C2 Question help
Hey stuck on a few questions, wondered if anyone could explain what is going on in each individual step:
QUESTION 1:
Expand (1 + 3x)^8
in ascending powers of x up to and including the term in x^3
You should simplify each coefficient in your expansion. (4)
My answer: 1 + 24x + 252x^2 + 1512x^3
(b) Use your series, together with a suitable value of x which you should state, to estimate the value of (1.003)^8, giving your answer to 8 significant figures.
*don't understand :/ ???
QUESTION 2:
The circle C has centre (−3, 2) and passes through the point (2, 1).
(a) Find an equation for C.
(b) Show that the point with coordinates (−4, 7) lies on C
(c) Find an equation for the tangent to C at the point (−4, 7). Give your answer in the form: ax + by + c = 0, where a, b and c are integers.
*I understand it, just don't know how to apply it.
QUESTION 3:
A geometric series has first term a and common ratio r where r > 1.
The sum of the first n terms of the series is denoted by Sn.
Given that S4 = 10 × S2,
(a) find the value of r.
Given also that S3 = 26,
(b) find the value of a,
(c) show that S6 = 728.
*Need help on this too :/
Many Thanks guys!
QUESTION 1:
Expand (1 + 3x)^8
in ascending powers of x up to and including the term in x^3
You should simplify each coefficient in your expansion. (4)
My answer: 1 + 24x + 252x^2 + 1512x^3
(b) Use your series, together with a suitable value of x which you should state, to estimate the value of (1.003)^8, giving your answer to 8 significant figures.
*don't understand :/ ???
QUESTION 2:
The circle C has centre (−3, 2) and passes through the point (2, 1).
(a) Find an equation for C.
(b) Show that the point with coordinates (−4, 7) lies on C
(c) Find an equation for the tangent to C at the point (−4, 7). Give your answer in the form: ax + by + c = 0, where a, b and c are integers.
*I understand it, just don't know how to apply it.
QUESTION 3:
A geometric series has first term a and common ratio r where r > 1.
The sum of the first n terms of the series is denoted by Sn.
Given that S4 = 10 × S2,
(a) find the value of r.
Given also that S3 = 26,
(b) find the value of a,
(c) show that S6 = 728.
*Need help on this too :/
Many Thanks guys!
Scroll to see replies
Original post by Reety
Hey stuck on a few questions, wondered if anyone could explain what is going on in each individual step:
QUESTION 1:
Expand (1 + 3x)^8
in ascending powers of x up to and including the term in x^3
You should simplify each coefficient in your expansion. (4)
My answer: 1 + 24x + 252x^2 + 1512x^3
(b) Use your series, together with a suitable value of x which you should state, to estimate the value of (1.003)^8, giving your answer to 8 significant figures.
*don't understand :/ ???
Many Thanks guys!
QUESTION 1:
Expand (1 + 3x)^8
in ascending powers of x up to and including the term in x^3
You should simplify each coefficient in your expansion. (4)
My answer: 1 + 24x + 252x^2 + 1512x^3
(b) Use your series, together with a suitable value of x which you should state, to estimate the value of (1.003)^8, giving your answer to 8 significant figures.
*don't understand :/ ???
Many Thanks guys!
(1 + 3x)=1.003
Because both have powers of 8. Solve here for x.
Subsititute x value to the sequence.
I get x= 1.0242535
Original post by Reety
Hey stuck on a few questions, wondered if anyone could explain what is going on in each individual step:
QUESTION 2:
The circle C has centre (−3, 2) and passes through the point (2, 1).
(a) Find an equation for C.
(b) Show that the point with coordinates (−4, 7) lies on C
(c) Find an equation for the tangent to C at the point (−4, 7). Give your answer in the form: ax + by + c = 0, where a, b and c are integers.
*I understand it, just don't know how to apply it.
Many Thanks guys!
QUESTION 2:
The circle C has centre (−3, 2) and passes through the point (2, 1).
(a) Find an equation for C.
