October,November,2012 Maths Examination.

[123dawg]

wats about the minimum value of gradient how to show that

Albert Einstein i would expect more from you

well they gave you dy/dx so that is the gradient so you do the second derivative and make it equal to zero and then find the x value at this point.
then plug this x value back into the first derivative ( the equation of the gradient ) and get the minimum gradient which i got to be 5?
hope that helps.

 

[makorpokey]

I think you are talking about 2 different questions, but the vector one :
they gave OA and OB so you could find AB in terms of k by using OB - OA= AB

And then once you got that you said the magnitude of that must equal 1 cuss the definition of a unit vector is hat its length is = 1
so then you could find k quite simply.

For the other question with 2cosx=3tanx
you expressit as a quadratic in sin

2 ( cosx)^2 = 3sinx
2( 1- ( sinx)^2) -3sinx=0
2(sinx)^2+3sinx - 2=0

then for the second part you just sub x with 2y and get
2(sin2y)^2+3sin2y-2=0
so sin2y= 0.5 and sin2y cant = -2
so sin2y = 0.5 only
then 2y=30
2y= 150

so final ans y=15 or y=75
hope this helps

thanks...

 

[123dawg]

thanks...

 

[Hassan Hijazi]

yes it was so easy......

wei

I think you are talking about 2 different questions, but the vector one :
they gave OA and OB so you could find AB in terms of k by using OB - OA= AB

And then once you got that you said the magnitude of that must equal 1 cuss the definition of a unit vector is hat its length is = 1
so then you could find k quite simply.

For the other question with 2cosx=3tanx
you expressit as a quadratic in sin

2 ( cosx)^2 = 3sinx
2( 1- ( sinx)^2) -3sinx=0
2(sinx)^2+3sinx - 2=0

then for the second part you just sub x with 2y and get
2(sin2y)^2+3sin2y-2=0
so sin2y= 0.5 and sin2y cant = -2
so sin2y = 0.5 only
then 2y=30
2y= 150

so final ans y=15 or y=75
hope this helps

are you sure about the vector thing???

 

[Hassan Hijazi]

guyz did anyone solve the question which says:

show that the gradient is min for x=2 and hence find this min value

dy/dx= 1+ 4/x^2

 

[Hassan Hijazi]

So you got the answer? What about the third part of the same question? Did you get 35.something? And..the 4th question... the curve 4y=x^2 and the line y=(k/x)+k that is tanget to it at point P, find k and P. How do you do that? did you use the X-intercept and Y-intercept to find gradient,then x and then k?

for finding k, you have to equate y=y and then use b^2-4ac=0 anyway k=-1 and p i forgot it!!!!

 

[Hassan Hijazi]

Please tell me you are being sarcastic !

easy!!!! i think ur kidding for sure!!
dude it was very tough!

 

[Hassan Hijazi]

volume for question 9 ??

was it 9*pie

 

[Hassan Hijazi]

it was

Yea you get 35.something for the third part of the question.
For the 4th question wasn't the question with curve 4y=x^2 and line y=(x/k) +k?

37.something!!!

 

[123dawg]

wei
are you sure about the vector thing???

Yup i even checked with like the brainiack in our grade, its right

 

[123dawg]

it was
37.something!!!

nope 35.something, check calculator settings must be in rad

 

[123dawg]

volume for question 9 ??

was it 9*pie

Thats what I got ​

 

[Hassan Hijazi]

nope 35.something, check calculator settings must be in rad

can u tell me hw did u do it?

 

[Hassan Hijazi]

Questions were easy but some were quite tricky which obviously made the paper tough. You can take the first part of the last question, the solution was very simple, it would just take a simple logic to solve it, but many of us failed to do it. Exam pressure could also be the reason behind this.

yes exactly.....if there was no pressure we could of done the paper easilyy

 

[Hassan Hijazi]

I think you are talking about 2 different questions, but the vector one :
they gave OA and OB so you could find AB in terms of k by using OB - OA= AB

And then once you got that you said the magnitude of that must equal 1 cuss the definition of a unit vector is hat its length is = 1
so then you could find k quite simply.

For the other question with 2cosx=3tanx
you expressit as a quadratic in sin

2 ( cosx)^2 = 3sinx
2( 1- ( sinx)^2) -3sinx=0
2(sinx)^2+3sinx - 2=0

then for the second part you just sub x with 2y and get
2(sin2y)^2+3sin2y-2=0
so sin2y= 0.5 and sin2y cant = -2
so sin2y = 0.5 only
then 2y=30
2y= 150

so final ans y=15 or y=75
hope this helps

i got final ans as 105, and 165!!

 

[cheemaboyz]

Hey well i am not really a bro , more like a bro"det (girl), and i would say if the question was out of 4 then minimum number of marks they would give you is 2 but hey just a student so i wouldn't really know. But you got the main part of the question right

they should give at least 2 so how many marks u think u will get in dis paper....

 

[123dawg]

i got final ans as 105, and 165!!

if you plug that back into 2(sin2y)^2+3sin2y-2=o you don't get=o

 

[123dawg]

they should give at least 2 so how many marks u think u will get in dis paper....

i dont know i hope like crazy i get an A but i think i messed u the first question i got a=2 and i think that is wrong and i got a lot of the last question wrong , but all we can do is wait and see

 

[cheemaboyz]

yeah we have to w8 and see but i think in the first ques i also got a=2 ....... and most probably it is right

 

[123dawg]

yeah we have to w8 and see but i think in the first ques i also got a=2 ....... and most probably it is right

yea a = 2 is right i just tried it again haha few at least we have got 3 marks