Area of Rhombus
= 0.125pq * 4
= 0.5pq

[yama]

ops... sorry for my careless

actually the questions is requiring proving


2) prove that sin 3x + sin 6x = sin 9x

sorry and thank you for ur help, kimsiang!

[kimsiang]

2) prove that sin 3x + sin 6x = sin 9x

sorry and thank you for ur help, kimsiang!

But sin 3x + sin 6x = sin 9x is not an equation...
for example,when x = 15 ,
sin 3x + sin 6x = sin 45 + sin 90 = (sqrt 2)/2 + 1
but sin 9x = sin 135 = sin 45 = (sqrt 2)/2
means (sin 3x + sin 6x) not = sin 9x
So,the original question that u have typed shd be ok..

[confused-freaker]

Hi there, i got some problems with this q:

11. (b) A large quantity of shoes from brand B is inspected before been distributed to slaes outlets. It is found that 1% of the left side shoes are defective while 2% of the right side shoes are defective. By assuming that the defects occur independently and using suitable approximation, calculate the probability that in a random sample of 50 pairs of shoes,

(i) there are two pairs of shoes with defects on their left side.
(ii) there is a pair of shoes with defects on both sides.

[stpm 1999]


answers from the 5-year series:

11. (b) (i) 0.076 (ii) 0.251

[kimsiang]

Hi there, i got some problems with this q:

11. (b) A large quantity of shoes from brand B is inspected before been distributed to slaes outlets. It is found that 1% of the left side shoes are defective while 2% of the right side shoes are defective. By assuming that the defects occur independently and using suitable approximation, calculate the probability that in a random sample of 50 pairs of shoes,

(i) there are two pairs of shoes with defects on their left side.
(ii) there is a pair of shoes with defects on both sides.

[stpm 1999]


(i)For left side shoes : n=50,p(defective)=0.01 => np=0.5...then we can use poisson distribution as approximation...

P(X=2)=[e^(-0.5)]*[(0.5)^2]*/2 = 0.075816332
= 0.076
(#) Here,n means number of left side shoe,X means number of left side shoe which is(are) defective.

(ii)Lets consider each pair of shoes...
p(both defective)=0.01*0.02=0.0002
p(not both defective)= 1 - 0.0002 = 0.9998 (or u can add up 0.01*0.98,0.99*0.02,0.98*0.99)

m=50,p=0.0002 =>mp=0.01
Poisson again......
P(Y=1)=[e^(-0.01)]*[0.01] = 0.0099

(#) Here,m means number of pairs,Y means number of pairs which is(are) both defective.
There r maybe other ways to solve (ii)......maybe there r better n more accurate solution....but this is wat i can get
since i din study that for a long time...Btw,The answer given is definitely strange....dun u feel that?

[confused-freaker]

Yea, the part (i) i can do as normal....comes to part (ii)....i'm kinda blur why their working diff d........let me type out the whole thing for you:

Let X be the number of pairs of brand B shoes, out of 50 pairs, which have defects on the left sides, and that Y the number of pairs which have defects on the right side, then

X~B(50,0.01)
Y~B(50, 0.02)

(i) Since n =50 is large and np = 0.5 < 5, and so

X~P(0.5) approximately,
P(X=2) = e^(-0.5) * 0.5^2 / 2! = 0.0758 (this is ok)

(ii) Since n = 50 is large and np = 1 < 5, hence

Y~P(1) approximately
Z = X + Y
so Z~P(1.5)
P(Z=2) = e^(-1.5) * 1.5^2 / 2! = 0.251

I dont really get why they go and plus them...never consider the 0.01x0.02 thing......sigh

Anyway, if i dont use this poisson approx, can i use the binomial approx?

[confused-freaker]

another q:

Given two parallel lines l1 and l2 passing through (5, 0) and (-5, 0) respectively, and meet the line 4x + 3y = 25 respectively at P and Q. If PQ equals to 5 units, find the possible slopes of l1 and l2.

[stpm 2001 p1 q5]

sorry for flooding so many q =.= but thanks in advance.
The answer/working uses the angle n tangent thing....but is there any other easier to understand way?

[kimsiang]

another q:

Given two parallel lines l1 and l2 passing through (5, 0) and (-5, 0) respectively, and meet the line 4x + 3y = 25 respectively at P and Q. If PQ equals to 5 units, find the possible slopes of l1 and l2.

[stpm 2001 p1 q5]

sorry for flooding so many q =.= but thanks in advance.
The answer/working uses the angle n tangent thing....but is there any other easier to understand way?

I think using the method using angle wil be easier compare to others...
For the questions regarding statistic above,Z is obviously not equal to X + Y.So,juz ignore it.
When talking about approximation,we r approximating binomial to others...means from discrete to continous.......but Why?? for example.when n = 500 ....i dun think our calculator can give us the value of 500 C 25 in binomial.......but by using poisson,when the p is 0.01,we can calculate the probability easily...isn`t?

[confused-freaker]

aah i see, thanks for the explaination :) ...

for the coordinate geometry q above, i find their working seems a little weird....i shall paste in the solution for you to spot if there is any probs ~

[confused-freaker]

books solution :
http://server2.uploadit.org/files/chickens-maths2001q5soln.jpg

worked out solution:
http://server3.uploadit.org/files/chickens-maths2001q5mysoln.jpg

need judgement, thx in advance kimsiang

[kimsiang]

oups...sorry for late reply......
yr solution is ok...but for the sample solution,1 of the mistakes is that it juz simply state that since theta can be positive n negative,n then come out with 2 solutions.Instead of doing that,1 shd draw another line where the slope is negative,n then start to evaluate the theta,beta,etc...shd come out with the same answer.
The way u used is a normal way,which is *secure*.But sometimes,do do it in other alternative ways.It helps to improve yr math... :wink:

[yama]

these are the questions tat i face problems , hope someone can help me to find the solution soon . thank you very muz !!

1. Find the points of intersection of the curve y^2 = 3ax and x^2 = 3ay. Calculate the are of the region bounded by the curves .

2. Find the set of the values of k such that the roots of the equation x^2 + (k+1)x +k + 4 = 0 are positive.

3. Expand (2+x)^-2 in ascending powers of x^-1 until and including the term in x^-4. state the range of values of x for which this expansion is valid. Find the coefficient of x^-18 in the expansion of (3x+x^-2)^12.

4. If ( x+iy )^3 = x + iy, find the real values of x and y .

5. Prove that x^2 + y^2 > 2xy if x, y E R and x is not equal to y. if x^2 + y^2 = 1 and w^2 + z^2 = 1 , show that xw + yz <1.

6. Show that (cosA + cos B)^2 + (sinA + sinB)^2 = 4cos^2 ? (A-B) . Hence, show that cos 15 = ? (route 2 + route6)

7. In a chemical process, substance A is continuously changing into the substance B at a rate which is proportional to the mass of A and inversely proportional to the mass of B , at any time t. the total mass of A and B remains constant and equals to M. the mass pf the B at time is m.
Obtain a differential equation relating m and t.
Given that M = 120 grams, 20 grams of substance B remains after 2 minutes, find the time taken for which the masses of both substances are equal.


8. If a and b are real roots of the equation x^2 + (k+2)x + (k+5) = 0 and k ER, find the ranges of k such that
(i) both a and b are positive,
(ii) both a and b are positive and less than 7

9. The parametric equations of a curve is given by x= t(t-2) and y= 2(t-1). If P is a point on the curve and O is the origin, show that the equation of the locus of Q which divides OP internally in the ratio 1:2 is given by 9y^2 = 4(3x+1). If the normal at P passes through the point A ( 1,8 ) , find the coordinates of P.

[reign226]

I have been thinking over this problem for a VERY long time and the answers keep contradicting. It refers to linear combination of two independant normal random variables.

X=N(10, 0.1)

Let's say X is distributed normally with mean = 10 and SD = 0.1

Now let's say random variable Y = 10X.