(b) Show that the point with coordinates (−4, 7) lies on C
(c) Find an equation for the tangent to C at the point (−4, 7). Give your answer in the form: ax + by + c = 0, where a, b and c are integers.
*I understand it, just don't know how to apply it.
Many Thanks guys!
A) First find the distance between both coordinates using the equation
You get the radius. You have the centre.
Apply it to the equation of circle. Dont leave the radius in square format. Square it in your final answer.
B)sUBstitute these corrdinates(-4,7) to the equation of the circle and get your answer
C)gET the gradient between coordinates (-4,7) and centre (-3,2)
You know any perpendicular gradients equals -1
So gradient os these points are -5
So when multiplying as follows : -5 x m = -1
so m of tangent is 1/5
Now you have the gradient, and the coordinates (-4,7)
You can use y-y1=m(x-x1)
You can subsitiute the coordinates in x1 and y1 and substitute m here . Expand and you get an equation.
Alternative use y=mx+c
Substitute y and coordinates and m and find c.Then re-write the equation.
OK so let's go through each one:
Question 1
Part a looks good.
Part b, is asking you to find (1.003)^8
Looking at part a, we already found the first 4 terms of (1+3x)^8 so it makes sense to use it. Now if we want to equate (1+3x) and (1.003) we will find that 3x=0.003 and x=0.001.
So now we know that we can find (1.003)^8 with our binomial expansion. Subbing in 0.001 to the above answer gets us 1.02425351 (9sf).
Now I rounded that to 9sf to show that, even though we only found the first three terms, this is an appropriate estimate.
As, as we raise 0.001 by increasing powers it gets smaller, therefore our answer will not change by any large degree.
To prove this i just added the term 5670x^4 to get 1.02425352 (9sf) notice that - whilst our answer chnaged) it is still the same to 8sf therefore our final answer is:
1.0242535
checking on a calculator reveals this is (approximately) correct.
Question 2
We should know the equation of a circle is:
(x-a)^2+(y-b)^2=r^2
where (a,b) is the centre and r is the radius.
We can sub in our values for (a,b): (-3,2) to get
(x+3)^2+(y-2)^2=r^2
We still need a value for r and, as we know the point (2,1) lies on the circle we can sub that in to (x,y)
5^2+(-1)^2=r^2
26=r^2
Therefore the equation of our circle is:
(x+3)^2+(y-2)^2=26
The next one just require subbing in (-4,7) into our equation and seeing whether it works out, lets try:
(-1)^2+5^2=1+25+26
So yes it lies on the circle
First we work out the gradient of the radius at the point (-4,7):
The centre of the circle is (-3,2) therefore the change in y is: 2-7=-5
and the change in x is: -3-(-4)=1
As gradient = change in y/change in x = -5/1
therefore the gradient of the radius is -5
The tangent is perpendicular to the radius, ergo, -5*m=-1 where m is the gradient of the tangent.
m= 0.2
Now we have the gradient and a point we can sub it into the equation of a straight line (either y=mx+c or (y-y1)=m(x-x1) )
y=mx+c
7=0.2*-4+c
c=7.8
y=0.2x+7.8
5y=x+39
x-5y+39=0
(y-y1)=m(x-x1)
y-7=0.2x+0.8
y=0.2x+7.8
5y=x+7.8
x-5y+7.8=0
Question 3
If S4=10*S2 then, subbing in the formula Sn from formulae book,
(a(1-r^4))/(1-r) = (10a(1-r^2))/(1-r)
rearranging that gives us
10(1-r^2)=(1-r^4)
r^4-10r^2+9
this factorises as:
(r^2-9)(r^2-1)
which means r=root(9)=+-3
or r=root(1)=1
as r>1
r=3
Now, again subbing values into the formula to Sn
26=(a(1-3^3))/(1-3)
26=(-26a)/(-2)
26=13a
a=2
Subbing in again with S6 into Sn
S6=(2(1-3^6))/(1-3)
S6=728
Hope this helped
Alex
Question 1
Part a looks good.
