STPM Maths T help

[ChristopherCWL]

The question is : If A = 2-i and B = 1+3i, simplify each of the following in x+yi from where necessary.
1) A + B
2) lABl
3) A-iB over 1+iB

[lawteoh]

The question is : If A = 2-i and B = 1+3i, simplify each of the following in x+yi from where necessary.
1) A + B
2) lABl
3) A-iB over 1+iB

This look quite easy honestly..

1. Using my brain, the ans is 3 + 2i . Jz do it like algebra.

2. AB = 2 +5i + 3 (note: i square wouuld be -1 as i = sq root -1 )
AB = 5+5i
Thus, the modulus would be sq root (5 square +5 square), which is sq root 50

3. substitute..

Then, u will get until here.. (5-2i)/(4i-2) ..Remember, u need to multiply by the conjugate to get rid of the denominator.. In this case, multiply by 4i+2 ..

When u solve it, the final answer would appear as (8i-9)/10 .. hope I din do any careless mistakes in this ques.. basically, the concept is like that. ;)

[ChristopherCWL]

Buddy, the answer given by my math teacher is totally diff with what u have calculated...now i wonder which side do i have to look at...swt...

[markwongsk]

The question is : If A = 2-i and B = 1+3i, simplify each of the following in x+yi from where necessary.
1) A + B
2) lABl
3) A-iB over 1+iB

1) A + B = 2+1 + (3-1)i = 3 + 2i
2) AB = (2-i)(1+3i) = 5 + 5i

modulus = (25+25)^1/2 = (50)^1/2 simplify

3) A-iB = 2-i-i+3 = 5-2i
1+iB = 1+i-3 = -2+i

therefore A-iB over 1+iB = (5-2i) over (-2+i) solve it

[nickvl]

The question is : If A = 2-i and B = 1+3i, simplify each of the following in x+yi from where necessary.
1) A + B
2) lABl
3) A-iB over 1+iB

for ques
1) (2-i)+(1+3i) = (2+1) + (-i+3i)
= 3 + 2i

2) dunno...havent reach that part yet

3) -12 - i over 5

[lawteoh]

Buddy, the answer given by my math teacher is totally diff with what u have calculated...now i wonder which side do i have to look at...swt...

Oops.. i calculate wrongly for part c.. a bit careless when substituting.. lol.. see mark wong's answer for part c la.. cz using my brain to do it that time.. so, there's a possibility in being careless.. =)

Continuing from Mark Wong's step,

therefore A-iB over 1+iB = (5-2i) over (-2+i) solve it

so, just multiply by -2-i/-2-i to destroy the denominator..

U should get (-12-i)/ 5 , as what nickvl said..

and for part b, after simplifying it, u will get 5 sq root 2.

[ChristopherCWL]

Thx for the answers from all of u...THx...i solved it now ^^

Hey, i got new question to ask again...pls help me answer it...

1) If (3^x+1)(4^x+2) = 5^x+3, find the value of x in 2 significant figure

2) Find, correct to 2 significant figures, the value of x such that
e^x - 2e^-x
------------- = 1 / 5
2e^x + e^-x

3) Find the values if x in surd form if

log3 x^2 + log3 x = log9 27

4) Determine the value of a if 2^1/2 + a i over 1+ 2^1/2 i is a real number
and find this real number....(complex number question)

[Johnivan]

Thx for the answers from all of u...THx...i solved it now ^^

Hey, i got new question to ask again...pls help me answer it...

1) If (3^x+1)(4^x+2) = 5^x+3, find the value of x in 2 significant figure


(3^x+1)(4^x+2) = 5^x+3
[3(3^x)][4^2(4^x)] = 5^3(5^x)
48(12)^x = 125(5^x)
48 = (5 )^x
125 (12)

Then solve using lg (48/125) / lg (5/12) = x

[tecklee89]

Third question is quite easy:
log3 x^2 + log3 x = log9 27
3log3 X = (log3 27)/(log3 9)
= 3/2
log3 X = 3/6
X= sq rt (3)
:P

[lawteoh]

Thx for the answers from all of u...THx...i solved it now ^^

Hey, i got new question to ask again...pls help me answer it...



