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omg i found that soooo hard!!! i mean most of my answers sounded right but i got a bad feeling, then again i always have a bad feeling
Reply 2
C2 Answers which I put.

1) 4rt2 - 2
2) -16 and (x +2)(3x-2)(x-3)
3) 6K + 15K^2 + 20K^2... k = 2/5 and coeff of x^3 was 32/25
4) rt7/4
5) 4.04, and 1.96
6) 1.07, x=21
7) y = -2/3x + 3.... (x-6)^2 + (y+1)^2 = 34
8) £760,000, and exceeded in the 17th year (could be 18th, but i think 17th)
9) (5pi/6,0) (11pi/6,0) and (0, 1/2) 0.183c, and 1.91c (missing 2 solutions)
10) 7cm, and -40rt2 < 0 so maximum
same here aminalia89, but it never works out for me :frown: . Well i found C2 hard compared to jan 2007. Here was me hoping to get an A after my b in the jan. C1 i thought was pretty good though. I shall see from my results.
for q3) i got 5/16 or something like that and q8) i got 2024 i.e 18 years oh and peaceseeker i hope you do well :smile:
Reply 5
Guitarman5
C2 Answers which I put.

1) 4rt2 - 2
2) -16 and (x +2)(3x-2)(x-3)
3) 6K + 15K^2 + 20K^2... k = 2/5 and coeff of x^3 was 32/25
4) rt7/4
5) 4.04, and 1.96
6) 1.07, x=21
7) y = -2/3x + 3.... (x-6)^2 + (y+1)^2 = 34
8) £760,000, and exceeded in the 17th year (could be 18th, but i think 17th)
9) (5pi/6,0) (11pi/6,0) and (0, 1/2) 0.183c, and 1.91c (missing 2 solutions)
10) 7cm, and -40rt2 < 0 so maximum


for Q10 did you forget to put your value of x back into the equation? (I think it asked for the maximum volume didn't it?)
for the year in 8) it was 17.something (I think), does that mean it's in the 18th year?
and for 7) I think you made the same mistake as me, iirc the y's were -1 and -2 which would give 1^2, not 3^2, giving 26 not 34 - can't guarentee I remembered the y's though
yeah i did too!!! i was so annoyed with myself
Reply 7
Jsuper
for Q10 did you forget to put your value of x back into the equation?


Eh? As in put the value from dy/dx into the second diff? Were we meant to work out the smallest possible volume? :s-smilie:

Damnit! another mark lost!
i just realised :smile: 6/15 = 2/5 man i'm dumb!!
Reply 9
any C1 ppl here. i did it and missed the last question :frown: just thought Q10 was the end of the paper. how many marks was Q11.
Reply 10
Guitarman5
Eh? As in put the value from dy/dx into the second diff? Were we meant to work out the smallest possible volume? :s-smilie:

Damnit! another mark lost!


you differentiate and put to 0 to find the stationary point of the curve (in this case a max).
you then put the value for x back into the original volume equation to find the maximum volume.
umm c1 last question around 10 ish - sorry
Reply 12
Ok, I got bored of waiting for the whole Edexcel rules thing so I'm going to post my take on the C2 exam. I'd like to stress that my answers aren't necessarily correct and I'm doing this all from memory after 12 hours so I may have forgotten the questions and so I might be posting complete crap. If you do get something different (and you think you're right) please let me know.

But here it is:

1) The definite integration question.

You had to find the integral giving your answer in the form a+b2a + b\sqrt{2}

18(1/x)dx=2+42\int^8_1 (1/\sqrt{x}) \, \mathrm{d}x = -2 + 4\sqrt{2}

2) The polynomial question with the factor theorem.

a) They gave you some function and wanted you to find the remainder when you divided it by (x-2) I think. So all you had to do was:

f(2)=16f(2) = -16

b) They told you (x+2) is a factor of the function and wanted you to hence show that the function factorises like (x+2)(3x2)(linear)(x+2)(3x-2)(linear)

I would show how it's done but I don't remember the function well. I'm not entirely sure whether the factor was (x-2) or (x+2).