Part b, is asking you to find (1.003)^8
Looking at part a, we already found the first 4 terms of (1+3x)^8 so it makes sense to use it. Now if we want to equate (1+3x) and (1.003) we will find that 3x=0.003 and x=0.001.
So now we know that we can find (1.003)^8 with our binomial expansion. Subbing in 0.001 to the above answer gets us 1.02425351 (9sf).
Now I rounded that to 9sf to show that, even though we only found the first three terms, this is an appropriate estimate.
As, as we raise 0.001 by increasing powers it gets smaller, therefore our answer will not change by any large degree.
To prove this i just added the term 5670x^4 to get 1.02425352 (9sf) notice that - whilst our answer chnaged) it is still the same to 8sf therefore our final answer is:
1.0242535
checking on a calculator reveals this is (approximately) correct.
Question 2
We should know the equation of a circle is:
(x-a)^2+(y-b)^2=r^2
where (a,b) is the centre and r is the radius.
We can sub in our values for (a,b): (-3,2) to get
(x+3)^2+(y-2)^2=r^2
We still need a value for r and, as we know the point (2,1) lies on the circle we can sub that in to (x,y)
5^2+(-1)^2=r^2
26=r^2
Therefore the equation of our circle is:
(x+3)^2+(y-2)^2=26
The next one just require subbing in (-4,7) into our equation and seeing whether it works out, lets try:
(-1)^2+5^2=1+25+26
So yes it lies on the circle
First we work out the gradient of the radius at the point (-4,7):
The centre of the circle is (-3,2) therefore the change in y is: 2-7=-5
and the change in x is: -3-(-4)=1
As gradient = change in y/change in x = -5/1
therefore the gradient of the radius is -5
The tangent is perpendicular to the radius, ergo, -5*m=-1 where m is the gradient of the tangent.
m= 0.2
Now we have the gradient and a point we can sub it into the equation of a straight line (either y=mx+c or (y-y1)=m(x-x1) )
y=mx+c
7=0.2*-4+c
c=7.8
y=0.2x+7.8
5y=x+39
x-5y+39=0
(y-y1)=m(x-x1)
y-7=0.2x+0.8
y=0.2x+7.8
5y=x+7.8
x-5y+7.8=0
Question 3
If S4=10*S2 then, subbing in the formula Sn from formulae book,
(a(1-r^4))/(1-r) = (10a(1-r^2))/(1-r)
rearranging that gives us
10(1-r^2)=(1-r^4)
r^4-10r^2+9
this factorises as:
(r^2-9)(r^2-1)
which means r=root(9)=+-3
or r=root(1)=1
as r>1
r=3
Now, again subbing values into the formula to Sn
26=(a(1-3^3))/(1-3)
26=(-26a)/(-2)
26=13a
a=2
Subbing in again with S6 into Sn
S6=(2(1-3^6))/(1-3)
S6=728
Hope this helped
Alex
Original post by Reety
Hey stuck on a few questions, wondered if anyone could explain what is going on in each individual step:
QUESTION 3:
A geometric series has first term a and common ratio r where r > 1.
The sum of the first n terms of the series is denoted by Sn.
Given that S4 = 10 × S2,
(a) find the value of r.
Given also that S3 = 26,
(b) find the value of a,
(c) show that S6 = 728.
*Need help on this too :/
Many Thanks guys!
QUESTION 3:
A geometric series has first term a and common ratio r where r > 1.
The sum of the first n terms of the series is denoted by Sn.
Given that S4 = 10 × S2,
(a) find the value of r.
Given also that S3 = 26,
(b) find the value of a,
(c) show that S6 = 728.
*Need help on this too :/
Many Thanks guys!
Which year is this?
Original post by kers123
(1 + 3x)=1.003
Because both have powers of 8. Solve here for x.
Subsititute x value to the sequence.
I get x= 1.0242535
Because both have powers of 8. Solve here for x.
Subsititute x value to the sequence.
I get x= 1.0242535
Yup, got this too ^!
