Post by: Vin on April 08, 2010, 10:08:53 am
i was wondering that are there ny math papers/specimen papers available for the new format of maths paper4..??
i've seen the format from may/june 2009 paper 4 is supposed to be written in the paper itself (earlier papers had the system of writing on a separate sheet...we've been givn m/j 09 n oct/nov 09 as our preliminary exam practice..)
i jus' thought there might be more papers available of this format of the paper...but couldn't find them
Post by: holtadit on April 08, 2010, 10:29:03 am
hi
i was wondering that are there ny math papers/specimen papers available for the new format of maths paper4..??
i've seen the format from may/june 2009 paper 4 is supposed to be written in the paper itself (earlier papers had the system of writing on a separate sheet...we've been givn m/j 09 n oct/nov 09 as our preliminary exam practice..)
i jus' thought there might be more papers available of this format of the paper...but couldn't find them
It doesnt make a difference whether you do it in the paper or not. Honestly the new format has been introduced to improve presentation. On the CIE main website there is a specimen paper (new format) of the paper 4 that was used in 2007.
Post by: Vin on April 10, 2010, 12:03:30 pm
Post by: Dark Prince on April 10, 2010, 08:24:54 pm
value of n.
Post by: holtadit on April 18, 2010, 02:27:38 pm
Q..Find any expression, using 1, 2, 3 and n exactly once, which will always be greater than 1 for any
value of n.
n+1(3-2)
SImple
Post by: Alpha on April 18, 2010, 03:28:35 pm
Q..Find any expression, using 1, 2, 3 and n exactly once, which will always be greater than 1 for any
value of n.
If you can use n^2,
(n - 3)^2,
n^2 + 3*1,
etc.
n+1(3-2)
SImple
When n <= -1?
Post by: Saladin on April 18, 2010, 03:37:17 pm
The paper (http://www.cie.org.uk/docs/dynamic/31237.pdf)
Post by: holtadit on April 18, 2010, 04:59:29 pm
If you can use n^2,
(n - 3)^2,
n^2 + 3*1,
etc.
When n <= -1?
-1+1(3-2) = 0 Hence 0 is one greater than -1 . Whats the problem ?
Post by: Saladin on April 18, 2010, 05:06:10 pm
-1+1(3-2) = 0 Hence 0 is one greater than -1 . Whats the problem ?
it said greater than 1, not negative 1.
Try it with all numbers, it will work.
Post by: holtadit on April 18, 2010, 05:10:27 pm
I thought he wanted a formula that would give an answer 1 greater than the value of n
Sorry people, Ari just messed up.... my bad :-[ :-\
Post by: Alpha on April 21, 2010, 04:19:11 pm
I just realised he wanted a formula that would give an answer tht is greater than 1
I thought he wanted a formula that would give an answer 1 greater than the value of n
Sorry people, Ari just messed up.... my bad :-[ :-\
No worries....... Didn't mind, our good. ;)
Post by: Vin on May 01, 2010, 10:54:59 am
i kno its quite silly n its kinda embarrassing for me to evn ask.. :-[ :-[
will some1 pls teach me the logic begind this ques.. ???
Post by: holtadit on May 01, 2010, 10:57:43 am
hey people attached to this post is my doubtSorry i cant answer now, I have to go for lunch.
i kno its quite silly n its kinda embarrassing for me to evn ask.. :-[ :-[
will some1 pls teach me the logic begind this ques.. ???
Just hang in there someone will help you.
Post by: Vin on May 01, 2010, 10:59:40 am
Post by: holtadit on May 01, 2010, 11:12:06 am
a = all trig functions are positive in this quadrant
s = only sine is positive in this quadrant
t= only sine is positive in this quadrant
c= only sine is positive in this quadrant
so if sine and cos are negative they must be in between 180 and 270.
Post by: holtadit on May 01, 2010, 11:16:57 am
Post by: Vin on May 01, 2010, 11:18:29 am
sorry i forgot to tell u so stupid of me i already figured the a) ques..
i wanted b) thanks fo ya time.. :)
sorry again!!
but i liked this method too!!
Post by: holtadit on May 01, 2010, 11:31:09 am
zero since its zero hours after midnight. Asnwer is 5 m
Okay for ii) 5 + 4 sin (30*10) = 1.54m
iii) 5 + 4*sin 360 since 12 hours is the max after midnight
Post by: Vin on May 01, 2010, 12:00:04 pm
but for iii) 5 + 4*sin360 equals 5 thts wat i thought..
but the ans is 9m in ms!!
confused ???
they might hav taken 5 + 4* sin(30 * 3) 3 am/pm
ans to the next is also 3 am n 3 pm..
so basically its trial n error..try till u get the greatest value!
Post by: Vin on May 01, 2010, 12:02:05 pm
Post by: Ghost Of Highbury on May 01, 2010, 12:45:18 pm
Post by: Ghost Of Highbury on May 01, 2010, 01:22:11 pm
(1) + (2) = (3)
Therefore the answer is (5,3)
7)a)ii) Now, MT(A) means to apply the transformation T first and then apply M to the new image.
After applying T , the image's co-ordinates are (5,3)
M is reflection in the line y=x. First, draw the line y=x, reflect the point. The new co-ordinates shud be (3,5).
PS : Wenever a point on the cartesian plane is reflected in the line y=x, the image co-ordinates are actually the inverse of the object co-ordinates.
7)b) This is a direct question . Answer : (0 1)
(1 0)
U can find the list here : https://studentforums.biz/index.php/topic,5575.msg167872.html#msg167872 (3rd message)
7)c) For this u have to do the transformation mentioned in the q i.e TM(Q)
So, M(Q) = (k-3, k-2) (Remember the reversing method)
T(k-3, k-2) = ((k-3) + 3, (k-2) + 2) = (k,k)
The y-co-ordinate of the image is k, and the x-co-ordinate of the image is also k. Thus its always on the line y=x..
7)d) The inverse of the identity matrix remains the same,
7)e)i) N = (0 4) - (0 3) = (0 1)
(0 0) (1 0) (-1 0)
7)e)ii) Check the list : https://studentforums.biz/index.php/topic,5575.msg167872.html#msg167872
rotation 90 degrees clockwise around centre (0,0)
Post by: Vin on May 01, 2010, 01:41:24 pm
really u've made my day! :) :P
awesome didnt expect ny1 to do this for me
thanks again!!:):)
By the way the links are really good.. ;D
Post by: Ghost Of Highbury on May 01, 2010, 03:49:56 pm
Post by: 6394 on May 10, 2010, 03:16:23 pm
ALL: all
SILVER: only sine
TEA : only tan
CUPS: only cos
:) :) :) :) :) :) :)
Post by: ankitd_1994 on May 15, 2010, 02:30:36 pm
Use the pic.
a = all trig functions are positive in this quadrant
s = only sine is positive in this quadrant
t= only sine is positive in this quadrant
c= only sine is positive in this quadrant
so if sine and cos are negative they must be in between 180 and 270.
u cn also use the sentence 'all silver tea cups' (ASTC) to remembr the quadrant thing...it actually helpd me in nov 09 add maths