That means:
E(Y) = 10E(X)

No problems so far. But:
Var(Y) = Var(10X)
= 100Var(X)

Right? But in some books, I the way they get the variance for Y, Var(Y) is to simply multiply Var(X) by 10, as there are ten X in Y. Which do you think is correct? Personally I go with 100Var(X) as it's mathematically proven. But I still have my doubt. Somebody help me!!

[yama]

I have been thinking over this problem for a VERY long time and the answers keep contradicting. It refers to linear combination of two independant normal random variables.

X=N(10, 0.1)

Let's say X is distributed normally with mean = 10 and SD = 0.1

Now let's say random variable Y = 10X.

That means:
E(Y) = 10E(X)

No problems so far. But:
Var(Y) = Var(10X)
= 100Var(X)

Right? But in some books, I the way they get the variance for Y, Var(Y) is to simply multiply Var(X) by 10, as there are ten X in Y. Which do you think is correct? Personally I go with 100Var(X) as it's mathematically proven. But I still have my doubt. Somebody help me!!




in this type of question

if the Q states that ten of x,
Y = X1+X2+X3+X4+....+X10
that mean Var(Y)= 10Var(X)

but

if the Q states that x is repeated ten times ,
Y = 10X
that means Var(Y)= Var(10X) = 100Var(X)

hope that u can understand that i trying to say

[reign226]

The Q was: "Random variable X = such and such, which represents the weight of a chocolate bar. Random variable Y is the weight of 4 chocolate bars."

So what is the variance for Y? Isn't Y = 10x?

Hmm, then again i think i understand what you're trying to say. I just have to get into my brain that

Y = X1 + X2 + X3 + X4

is not equal to

Y = 4X.

*sigh* Thanks for the help!

[reign226]

Here's another little devil:

if y = x + (1/x), show that x^4 - 2x^3 - 6x^2 - 2x + 1 = 0 can be transformed into a quadratic in terms of y

i got until this far only:
y^2(x^2 - 2x) + y(2-8x) + 8 = 0

(by long division with X + (1/X)) And now i'm stuck. Help!

[kimsiang]

Here's another little devil:

if y = x + (1/x), show that x^4 - 2x^3 - 6x^2 - 2x + 1 = 0 can be transformed into a quadratic in terms of y

i got until this far only:
y^2(x^2 - 2x) + y(2-8x) + 8 = 0

Since y = x + (1/x) ------(1)
we can get y^2 = x^2 + (1/x^2) + 2
==>y^2 - 2 = x^2 + (1/x^2) ------(2)
then,x^4-2x^3-6x^2-2x+1 = x^2[x^2 - 2x - 6 - (2/x) + (1/x^2)]=0
Since y = x + (1/x),we can deduce that x not = 0.
then we can get [x^2 - 2x - 6 - (2/x) + (1/x^2)]=0
<=> [x^2 + (1/x^2) - 2x - (2/x) - 6]=0
<=> [x^2 + (1/x^2) - 2(x+(1/x)) - 6]=0
Substitute (1)and (2) into it,
we can get <=> [y^2 - 2 - 2y - 6]=0
<=> [y^2 - 2y - 8]=0
<=>(y-4)(y+2)=0
<=> y=4 or y = -2

[reign226]

Brilliant! Thanks a lot kimsang. Unfortunately I could never have thought of this way to do it...ever. *Sigh* Thanks a lot dude!

[kimsiang]

You are welcome.......... :D

[Nelson]

2. Find the set of the values of k such that the roots of the equation x^2 + (k+1)x +k + 4 = 0 are positive.


I think the solution is as follows:

Given that the equation x^2 + (k+1)x +k + 4 = 0 has roots, this means that b^2-4ac≥0.
By taking a = 1, b = (k+1) and c = k+4
we have
(k+1)^2-4(1)(k+4)≥0
k^2+2k+1-4k-16≥0
k^2-2k-15≥0
(k+3)(k-5)≥0
Hence k≤-3 or k≥5

Let α and β be the roots of this equation
Given that those roots must be positive
Therefore α>0 and β>0
α+β = -k-1>0
k<-1
αβ=k+4>0
k>-4
The solution is -4<k<-1

By representing these solutions on a real number line,
we have the set of values of k is {k:-4<k≤-3}.

[DecentMerson]

I have been thinking over this problem for a VERY long time and the answers keep contradicting. It refers to linear combination of two independant normal random variables.

X=N(10, 0.1)

Let's say X is distributed normally with mean = 10 and SD = 0.1

Now let's say random variable Y = 10X.

That means:
E(Y) = 10E(X)

No problems so far. But:
Var(Y) = Var(10X)
= 100Var(X)

Right? But in some books, I the way they get the variance for Y, Var(Y) is to simply multiply Var(X) by 10, as there are ten X in Y. Which do you think is correct? Personally I go with 100Var(X) as it's mathematically proven. But I still have my doubt. Somebody help me!!

about the Var(Y) = Var (10X) question... it is definitely
Var(Y) = Var(10X)
Var(Y) = 100Var(X)

but if Var(Y) = Var (X1 + X2 + X3 + ... + X10), assuming that Xi's are independent and identically distributed(i.i.d) random variable... then, the variance of the sum = sum of the variance because covariance = 0 when the random variables are independent...

so, Var (Y) = Var (X1) + Var (X2) + ... + Var (X10)
= 10 Var (X) (since that they are iid)

[DecentMerson]

6. Show that (cosA + cos B)^2 + (sinA + sinB)^2 = 4cos^2 ? (A-B) . Hence, show that cos 15 = ? (route 2 + route6)

by using the identity
cos a + cos b = 2cos(a+b/2)cos(a-b/2)
sin a + sin b = 2sin(a+b/2)cos(a-b/2)

u get (cosA + cos B)^2 + (sinA + sinB)^2 = 4 [ cos^2 (A+B/2) cos^2(A-B/2) + sin^2(A+B/2)cos^2(A-B/2)]

as sin^2 X = 1 - cos^2 X... by applying this identity... and expanding it a little, u will come to 4cos^2 ? (A-B)

hence, using the equation u just proved... A-B/2 = 15... A-B = 30...
so, u can let A= 90 B=60, or A=30, B=0... and substitute the value into the equation... and u shld be able to get the answer...

[reign226]

About the variance question again:

So what does this statement mean: "X = random variable representing mean number of hours studying by Males"

Y= random variable representing mean number of studies by 10 males".

Is Y = 10X or Y = X1 + X2...+ X10?

I saw this question in our textbook and they say Y= 10X. But the thing is, in another book, it's practically the same question (replace mean number of hours studying with weight of chocolate) and Y = random variable representing 10 chocolate bars and they use Y = X1 + X2 etc.

@<hidden>@<hidden>

It would be best if the question just state Y = what. But I'm not counting on it. Arrgh!

[DecentMerson]

About the variance question again:

So what does this statement mean: "X = random variable representing mean number of hours studying by Males"

Y= random variable representing mean number of studies by 10 males".

Is Y = 10X or Y = X1 + X2...+ X10?

I saw this question in our textbook and they say Y= 10X. But the thing is, in another book, it's practically the same question (replace mean number of hours studying with weight of chocolate) and Y = random variable representing 10 chocolate bars and they use Y = X1 + X2 etc.

@<hidden>@<hidden>

It would be best if the question just state Y = what. But I'm not counting on it. Arrgh!

i would say the questions are not good questions... unclear... but, the first question, looking at the sentence structure, i would guess Y = 10X...

and about the chocolate bar questions, try scrutinizing the words... did they mention anything about taking 10 different chocolate bars or wat??

[Gar87field]

SOS! A few problems here... Can someone help?

