Author Topic: Math paper4 Jun. 2009 help  (Read 6495 times)

Offline @d!_†oX!©

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Re: Math paper4 Jun. 2009 help
« Reply #45 on: November 03, 2009, 04:30:09 pm »
there u go:

a)iii) K(-1,0) and L(3,0)
       M lies on the line of symmetry so the the point on the x-axis along this line of symmetry will be the mid-point of the line KL
       Mid point = ((x2+x1)/2 , (y2+y1)/2)
       Mid point = (1,0)
As the point M lies on the same line of symmetry thus its x-coordinate is 1
by using the value of x(as 1) and the values of p, q and r given find y
y= -4

b)ii)The graph will be the same figurebut its base will not go below the origin(0,0)

c)ii)
put the values for each coordinate in the equation forming 2 equations nd then solve them simultaneously
for (3,0)
0=a(3)2+b(3) +0
=>9a+3b=0
for (4,8)
8=a(4)2+b(4)+o
=>16a+4b=8

solve them simultaneously to get a=2, and b=-6
AAL IZZ WELL!!! ;)

Offline @d!_†oX!©

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Re: Math paper4 Jun. 2009 help
« Reply #46 on: November 03, 2009, 04:37:24 pm »
for aiii

best wud be to draw the graph, and see where the minimum point lies. write its cordinates

for confirmation u can use differentiation

dy/dx = 2x - 2

2x - 2 =0

2x = 2

x = 1 substituting y = 4

dont bother if u dont understand this

for bii

u know then the value are applied in the equation

u r left with only y = x2 which is a parabola and passes through the origin (0,0)

for cii

it says it passes through the origin
which means, c =0

u r left with

y= ax^2 + bx

substitute the cordinates (3, 0) and (4, 8).

to get

0 = 9a + 3b

and

8 = 16a + 4b

from the second we can get

a = 8-4b / 16
substitute this in the first to get
72 - 36b
--------  = -3b
     16

72 - 36b = -48b

solve to get

b = -6 and a = 2

@wassup - for aiii - it is not they-axis, its a line of symmetry

aadi won't it be tooo time consuming to draw the graph and do the differentiation(a part)....just see wat i did....i think dts the right method!!
AAL IZZ WELL!!! ;)

Offline Ghost Of Highbury

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Re: Math paper4 Jun. 2009 help
« Reply #47 on: November 03, 2009, 04:38:16 pm »
aadi won't it be tooo time consuming to draw the graph and do the differentiation(a part)....just see wat i did....i think dts the right method!!

yes. that would be better. i considered that and thus didnt comment. u wanted to hear this. lol
divine intervention!

Offline moon

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Re: Math paper4 Jun. 2009 help
« Reply #48 on: November 03, 2009, 04:47:56 pm »

iii) On M x=0
so, y=r
     y=-3
M(0,-3)


This answer isn't like that in the mark scheme it should b (1, -4)
Thanks u all for ur help & I think that it is not that much difficult.

Offline wassup

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Re: Math paper4 Jun. 2009 help
« Reply #49 on: November 03, 2009, 04:57:19 pm »
This answer isn't like that in the mark scheme it should b (1, -4)
Thanks u all for ur help & I think that it is not that much difficult.

sorry i thot it was d y axis.

Offline @d!_†oX!©

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Re: Math paper4 Jun. 2009 help
« Reply #50 on: November 03, 2009, 04:58:43 pm »
yes. that would be better. i considered that and thus didnt comment. u wanted to hear this. lol
hey nthng like that dude...just wanted to let u know....
AAL IZZ WELL!!! ;)

Offline ~~~~shreyapril~~~~

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Re: Math paper4 Jun. 2009 help
« Reply #51 on: November 04, 2009, 09:56:12 am »
abbe kitan jhagde ho tum dono!!!!
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