Mathematics: Post your doubts here!

[zain786]

can you please help me in how to answer questions such as exam 9709/32/O/N10 the question 8 part b

its basically after making partial fractions...expanding it to ascending powers upto x^3 etc

how do you do these? plz help

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_3.pdf
question 10 part 2
anyone ?
PhyZac

See, can u help me in (a) , i cant use the r-a.n equation , sorry.

for finding angle,
u take normal,
2i−3j+6k
and direction vector,
i − 2j + 2k

take scalar product
2+6+12 = 20

now divide with the mod of both vectors
and u get
20/21

20/21 normally =cos (angle)
but when plane and line
it is=sin (angle)
so sin^-1 (20/21) = 72.2

 

[ElvinJc]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_32.pdf
can anybody plz tell me how to solve Q2(i)&Q8...

 

[PhyZac]

can you please help me in how to answer questions such as exam 9709/32/O/N10 the question 8 part b

its basically after making partial fractions...expanding it to ascending powers upto x^3 etc

how do you do these? plz help

share link next time plz.


In the formula book u have the formula under binomial expansion

and use this video and then try resolving hope u will get it.
http://www.examsolutions.net/maths-...uences-series/binomial/formula/tutorial-3.php

 

[ElvinJc]

share link next time plz.


In the formula book u have the formula under binomial expansion

and use this video and then try resolving hope u will get it.
http://www.examsolutions.net/maths-...uences-series/binomial/formula/tutorial-3.php

i sorry about that....actually i have tried to solve it....but in Q8(i），my answer is C=-1/3，but the ms is 1/3...and i can‘t find any way to get “-”

 

[PhyZac]

i sorry about that....actually i have tried to solve it....but in Q8(i），my answer is C=-1/3，but the ms is 1/3...and i can‘t find any way to get “-”

I am afraid i am confused.

See, i quoted a post of different member, not u.

And what question are you referring to?

 

[applepie1996]

See, can u help me in (a) , i cant use the r-a.n equation , sorry.

for finding angle,
u take normal,
2i−3j+6k
and direction vector,
i − 2j + 2k

take scalar product
2+6+12 = 20

now divide with the mod of both vectors
and u get
20/21

20/21 normally =cos (angle)
but when plane and line
it is=sin (angle)
so sin^-1 (20/21) = 72.2

SURE
part a right ?

 

[PhyZac]

SURE
part a right ?

Yes part a.....thanks! Jazaki Allah khairan.

 

[Dudu]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_1.pdf

Q)4) help appreciated.


Anyone have trigonometry notes? Clearly my worst topic from P1.

 

[minie23]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w09_qp_31.pdf

Please help for no. 10 (ii) & (iii)

 

[sma786]

for stationary point, take the dy/dx equation and equate to 0.
for nature of stationary point use d2y/dx2 and is -ve number comes means max point if +ve number comes means minimum.

thankss helped alot

 

[applepie1996]

Yes part a.....thanks! Jazaki Allah khairan.

ok so here goes
(r-3i) . (2i - 3j + 6k)=0
and r=((1+s) , (6-2s) , (-3+2s))
now you have to subtract r from 3i + oj + 0k
and then you get (-2+s , 6-2s , -3+2s)
now the dot product with (2i - 3j + 6k)
sooo (-2+s , 6-2s , -3+2s). (2 , - 3 , 6)=o
you get 2(-2+s) + -3(6-2s) + 6(-3+2s) = 0
-4 + 2s + 6s -18 -18 +12s=0
-40=-20s
s=2
replace s=2 in r=((1+s) , (6-2s) , (-3+2s))
and you get 3i + 2j + k
hope you got it

 

[Scafalon40]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf
Q10 part (iii) please!

 

[PhyZac]

ok so here goes
(r-3i) . (2i - 3j + 6k)=0
and r=((1+s) , (6-2s) , (-3+2s))
now you have to subtract r from 3i + oj + 0k
and then you get (-2+s , 6-2s , -3+2s)
now the dot product with (2i - 3j + 6k)
sooo (-2+s , 6-2s , -3+2s). (2 , - 3 , 6)=o
you get 2(-2+s) + -3(6-2s) + 6(-3+2s) = 0
-4 + 2s + 6s -18 -18 +12s=0
-40=-20s
s=2
replace s=2 in r=((1+s) , (6-2s) , (-3+2s))
and you get 3i + 2j + k
hope you got it

Thank You sooooooo much !!!!!!! And yes i get it very well Alhamdulilah...!! Jazaki Allah khairan...May Allah S.W.T reward you and your family with highest grades in the world and hereafter. May Allah S.W.T bless you, Aameen.!!! Thank ALOT

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf
Q10 part (iii) please!

I solved this before, very recently.!​

We have to find the point p where the perpendicular distance to the two planes is same.


first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]​

 

[PhyZac]

Some question are already solved in thread.
Try searching before posting.
It works for me sometime!

Here is how I do it.

For example my doubt is in this link
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

You copy the last bit in this case "9709_s12_qp_32"

Now paste this in google and add the word doubt.

so you get

9707_s12_qp_32 doubt.

You might find many pages from xtremepapers, open all of them in new tabs (right click, newtab) and then look in each if it has same question, if yes, then follow up to see if anyone replied.

 

[Scafalon40]

I solved this before, very recently.!​

We have to find the point p where the perpendicular distance to the two planes is same.​

​

​

first we will for an equation the way the paper gave.​

therefore​

for plane m​

|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]​

for plane n​

|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3​

​

​

since you are finding a point where both distance is same , therefore.​

|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)​

=​

|x+2y-2z-1| = |2x-2y+z-7|​

​

now we will sub the line values in the equation.​

the x component (1+2t) [P.S, t is lamda ]​

the y comp. (1+t)​

the z comp. (-1+2t)​

​

therefore.​

| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |​

simplifying it you get​

|4| = |-8+4t|​

now to solve modulus, we do squaring method​

16 = 64-64t+16t^2​

simplify​

2 = 8 - 8t +2t^2​

2t^2 - 8t + 6 = 0​

solve it and get​

t = 3 or t= 1​

when t=3 the position of point is [P.S, u do this by sub t value in line equation)​

(7, 4, 5) => OA​

when t = 1 the position of point is​

(3, 2, 1) => OB​

​

BA (or AB, same thing) = OA - OB​

= (4 , 2, 4)​

​

now find the mod​

sqrt(4^2 + 2^2 + 4^2)​

= 6​

​

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]​

Dude you're a life saver! Thanks!

 

[applepie1996]

Thank You sooooooo much !!!!!!! And yes i get it very well Alhamdulilah...!! Jazaki Allah khairan...May Allah S.W.T reward you and your family with highest grades in the world and hereafter. May Allah S.W.T bless you, Aameen.!!! Thank ALOT

Jazak Allah to you too
for helping me
InshaAllah i will remember you in my prayers

 

[Rutzaba]

this time i be the punjab police... entering wen there are no doubts!

 

[applepie1996]

this time i be the punjab police... entering wen there are no doubts!

LOL xD