2) Find, correct to 2 significant figures, the value of x such that
e^x - 2e^-x
------------- = 1 / 5
2e^x + e^-x



For ques 2, answer should be 0.65

working.. Just take out e^-x

[e^-x ( e^2x -2)] / [e^-x (2e^2x +1) ] = 1/5

Cancel the e^-x. you will get

(e^2x -2)/(2e^2x +1) = 1/5
a few more steps later, u will get this..

e^2x = 11/3 .. and im sure u can solve that. Just ln both sides.

[ChristopherCWL]

New question !!!

1) If lg a N = x and lg b N = y, prove that lg ab N = xy over x + y

[tecklee89]

http://farm4.static.flickr.com/3572/3569573075_91b69e8ce8_b.jpg

Check it out some of the examination paper on my blog..
http://teckleewee.blogspot.com/2009/04/stpm-notes-trial-stpm-and-so-on.html

good luck

[nickvl]

New question !!!

1) If lg a N = x and lg b N = y, prove that lg ab N = xy over x + y

oh ho ho...i did this ques twice... Let's see if i remember correctly...

if lg a N =x, a^x=N and also a=N^(1/x)
lg b N = y, then b^y=N and b=N^(1/y) <---gotten by log laws

so ab = (N^(1/x) )(N^1/y)
= N^(1/x+1/y)
ab =N^((y+x)/xy)

take log to base ab
so log ab ab = log ab N^((y+x)/xy)
1 = (y+x)/xy log ab N
1/(y+x)/xy = log ab N
therefore, log ab N = xy/(x+y)

[anbbly]

help me pls~~
4^x-3 . 5^2-3x = 20
wat is x??
y i cant use 1=x-3 or 1= 2-3x to solve tis question?

[extreme]

help me pls~~
4^x-3 . 5^2-3x = 20
wat is x??
y i cant use 1=x-3 or 1= 2-3x to solve tis question?


the soln:
4^(x-3).5^(2-3x)=20
(x-3)ln 4.(2-3x)ln 5=20 assume ln4.ln5=J
(xln 4 -3ln 4).(2 ln 5 -3x ln5)=20
2x J-(3x^2) J- 6 J+ 9x J =20
J(11x-3x^2-6)=20

thus
(-3)(x^2)+11x-6=(20)/(J)=(20)/(ln4 ln5)

solve this and u get x


this is the way I'm gonna approach this type of question.Hope this helps you.


the soln:
4^(x-3).5^(2-3x)=20
(x-3)ln 4.(2-3x)ln 5=20 assume ln4.ln5=J
(xln 4 -3ln 4).(2 ln 5 -3x ln5)=20
2x J-(3x^2) J- 6 J+ 9x J =20
J(11x-3x^2-6)=20

editing my original working to the correct method
(x-3) ln 4+(2-3x) ln 5=20
x(ln 4 -3 ln5)+(2 ln 5 -3 ln4)=2
find x which is
x=(2- (2 ln 5 -3 ln4))/(ln 4 -3 ln5)

forgotten about the basic rule of log sometimes
XDXD


:)):))

[markwongsk]

New question !!!

1) If lg a N = x and lg b N = y, prove that lg ab N = xy over x + y

You can use Nick's method or mine, which is thus:

lg N base a = x therefore lg a base N = 1/x
lg N base b = y therefore lg b base N = 1/y

adding both logarithms = lg ab base N (logarithm laws) = 1/x + 1/y = (x+y)/(xy)
proven Quod ed demonstratum.

4^x-3 . 5^2-3x = 20
wat is x??
y i cant use 1=x-3 or 1= 2-3x to solve tis question?