1. Find the square root of 11-6(2)^(1/2)

2. Prove that a^2 + b^2 =14ab, then log((a+b)/4)= 1/2 (log a + log b)

3. Write the square roots of a^(1/2) +b(c)^(1/2) in the form p+q(r)^(1/2)

Too much maths.. Blur... Thanx in advance~! :D

[kimsiang]

1. Find the square root of 11-6(2)^(1/2)

2. Prove that a^2 + b^2 =14ab, then log((a+b)/4)= 1/2 (log a + log b)

3. Write the square roots of a^(1/2) +b(c)^(1/2) in the form p+q(r)^(1/2)


Here`s my solution....
1.let sqrt[11-6(2)^(1/2)]= a + b(sqrt2)....where a,b is rational no.
then,by squaring both side,we can get
11-6(2)^(1/2) = [a + b(sqrt2)]^2
11-6(sqrt2) = a^2 + 2b^2 + 2ab(sqrt2)
by equaling real number part n irrational number (sqrt2) part,
we can get 11 = a^2 + 2b^2 , -6 = 2ab ==>ab = -3
By substiltute a=-3/b into 11 = a^2 + 2b^2,
we can get 2b^4 - 11b^2 +9 = 0
thus , b = 1 or -1
a = -3 or 3
==> a+b(sqrt2) = -3 + (sqrt2) or 3 - (sqrt2)
but,since sqrt(x) > 0 , for every x > 0,
hence the solution is 3 - (sqrt2) #

2. a^2 + b^2 =14ab <==> a^2 + b^2 +2ab = 14ab + 2ab
<==> ( a+b )^2 = 16ab
<==> (a+b)^2/16 = ab
<==> [(a+b)/4]^2 = ab
take log for both side, <==> log [(a+b)/4]^2 = log (ab)
<==> 2 log [(a+b)/4] = log a + log b
<==> log [(a+b)/4] = (1/2)*(log a + log b)#

3.It is too long to be written here...SO,i wil only explain briefly.

1st,let sqrt[a^(1/2) +b(c)^(1/2)] = p+q(r)^(1/2)
then (sqrt a) + b(sqrt c) = [p+q(sqrt r)]^2
(sqrt a) + b(sqrt c) = p^2 + rq^2 + 2pq(sqrt r)

squaring both side again,letting A =p^2 + rq^2 and B = 2pq
Hence , a + cb^2 + 2b(sqrt ac) = A^2 + rB^2 + 2AB(sqrt r)
From here,we can deduce that r = ac.

Then we have to find p and q,which requires quite many steps.
Please try to slove a + cb^2 = A^2 + rB^2 and b = AB at yr own.
At some step,u may get 2pq = 1/(sqrt c) or b/(sqrt a)
Good luck.

p.s: if u mistyped the question,which the correct 1 is a + b(sqrt c),
u can use the method in question no.1.

[reign226]

Eh, I thought STPM Maths 1 over oredi? Those are questions for maths t 2 meh?

[Gar87field]

Thanx Kimsiang.. Ur maths is great! I get ur solution but still have to work out Q3 because it's a bit blur to me now. Thanx thanx thanx. Btw, i'm in lower 6 now , reign 226. :wink:

[Gar87field]

Another simple question which i cant solve.. No idea where to start...

Find the equatic equation with the roots -3+4i and -3-4i, giving ur answer in the form x^2+px+q=0 where p and q are real numbers.

[kimsiang]

Find the equatic equation with the roots -3+4i and -3-4i, giving ur answer in the form x^2+px+q=0 where p and q are real numbers.

At 1st,let a=-3+4i and b=-3-4i.
==> a+b = -6
==> ab = (-3+4i)(-3-4i)
= 9 + 16 = 25
Then,since a,b are roots of a quadratic equation,using the fact that x^2 - (a+b)x + ab = 0
We can get x^2 - (-6)x + 25 = 0
==>x^2 + 6x + 25 = 0

[Gar87field]

The complex number z is given by z= 1+ cos x + i sin x

1)a) Find the modulus and argument of z and 1/z when x=pie/3

b) Show that for all values of x, the point representing z in a argand diagram is located on a circle. Find the centre and the radius of the circle.

c) Prove that the real part of 1/z is 1/2 for all values of x.


p.s: I can solve the a part only...

2)Find the constant a such that for all real values of b, one of the roots of the equation 2x^3+ax+4 =b(x-2) is 2.

a) When a takes this value, find the set of values of b where the given equation has three real and distinct roots. ( I got the answer but dont understand why b cannot be 14)

b) When a=-10 and b=-12, find the real n comple roots of the above equation.

Thanx kim siang. :D

[kimsiang]

The complex number z is given by z= 1+ cos x + i sin x

1)a) Find the modulus and argument of z and 1/z when x=pie/3

b) Show that for all values of x, the point representing z in a argand diagram is located on a circle. Find the centre and the radius of the circle.

c) Prove that the real part of 1/z is 1/2 for all values of x.


p.s: I can solve the a part only...

2)Find the constant a such that for all real values of b, one of the roots of the equation 2x^3+ax+4 =b(x-2) is 2.

a) When a takes this value, find the set of values of b where the given equation has three real and distinct roots. ( I got the answer but dont understand why b cannot be 14)

b) When a=-10 and b=-12, find the real n comple roots of the above equation.


1.b) z = 1+ cos x + i sin x
<==> z - 1 = cos x + i sin x
<==> modulus [z - 1] = modulus [cos x + i sin x]
<==> modulus [z - 1] = sqrt [(cos x)^2+(sin x)^2]
<==> modulus [z - 1] = sqrt [1] = 1
Thus,the radius is 1 and the center is (1,0)#

c) z = 1+ cos x + i sin x
1/z = 1/(1+ cos x + i sin x)
= (1+ cos x - i sin x)
(1+ cos x + i sin x)(1+ cos x - i sin x)
= (1+ cos x - i sin x)
(1+cos x)^2+(sin x)^2
= (1+ cos x - i sin x)
1 + 2cos x + (cosx)^2+(sinx)^2
= (1+ cos x - i sin x)
(2+ 2cos x)
= 1/2 - i[(sin x) /(2+ 2cos x)]
Thus,the real part of 1/z is alwaz 1/2#

2.a)a=-10,then 2x^3 -10x + 4 = b(x-2)
<==> 2x^3 -(10+b)x + 4+2b = 0
<==> (x-2)[2x^2 + 4x - (2+b)] = 0
Now,other than 2,we need 2 more different roots.
Since for [2x^2 + 4x - (2+b)]=0 , x cant be 2(coz we wil have two 2 as root) , so , when x= 2 , [2x^2 + 4x - (2+b)] not equal to 0.(means 2 is not the root of [2x^2 + 4x - (2+b)])
Thus,we can get 2(2)^2 + 4(2) - (2+b) not 0
==> 16 - (2+b) not 0
==> 14 - b not 0
==> b not 14 #

b)when a=-10 and b=-12,
[2x^2 + 4x - (2+b)] = 0 ==> [2x^2 + 4x + 10] = 0
==> x = [- 4 + sqrt(16-4(2)(10))]/4 or [- 4 - sqrt(16-4(2)(10))]/4
==> x = [- 4 + sqrt(-64)]/4 or [- 4 - sqrt(-64)]/4
==> x = [- 4 + 8i]/4 or [- 4 - 8i]/4
==> x = -1 + 2i or -1 - 2i and also 2.....#

[Gar87field]

1.b) z = 1+ cos x + i sin x
<==> z - 1 = cos x + i sin x
<==> modulus [z - 1] = modulus [cos x + i sin x]
<==> modulus [z - 1] = sqrt [(cos x)^2+(sin x)^2]
<==> modulus [z - 1] = sqrt [1] = 1
Thus,the radius is 1 and the center is (1,0)#

Sorry but i dont get this clear.. Mind to clarify?

[kimsiang]

1.b) z = 1+ cos x + i sin x
<==> z - 1 = cos x + i sin x
<==> modulus [z - 1] = modulus [cos x + i sin x]
<==> modulus [z - 1] = sqrt [(cos x)^2+(sin x)^2]
<==> modulus [z - 1] = sqrt [1] = 1
Thus,the radius is 1 and the center is (1,0)#

Sorry but i dont get this clear.. Mind to clarify?

modulus [z - 1] means distance of z from 1.
So,if the distance is constant,means that the locus is a circle.

[Gar87field]

1. If (x+iy)^2 = x+yi, where x n y are real numbers, find the possible values of x and y.

2. If sqrt(3-i)/ (1+i) = x +yi where x and y are real numbers, find the values of x and y.

[Gar87field]

Kim siang, can help with this?

