Math paper4 Jun. 2009 help

[Ghost Of Highbury]

Yep IGCSE 0580 Maths Paper 4 June 2009

Yea I could do it that way, but I just want to check using differentiation because i'm going to need to be able to do that way later. Also sometimes my tangents aren't very accurate so I just want to be able to use differentiation to check it.

Thanks

dy/dx (x/2 - 2/x) = 1/2 +  2/x² => substituing x=2 --> m = 1

[@d!_†oX!©]

hey aadi cud u explain me differentiation more briefly...i wud want to learn to check my answer if possible...pm me or better come on gmail...

[Ghost Of Highbury]

its difficult to learn differentiation in such a short span of..there are different methods for different type

for e.g here...u have to apply the u/v rule..

so yea..maybe later..

[lol]

Thanks but i don't understand why it is x^2, isn't it x^-2 because the original power is -1 and we reduce it by 1 we get -2

[~~~~shreyapril~~~~]

ya ure rite!!
but x^-2 becomes 1/x^2!!!

[Ghost Of Highbury]

n ono...its not that simple

its the u/v method

here

d/dx (u/v) = (vdu-udv)/v²
break the expressin into two halves

1. x/2

2. 2/x

first..

u = x
v = 2
du = 1
dv = 0

->(vdu-udv)/v²

= 2/4

second..

u = 2
v =x
du = 0
dv = 1

(vdu-udv)/v²

=> (0-2)x² = -2/x²

so the final differentiation = 1/2 - (-2/x²) = 1/2 + 2/x²

[@d!_†oX!©]

its difficult to learn differentiation in such a short span of..there are different methods for different type

for e.g here...u have to apply the u/v rule..

so yea..maybe later..

ya ya...kk..
u teach me after the ig'z!!

[lol]

OMG I'm such an idiot. This is exactly the kind of thing i'm going to do tomorrow. I can do the work but I just make so many errors. After substituting x=2 into the equation i forgot to multiply by 2 before adding a half.

Thank you all so much and sorry about the hassle

[Ghost Of Highbury]

OMG I'm such an idiot. This is exactly the kind of thing i'm going to do tomorrow. I can do the work but I just make so many errors. After substituting x=2 into the equation i forgot to multiply by 2 before adding a half.

Thank you all so much and sorry about the hassle

thats k..

[~~~~shreyapril~~~~]

baapre aadi yuo tuped so much just to explain him differentiation!!

i guess adi toxic is now scared!!

By the way there is one trick by differentiation you can find the equation of the normal and tangnt in maths !!

i ahve used it many times!!
although this is not correct method!!

[moon]

I need help in NOV 2004 Q7 Please.It is diffiicult.

[Ghost Of Highbury]

I need help in NOV 2004 Q7 Please.It is diffiicult.

the whole q?

[moon]

please I need a3, bii and c2

[wassup]

I need help in NOV 2004 Q7 Please.It is diffiicult.

7 a)i) y = px + qx + r
      y = 4^2+(-2*4)+(-3)
      y = 5

ii) On K and L y=0
so, px + qx + r=0
     x-2x-3=0
     Solve this n u'll get x=-1, 3
K(-1,0) L(3,0)

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

b)i) Upside down
ii) Eqn, will be x^2=y, so just put some values n get the shape of graph

c)i) c=0 because it passes through d origin
ii) (3,0) (4,8)
put these in the eqn.

3^2a + 3b = 0
4^2a + 4b = 8

solve these n u'll get a = 2, b = –6

[Ghost Of Highbury]

please I need a3, bii and c2

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, .

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