[STPM] Mx T

[sunshine88]

hehe..this is my first question posted in this thread...I'm slightly building up the interest for maths. somehow... :P Alrite..can anyone solve this question for me...
It's from the topic trigonometry.This question asks us to prove.
1 - sin θ/1 + sin θ = sec θ - tan θ

I have a slight twitching here.....I know that 1/cos=sec and sin/cos = tan...and so when i wanna solve LHS

i write .....1 - sin θ / cos θ.....then?????....maybe my step could be wrong.Can anyone correct me?

[ccyik]

1 - sin θ/1 + sin θ = sec θ - tan θ


1 - sin θ/1...divided by 1??

[sunshine88]

it's 1 sin θ [over] 1 + sin θ ...


Actually,i solved it already,thanks..hehe..Just sort it out today with my teacher.

[allessioz]

Is it possible to find a polynomial of degree 3 that have no real zeros? Give an
example of such polynomial if you think it exists, or explain why it is impossible
to find such a polynomial...

sry..guys may i ask for your help?? thx 1st

[chiunlin]

Is it possible to find a polynomial of degree 3 that have no real zeros? Give an
example of such polynomial if you think it exists, or explain why it is impossible
to find such a polynomial...

sry..guys may i ask for your help?? thx 1st

Hint: let's say the polynomial is ax^3+bx^2+cx+d, what happens when x = +- infinity? Then apply Intermediate Value Theorem.

[chong20]

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 question2,
substitute z=x+iy into |z+2|=2|z-i| and do some algebraic and you will reach the equation 3x^2 + 3y^2 -4x -8y = 0

For question 3,
substitute z=x+iy into the equation and you can solve for x.

[terence_lim]

Is it possible to find a polynomial of degree 3 that have no real zeros? Give an
example of such polynomial if you think it exists, or explain why it is impossible
to find such a polynomial...

sry..guys may i ask for your help?? thx 1st

polynomial of degree 3 that has no real roots/ zeros does not exist.. polynomials of degree 3 will have at least one real zero.. alright, this can be illustrated by sketching 3 possible cubic curves:
(i) with only one real root (the curve intersects the x-axis only once with the turning point/ stationary point of the cubic curve hanging either above or below the x-axis)
(ii) with three real roots where one pair of the roots are equal (the cubic curve intersects the x-axis at only one point with its turning point/ stationary point touches the x-axis)
(iii) with three real and distinct roots (the cubic curve intersects the x-axis at 3 different points)

furthermore, according to the conjugate pair theorem,
if P(x) = 0 is a polynomial equation with real coeffifients, then when a+bi is a root, a-bi is also a root

IF a polynomial with degree 3 has no real roots, means this polynomial is expected to has only complex roots: (a+bi), (a-bi), (c+di), (c-di) (4 roots in total)
this is impossible since polynomials with degree 3 have three roots..

an example of polynomial with only one real root is
P(x)=x^3-x^2-7x+15
to find the roots of this polynomial:
by trial & error(in accordance with factor theorem) or calculator, you can easily determine the real root
it is advisable to substitute (1,-1,3,-3,5,-5..factors of 15) into P(x)
in this case,
P(-3) = 0, hence x=-3 is a zero of P(x) and (x+3) is a factor of P(x)
then divide P(x) by (x+3) by using long division, you will get (x^2-4x+5)as quotient
hence, P(x)=(x+3)(x^2-4x+5)
find the complex roots by solving (x^2-4x+5) (by completing the square or by using quadratic formula)

hence, the roots of P(x) are x=-3, x=(2+i), x=(2-i)

similarly, question may ask you to find the cubic equation with roots -3 and (2+i)
in this case, you are expected to know the other root is (2-i) in accordance with conjugate pair theorem.. this is a tricky question, teachers used to ask this type of question in exams..

[terence_lim]

question on sequences&series:

given that C1 is a circle of radius 1 unit. equilateral triangle T1 is inscribed in C1. C2 is inscribed in T1, and so on. show that the areas of the triangles form a G.P. hence, compute Sigma (from k=1 to infinity) area(Tk)

answer: must prove that the common ratio is 1/4
area: surd 3 unit^2

this question came out in our mid year exam (maths T paper 1)
it is a modified malaysia mathematical olympiad past yr question..

[terence_lim]

Hint: let's say the polynomial is ax^3+bx^2+cx+d, what happens when x = +- infinity? Then apply Intermediate Value Theorem.

chiunlin, could you pls explain how to solve this question by using this theorem?
thanks in advance.

[holyvin]

It looks so easy but i cant seem to find the answer when im trying to equate x into a factor of zero.

2
--------------------------
(1-x)^2(1+x^2)

[withoutwax]

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 Q1,

(x^1/3 + y^1/3)^3 = x+y+3(x^2/3)(y^1/3)+3(x^1/3)(y^2/3)
= x+y+3(x^1/3)(y^1/3)[x^1/3 + y^1/3]

x^1/3 + y^1/3=z^1/3
(x^1/3 + y^1/3)^3=z
x+y+3(x^1/3)(y^1/3)[x^1/3 + y^1/3]=z
x+y-z+3(x^1/3)(y^1/3)[x^1/3 + y^1/3]=0
x+y-z=-3(x^1/3)(y^1/3)[x^1/3 + y^1/3]
{x+y-z=-3(x^1/3)(y^1/3)[x^1/3 + y^1/3]}^3
(x+y-z)^3=-27xyz
(x+y-z)^3+27xyz=0 -proved



hehe..hope dat helps..
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[lawteoh]