For this one, noticing that 20 = 4X5 will help a lot eg:

4^(x-3) . 5^(2-3x) = 4X5

regrouping the indices

4^(x-4) = 5^(3x-1)

Taking lg or ln on both sides:

(x-4)lg 4 = (3x-1)lg5

easy to solve from here =)

the soln:
4^(x-3).5^(2-3x)=20
(x-3)ln 4.(2-3x)ln 5=20 assume ln4.ln5=J
(xln 4 -3ln 4).(2 ln 5 -3x ln5)=20
2x J-(3x^2) J- 6 J+ 9x J =20
J(11x-3x^2-6)=20

thus
(-3)(x^2)+11x-6=(20)/(J)=(20)/(ln4 ln5)

solve this and u get x


this is the way I'm gonna approach this type of question.Hope this helps you.


:)):))

Uhm... this method isn't accurate because in the second step when you took the logarithm, your left hand side is wrong. When you take a logarithm, you must take the whole side ie:

lg [4^(x-3).5^(2-3x)]

but according to the laws of logarithm, this is equivalent to

lg [4^(x-3)] + lg[5^(2-3x)]

which isn't the same as (x-3)ln 4.(2-3x)ln 5

A simpler example is this:

lg xy =???

Thus, the reasoning should go lg xy = lg x + lg y and not (lg x) X (lg y)

help me pls~~
4^x-3 . 5^2-3x = 20
wat is x??
y i cant use 1=x-3 or 1= 2-3x to solve tis question?

The reason why you cannot use an unknown to substitute the indices is because firstly, the base is not included in your substitution... Even if you did make a substitution of for example, 4^x = y, you wont get an answer because 5^x and 4^x have nothing in common (except they are to the power x)... and thus are not related easily in terms of y... you only use substitution when the equation contains a base shareable between the two numbers. eg:

4^x + 2^x = 10

notice that 4^x = 2^2x + 2^x = 6... therefore 2 is the related base... we substitute 2^x = y and therefore

y^2 + y - 6 = 0

solving yields y = 2 (not y = -3?? as 2^x cannot be negative)

Don't forget that we want to find x

Therefore 2^x = 2, x = 1 :laugh

Hope this helps hehe

[daisylady]

sorry.. i've been thinking all night for this lol.. please help please? help me to solve this maths question.. i'm stupid so please help meee dear clever people! heehe

9/25x-5/3x^-2=0

i just want to know that so i can get the idea so i can move on to other ques.. plz?

[BGT]

sorry.. i've been thinking all night for this lol.. please help please? help me to solve this maths question.. i'm stupid so please help meee dear clever people! heehe

9/25x-5/3x^-2=0

i just want to know that so i can get the idea so i can move on to other ques.. plz?

The answer is 3/5. I suppose u got stucked with 5/(3x^-2)? This is equal to (5x^2)/3. Think over about it. :)

[chikugunya]

sorry.. i've been thinking all night for this lol.. please help please? help me to solve this maths question.. i'm stupid so please help meee dear clever people! heehe

9/25x-5/3x^-2=0

i just want to know that so i can get the idea so i can move on to other ques.. plz?
juz like wats BGT had told earlier...5/(3x^-2) is equal to (5x^2)/3..
so u can simply bring this to the right hand side...
n then u can simplify both side...
n the final answer will be 3/5.....:P

[RheBA]

sorry.. i've been thinking all night for this lol.. please help please? help me to solve this maths question.. i'm stupid so please help meee dear clever people! heehe

9/25x-5/3x^-2=0

i just want to know that so i can get the idea so i can move on to other ques.. plz?

First of all the questions is written in a rather ambiguos manner...
What does 9/25x mean?? do u mean 9/(25x) or do u mean (9/25)x??
Same for the second term...
Anyways...word of advice..do refrain from calling urself stupid..one must always think positively about themselves.. if u believe that ur smart, i'm sure u will be smart!
P/S: asking help from ppl doesn't make u stupid..My younger sister(f4) was able to answer one of my Form 6 Math problems..

