hi i need help with oct/Nov 2008 p3 question 8 (i)
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w08_qp_3.pdf
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w08_qp_3.pdf
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Thanks a lot!For paper 61 Q6 part ii
When using trigonometry, Use radian mode. I mean When the question includes pie and trig together, we have to use radian mode.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
for q8 in the past paper above, i was supposed to calculate in radians, while the whole time i am calculating in degree. And no hint is give in the question wether to use radians or degrees.
Well for such questions how am i supposed to know whether to calculate in degrees or radians, with no hint give.
I took me 30 minutes just to get this question, first i solved my self calculating in deg, which took me to a weird answer, When i opened mark scheme, it was completely different i was so confused, i used like 3 calculators to calculate it, same result and then i looked up on internet and most of them are calculating cos(pi in radians. So when i calculated in red i got correct answer. So i was confused with it.When using trigonometry, Use radian mode. I mean When the question includes pie and trig together, we have to use radian mode.
Btw, how did you solved it ? Tell me.
I don't know the whole part, I am studying in school, but I am not having a single teacher, School is at Nadir. I am in worse condition more than you are.I took me 30 minutes just to get this question, first i solved my self calculating in deg, which took me to a weird answer, When i opened mark scheme, it was completely different i was so confused, i used like 3 calculators to calculate it, same result and then i looked up on internet and most of them are calculating cos(pi in radians. So when i calculated in red i got correct answer. So i was confused with it.
Also where i study, is hell, teachers them selves don't know anything. So even having tutions (Private classes) it is like studying alone, all the schools for cie are closed in my area because a retarded fag has problems with all schools. I hope get out of this hell quickly, no good teachers are available, my exams for maths are starting in less than a week and its like i have just started leaning maths. I hope by some miracle i get good grades to get out this freaking hell.
And thank you for your help, i am currently solving this past paper and can't get through same question (q8, part ii).
Would you like to help me in this case too. I am stuck at cos2x=3/4. I am getting wrong answer after that. :/
Thaaaankyouan extremely idiotic atempt
Heres the deal
If you have an equation
Ax²-bx +c =0
Frst make the coefficient of x²=1
We ignore the c and see only the a and b part
To remove the a and make it one we take common a that wud bring us to
A (x²- (b/A)x) +c
Then we divide the b wali term by 2 ( so that we can justify the 2ab part when we expand (a -b)^2)
A(x² - (b/2A)x) +c
Add the (b/2) square and subtract b/2 square
[A(x²-(B/2A)- (B ²/4A) +(B ²/4A)]+C
Next we take out the +(b ²/4A) out of the bracket
Remember in the process it will be multiplied by A
b²/4A wen multiplied by A will give B ²/4
thus
A[(x²-(B/2A) +(B ²/4A)] -B ²/4 +c= 0
Can u see the statement inside the []?
Does it resemble a ² - 2ab +c ?
So A[x- (B/2A)] ²] +c-B ²/4
robinhoodmustafa
Still I am weak at those. Can you do it step by step all 3 parts please.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
Q8 part ii
I am stuck at cos(2x)=3/4
If i do something like 2x=cos^-1(3/4)
It get wrong answer any help! Mark Scheme has two answers how ?
im glad and surprisedThaaaankyou
Pooora Samajh agaya.
this is a given formu;a in the mf9Thanks a lot but, how did you do change cos2x to 2cos^2 x-1
I only didn't get that part and i haven't studied some thing like that?
Thanks for your help.You just need to input the formulas to the equation. For example, in the formula, tan (180-x) = - tan x. Then, if tan x =k, tan (180-x) = - tan k.
For the origin of the formulas, probably there's an explanation in the examsolutions.net? I use the site for my studies too but I don't know if they provide explanations for that or not. It's pretty hard to explain it by only writing it down.
If you memorize all the formulas, for that type of question you'll be fine.
For the 3rd picture of your question, remember that m = tan x and thus x = tan^-1 m. So, if they ask for tan^-1 k, they're just asking for m of the normal.
And what about that differentiation last part ?You just need to input the formulas to the equation. For example, in the formula, tan (180-x) = - tan x. Then, if tan x =k, tan (180-x) = - tan k.
For the origin of the formulas, probably there's an explanation in the examsolutions.net? I use the site for my studies too but I don't know if they provide explanations for that or not. It's pretty hard to explain it by only writing it down.
If you memorize all the formulas, for that type of question you'll be fine.
For the 3rd picture of your question, remember that m = tan x and thus x = tan^-1 m. So, if they ask for tan^-1 k, they're just asking for m of the normal.
Thanks for your help.
But I din't understood in vectors.. First part, how you know, What to take on i, j, k axis ? I know you explained that 3 in i so 3i cap, 6 in j so 6 j cap and -3 in k so -k cap. But how did you get those 3,6,-3 values Thanks in advance
And thanks for other question help.
And what about that differentiation last part ?
Thanks, I'll see to ithttp://www.examsolutions.net/maths-revision/core-maths/vectors/notation/tutorial-1.php
From minute 07:00
Sorry what do you mean? Which differentiation?
Oh, han!Oh, that one! I already gave the explanation to you in the previous post.
"For the 3rd picture of your question, remember that m = tan x and thus x = tan^-1 m. So, if they ask for tan^-1 k, they're just asking for m of the normal"
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