Be kind and check my answers for nov 2004 p2 Watch

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habosh
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ok for november 2004 I just wanna check my answers fermat or newton or any of you guyz please
if you are not bothered to read it all just scroll the last functions part because it's the one I'm not sure about it at all

2) a) find the first four terms in ascending powers of x, in the binomial expansion of (k + x/2)^5, where k is a constant. (2 marks)
(k^5)+2.5 K^4 x +2.5 k^3 x^2 +1.25 k^2 x^3

given that the third term of the series is 540x^2,

b) show that k=6 (2 marks)

540=2.5 K^3
540/2.5=216
quiped sqrt of 216=6


c)find the coefficient of x^3 (2 marks)
1.25*36=45

3) y=4x-6/x, x>0
the shaded region R is bounded by the curve, the x axis and the lines with equations x=2 and x=4. This region is rotated through 2pi radians about the x axis.
Find the exact value of the volume of the solid generated (8 marks)
635/3 (pi)

4) f(x)= (x^3)-2-1/x, x is not 0
a) show that the equation f(x)=0 hasa root between 1 and 2 (2 marks)
f(1)=-2<0
f(2)=5.5>0 hence change in sign proves that root lies between 1 and 2


an approximation for this root is found using the iteration formula
Xn+1= (2+ 1/Xn)^1/3, with Xo=1.5
b)By calculating the values of X1, X2, X3 and X4, find an approximationto this root, givingyou answer to 3 d.p (4 marks)
x1=1.387
x2=1.396
x3=1.395
root=1.395


c)By considering the change of sign of f(x) in a suitable interval, verify that your answerto part (b) is correct to 3 d.p (2 marks)
1.396^3-2-1/1.396= 4.214757777*10^-3
1.394^3-2-1/1.394=-8.489130778*10^-3
hence change in sign
1.394<x<1.396
so=1.395


5) f(x)=(2x+5/x+3) - (1/(x+3)(x+2)), x>-2
a) express f(x) as a single fraction in its simplest form (5 marks)
(2x+3)/(x+2)

b) Hence, show that f(x)= 2 - 1/x+2, x>-2 (2 marks)
I'm not sure about this but well I'm gonna try
2-1/(-2+2)=not in the range
(2x+3)/x+2) by subtituating with -2 also not in the range ,I cant find any other way


The curve y=1/x, x>0 is mapped onto the curve y=f(x), using three successive transformations T1, T2, T3 where T1 and T3 are translations
c) describe fully T1,T2,T3 (4 marks)
this is a hard one too but I tried and dunno it I'm right or wrong
transation (0,(-5/4)) translation ((8/3),0)
and strech with factor (2/7)through y axis
well that is all thanx alot to anybody who reads it through and just give me brief comment about my mistakes and erm specially the last question
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habosh
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so nobody is kind enough I get:p:
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El Stevo
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i'll have a look...
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habosh
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(Original post by El Stevo)
i'll have a look...
yay thanx alot hun
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davidreesjones
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I have just checked q5 -

a) right - but perhaps write x>-2 as well
b) this is much easier than it seems (i dont think ur method is right) - just

f(x) = (2x + 3)/(x+2)
= (2(x+2)-1)/(x+2) - factorise x+2 out of 2x +3 to get a remained of -1
= 2(x+2)/(x+2) - 1/(x+2) - split the fraction
= 2 - 1/(x+2) - cancel

c) This is a bit harder - think it through backwards and then reverse all the steps to get

T1 - to go from 1/x to 1/(x+2) translate "left" 2 units
T2 - from 1/(x+2) to -1/(x+2) reflect in x axis
T3 - from this to 2 - 1/(x+2) translate "up" 2 units

hope this helps
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El Stevo
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two looks good to me...

would 3) be y=(4x-6)/x or y=4x-(6/x)
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habosh
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(Original post by davidreesjones)
I have just checked q5 -

a) right - but perhaps write x>-2 as well
b) this is much easier than it seems (i dont think ur method is right) - just

f(x) = (2x + 3)/(x+2)
= (2(x+2)-1)/(x+2) - factorise x+2 out of 2x +3 to get a remained of -1
= 2(x+2)/(x+2) - 1/(x+2) - split the fraction
= 2 - 1/(x+2) - cancel

c) This is a bit harder - think it through backwards and then reverse all the steps to get

T1 - to go from 1/x to 1/(x+2) translate "left" 2 units
T2 - from 1/(x+2) to -1/(x+2) reflect in x axis
T3 - from this to 2 - 1/(x+2) translate "up" 2 units

hope this helps
oh my god...all this time I though the equation was (2-1)/(x+2)
I get the first part but I dont get any of the translationscan you explain more I tried to sketch a graph as it would be easier yet it didn't work out
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habosh
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(Original post by El Stevo)
two looks good to me...

