Mathematics: Post your doubts here!

[haha101]

You're right but they subtracted distance of P from distance of Q.. it has to be Sp-Sq=4.9

oh really ? I didnt seem to understand the method did the Q myself and posted what i thought was logical

 

[Pie-man]

because P is further away and has covered more distance than q . it it were the other way around then we would get a negative distance which is impossible ( I am talking about distance not displacement )

BTW i did it another way

how did you do it then?

 

[Esme]

all i know is if you have 3 different forces on an object you can use lami
i'm not exactly sure why i got CP2X it just made more sense to me and after getting all the other angles i was sure it was right but it will make more sense if you turn it into a triangle (well for most people i know)

Thanks I'll try to make sense of it

 

[Pie-man]

I took time taken by P to be 't' and time taken by Q to be (t-2). Then i formed the equations for the distance travelled by both the particles and subtracted distance of Q from distance of P. I got t=2.4. So the time taken by Q is t-2= 2.4-2= 0.4

why t-2 and not t+2 as it is 2 seconds late?

 

[Esme]

why t-2 and not t+2 as it is 2 seconds late?

See, P started before Q. So at any instant, P has been travelling for more time Q has. So if time taken by P is t, the time taken by Q to reach that point will be less by 2, hence (t-2)

 

[haha101]

how did you do it then?

First of all made equations for the dist covered by p and q :
Dist of p= o.5*1.75*(2+t)^2 (s=ut+0.5at^2) (u=0)
dist of Q = o.5*1.75*t^2

the difference was 4.9 => dist of P - Dist of Q
o.5*1.75*(2+t)^2 - o.5*1.75*t^2 = 4.9 (took o.5*1.75 common and divided it by 4.9 )

(2+t)^2 - t^2 =5.6

expanded the eq 4 +4t+t^2 -T^2 =5.6 ( T^2 is cancelled )

t =0.4

 

[Esme]

why t-2 and not t+2 as it is 2 seconds late?

You're right too. If the time taken by Q is t, then the time taken by P will be t+2.

 

[Pie-man]

See, P started before Q. So at any instant, P has been travelling for more time Q has. So if time taken by P is t, the time taken by Q to reach that point will be less by 2, hence (t-2)

ok thanks that really helped now i hope they get this on the exam its not as hard as it seems

 

[UnprepareD]

Dear, mathematicians I've hit rock bottom with two of the questions on this paper - Questions 2 and 8.The former I'm absolutely clueless how to do and the later, I'm struggling to substitute du for dx. Any help would be welcome. Thank you .
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

 

[Dug]

Dear, mathematicians I've hit rock bottom with two of the questions on this paper - Questions 2 and 8.The former I'm absolutely clueless how to do and the later, I'm struggling to substitute du for dx. Any help would be welcome. Thank you .
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf

Q2)
i) Area of sector = 1/3 (Area of triangle)
(1/2)r²Θ = (1/3)[(1/2)(atanΘ)(a)]
Simplify and rearrange to get tanΘ = 3Θ

ii) Taking initial value = π/4
Θ1 = 1.1694
Θ2 = 1.2391
Θ3 = 1.3185
Θ4 = 1.3232
Θ5 = 1.3240
Θ6 = 1.3242
Θ7 = 1.3242

Θ = 1.32

 

[Talhakhan]

Which Mechanics M1 paper has been the most difficult in the last decade.. I mean which had the lowest grade threshold. I want to attempt that paper plz if anyone have any idea plz share. ;-)

 

[Silent Hunter]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w06_qp_4.pdf

Pie-man

or anyone ....... how to do 7(iii) ?

Thank you

 

[prekshya]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_42.pdf
ques num 4 please!!