Solve the equations, for x and y,

x^2+xy=1/2(a)(a+b)

xy+y62=1/2(a)(a-b)

where a is not 0.

[kimsiang]

Solve the equations, for x and y,

x^2+xy=1/2(a)(a+b) ------- (1)

xy+y^2=1/2(a)(a-b) ------- (2)

where a is not 0.

(1)+(2),
we can get x^2+2xy+y^2=1/2(a)(a+b)+1/2(a)(a-b)
===> (x+y)^2=a^2
===> x+y = a ----- (3) or x+y= -a ----- (4)
then (1)-(2),
we can get x^2 - y^2 = b^2
====> (x+y)(x-y) = b^2 ------- (5)

1st,substitute x+y=a from (3) into (5)
we get (x-y)=b^2/a (a not 0) ------- (6)
solving (3) and (6),

==> x = (a^2+b^2)/2a , y = (a^2-b^2)/2a

Then, substitute x+y= -a from (4) into (5)
we get (x-y)= - b^2/a (a not 0) ------- (7)
solving (4) and (7),

==> x = - (a^2+b^2)/2a , y = - (a^2-b^2)/2a #

[kimsiang]

Besides STPM,STPM math,i would like to recommend u all to do some math from other countries such as Japan.Since i m in japan now,i have many questions for university entrance exam(high school level) in japan.Since I have translated it,i will post it here.For those who love math very much,have a try.I wil try to post the answer once per week.Good luck.

[J-Question 1]
The sum of some consecutive natural numbers is 1000.Find out what are those numbers.

[Yamagata University,Humanities Faculty 1989]

[Hint]:When their sum is 100,we can have 18+19+20+21+22=100.

[Tasslehoff]

I'll try this out
ofcourse with that hint, we could use a series of 5 numbers
198, 199, 200, 201, 202

^^ done by, 1000/5 == 200
since 200 is the middle number, the rest of the numbers must have a max (n-1)/2 deviation where n == total numbers in the series
thus, the max deviation is 2
with that, the series is 200-2, 200-1, 200, 200+1, 200+2
and we get
198, 199, 200, 201, 202

I tried to work out a working to solve it.. haha

for natural numbers, the series is an arithmetic one with a standard deviation of 1
let the first number of the series = x
d =1
total numbers in the series = n
Sn = (n/2)[2a+(n-1)d] = (n/2)[2x+n-1] = 1000

2xn+n^2-n = 2000
n^2+(2x-1)n-2000=0

dunno what the second equation was. or perhaps i started out wrong anyways :P

[kimsiang]

I'll try this out
ofcourse with that hint, we could use a series of 5 numbers
198, 199, 200, 201, 202

^^ done by, 1000/5 == 200
since 200 is the middle number, the rest of the numbers must have a max (n-1)/2 deviation where n == total numbers in the series
thus, the max deviation is 2
with that, the series is 200-2, 200-1, 200, 200+1, 200+2
and we get
198, 199, 200, 201, 202

I tried to work out a working to solve it.. haha

for natural numbers, the series is an arithmetic one with a standard deviation of 1
let the first number of the series = x
d =1
total numbers in the series = n
Sn = (n/2)[2a+(n-1)d] = (n/2)[2x+n-1] = 1000

2xn+n^2-n = 2000
n^2+(2x-1)n-2000=0

dunno what the second equation was. or perhaps i started out wrong anyways :P

Good work!!!
But there are stil 2 more sets of answers.
Good luck. :D

#if u let the total numbers in the series as n+1
u wil get (n+1)(2x+n) = 2000
which is easier to solve.

[kimsiang]

[J-Question 2]
Find out all natural numbers,n which satisfy the following equation.

[sqrt(n^2 - 7n + 11)]^(n^2 - 8n + 7) = 1

[Gakushuin University,Science Stream 1998]

[DecentMerson]

[J-Question 2]
Find out all natural numbers,n which satisfy the following equation.

[sqrt(n^2 - 7n + 11)]^(n^2 - 8n + 7) = 1

[Gakushuin University,Science Stream 1998]

tricky...

[sqrt(n^2 - 7n + 11)]^(n^2 - 8n + 7) = 1
=> (n^2 - 7n + 11)^1/2(n^2 - 8n + 7) = 1

take log both sides
=>1/2(n^2 - 8n + 7) lg(n^2 - 7n + 11) = lg 1 = 0

multiply by 2
=>(n^2 - 8n + 7) lg(n^2 - 7n + 11) = 0

so... the equation will only be zero if (n^2 - 8n + 7) = 0 and/or (n^2 - 7n + 11) = 1

so, solve both equations and u will get n = 1, 2, 5, 7
(i guess)

[kimsiang]

[sqrt(n^2 - 7n + 11)]^(n^2 - 8n + 7) = 1
=> (n^2 - 7n + 11)^1/2(n^2 - 8n + 7) = 1

take log both sides
=>1/2(n^2 - 8n + 7) lg(n^2 - 7n + 11) = lg 1 = 0

multiply by 2
=>(n^2 - 8n + 7) lg(n^2 - 7n + 11) = 0

so... the equation will only be zero if (n^2 - 8n + 7) = 0 and/or (n^2 - 7n + 11) = 1

so, solve both equations and u will get n = 1, 2, 5, 7
(i guess)

Good attempt.
but it is quite dangerous to take log in this situation since u duno the value inside is positive or negative.[Hint:inside can be negative]
there are stil 1 more value for n.Good luck.

[kimsiang]

[J-Question 3]

When x = (1-sqrt3)/(1+sqrt3) and y = (1+sqrt3)/(1-sqrt3)
Find out the value of sqrt[(x^y)/(y^x)].

[Tsukuba International University , 1998]

[Gar87field]

[J-Question 2]
Find out all natural numbers,n which satisfy the following equation.

[sqrt(n^2 - 7n + 11)]^(n^2 - 8n + 7) = 1

[Gakushuin University,Science Stream 1998]



Solution: [sqrt(n^2 - 7n + 11)]^(n^2 - 8n + 7) = 1
n^2 - 7n + 11)]^(n^2 - 8n + 7)=1
n^2-7n+11=1^1/(n^2-8n+7)
n^2-7n+11=1
n^2-7n+10=0
(n-5)(n-2)=0
n=5, n=2

n^2-8n+7 has to be zero,
(n-7)(n-1)=0
n=7, n=1


* Solve like this can? Where is another value of n hor :?:

[kimsiang]

Solution: [sqrt(n^2 - 7n + 11)]^(n^2 - 8n + 7) = 1
==> n^2 - 7n + 11)]^(n^2 - 8n + 7)=1
==> n^2-7n+11=1^1/(n^2-8n+7)
n^2-7n+11=1
n^2-7n+10=0
(n-5)(n-2)=0
n=5, n=2

==> n^2-8n+7 has to be zero,
(n-7)(n-1)=0
n=7, n=1


* Solve like this can? Where is another value of n hor :?:

*Some advise about writing solution in math:
for the part with ==>,it is better to write out what u did to those numbers.for example,take square for both side,take log for both side,etc....So,that other ppl wil could see clearly each step. :wink:

Since [sqrt(n^2 - 7n + 11)]^(n^2 - 8n + 7) = 1
For this to be true,

when (n^2 - 7n + 11) > 0 ,
we must have either (n^2 - 7n + 11) = 1 or (n^2 - 8n + 7) = 0
which we wil get n = 1,2,5,7

But,when (n^2 - 7n + 11) < 0,
sqrt(n^2 - 7n + 11) = sqrt[(-1)(-n^2 + 7n - 11)]
= isqrt(-n^2 + 7n - 11)
where i is sqrt(-1).

Since the right hand side is 1,the modulus of isqrt(-n^2 + 7n - 11)
must 1 also.Hence,(-n^2 + 7n - 11) = 1
==>n^2 - 7n + 12 = 0
==>n = 3 , 4

So,now,we hv [isqrt(-n^2 + 7n - 11)]^(n^2 - 8n + 7) = 1

When n = 3,
left hand side = [i]^(3^2 - 8*3 + 7)
= [i]^8=[i^4]^2=1=right hand side
so,n=3 also satisfy the equation.

