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Be kind and check my answers for nov 2004 p2
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
(k^5)+2.5 K^4 x +2.5 k^3 x^2 +1.25 k^2 x^3
1.25*36=45
635/3 (pi)
f(1)=-2<0
f(2)=5.5>0 hence change in sign proves that root lies between 1 and 2
x1=1.387
x2=1.396
x3=1.395
root=1.395
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
(2x+3)/(x+2)
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
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
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)
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
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)
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)
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)
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)
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)
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
Scroll to see replies
Reply 4
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
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
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
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 translations
can you explain more I tried to sketch a graph as it would be easier yet it didn't work out 
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
Reply 10
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
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
Reply 15
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)
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)
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)
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???
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
= 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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