Mathematics: Post your doubts here!

[robinhoodmustafa]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
Thought blocker

Q-1

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
Thought blocker

Q-1

You can go to examsolution site for detail understanding.
Use the formula :
Here a is 2x, b is -1/x and n is 5
so put this values in formula, and solve it until you get the coff of 1/x like this :¬

 

[RadzMau]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_31.pdf
Q6 please!!!

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_31.pdf
Q6 please!!!

Oh, so sorry, I didn't notice any alerts. Well, here you go :¬
Q6 :
i)
Sector area OABC : 0.5 x r^2 x (2π x 4θ) = πr^2 - 2θr^2
Area AOB : 0.5 x r^2 x sin( π - 2θ ) = 0.5r^2 x sin2θ
2 triangles : r^2 x sin2θ
Sector area ABOC : 0.5 x 2θ x (2rcosθ)^2 = 4θr^(2)cos^(2)θ

4θr^(2)cos^(2)θ - r^(2)sin2θ + πr^2 - 2θr^(2) = (πr^2)/2
2θr^(2) x (2cos^(2)θ - 1) - r^(2)sin2θ + πr^2 = (πr^2)/2
2θr^(2) x cos2θ - r^(2)sin2θ + πr^2 = (πr^2)/2
4θr^(2) x cos2θ - 2 r^(2)sin2θ + 2πr^2 = πr^2
4θcos2θ = 2sin2θ - π
cos2θ = (2sin2θ - π)/4θ.

ii)
I was confused without my calculator, was getting random answers, then chistguy helped me and said, put your calculator in radian mode, I was really frustrated at that moment. But I finally got answer, and yes verified with two people, ms, and calculator too.
So, here just use that formula : iterative formula
1st put θ value as 1 in that formula, after solving it you'll get answer as 0.953 now use θ as 0.953 in that formula, and repeat this process until two answers are same and that will be your answer in 2dp that is 0.95

Once again sorry for the delay, I was bit confused in this one. Thanks to my friends

 

[Boss201]

solve the equation x^3.9=11x^3.2 , where x not equal to 0 (ans :30.7)


the question paper is Oct/NOV 2004 paper 2 number 2


any help how to solve this

 

[Thought blocker]

solve the equation x^3.9=11x^3.2 , where x not equal to 0 (ans :30.7)


the question paper is Oct/NOV paper 2 number 2


any help how to solve this

I dont get the equation, tell me which paper ? You said oct-nov Paper 2 but which year ?

 

[Thought blocker]

solve the equation x^3.9=11x^3.2 , where x not equal to 0 (ans :30.7)


the question paper is Oct/NOV paper 2 number 2


any help how to solve this

Till I have solved the papers, I know it was in 9709_w04_qp_2, so here you go if it was the question :¬
Q. x^(3.9) = 11x^(3.2)
A. [{x^(3.9)}/{x^(3.2)}] = 11
--> x^(0.7) = 11
TAKE LOG TO BOTH THE SIDES.
--> 0.7log(x) = log11
--> x = 30.7
EDIT as you asked in convo: to remove log we do 10 ^ like for here, x = 10 ^ [log(11)/0.7] = 30.7

All the best.

 

[M.Shahzaib Shoaib]

http://spcmaths.yolasite.com/resources/Nov' 01 P1.pdf

Question 3 & 7(1) ..

 

[M.Shahzaib Shoaib]

Would really appreciate it if someone gives detailed solutions ..

 

[Thought blocker]

http://spcmaths.yolasite.com/resources/Nov' 01 P1.pdf

Question 3 & 7(1) ..

Would really appreciate it if someone gives detailed solutions ..

Q3 :¬
Whats the problem here ?
Q7 :¬
i)
Okay, so here we just have to square a and b, after that we have to add them.
It follows as :
a² = 4sin²Θ + cos²Θ + 4sinΘcosΘ
b² = 4cos²Θ + sin²Θ - 4sinΘcosΘ
a² + b² = 5sin²Θ + 5cos²Θ
We know that sin²Θ + cos²Θ = 1
so the answer will be 5.
We get to know that a² + b² is constant for all value of Θ

 

[M.Shahzaib Shoaib]

Well I do not know how to do the whole question 3.

If you could do it on a paper & post the picture. Would be really helpful.

THANKS FOR QUESTION 7 ..

 

[Thought blocker]

Well I do not know how to do the whole question 3.

If you could do it on a paper & post the picture. Would be really helpful.

THANKS FOR QUESTION 7 ..

So, as on your request here you go :¬
NOTE : I made two separate graphs so that to make sure how their graphs actually look like, now try them in one graph, do label it correctly, look at the points on cos 3x and cos x. You can also label it as 60 90 120 180 240 270 300 360 and also in radians, its our own wish, but I do use radians, as its related to trigonometry



If you still are not cleared. Do ask.

 

[M.Shahzaib Shoaib]

Thank you soo much.

This was very helpful.

 

[M.Shahzaib Shoaib]

http://www.a-level365.com/upload/2011/3/28191047966.pdf

Question 5 (1) ...

The vector MC=OC-OM ..

Now I get OM correct but have a problem with the unit vector K of OC..
How is it 12?

 

[Thought blocker]

http://www.a-level365.com/upload/2011/3/28191047966.pdf

Question 5 (1) ...

The vector MC=OC-OM ..

Now I get OM correct but have a problem with the unit vector K of OC..
How is it 12?

examsolutions - This very good site, and it helped me even to finish my math P1 in 1 month and P4 in 1 month. But it need practice, if you won't, then Math is useless. All the best.
Here is your answer in my way :¬

 

[Rockstar RK]

Integrate cosec4xcot4x dx (indefinite)

 

[Thought blocker]

Integrate cosec4xcot4x dx (indefinite)

Is it -cosec(4x)/4 ?
Check my answer below.

 

[Manoj Panigrahi]

Please Some body help me to solve Oct/Nov2011 Paper 4(Extended) 0580/42 Question No.4 (a,b,c)

 

[Manoj Panigrahi]

Please Some body help me to solve Oct/Nov2011 Paper 4(Extended) 0580/42 Question No.4 (a,b,c)

 

[Manoj Panigrahi]

Please help me to solve Oct/Nov2011 Paper 4(Extended) 0580/42 Question No.4 (a,b,c)