Is there anyway to post X^3 into x3? (the 3 ought to be at the top).. I had a great headache by seeing what u guys post..

btw, I thought Daniel's question is taken from the Pelangi Book? the question looked familiar.. but its easy.. Miscellaneous exercise 3 is tough...... need some time in understanding it.. since the teacher in school mainly copies out the examples on the board.. :x

[Sillyboy]

Is there anyway to post X^3 into x3? (the 3 ought to be at the top).. I had a great headache by seeing what u guys post..

btw, I thought Daniel's question is taken from the Pelangi Book? the question looked familiar.. but its easy.. Miscellaneous exercise 3 is tough...... need some time in understanding it.. since the teacher in school mainly copies out the examples on the board.. :x

'Copy paste teacher'...... :P

[day_dreamer]

Is there anyway to post X^3 into x3? (the 3 ought to be at the top).. I had a great headache by seeing what u guys post..

btw, I thought Daniel's question is taken from the Pelangi Book? the question looked familiar.. but its easy.. Miscellaneous exercise 3 is tough...... need some time in understanding it.. since the teacher in school mainly copies out the examples on the board.. :x

'Copy paste teacher'...... :P

Don't rely on teacher for Maths T , do your own studies .. It's better ~

[hercules]

Anybody knows how to integrate


(x^2)*(1+x^2)^(1/2) ?

I have tried integration by parts, trigonometric substitution and all the tecniques. It didnt go well...................................................


HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[Sillyboy]

Anybody knows how to integrate


(x^2)*(1+x^2)^(1/2) ?

I have tried integration by parts, trigonometric substitution and all the tecniques. It didnt go well...................................................


HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Is it X power 2 multiply (1+x^2)^(1/2)??

If the function is as above, you might have to consider hyperbolic functions. Substitute X with Sinh U and you will get the answer.

Let X=Sinh u
dx/du= Cosh u

after substitution, function is as below:

Integrate sinh^2(u)*cosh^2(u) du

=1/4 integrate sinh2u du
=(cosh2u)/2

[holyvin]

Pelangi m/s 215 Exercise 5.7

7.Find the equation of the tangent and normal to the parabola y^2=4x at the point (t^2,2t). If P is the point with t = 2^1/2 , find the coordinates of the point Q where the normal at the point P meets the parabola again. If O is the origin , show that OP is perpendicular to OQ.

I tried intersecting the normal equation with the parabola and substitute t with 2^1/2 after that but i cant seem to find Q...ahhh u muist help me.......so long still i cant solve it XD! ty...

[withoutwax]

Is your normal equation: y-2t = -tx + t^3 ?
So normal at P would be
y=4[(2)^1/2] - x(2)^1/2 as you substitute t= (2)^1/2

parabola equation:
y^2 = 4x
sub normal at P into it
{4[(2)^1/2] - x(2)^1/2}^2 = 4x
.
.
.
x^2 -10x +16=0
.
.
x=8 or x=2
y= -4(2)^1/2 y= 2(2)^1/2
Q(8, -4(2)^1/2) P(2, 2(2)^1/2)
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[holyvin]

yes that its the correct equation and answer...and its really fast som1 replied ah ty for the answer ^^

[withoutwax]

lol..no prob..post more..next time :P
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[lawteoh]

Without wax, the most expert in maths here.. haha.. just to ask.. currently, ur teacher in school is teaching what chapter? Mine until half of differentiation ady.. but we as students can't understand what he is talking most of the time.. lol

[withoutwax]

wow..ur tcher is a he..mine is a she...she sucks too..well..we also dun bother bout wad she say..when she walk around..we'll juz pretend like we are hard working..and try to ask her some dumb question..which she'll be thinking that she is the greatest..haha..well..actually we are all chatting around..Future Oscar Winner~

She juz started differentiation..
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[fullerene]

differentiation is fun!

so is integration!

=D

[holyvin]

wow thats so fast....i only reach chapter 6 by now XD....or we r too slow haha :lol:

[lawteoh]

Holyvin: You are not slow.. Just that my teaching is rushing to finish syllabus.. and because we didn't ask him any question, he speed up.. haiz..

Withoutwax: Try asking her tough question and see whether see could solve or not.. haha..

[withoutwax]

Yea..yea..my tcher speeding too~ rushing for exam..her reason

Well..i asked her tough ques b4..like the one on pg 134 Qno.5b..
guess wad she told me the next day i asked her.."this ques i think got problem..you better don't try"..guess wad i shot her bak.."sry..tcher..i've found the solution.." haha..
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[lawteoh]

Haha! ur teacher looked the same like my teacher i guess..For me, if my teacher don't know hor. he will said this question is out of syllabus.. Won't come out in STPM.. and guess what, he is setting the paper.. shouldn't be that difficult if he said that.. Right? hehe.

Hmm.. seems like you love tough questions.. must find one that is hard for you.. but I guess you know all.. lol..

Btw, pg 134, question b? From pelangi ah? or Fajar Bakti? I only use one book, which is Fajar Bakti.. And well, got a bit of problem in their graph sketching answer.. some of them are wrong.. (not all)

[withoutwax]

Nah..i'm using pelangi..i got fajar bakti oso..i used them onli once in a blue moon..haha..

When is ur exam?
________
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[lawteoh]

Actually, my exam is this week.. that's why u din see me onlining this few days.. haha.. interested to trade our maths questions after exam ah? when's yours?

[withoutwax]

after raya holidays
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