Mathematics: Post your doubts here!

[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

 

[Scafalon40]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_33.pdf
Q9 (iii) and 10 part b ii PhyZac
last question for the day

 

[Shreeram]

Hi!! Can anyone help me to give a detailed explanation for june 12 p 42 q7 thanks

 

[PhyZac]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_31.pdf

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

okay first, move value to make two integration
(let ! be the integration sign)
dr/dt = 0.08r^2
1/r^2 dr = 0.08 dt
! 1/r^2 dr = !0.08 dt
! r^-2 dr = 0.08t +c
-1/r = 0.08t + c
sub values given in question
-1/5 = 0.08(0) + c
c = -1/5

-1/r = 0.08t - 1/5
r = -1 / (0.08 - 1/5)
r = -5 /0.4t - 1


Now for part iii
i am not sure
but what i think is the answer must be positive.
to do this the value shudnt be smaller than 0, (try calculating)
and even not more than 2.5
i got 2.5 by
0.4t-1 =0
0.4t =1
t= 2.5

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_33.pdf
Q9 (iii) and 10 part b ii PhyZac
last question for the day

Okay i have the answering, but i dont know Q9 (i) if u can explain plzz.

so Q9 (iii)

If u got the a b values, 3 and 2

u sub them in the components.

3-λ = 4 + aµ
-2+2λ= 4 + bµ
1+λ = 2 - µ

so u get
3-λ = 4 + 3µ
-2+2λ= 4 + 2µ
1+λ = 2 - µ

choose any two and solve simultaneously

and u get µ = -1

no need to find λ or vice versa

take the value u got, i my case µ = -1
into its line equation and you willget
the i j k values.

about Q10 sorry i just saw it, will look into it, EDIT: i looked into it, i need to solve full question in order to answer tomorrow In Sha Allah

 

[Adorkableme]

I urgently need someone to solve Qs 1 from
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w08_qp_3.pdf

 

[PhyZac]

I urgently need someone to solve Qs 1 from
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_3.pdf

ln(x+2) = 2+lnx
ln(x+2) - lnx = 2
ln ((x+2)/x) = 2
(x+2)/x = e^2
x+2 = xe^2
xe^2 - x = 2
x(e^2 - 1) =2
x = 2/ (e^2 -1)
using calc.
x = 0.313

 

[immie.rose]

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

Q)4) help appreciated.


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

You could try the trigonometry notes linked on the first page of this thread.
Here's what my guide says about the ques:

a: indicates the amplitude of the curve which is the max or min range of the curve from the mean position.
so, a=9-3=6

b: indicates the periodicity of the curve. we see that in the given figure there are two main cycles in a normal period of sine.
therefore, b=2

c: indicates the vertical shift in the graph from the horizontal axis, which in this case is 3 units.
thence, c=3 units

 

[unseen95]

please help me with question number 5(iii), shouldn't the two equations be
"0.1*1o - T = 0.1*a" and
"T - 0.8*0.5*10 = 0.5a"
if not what should the equations be and why? link to the question paper is http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_4.pdf

 

[unseen95]

please help me with question number 6(i). Link to the question is http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_4.pdf