October,November,2012 Maths Examination.

[Khunkar]

Are you serious?! No fair...

yeah!! I was like . At first they stopped the examination for about 15 minutes, and they gave an additional time of 1/2 an hour. You can imagine the situation...... NO ANSWER SHEET IN MATHS EXAM!!

 

[Firdo Mino]

I will take P1 exam on 17th oct. and S1 on 5th Nov. anybody will take the same exam? what's your schedule guys?

Well maybe u're doing another variant because in different countries and variants differ and so does the dates for the exams...

 

[Ruman Tiwari]

hey guyz how was your paper??
did anyone solve the last question first part????

yes it was so easy......

 

[Ruman Tiwari]

quite tough.... but luckily we got 1/2 hour extra time since the british council ran out of paper.... funny!!

same here..i guess we were in same hall...

 

[123dawg]

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 tangent 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?

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?

 

[123dawg]

yes it was so easy......

Please tell me you are being sarcastic !

 

[Ruman Tiwari]

Please tell me you are being sarcastic !

no really..i mean it...

 

[123dawg]

no really..i mean it...

well awesome for you well done

 

[Ruman Tiwari]

well awesome for you well done

tx dude......bst of luk for rest of ur papers..

 

[Khunkar]

Please tell me you are being sarcastic !

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.

 

[Firdo Mino]

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.

I totally agree with u !! coz it's true!!

 

[shikha95]

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?

Yes i was....sorry, human error

 

[shikha95]

yeah!! I was like . At first they stopped the examination for about 15 minutes, and they gave an additional time of 1/2 an hour. You can imagine the situation...... NO ANSWER SHEET IN MATHS EXAM!!

We got 10 minutes extra, but the teacher kept shouting,“Oh my God! Give the papers! The time is up!“ lol XP

 

[shikha95]

yes it was so easy......

I still haven‘t understoo how to do it, can you explain?

 

[123dawg]

Yes i was....sorry, human error

I think i did something like this :

(x^2)/4= x/k + k
k(x^2)= 4x + 4k^2

k(x^2) - 4x - 4k^2=0

the discriminant of his should equal zero cus a tangent intersects the curve at one spot therefore

(-4)^2 - 4(k)(-4k^2) = 0
16+16k^3=0
k^3 = -16/16
k^3= -1
so k= -1

something like that

 

[cheemaboyz]

I think i did something like this :

(x^2)/4= x/k + k
k(x^2)= 4x + 4k^2

k(x^2) - 4x - 4k^2=0

the discriminant of his should equal zero cus a tangent intersects the curve at one spot therefore

(-4)^2 - 4(k)(-4k^2) = 0
16+16k^3=0
k^3 = -16/16
k^3= -1
so k= -1

something like that

hey bro how many marks will i get if have done
(x^2)/4= x/k + k
k(x^2)= 4x + 4k^2

k(x^2) - 4x - 4k^2=0

the discriminant of his should equal zero cus a tangent intersects the curve at one spot therefore

(-4)^2 - 4(k)(-4k^2) = 0
correct and den did some mistake and my value of k is wrong

 

[makorpokey]

Yea I got how to do the last one as soon as i left the exam room < soooooo annoying. The line of the sector was a tangent to the circle so the radius of the circle (x) would make a 90 degree angle with the radius of the sector ( that was 20 cm in length ) so you could use sin of (1.2/2) =(x/20-x) and then if you went from there you got x = 7.218 and then after that the rest of the question would have been easy. hate that my brain freezes under pressure . But other than that question how did people find the paper ?

what about the one where they asked you to find k where AB is a unit vector?
(ii) 2cos2y=3tan2y...did u use the formula 2cos2y=1-2sin^2 y ..or use the earlier quadratic equ in (i)?

 

[123dawg]

hey bro how many marks will i get if have done
(x^2)/4= x/k + k
k(x^2)= 4x + 4k^2

k(x^2) - 4x - 4k^2=0

the discriminant of his should equal zero cus a tangent intersects the curve at one spot therefore

(-4)^2 - 4(k)(-4k^2) = 0
correct and den did some mistake and my value of k is wrong

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

 

[Albert Einstein]

wats about the minimum value of gradient how to show that

 

[123dawg]

what about the one where they asked you to find k where AB is a unit vector?
(ii) 2cos2y=3tan2y...did u use the formula 2cos2y=1-2sin^2 y ..or use the earlier quadratic equ in (i)?

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