Mathematics: Post your doubts here!

[MiniSacBall]

http://papers.xtremepapers.com/CIE/...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.

 

[Thought blocker]

 

[Thought blocker]

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.

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.

 

[MiniSacBall]

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 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. :/

 

[Thought blocker]

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. :/

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.
May this site helps you --> http://www.examsolutions.net/maths-revision/syllabuses/CIE/period-1/specification.php
I'am preparing my math from this site, and believe me, without any help, I can score more than 75%. I still remember past week I was just stuck at 20% but this site helped me a lot. HOPE IT HELPS YOU TOO! Still I have to look at trigonometry. So I am not able to solve your doubt. So sorry. :'(

 

[robinhoodmustafa]

an 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

Thaaaankyou
Pooora Samajh agaya.

 

[MiniSacBall]

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 ?

 

[SitiPutri]

Still I am weak at those. Can you do it step by step all 3 parts please.

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.

View attachment 40048 View attachment 40049 View attachment 40050 View attachment 40052

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.

 

[SitiPutri]

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 ?

 

[Rutzaba]

Thaaaankyou
Pooora Samajh agaya.

im glad and surprised

 

[MiniSacBall]

Thanks 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?

 

[Rutzaba]

Thanks 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?

this is a given formu;a in the mf9
cos^2 x= (1+cos 2x)/2

 

[Thought blocker]

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.

 

[Thought blocker]

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 ?

 

[SitiPutri]

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.

http://www.examsolutions.net/maths-revision/core-maths/vectors/notation/tutorial-1.php

From minute 07:00

And what about that differentiation last part ?

Sorry what do you mean? Which differentiation?

 

[Thought blocker]

http://www.examsolutions.net/maths-revision/core-maths/vectors/notation/tutorial-1.php

From minute 07:00



Sorry what do you mean? Which differentiation?

Thanks, I'll see to it

https://www.xtremepapers.com/community/attachments/db4-png.40050/ This one.

 

[SitiPutri]

Thanks, I'll see to it

https://www.xtremepapers.com/community/attachments/db4-png.40050/ This one.

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"

 

[Thought blocker]

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"

Oh, han!

 

[muzzaml]

Hi everyone, just a question
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_61.pdf
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20%289709%29/9709_s11_ms_61.pdf


In question 6, when we are asked to estimate the mean, the mark scheme uses 50.5 as the midpoint for the first class. Can anyone explain why that is? In addition, if anyone has a good article on the different cases of finding the midpoint, that would be much appreciated. Most I've seen only tell you to take the higher and lower limit, which doesn't always seem to work for me.

Thanks

 

[Ayeshak93]

NOTE i am considering @ as thita

as u found out in part i ..
4Cos@ + 3Sin@ = 5Cos(@-0.6435)

5Cos(@-0.6435) = 2
Cos(@-0.6435) = 2/5
@ = Cos^-1(2/5) + 0.6435

@= 1.8028 radians
and
2(pie)-1.8028 = 4.48 radians

can't seem to grasp onto the last part tho

Thanks