When n = 4,
left hand side = [i]^(4^2 - 8*4 + 7)
=[i]^(-9)
=1/[i]^9
=1/ i (from above [i]^8 = 1)
= i/i^2 (multiply i on upper n lower side)
= - i not = right hand side
So,n=4 is not the solution.

As a result , we have n = 1,2,3,5,7

[Gar87field]

[J-Question 3]

When x = (1-sqrt3)/(1+sqrt3) and y = (1+sqrt3)/(1-sqrt3)
Find out the value of sqrt[(x^y)/(y^x)].

[Tsukuba International University , 1998]

The solution very long... Got the answer (1+sqrt3/1-sqrt3)^2.. dunno correct or not.

[kimsiang]

When x = (1-sqrt3)/(1+sqrt3) and y = (1+sqrt3)/(1-sqrt3)
Find out the value of sqrt[(x^y)/(y^x)].

[Tsukuba International University , 1998]
The solution very long... Got the answer (1+sqrt3/1-sqrt3)^2.. dunno correct or not.

nope.But good try. :wink: .Happy to have math fan like you,hehe :D .

[Hint]:For this question,you have to understand clearly the meaning of x^y and y^x .

[kimsiang]

[J-Question 3]

When x = (1-sqrt3)/(1+sqrt3) and y = (1+sqrt3)/(1-sqrt3)
Find out the value of sqrt[(x^y)/(y^x)].

[Tsukuba International University , 1998]

From above,we can see that y = 1/x

Then sqrt[(x^y)/(y^x)] = sqrt[(x^(1/x))/((1/x)^x)]
= sqrt [(x^(1/x))/(x^-x)]
= sqrt [(x^((1/x)+x))]

But,(1/x)+x = (1-sqrt3)/(1+sqrt3) + (1+sqrt3)/(1-sqrt3)
= -4

Hence, sqrt[(x^y)/(y^x)] = sqrt (x^-4)
= x^-2
= [(1+sqrt3)/(1-sqrt3)]^2
= 7+4(sqrt3)

#actually answer from Gar87field was nearly correct.juz she havent simplify it. :wink: .Remember to simplify the answer next time.

[kimsiang]

[J-Question 1]
The sum of some consecutive natural numbers is 1000.Find out what are those numbers.

[Yamagata University,Humanities Faculty 1989]

Solution:
Let the 1st no. as m,and is made up of n+1 numbers,we hv

m+(m+1)+(m+2)+(m+3)+.....+(m+n) = 1000
==> (2m+n)(n+1)/2=1000
==> (2m+n)(n+1) = 2000 = (2^4)(5^3)

But,(2m+n)+(n+1)=2(m+n)+1=odd number
Hence,one of the (2m+n),(n+1) must be odd n the other must be even.
*[odd number + odd number = even number
odd number + even number = odd number
even number + even number = even number]

Since 2^4 is always even,2^4 will always on 1 side.
Thus the possible combinations are {16,125},{16*5,25},{16*5*5,5}={16,125}{80,25}{400,5}

Since we know that (2m+n)>(n+1),we can get
{(2m+n),(n+1)}={125,16}{80,25}{400,5}
solving it,
we get {n,m}={15,55}{24,28}{4,198}

Thus,the solutions are 55+56+57+....+70 , 28+29+....+52 and 198+199+200+201+202.

[kimsiang]

[J-Question 4]

Prove that pi is greater than 3.05.

[Tokyo University,Science 2003]

[kimsiang]

[J-Question 5]

Find out all polynomials f(x) which satisfied f[f(x)]=[f(x)]^2 for all real x.

[Meiji University,199?]

[kimsiang]

[J-Question 6]

A(a,a^2) , B(b,b^2) are 2 different points on the parabola y=x^2.(a>b).Find out the range of a,b such that point C which satisfy [angle ACB = 90] exist on the same parabola.
[Kinki University,Commerce 1984]

[kimsiang]

[J-Question 4]

Prove that pie is greater than 3.05.

[Tokyo University,Science 2003]

Solution

Consider a circle with unit radius.Now,draw a n-poligon inside it which the edges of the poligon touch the circle.Combine the center of the circle to its edges.Now u wil hv n triangle inside a circle.

Hence,the area of the circle is greater than of the poligon.

==> Area of unit radius circle > Area of the n sides poligon
==> pie(1)^2 > n x area of a triangle
==> pie > n x 0.5 x (1)^2 x sin(360/n)
==> pie > [nsin(360/n)]/2

When n=12,
we have pie > [12sin(360/12)]/2
= 6sin30=3 (not enough)
When n=24,
we have pie > [24sin(360/24)]/2
= 12sin15
= 12sin(45-30)
= 12[sin45cos30-sin30cos45]
= 12[(sqrt2*sqrt3/4)-(sqrt2/4)]
= 3(sqrt2)(sqrt3-1)
>3(1.41)(1.73-1)
=3.0879
>3.05
u can try larger n for more accurate value for pie.....this was wat archimedes did...hehe...

[kimsiang]

[J-Question 5]

Find out all polynomials f(x) which satisfied f[f(x)]=[f(x)]^2 for all real x.

[Meiji University,199?]

At 1st,let f(x) not = 0.
f(x)=ax^n+....... (a not =0,n is 0 or positive integer)

Then,f[f(x)] = a(f(x))^n+.......
= a(ax^n+.....)^n+......
= [a^(n+1)][x^(n^2)]+........

and [f(x)]^2= [ax^n+.......]^2
= [a^2][x^2n]+........
From f[f(x)]=[f(x)]^2,we compare the term with the highest degree,
then we can get n^2 = 2n ------------(1)
a^(n+1) = a^2 ----------(2)
From (1),we can get n=0 or n=2.

[Case 1] n=0
substitute it into (2),we have a=a^2.
Since a is not 0.we have a=1.
Thus f(x)=1

[Case 1] n=2
substitute it into (2),we have a^3=a^2.
Since a is not 0.we have a=1.
Thus,we can rewrite f(x) as f(x)=x^2+bx+c.(where b,c are constants)

Then substitute f(x)=x^2+bx+c into f[f(x)]=[f(x)]^2,
we get (x^2+bx+c)^2+b(x^2+bx+c)+c = (x^2+bx+c)^2.

Rearranging it,
we can get bx^2+(b^2)x+c(b+1)=0
Since it is always true every value of x,
we can conclude that b=0 , b^2=0 , c(b+1) =0
Solving it,==> b=0 , c=0 .
Thus, f(x)=x^2

Other than the 2 polynomials found,f(x)=0 is undoubtedly the solution too.

Hence,the answers are f(x)=0 , f(x)=1 , f(x)=x^2 #

[kimsiang]

[J-Question 7]

A,B are 2x2 matrixs.When AB-BA=A,prove that A^2=0.

[Aichi University,Literature 1984]

[IchimaruGin]

integrate :
(e^x)(ln x)

Help me solve this question.........PLease.....

[bush]

integrate :
(e^x)(ln x)

Help me solve this question.........PLease.....

If i am not mistaken.....integrate it by parts....

[IchimaruGin]

integrate :
(e^x)(ln x)

Help me solve this question.........PLease.....

If i am not mistaken.....integrate it by parts....

ya......can u show me the steps....
i do it this way:
1. (fix ln x , integrate e^x) - (integrate (the integral of e^x multiply
with the differentiation of ln x)
=(e^x)(ln x) - (integrate [(e^x)/x])
2. then i do integrate by part again by fixing (e^x)

SO....is this the correct method?

[bush]

I think.....by parts then substitution

[kimsiang]

integrate :
(e^x)(ln x)

Here`s my solution:

At 1st,we consider int [e^x/x] dx.
From the series of e^x(e^x=1+x+x^2/2!+x^3/3!+......+x^n/n!+....),

Int [e^x/x] dx = Int [(1+x+x^2/2!+x^3/3!+......+x^n/n!+....)/x] dx
= Int [1/x+1+x/2!+x^2/3!+......+x^(n-1)/n!+.....] dx
= ln x + x + x^2/(2)2! + x^3/(3)3! + .... + x^n/(n)n!
+...