[bcf1991]

using algebraic law of sets,prove that
(B-C) U (C-B)=(B U C)n(B n C)'

[nickvl]

using algebraic law of sets,prove that
(B-C) U (C-B)=(B U C)n(B n C)'

LHS : (B-C) U (C-B)

(B n C') U (C n B')------> difference of set

[( B n C') U C)] n [(B n C') U B') -----> distributive law (expand)

[ ( B U C) n (C' U C) ] n [(B U B') n (C' U B')]--->distributive law (expand again)

(B U C) n E ) n ( E n (B' U C')----> complement law

(B U C) n (B' U C')-----> identity law

(B U C ) n (B n C)' ----> de Morgan's law (shown)/RHS

[RheBA]

The polynomial P(x) = a(x^4)+b(x^2)+x+1 gives a remainder x+3 when divided by (x^2-1). Find the values of a and b.
Actually i undrestand how to get the answer..but..I am unsure on how to express it because a+b will always equal to 2 because they are of the same value...

[markwongsk]

notice that when the polynomial is divided by http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=x%5E2-1 you get a remainder of x+3... This implies that either when x = 1 or x = -1, you get a remainder of x+3... substituting x = 1 and x = -1 into the polynomial yields you two equations, which is sufficient for you to solve a and b... hope that helps :laugh

[Lil_live]

The polynomial P(x) = a(x^4)+b(x^2)+x+1 gives a remainder x+3 when divided by (x^2-1). Find the values of a and b.
Actually i undrestand how to get the answer..but..I am unsure on how to express it because a+b will always equal to 2 because they are of the same value...

goodness Rhe, You are doing polynomial already..? Paper 1 or 2 book?

[RheBA]

notice that when the polynomial is divided by http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=x%5E2-1 you get a remainder of x+3... This implies that either when x = 1 or x = -1, you get a remainder of x+3... substituting x = 1 and x = -1 into the polynomial yields you two equations, which is sufficient for you to solve a and b... hope that helps :laugh

Hehe...i do know that...bt notice too that when u do substitute both values of x into the equation P(x) u end up with similar equations which is a+b=2..Can u please explain further?? ^_^

Edit: Upon further inspection of the question, i realised that it was indeed a poor question. The second part of the question asks for the value of 100010101 divided by 9999. The value of a and b can be obtained when x=100 is substituted P(x) equals to 100010101. But since the question actually asks for the value of a and b without any relation to the second part of the question it is evident that this is a poor and lousy question and should not be attempted by anyone!

[PsyMaths]

If 3(4^p) = 5(2^q) and 9(8^p) = 10(4^q), show that 2^(p+1)=5.

Dunno how to continue after converting all the value in (bracket) to base 2.

[nicodemus]

If 3(4^p) = 5(2^q) and 9(8^p) = 10(4^q), show that 2^(p+1)=5.

Dunno how to continue after converting all the value in (bracket) to base 2.

3(4^p) = 5(2^q)

3(2^2p) = 5(2^q) ------- Eq 1

9(8^p) = 10(4^q)

9(2^3p) = 10 (2^2q) -------- Eq 2


Eq 2 divide Eq 1,

3(2^p) = 2(2^q)

Rearrange,

(2^q) = 3/2 (2^p) -------- Eq 3

Substitute Eq 3 into Eq 1,

3(2^2p) = 15/2 (2^p)

6(2^2p) = 15(2^p)

(2^2p)/(2^p) = 15/6

(2^p) = 5/2

2(2^p) = 5

2^(p+1) = 5

QED.... :-)

[Aced]

can anyone tell me the most effective way of getting high marks in Mathematics T?

[DonFrank]

can anyone tell me the most effective way of getting high marks in Mathematics T?