Original post by theJdog
Yup, got this too ^!
Understood?
Original post by alexjackson1711
OK so let's go through each one:
Question 1
Part a looks good.
Part b, is asking you to find (1.003)^8
Looking at part a, we already found the first 4 terms of (1+3x)^8 so it makes sense to use it. Now if we want to equate (1+3x) and (1.003) we will find that 3x=0.003 and x=0.001.
So now we know that we can find (1.003)^8 with our binomial expansion. Subbing in 0.001 to the above answer gets us 1.02425351 (9sf).
Now I rounded that to 9sf to show that, even though we only found the first three terms, this is an appropriate estimate.
As, as we raise 0.001 by increasing powers it gets smaller, therefore our answer will not change by any large degree.
To prove this i just added the term 5670x^4 to get 1.02425352 (9sf) notice that - whilst our answer chnaged) it is still the same to 8sf therefore our final answer is:
1.0242535
checking on a calculator reveals this is (approximately) correct.
Question 2
We should know the equation of a circle is:
(x-a)^2+(y-b)^2=r^2
where (a,b) is the centre and r is the radius.
We can sub in our values for (a,b): (-3,2) to get
(x+3)^2+(y-2)^2=r^2
We still need a value for r and, as we know the point (2,1) lies on the circle we can sub that in to (x,y)
5^2+(-1)^2=r^2
26=r^2
Therefore the equation of our circle is:
(x+3)^2+(y-2)^2=26
The next one just require subbing in (-4,7) into our equation and seeing whether it works out, lets try:
(-1)^2+5^2=1+25+26
So yes it lies on the circle
First we work out the gradient of the radius at the point (-4,7):
The centre of the circle is (-3,2) therefore the change in y is: 2-7=-5
and the change in x is: -3-(-4)=1
As gradient = change in y/change in x = -5/1
therefore the gradient of the radius is -5
The tangent is perpendicular to the radius, ergo, -5*m=-1 where m is the gradient of the tangent.
m= 0.2
Now we have the gradient and a point we can sub it into the equation of a straight line (either y=mx+c or (y-y1)=m(x-x1) )
y=mx+c
7=0.2*-4+c
c=7.8
y=0.2x+7.8
5y=x+39
x-5y+39=0
(y-y1)=m(x-x1)
y-7=0.2x+0.8
y=0.2x+7.8
5y=x+7.8
x-5y+7.8=0
Question 3
If S4=10*S2 then, subbing in the formula Sn from formulae book,
(a(1-r^4))/(1-r) = (10a(1-r^2))/(1-r)
rearranging that gives us
10(1-r^2)=(1-r^4)
r^4-10r^2+9
this factorises as:
(r^2-9)(r^2-1)
which means r=root(9)=+-3
or r=root(1)=1
as r>1
r=3
Now, again subbing values into the formula to Sn
26=(a(1-3^3))/(1-3)
26=(-26a)/(-2)
26=13a
a=2
Subbing in again with S6 into Sn
S6=(2(1-3^6))/(1-3)
S6=728
Hope this helped
Alex
Question 1
Part a looks good.
Part b, is asking you to find (1.003)^8
Looking at part a, we already found the first 4 terms of (1+3x)^8 so it makes sense to use it. Now if we want to equate (1+3x) and (1.003) we will find that 3x=0.003 and x=0.001.
So now we know that we can find (1.003)^8 with our binomial expansion. Subbing in 0.001 to the above answer gets us 1.02425351 (9sf).
Now I rounded that to 9sf to show that, even though we only found the first three terms, this is an appropriate estimate.
As, as we raise 0.001 by increasing powers it gets smaller, therefore our answer will not change by any large degree.
To prove this i just added the term 5670x^4 to get 1.02425352 (9sf) notice that - whilst our answer chnaged) it is still the same to 8sf therefore our final answer is:
1.0242535
checking on a calculator reveals this is (approximately) correct.
Question 2
We should know the equation of a circle is:
(x-a)^2+(y-b)^2=r^2
where (a,b) is the centre and r is the radius.