Then,
Int [(e^x)(ln x)] dx = (ln x)(e^x) - Int [e^x/x] dx
= (ln x)(e^x) - (ln x + x + x^2/(2)2! + x^3/(3)3!
+ .... + x^n/(n)n!+.....) #

I hope it is correct :D ...if not,pls correct me....thanks

[IchimaruGin]

integrate :
(e^x)(ln x)

Here`s my solution:

At 1st,we consider int [e^x/x] dx.
From the series of e^x(e^x=1+x+x^2/2!+x^3/3!+......+x^n/n!+....),

Int [e^x/x] dx = Int [(1+x+x^2/2!+x^3/3!+......+x^n/n!+....)/x] dx
= Int [1/x+1+x/2!+x^2/3!+......+x^(n-1)/n!+.....] dx
= ln x + x + x^2/(2)2! + x^3/(3)3! + .... + x^n/(n)n!
+...

Then,
Int [(e^x)(ln x)] dx = (ln x)(e^x) - Int [e^x/x] dx
= (ln x)(e^x) - (ln x + x + x^2/(2)2! + x^3/(3)3!
+ .... + x^n/(n)n!+.....) #

I hope it is correct :D ...if not,pls correct me....thanks

sry but i still dun understand....
i do it like this.....
int [(e^x)(ln x)] dx = (ln x)(e^x) - int [(e^x)/x] dx
= (ln x)(e^x) - {(e^x)(ln x) - int [(ln x)(e^x)] dx}

and everything is balanced out and i left with nothing....dunno why
did i do anything wrong?

[kimsiang]

Oh,it is becoz u do the step 1 to 2 , and then do step 2 to 1.Of coz u wil get back the same thing n everything cancel out.
i.e.
int[ab]= [int(a) . b] - int[int(a) . diff(b)] (1)
= [int(a) . b] - {int(a).int[diff(b)] - int[int[diff(b)].diff[int(a)]}(2)
= [int(a) . b] - {int(a).(b)] - int[(b).(a)]} (3)
= [int(a) . b] - [int(a).(b)] + int[(b).(a)] (4)
= int[(b).(a)] = int[ab]

From step(1) to step(2),u juz applied the integration by parts in the reverse.Am i clear?

About my method,it may look strange to u,but that is the only way that i found.SOrry.

[IchimaruGin]

ooo....so that's the reason i din have anything in the end.....ok thanks a lot kimsiang....i will try out your method.....Thanx ya

[kimsiang]

[J-Question 7]

A,B are 2x2 matrixs.When AB-BA=A,prove that A^2=0.

[Aichi University,Literature 1984]

Some warming up b4 the solution

Cayley-Hamilton's theorem

When A is a 2x2 matrix , A = [ a b ] ,
[ c d ]
the following equation hold:

A^2-(a+d)A+(ad-bc)I=0 -------------- (1)

Trace A = Tr(A) = a+d
Tr(AB)=Tr(BA)
--------------------------------------------------------------------------------------
Solution:
Since AB-BA=A,we can get Tr(AB-BA)=Tr(A)
<==>Tr(AB)-Tr(BA)=Tr(A)
<==>Tr(AB)-Tr(AB)=Tr(A)
<==>Tr(A)=0
<==>a+d=0 ----------- (2)
multiply AB-BA=A by A from left-hand side,
we get A^2B-ABA=A^2 ------------------ (3)
Now,multiply AB-BA=A by A from right-hand side,
we get ABA-BA^2=A^2 ------------------ (4)
adding both, A^2 = A^2B - BA^2 ------------ (5)
Substitute (2) into (1),
we have A^2=-(ad-bc)I
=-det(A)I ------------ (6)
Substitute (6) into right-hand side of(5),
we get A^2=-det(A)IB+det(A)BI
=-det(A)B+det(A)B
=0

[ohmygod]

[J-Question 7]

A,B are 2x2 matrixs.When AB-BA=A,prove that A^2=0.

[Aichi University,Literature 1984]

Some warming up b4 the solution

Cayley-Hamilton's theorem




hey, i'm new here...
Is Cayley-Hamilton's theorem inside the STPM syllabus???

[kimsiang]

hey, i'm new here...
Is Cayley-Hamilton's theorem inside the STPM syllabus???

er...it is inside the further math syllabus but not in math.....

[Seiryu]

further maths? hey can u tell me a lil more bout stpm further maths? i went to mph to find the text book but the girl told me that she couldn't find any stpm further maths textbooks in many mph bookstores around klang valley... i want to take on further maths as another extra subject for stpm coz i think it would help me to further my studies in actuarial science... (General Studies, physics, chemistry, maths t, maths L. :? . is that combi good? or is it better to take off maths L and put in Econs? pls enlighten me on that too :roll: )

thx and god bless.
btw this is my second post on recom.. i post two posts in one day :lol:

[bush]

There are a few things One would like to point out. While there is nothing wrong to take physics, chemistry, maths and FM, it wouldn't be the best choice as STPM chemistry isn't a nice subject to take if you do not have the heart for it. If you do not have the heart for it, you might have a high chance to underperform hence affecting your overall grades ie AAAB or AAAC. It won't look nice if you are looking for a scholarship after STPM.

Economics is good and is encouraged if one wants to take actuarial science but it is not essential. You can still pick it up once you're in university.

Not many people take STPM FM as it is simply horrendous. For those who took it and excelled in it(kimsiang), they will most likely be comfortable with advanced figures. Not many schools offer this subject, TARC offered it but scraped itj ust two years ago. Even if you manage to get into a school which offers this subject, you may find yourself be taught by a clueless nut. You can of course learn it by yourself with minimal guidance provided you have that sort of mathematical intuition and flair.

It is the total opposite for A-levels. There are comparatively many students and institutions offering this subject but the fees don't come cheap for A-levels. Furthermore, it might not be possible to pursue actuarial science at one of the local public universities if your scholarship quest fails.

I would suggest this, take STPM if you are willing to settle for a possible place at one of our local public universities if you fail to secure a scholarship. Take A-levels if you are willing to gamble a bit. It is fairly simple to gain a place in schools like LSE, macquarie, waterloo if you have the necessary grades. Since you have the grades, you might get a scholarship offer(no promises). Now this is where the catch comes in, what if you fail to get any funding? You won't be going anywhere, public universities would be out of the question. This is what makes A-levels risky.

[kimsiang]

further maths? hey can u tell me a lil more bout stpm further maths? i went to mph to find the text book but the girl told me that she couldn't find any stpm further maths textbooks in many mph bookstores around klang valley... i want to take on further maths as another extra subject for stpm coz i think it would help me to further my studies in actuarial science... (General Studies, physics, chemistry, maths t, maths L. :? . is that combi good? or is it better to take off maths L and put in Econs? pls enlighten me on that too :roll: )

thx and god bless.
btw this is my second post on recom.. i post two posts in one day :lol:

ok,of coz.Since there r less n less students taking f.math,textbooks were "disappeared" many years ago....n the textbooks r quite old n were for stpm long time ago...there r a few new topics like number theory,graph theory in the new syllabus(started from year 2002).
i took it in year 2003,4/4 students in my school got A for that.We were quite lucky becoz teachers r willing to help us n we hv some seniors who also took it the year b4 us.
If u really like math n wont give up easily,of coz u can try it.i wil tell u my experience here.
2002(lower 6):form6 started on april n there are 10 students who wanted to study f.math in my school.Since f.math is an advanced subject after we hv studied math t,our study plan is to finish math t in lower 6.Fortunately,we able to finish math T in Sep.Due to JPA scholarship n UTM intake n some other reasons,there 4 students left when we enter upper 6.
2003(upper 6):We started our f.math.Since my teachers hv experience teaching f.math,their explanation were quite good.But when it comes to problem solving,we alwaz cant finish our papers in exam becoz there r more things to be considered n the solutions are quite long.So,i did past year questions from 1980++ till 2002 for stpm , A-level , n also doing some math from IMO in order to get used to it.
But then,i do hv some frens from other states(i m from penang) who studied on theirselves got A also in stpm. So,dun hv to worry that u wil fail if u study on yr own. :)
I hope that my experience wil hv some help on u.... :D

[ohmygod]

i want to take on further maths as another extra subject for stpm coz i think it would help me to further my studies in actuarial science... (General Studies, physics, chemistry, maths t, maths L. :? . is that combi good? or is it better to take off maths L and put in Econs? pls enlighten me on that too :roll: )




I saw your post under Actuarial Science topic in Education Forum.
Since you're planning to do actuarial science, I think it's better for you to take further maths. Coz further maths will introduce you to more advanced mathematics, and it inderectly helps you in passing the actuarial exams. And about the econ, you'll take it in your future time (i mean university time)...but now don't worry about it, it's better for you to be equipped with higher mathematical knowledge as early as possible, then for your university year, you'll feel more easy in the studies of mathematics.