1)Understand the problems first
2)Think of how to solve it first before writing your solution
3)Be logical in answering; once you have the notion that your solution is illogical, change it and present a logical one.
4)Make sure you are clear in presenting your answers; all the notations and symbols should be correctly used.
5)Don't do shortcuts and don't assume that your examiners know what you are writing most of the time; show everything clearly to them.

[Equus]

At this time, is it better to do more exercises(excluding past year ques.) or go through all the notes once again? STPM maths is so different from SPM's, so does more practice help?

[teenjay]

more practice will help....try to do exercises from purple book ( dunno which company liao), some question will come out from there because the book content past year question ( 1980-an) which is quite difficult.....

[dkpit]

Can someone help me out with these questions:
1.Given that 2^a=3^b=18^c, show that ab=c(b+2a)
2.Show that 2^n + 2^(n+1) + 2^(n+2) is divisible by 7 where n is a positive integer.

[PsyMaths]

Can someone help me out with these questions:
1.Given that 2^a=3^b=18^c, show that ab=c(b+2a)
2.Show that 2^n + 2^(n+1) + 2^(n+2) is divisible by 7 where n is a positive integer.

2^n + 2^(n+1) + 2^(n+2)
= [(2^n) + 1] + [(2^n)+ 2] + [(2^n) + 4]
= (2^n) (1 + 2 + 4)
= 7 (2^n)

which is divisible by 7

[jiacheng91]

1.
start from 2 equations

2^a=3^b
3^b=18^c

add lg to both sides. then notice that 18=2x3x3

then try to solve it

[dkpit]

can someone help me at these questions?
1. The tangent to the curve y=e^(2x) at pt P whr x=1 meets axis at A. The normal at P meets the y axis at B. Find the area of triangle APB.
2. If y=(ax+b)e^(-2x) whr a and b are constants, find dy/dx and d2y/dx2. Find the values of the constants p and q such that d2y/dx2 + p dy/dx +qy=0 for all values a and b.
3.how to show somethg is a subset by proving?

[Feni2008]

1.clues:you nd to differentiate y=e^(2x),then put x=1 into it,it would become ur gradient o the tangent at pt P,bt seriously u may miss some word for what axis at A.the gradient of normal can be obtained from gradient of tangent.touching axis meaning one of the value(either x or y) must be 0.
2.clues:try product rule of differentiation.
3.you need to learn no only set law,but oso set definition like wat is definition of 'and' and 'or',check with pelangi reference book the correct method.

if u wan more detail i can help u.

[dkpit]

1. A meets at x axis. I have prob determining whether A and P have the same eqn o not. Thn the area normally use the coordinates method? The one that 1/2 X modulus coordinates?
2. For this question, do I have to solve simultaneously? If yes, which two eqns do I use to solve?
3. can i know which pelangi book you mean? the small one, the top is yellow?

[Feni2008]

1. A meets at x axis. I have prob determining whether A and P have the same eqn o not. Thn the area normally use the coordinates method? The one that 1/2 X modulus coordinates?
2. For this question, do I have to solve simultaneously? If yes, which two eqns do I use to solve?
3. can i know which pelangi book you mean? the small one, the top is yellow?

1.here have 2 condition, if u r using gradient of tangent at P,then ur straight line with gradient of tangent will touch/passing through pt A at x-axis.what's the meaning? think about it.
2.u can 1st differentiate 1st time for the eqn,if some part can be subtituted with y=(ax+b)e^(-2x),u can substitute it.try to think what will gonna happen if you differentiate y in corresponding to x ,for example,if y^2,it would differentiate into 2y(dy/dx) etc.it might help u to simply all ur working when u faced same type of question.
3.i mean the small one,front page might have red colour sumthing.

[dkpit]

Feni 2008, thanks for giving me some directions on the questions. I think I understood some already. I'll try them out again and I'll go find that book.