How was today's add-maths paper 2 ?

[osama]

my answer to 11 was.....0.14....i guess i did it directly rather than separating the area of triangle separately....will that be correct ???

 

[maaz007]

I guess so! But u had to minus the Area under the curve from the triangle! so -1! u could see it in the Marking Scheme!

 

[maaz007]

The limits were supposed to be 3/2pi and 2pi

Sorry it was 2pie and 3pie/2

 

[maaz007]

How did u do the dimensions ?

Dimensions were confusing first u have to make a volume equation
x*x*h=256cm3
h=256cm3/x^2 H in terms of X
S.A=(x*x)+4(x*h), put h
dA/dx=0 gives u x=8
Find the height dimensions 8 by 8 by 4

 

[HarisLatif]

my answer to 11 was.....0.14....i guess i did it directly rather than separating the area of triangle separately....will that be correct ???

so you took the limits of shaded region and solved it?

 

[Tharindu]

Wot was the answer for that co ordinate question in paper 1 and how the hell is that vector represented as mew, a and b only

 

[Tharindu]

The nos for permutation was 1,2,3,4,5,6

 

[Tharindu]

P1 has been pretty much washed frm my mind now

Ya but rough values

 

[osama]

so you took the limits of shaded region and solved it?

of course yaar....i took 2.37 and 3pi/2 i guess

Wot was the answer for that co ordinate question in paper 1 and how the hell is that vector represented as mew, a and b only

for the coordinate question an easier method to find the co-ordinates was to use the position vector method...BTW answer was p= (3,1)....for the last question i have no idea i messed up that question .....and value of mu was coming 4/15 and -10/11 for lambda....i expressed the second part in terms of mu, lambda a, b ;(

 

[MustafaMotani]

of course yaar....i took 2.37 and 3pi/2 i guess

for the coordinate question an easier method to find the co-ordinates was to use the position vector method...BTW answer was p= (3,1)....for the last question i have no idea i messed up that question .....and value of mu was coming 4/15 and -10/11 for lambda....i expressed the second part in terms of mu, lambda a, b ;(

lambda cant be negative I got 10 / 11...
and did everyone got indepndent term as 135, I got it as 180.. I cant believe I did that wrong..

 

[syedhamza]

syedhamza and Syed Zafar answer to Q-11 was exact o.1213 units
With base of triangle equals to 2 and height equals to underroot 2/2 and area under the graph with limits 2pie and 3pie/2 where curve intersection was to be calculated through quadrants. While Q1 E symbol states (element of) u might need to check it on internet!
and Q10 stated divisble by five so ur and mine answer still remains 60! Hope u get an A*! and best luck for Maths!

Thanks now i remember Mine ans was also .12 sth . were ur coordinates of Q (6.7,0) ? and in paper 1 that vector Qs.. Were u able to find QR in terms of a, b and meu (as my teachers calls it ) the first part was easy that lamda one but in da 2 part I found QR in terms of lamda,meu,a and b! one of the values were -3/2?

 

[Buzzy]

w8 a sec guys...

this is important...is what maaz007 syin true?

Confirmation plz!

 

[Tharindu]

of course yaar....i took 2.37 and 3pi/2 i guess

for the coordinate question an easier method to find the co-ordinates was to use the position vector method...BTW answer was p= (3,1)....for the last question i have no idea i messed up that question .....and value of mu was coming 4/15 and -10/11 for lambda....i expressed the second part in terms of mu, lambda a, b ;(

of course yaar....i took 2.37 and 3pi/2 i guess

for the coordinate question an easier method to find the co-ordinates was to use the position vector method...BTW answer was p= (3,1)....for the last question i have no idea i messed up that question .....and value of mu was coming 4/15 and -10/11 for lambda....i expressed the second part in terms of mu, lambda a, b ;(

Ok so wots the area of that co ordinate thing

 

[MustafaMotani]

Thanks now i remember Mine ans was also .12 sth . were ur coordinates of Q (6.7,0) ? and in paper 1 that vector Qs.. Were u able to find QR in terms of a, b and meu (as my teachers calls it ) the first part was easy that lamda one but in da 2 part I found QR in terms of lamda,meu,a and b! one of the values were -3/2?

u are correct about the area and Q cordinates

 

[MustafaMotani]

People I too got QR in terms of mu, lambda, a and b.
But someone told me that it was possible to get that term widout lambda if u wud have used similar triangle.
Now who would have thouht about that

 

[Tharindu]

What might be the min mark for A* this time..Any guesses

 

[scouserlfc]

What might be the min mark for A* this time..Any guesses

i guess close to 90 abt 88 or 87 !

 

[Tharindu]

i guess close to 90 abt 88 or 87 !

Will it ever go more than 90...like 93%....but compared to last years isnt it easier

 

[maaz007]

For vectors answer could be in negative! i also proved that answer several times! And again no.s were 2,3,4,5,6,7 i highlighted every no.s and rechecked it too! For Area of Triangle in P1 it was exact 156units square! I got it through vertice multiplying! QR was bieng expressed in mu lamda a and b due to OR being Mu a or b! and it was of 3 Marks so OR couldnt be expressed betterly!

 

[MustafaMotani]

For vectors answer could be in negative! i also proved that answer several times! And again no.s were 2,3,4,5,6,7 i highlighted every no.s and rechecked it too! For Area of Triangle in P1 it was exact 156units square! I got it through vertice multiplying! QR was bieng expressed in mu lamda a and b due to OR being Mu a or b! and it was of 3 Marks so OR couldnt be expressed betterly!

:S Could u prove how answers be negative...
u see it was OR = lambda OA
thus lambda can never be negative or elso its direction woul be change .. :S so please dont make fuss of it, and wats 2,3,4,5,6,7 . ??