We can sub in our values for (a,b): (-3,2) to get
(x+3)^2+(y-2)^2=r^2
We still need a value for r and, as we know the point (2,1) lies on the circle we can sub that in to (x,y)
5^2+(-1)^2=r^2
26=r^2
Therefore the equation of our circle is:
(x+3)^2+(y-2)^2=26
The next one just require subbing in (-4,7) into our equation and seeing whether it works out, lets try:
(-1)^2+5^2=1+25+26
So yes it lies on the circle
First we work out the gradient of the radius at the point (-4,7):
The centre of the circle is (-3,2) therefore the change in y is: 2-7=-5
and the change in x is: -3-(-4)=1
As gradient = change in y/change in x = -5/1
therefore the gradient of the radius is -5
The tangent is perpendicular to the radius, ergo, -5*m=-1 where m is the gradient of the tangent.
m= 0.2
Now we have the gradient and a point we can sub it into the equation of a straight line (either y=mx+c or (y-y1)=m(x-x1) )
y=mx+c
7=0.2*-4+c
c=7.8
y=0.2x+7.8
5y=x+39
x-5y+39=0
(y-y1)=m(x-x1)
y-7=0.2x+0.8
y=0.2x+7.8
5y=x+7.8
x-5y+7.8=0
Question 3
If S4=10*S2 then, subbing in the formula Sn from formulae book,
(a(1-r^4))/(1-r) = (10a(1-r^2))/(1-r)
rearranging that gives us
10(1-r^2)=(1-r^4)
r^4-10r^2+9
this factorises as:
(r^2-9)(r^2-1)
which means r=root(9)=+-3
or r=root(1)=1
as r>1
r=3
Now, again subbing values into the formula to Sn
26=(a(1-3^3))/(1-3)
26=(-26a)/(-2)
26=13a
a=2
Subbing in again with S6 into Sn
S6=(2(1-3^6))/(1-3)
S6=728
Hope this helped
Alex
Oh my lord!Thanks alot,buddy! I got it now. I got terrified when working out the 3rd question. Thanks you helped me out
Original post by kers123
Understood?
Oh, I already understood it. I was just going to help the OP when I saw someone beat me to it
But thanks anyway! You doing c2 tomorrow?Original post by theJdog
Oh, I already understood it. I was just going to help the OP when I saw someone beat me to it But thanks anyway! You doing c2 tomorrow?
But thanks anyway! You doing c2 tomorrow?Oh!I see. It's ok. If you post, it will also be helpful to them. You might have a bette way of presenting than I. Yes, I have exam tomorrow. Thanks!
Original post by kers123
Oh!I see. It's ok. If you post, it will also be helpful to them. You might have a bette way of presenting than I. Yes, I have exam tomorrow. Thanks!
Aww GOOD LUCK! You prepared? and me too
Original post by kers123
Oh my lord!Thanks alot,buddy! I got it now. I got terrified when working out the 3rd question. Thanks you helped me out
You're welcome, I just realised I made a small mistake on Q2 when showing the equation with (y-y1)... I forgot to multiply 7.8 by 5. It is correct on the y=mx+c one, however.
Basically, I have a question from the C2 internalional jan 2014 paper and I need help with question 3 b) :https://googledrive.com/host/0B9YYVK...e_20140113.pdf
I got up to the differentiation bit of x^(-1/2)-9x^(-3/2), how do i take it from there?
Thanks guys!!!!!!!
I got up to the differentiation bit of x^(-1/2)-9x^(-3/2), how do i take it from there?
Thanks guys!!!!!!!
Original post by theJdog
Aww GOOD LUCK! You prepared? and me too
good luck!!!
prepared and good!!!thanksOriginal post by alexjackson1711
You're welcome, I just realised I made a small mistake on Q2 when showing the equation with (y-y1)... I forgot to multiply 7.8 by 5. It is correct on the y=mx+c one, however.
oh ok! going to get ready to exam!
Quick Reply
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