[Seiryu]

I would suggest this, take STPM if you are willing to settle for a possible place at one of our local public universities if you fail to secure a scholarship. Take A-levels if you are willing to gamble a bit. It is fairly simple to gain a place in schools like LSE, macquarie, waterloo if you have the necessary grades. Since you have the grades, you might get a scholarship offer(no promises). Now this is where the catch comes in, what if you fail to get any funding? You won't be going anywhere, public universities would be out of the question. This is what makes A-levels risky.

Yes i am willing to enroll local Us.. If only i get the course i want... because i think ... if u are a fellow of any acturial science board, it would cover up what unis u come from (pls correct me if i am wrong.... ) As for A lvls.. i thought it's harder because SOOO many ppl taking A lvls worldwide compared to STPM so i thought STPM is a safer way of securing a scholarship.....

rite?

And sry for the interuption i know this forum should be bout mathematics T but pls help me to clear my doubts. Thank u and God bless.

[Seiryu]

Since you're planning to do actuarial science, I think it's better for you to take further maths. Coz further maths will introduce you to more advanced mathematics, and it inderectly helps you in passing the actuarial exams. And about the econ, you'll take it in your future time (i mean university time)...but now don't worry about it, it's better for you to be equipped with higher mathematical knowledge as early as possible, then for your university year, you'll feel more easy in the studies of mathematics.

THX A LOT!!!! another thing is stpm econs is in Malay so.... :? it's some sort of prejudice but who cares..

[Seiryu]

Since you're planning to do actuarial science, I think it's better for you to take further maths. Coz further maths will introduce you to more advanced mathematics, and it inderectly helps you in passing the actuarial exams. And about the econ, you'll take it in your future time (i mean university time)...but now don't worry about it, it's better for you to be equipped with higher mathematical knowledge as early as possible, then for your university year, you'll feel more easy in the studies of mathematics.

THX A LOT!!!! another thing is stpm econs is in Malay so.... :? prejudice prejudice... but who cares..

[bush]

I would suggest this, take STPM if you are willing to settle for a possible place at one of our local public universities if you fail to secure a scholarship. Take A-levels if you are willing to gamble a bit. It is fairly simple to gain a place in schools like LSE, macquarie, waterloo if you have the necessary grades. Since you have the grades, you might get a scholarship offer(no promises). Now this is where the catch comes in, what if you fail to get any funding? You won't be going anywhere, public universities would be out of the question. This is what makes A-levels risky.

Yes i am willing to enroll local Us.. If only i get the course i want... because i think ... if u are a fellow of any acturial science board, it would cover up what unis u come from (pls correct me if i am wrong.... ) As for A lvls.. i thought it's harder because SOOO many ppl taking A lvls worldwide compared to STPM so i thought STPM is a safer way of securing a scholarship.....

rite?

And sry for the interuption i know this forum should be bout mathematics T but pls help me to clear my doubts. Thank u and God bless.

Yes, correct. The degree is just to help take the professional papers. But I believe that the exposure will be very different if you get to study in lets say UM and LSE..........

Scholarships at post STPM/A-levels are not so much on grades, the offer from a university(preferably a reputable one) plays a larger role to a certain extend. Having said that, it doesn't really matter if you did A-levels or STPM.

[Seiryu]

Having said that, it doesn't really matter if you did A-levels or STPM.

Thx bush.. at least i am more confident that foreign Unis recognise STPM.

[bush]

Having said that, it doesn't really matter if you did A-levels or STPM.

Thx bush.. at least i am more confident that foreign Unis recognise STPM.

It doesn't matter if you have the grades but the difficulty to score these grades are a different story altogether. That sets these two exams apart.

[ohmygod]

I don't know whether it's a right place to post my problem...But I think the matrix is inside Mathematics T STPM...

Below is my question:


Let A be the nxn matrix whose entries are all 1. Show that

(A-nIn) ij = (-1) ^(n-2).n^(n-2) for all i, j, where (A-nIn)ij denotes the cofactor of the (i, j)- entry of A-nIn. #



Well, what i do is assigning a number for i & j and n, then I prove that it statisfies the equation. But I think it should be better and more correct to prove that for all i,j the equation is satisfied. However, I don't know how to prove that for all i, j and n.
Anyone can help me???

[kimsiang]

I don't know whether it's a right place to post my problem...But I think the matrix is inside Mathematics T STPM...

Below is my question:


Let A be the nxn matrix whose entries are all 1. Show that

(A-nIn) ij = (-1) ^(n-2).n^(n-2) for all i, j, where (A-nIn)ij denotes the cofactor of the (i, j)- entry of A-nIn. #



Well, what i do is assigning a number for i & j and n, then I prove that it statisfies the equation. But I think it should be better and more correct to prove that for all i,j the equation is satisfied. However, I don't know how to prove that for all i, j and n.
Anyone can help me???

Could you pls recheck the question?
maybe it is (A-nIn) ij = (-1) ^(n-1).n^(n-2) ?
and is it A-nIn means A-nI where I is nxn matrix?
Thanks!

[ohmygod]

[quote=ohmygod]
Could you pls recheck the question?
maybe it is (A-nIn) ij = (-1) ^(n-1).n^(n-2) ?
and is it A-nIn means A-nI where I is nxn matrix?
Thanks!


oh...ya..sorry... is actually (A-nIn) ij = (-1) ^(n-1).n^(n-2)...
I is identity matrix for the dimension of nxn...(n. In = nIn, coz i don't know how to type the lower case for In)


thanks.

[kimsiang]

oh...ya..sorry... is actually (A-nIn) ij = (-1) ^(n-1).n^(n-2)...
I is identity matrix for the dimension of nxn...(n. In = nIn, coz i don't know how to type the lower case for In)


Oh,thankyou.It is obviously the level of this problem is higher than STPM.But i will try to work out the solution when i am free.
What you did(substitute certain value for n,i,j) isn't the correct way since you only verified the equation is true only for some value but not all.I think we shd seperate the solution into 2 parts.
1st:when i=j (since all (A-nIn) ij are the same)
2nd:when i not = j ((A-nIn) ij looks different but what we hv to do is to change the shape and make all of them the same)

[Smilehoe]

I certainly are not in any position to answer those questions posted, but I have this link that may be useful. http://mathforum.org/dr.math/ask/agree.html

They reply quite fast for any maths-related questions. Have a nice day everybody.

[ohmygod]

Oh,thankyou.It is obviously the level of this problem is higher than STPM.But i will try to work out the solution when i am free.
What you did(substitute certain value for n,i,j) isn't the correct way since you only verified the equation is true only for some value but not all.I think we shd seperate the solution into 2 parts.
1st:when i=j (since all (A-nIn) ij are the same)
2nd:when i not = j ((A-nIn) ij looks different but what we hv to do is to change the shape and make all of them the same)


I certainly are not in any position to answer those questions posted, but I have this link that may be useful. http://mathforum.org/dr.math/ask/agree.html

They reply quite fast for any maths-related questions. Have a nice day everybody.