would 3) be y=(4x-6)/x or y=4x-(6/x)
I took it as 4x -(6/x) as no quatations were included :confused:
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El Stevo
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for iteration on q4b would work to 4 dp and give last answer to 3dp, answer is correct though...
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habosh
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(Original post by El Stevo)
for iteration on q4b would work to 4 dp and give last answer to 3dp, answer is correct though...
I usually write the full answer for iteration but i'll keep the four decimal places in mind from now on,...It's abit scary question 4 last bit the one with translations scared the hell out of me I admit I still have no idea about it yet i remember transformations quite clearly
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davidreesjones
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hope this helps - key in top right!
Attached files
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habosh
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(Original post by davidreesjones)
hope this helps - key in top right!
yeah it helped alot,but how are we supposed to know that we should sketch 1/(2+x) for example we have 2-1/(2+x0 but we dont have 1/(x+2)as an individual ....I mean how will I know I should sketch all of this ... :confused: :bawling:
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El Stevo
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split the question up and look for bits you know... q3 coming...
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habosh
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(Original post by El Stevo)
split the question up and look for bits you know... q3 coming...
you can the transformation thing all depends on 2-1/2+x and 1/2+x and -1/2+x you dont sketch these your doomed :hmpf:
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habosh
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(Original post by davidreesjones)
I have just checked q5 -

a) right - but perhaps write x>-2 as well
b) this is much easier than it seems (i dont think ur method is right) - just

f(x) = (2x + 3)/(x+2)
= (2(x+2)-1)/(x+2) - factorise x+2 out of 2x +3 to get a remained of -1
= 2(x+2)/(x+2) - 1/(x+2) - split the fraction
= 2 - 1/(x+2) - cancel

c) This is a bit harder - think it through backwards and then reverse all the steps to get

T1 - to go from 1/x to 1/(x+2) translate "left" 2 units
T2 - from 1/(x+2) to -1/(x+2) reflect in x axis
T3 - from this to 2 - 1/(x+2) translate "up" 2 units

hope this helps
please I need to know why did we drow 1/(2+x) how can we with out the 2 it's part of the whole equation and how did we drow -1/(2+x) based on what facts....please I need to know
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davidreesjones
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i think its best to start at the back and try to map f(x) onto 1/x.

So to remove the +2 we translate down (reverse this and get T3)
Then to remove the *-1 we reflect in the x axis (T2)
Then to get from this to 1/x tranlate two right using the rules (reverse for T1)
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habosh
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(Original post by davidreesjones)
i think its best to start at the back and try to map f(x) onto 1/x.

So to remove the +2 we translate down (reverse this and get T3)
Then to remove the *-1 we reflect in the x axis (T2)
Then to get from this to 1/x tranlate two right using the rules (reverse for T1)
oh I get it now.....we need to map 1/x over the original equation so we first translate it to the left 2 then refelct it then we translate it up 2 along the y axis is it like that???
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El Stevo
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= Pi Int [2,4] (4x-6/x)^2 dx
= Pi Int [2,4] [16(x^2) - 48 + 36(x^-2) dx]
= Pi [2,4] [(16/3)(x^3) - 48x - 36(x^-1)]
= Pi [(1024/3) - 192 - 9) - {(128/3) - 96 - 18}]
= Pi [(1024/3) - 192 - 9) - {(128/3) - 96 - 18}]
= Pi [(421/3) - (-214/3)
= 635/3 Pi
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habosh
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(Original post by El Stevo)
= Pi Int [2,4] (4x-6/x)^2 dx
= Pi Int [2,4] [16(x^2) - 48 + 36(x^-2) dx]
= Pi [2,4] [(16/3)(x^3) - 48x - 36(x^-1)]
= Pi [(1024/3) - 192 - 9) - {(128/3) - 96 - 18}]
= Pi [(1024/3) - 192 - 9) - {(128/3) - 96 - 18}]
= Pi [(421/3) - (-214/3)
= 635/3 Pi
yay,I feel abit reliefed after understanding the questions thanx both of you alot
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chats
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for habosh

p2 november 04

8) i)prove that tan theta+cot theta= 2 cosec 2theta

ii) Given that sin alpha= 5/13, 0<alpha< pi/2, find the exact value of

a) cos alpha

b)cos 2alpha

Given also that 13 cos(x+alpha)+5sinx=6, and 0<alpha<pi/2,

c)find the value of x
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