 

[Anika Raisa]

​

​

CIE PURE MATH 3:​

Can any one please help me with this sum's part iv below please!​

(I could do till iii but got badly stuck with part iv)​

​

Q.The complex number −2 + i is denoted by u.​

(i) Given that u is a root of the equation x​

3 − 11x − k = 0, where k is real, find the value of k. [3]​

(ii) Write down the other complex root of this equation. [1]​

(iii) Find the modulus and argument of u. [2]​

(iv) Sketch an Argand diagram showing the point representing u. Shade the region whose points​

represent the complex numbers z satisfying both the inequalities​

|z| < |z − 2| and 0 < arg(z − u) < 1/4π. ​

Please help asap am at door! And please draw n upload the diagram here too...pleaaase i m really confused abt the diagram dat i should draw for part iv​

Thank u! May u be blessed for helping!​

BESIDES here is the mark scheme below... to me it sounded completely gibberish so i couldnot work out the last part of the sum!​

(iv) Show point representing u in relatively correct position in an Argand diagram B1​

Show vertical line through z = 1 B1​

Show the correct half-lines from u of gradient zero and 1 B1​

Shade the relevant region B1 [4]​

[SR: For parts (i) and (ii) allow the following alternative method:
State that the other complex root is –2 – i B1
State quadratic factor x2 + 4x + 5 B1
Divide cubic by 3-term quadratic, equate remainder to zero and solve for k, or, using
3-term quadratic, factorise cubic and obtain k M1
Obtain k = 20 A1]

AND Also :

Can you please help me with 9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you

Link to Question: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf

Link to Mark Scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_ms_33.pdf
Anika Raisa said: ↑

Can you please help me with 9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you​

Pleaase help!

 

[Pie-man]

ok i'm too lazy to solve it right now so in about 15mins or so i'll attempt it

 

[Pie-man]

Can any one please help me with this sum's part iv below please!​

(I could do till iii but got badly stuck with part iv)​

​

Q.The complex number −2 + i is denoted by u.​

(i) Given that u is a root of the equation x​

3 − 11x − k = 0, where k is real, find the value of k. [3]​

(ii) Write down the other complex root of this equation. [1]​

(iii) Find the modulus and argument of u. [2]​

(iv) Sketch an Argand diagram showing the point representing u. Shade the region whose points​

represent the complex numbers z satisfying both the inequalities​

|z| < |z − 2| and 0 < arg(z − u) < 1/4π. ​

Please help asap am at door! And please draw n upload the diagram here too...pleaaase i m really confused abt the diagram dat i should draw for part iv​

Thank u! May u be blessed for helping!​

BESIDES here is the mark scheme below... to me it sounded completely gibberish so i couldnot work out the last part of the sum!​

(iv) Show point representing u in relatively correct position in an Argand diagram B1​

Show vertical line through z = 1 B1​

Show the correct half-lines from u of gradient zero and 1 B1​

Shade the relevant region B1 [4]​

[SR: For parts (i) and (ii) allow the following alternative method:
State that the other complex root is –2 – i B1
State quadratic factor x2 + 4x + 5 B1
Divide cubic by 3-term quadratic, equate remainder to zero and solve for k, or, using
3-term quadratic, factorise cubic and obtain k M1
Obtain k = 20 A1]

AND Also :

Can you please help me with 9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you

Link to Question: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf

Link to Mark Scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_ms_33.pdf
Anika Raisa said: ↑

Can you please help me with 9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you​

Pleaase help!

the whole thing sounded gibberish to me....then i realized its paper 3

 

[50sum]

First of all made equations for the dist covered by p and q :
Dist of p= o.5*1.75*(2+t)^2 (s=ut+0.5at^2) (u=0)
dist of Q = o.5*1.75*t^2

the difference was 4.9 => dist of P - Dist of Q
o.5*1.75*(2+t)^2 - o.5*1.75*t^2 = 4.9 (took o.5*1.75 common and divided it by 4.9 )

(2+t)^2 - t^2 =5.6

expanded the eq 4 +4t+t^2 -T^2 =5.6 ( T^2 is cancelled )

t =0.4

how did you know that the both particles have same acceleration?

 

[Anika Raisa]

the whole thing sounded gibberish to me....then i realized its paper 3

Haha...sorry i should have mentioned above!

 

[Anika Raisa]

CIE Puremath3: Some help here please!

Please help me with 9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you

Link to Question: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf

Link to Mark Scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_ms_33.pdf​

 

[Sk. Shahriar Hossain]

Plz help me solve this math.....i will be really grateful...