Hey, thanks kimsiang and smilehoe. Actually I tried quite few times but failed so somehow gave up this questions. Anyway, now kind of "reminded", erm...i 'll check up that website.

[lyt87]

i got question to ask.

1. the gradient of side BC of a rhombus ABCD is 2 and its coordinates A n C are (-3,-4) and (5,4) respectively. what is the area of the rhombus?

[kimsiang]

i got question to ask.

1. the gradient of side BC of a rhombus ABCD is 2 and its coordinates A n C are (-3,-4) and (5,4) respectively. what is the area of the rhombus?

http://www.imagestation.com/picture/sraid202/pa43a598f9f2ca3c218af8789f9266e9c/efd53c17.jpg

From the diagram,we can deduce that the side B is at the position below line AC.
Let B(x,y),since the gradient of BC = 2
==> (y-4)/(x-5) = 2
==> y = 2x - 6 ----------(1)
But then,since ABCD is a rhombus, AB = BC
==> sqrt[(x+3)^2+(y+4)^2] = sqrt[(x-5)^2+(y-4)^2]
==> [(x+3)^2+(y+4)^2] = [(x-5)^2+(y-4)^2]
==>(x^2+6x+9+y^2+8y+16) = (x^2-10x+25+y^2-8y+16)
==> 16x+16y = 16
==> x+y = 1 -----------(2)
solving (1) and (2) simultaneously,
we get x = 7/3 , y = - 4/3 respectively. ==> B( 7/3 , - 4/3 )

Midpoint of AC , M = ( [5-3]/2 , 0/2 ) = (1,0)
Hence , since M is also mid point of BD,
==> (1,0) = ( [h+(7/3)]/2 , [k-(4/3)]/2 )
==> 1 = [h+(7/3)]/2 , 0 = [k-(4/3)]/2
==> h = - 1/3 , k = 4/3
==> D( -1/3 , 4/3 )
now we have lenght of AC = 8(sqrt2) , length of BD = 8(sqrt2)/3
Since area of rhombus = 0.5(AC)(BD)
==> Area = 0.5[8(sqrt2)][8(sqrt2)/3]
= 64/3 #

OR
Using Heron's Theorem
When the length of sides of a given triangle are a,b,c respectively
the area of it , A = sqrt(s(s-a)(s-b)(s-c))
where s = (a+b+c)/2

Applying the law, we let AC=a = 8(sqrt2) , AB=b= 8(sqrt5)/3
and BC= c = 8(sqrt5)/3 .
Hence , s = 4(sqrt2)+8(sqrt5)/3
==> A = sqrt(s(s-a)(s-b)(s-c))
= 32/3
Since A = Area of triangle ABC,
Area of the rhombus = 2A = 64/3 #
Which agrees with the result above.

[youngyew]

Am just wondering whether it's Hero's formula or Heron's formula?

[kimsiang]

Am just wondering whether it's Hero's formula or Heron's formula?

Oups,I missed out the "n"..should be Heron's formula...thanks for reminding me

[caarzee]

Hey, i have a stats problem. the Q goes as follows:

The hardness (Rockwell hardness) of a metal specimen is determined by impressing the surface of the specimen with a
hardened point, and then measuring the depth of the penetration. The hardness of a certain alloy is normally distributed
with mean of 70 units and a standard deviation of 3 units.

If a specimen is acceptable only if its hardness is between 66 and 74 units, which option is the probability that a randomly
chosen specimen is acceptable?

My Answer is 0.682.

the formula goes like this;

p(66<x<74); min=70, sigma=3
p[66-70)/3 < (x-min) < (74-70)/3
p(-0.75 < z < 1.333)
1 - p(z > 1.333) - p(z > 0.75)
1 - 0.0913 - 0.2266
= 0.6821

but i was told the answer is not what i gave.

anyone with ideas?

thanks.

[kimsiang]

Hey, i have a stats problem. the Q goes as follows:

The hardness (Rockwell hardness) of a metal specimen is determined by impressing the surface of the specimen with a
hardened point, and then measuring the depth of the penetration. The hardness of a certain alloy is normally distributed
with mean of 70 units and a standard deviation of 3 units.

If a specimen is acceptable only if its hardness is between 66 and 74 units, which option is the probability that a randomly
chosen specimen is acceptable?

My Answer is 0.682.

the formula goes like this;

p(66<x<74); min=70, sigma=3
p[66-70)/3 < (x-min) < (74-70)/3
p(-0.75 < z < 1.333)
1 - p(z > 1.333) - p(z > 0.75)
1 - 0.0913 - 0.2266
= 0.6821

but i was told the answer is not what i gave.

anyone with ideas?

thanks.

Isn't it
p [66-70)/3 < (x-min)/3 < (74-70)/3]
= p [-1.333 < z < 1.333]
= 1 - 2p(z > 1.333)
= 0.8174 #

[caarzee]

hi kimsiang,

thanks for spotting my silly mistake!
appreciate it.

[doodle]

I don't know whether it's a right place to post my problem...But I think the matrix is inside Mathematics T STPM...

Below is my question:

Let A be the nxn matrix whose entries are all 1. Show that

(A-nIn) ij = (-1) ^(n-2).n^(n-2) for all i, j, where (A-nIn)ij denotes the cofactor of the (i, j)- entry of A-nIn. #

Well, what i do is assigning a number for i & j and n, then I prove that it statisfies the equation. But I think it should be better and more correct to prove that for all i,j the equation is satisfied. However, I don't know how to prove that for all i, j and n.
Anyone can help me???

A good question there, and here is my solution: 060319.pdf (http://www.geocities.com/thebigbagoftricks/recom/060319.pdf)

[ohmygod]

thank you doodle :o

I have a maths T question here. hope somebody can help.thanks.

By using the method of differences, find the sum of the first n terms of the series whose rth term, Ur , are as follows

Ur = 1/r(r+3)


does anybody know how to solve this?


Answer : 11n^3 + 48n^2 + 49n / 18(n+1)(n+2)(n+3)

[kimsiang]

I have a maths T question here. hope somebody can help.thanks.

By using the method of differences, find the sum of the first n terms of the series whose rth term, Ur , are as follows

Ur = 1/r(r+3)


does anybody know how to solve this?


Answer : 11n^3 + 48n^2 + 49n / 18(n+1)(n+2)(n+3)

Haha,didnt solve math problem here for a long time ....Here u go...

Let the sum of the 1st n-term of the series be S.
Hence,S = summation over r from 1 to n of Ur
= summation over r from 1 to n of 1/r(r+3)
= (1/3)summation over r from 1 to n of (1/r - 1/(r+3))
(Since 1/r(r+3)=(1/3)[1/r - 1/(r+3)])
= (1/3)[ 1/1 - 1/4
+ 1/2 - 1/5
+ 1/3 - 1/6
+ 1/4 - 1/7
+ 1/5 - 1/8
+ 1/6 - 1/9
.
.
.
.
+ 1/(n-3) - 1/n
+ 1/(n-2) - 1/(n+1)
+ 1/(n-1) - 1/(n+2)
+ 1/n - 1/(n+3)]
(As you can see ,the terms 1/4,1/5,1/6,.....1/n can be cancelled.)
Thus, S =(1/3)[1+1/2+1/3- 1/(n+1)- 1/(n+2)- 1/(n+3)]
=(1/3)[11/6 - {1/(n+1) + 1/(n+2) + 1/(n+3)}]
=11(n+1)(n+2)(n+3)-6{(n+2)(n+3)+(n+1)(n+3)+(n+1)(n+2)}
18(n+1)(n+2)(n+3)
=11(n^3+6n^2+11n+6)-6{3n^2+12n+11}
18(n+1)(n+2)(n+3)
=11n^3+66n^2+121n+66)-{18n^2+72n+66}
18(n+1)(n+2)(n+3)
=11n^3+48n^2+49n
18(n+1)(n+2)(n+3) #

[bush]

The only difficult part in this question is the realisation. The realisation that partial fractions must be performed to obtain the practical form of the expression in order to solve the problem. No '-', no solution.