3) The binomial expansion question.

a) Find the first four terms of (1+kx)6(1+kx)^6.

I used the 1+nx+n(n1)x22!...1 + nx + \frac{n(n-1)x^2}{2!}... formula to give:

1+6kx+15k2x2+20k3x31 + 6kx + 15k^2x^2 + 20k^3x^3

b) They told you the coefficient of x is the same as the coefficient of x^2. So:

6k=15k26k = 15k^2
15k26k=015k^2 - 6k = 0
k(15k6)=0k(15k - 6) = 0
k=6/15(k>0)k = 6/15 (k>0)

I actually really do not remember getting 6/15 at all but it seems right now. I could have the question wrong.

c) Find the coefficient of x^3 which was just plugging in k into 20k320k^3.

4) The question with the triangle + cosine rule.

a) They gave you a triangle with sides 4cm, 5cm, 6cm and an angle A between 5cm and 6cm. They asked you to show that cos A=3/4:

cosA=b2+c2a22bccos A = \frac{b^2 + c^2 - a^2}{2bc}
cosA=62+5242256cos A = \frac{6^2 + 5^2 - 4^2}{2*5*6}
cosA=3/4cos A = 3/4

b) Hence or otherwise, find the exact value of sin A.

In a right angled triangle, cos A = adjacent/hypoteneuse and sin A = opposite/hypoteneuse. So draw a right angled triangle, with the hypoteneuse as 4 and the adjacent as 3. Then the opposite can be worked out as:

a2+b2=c2(Pythagorastheorem)a^2 + b^2 = c^2 (Pythagoras' theorem)
32+b2=423^2 + b^2 = 4^2
b=7b = \sqrt{7}

sinA=opposite/hypoteneusesin A = opposite/hypoteneuse
sinA=7/4sin A = \sqrt{7}/4

5) The trapezium rule question.

a) Fill out the y values on the table. The x values on the table were 0, 0.5, 1, 1.5 and 2.

b) You were asked to work out an approximation for 02(xx3+1)dx\int^2_0 (x\sqrt{x^3 + 1}) \, \mathrm{d}x

Using the trapezium rule:

A=0.52(0+6+2(0.53+1.414+3.137))=4.04(3s.f)A = \frac{0.5}{2}(0 + 6 + 2(0.53 + 1.414 + 3.137)) = 4.04 (3 s.f)

c) They showed you a graph with a line going through the origin and the point (2,6). They also made it clear that the curve goes through the origin and the point (2,6) as well. Find the area between the line and curve, labelled R.

You had to realise that the area under the line was a triangle whose area you could work out doing:

Area=1/2baseheight=1/226=6Area = 1/2 * base * height = 1/2 * 2 * 6 = 6

This is because between the origin and the point (2,6) the x distance is 2, so the base is 2. And between the origin and the point (2,6), the y distance is 6, so the height was 6.

The area under the curve you worked out in part b as 4.04.

So the area of R was:

R = Area of line - area of curve = 6 - 4.04 = 1.96 (3 s.f)

6) The logarithms question.

a) 8^x = 0.8
x=log0.8log8=0.107x = \frac{\mathrm{log}0.8}{\mathrm{log}8} = -0.107


b) Solve 2log3xlog37x=12\mathrm{log}_3 x - \mathrm{log}_3 7x = 1.

log3x2log37x=1\mathrm{log}_3 x^2 - \mathrm{log}_3 7x = 1
log3x2/7x=1\mathrm{log}_3 x^2/7x = 1
3=x/73 = x/7
x=21x = 21

7) The circle question.

a) Working out the equation of line through M and some point was just y - b = m(x-a) and where m was dy/dx of the two points they gave you.

By a circle theorem, the equation of the line you worked out before was perpendicular to the equation of the line you're trying to work out in part a. Then it was just y - b = -1/m (x-a) (using the coordinates of the point M).

b) If the x coordinate of the centre is 6 (which they told you), you could work out its y coordinate using your answer in part a. I believe it was -1. So the point P, the centre, was (6,-1).

Equation of the circle is then:

(xa)2+(yb)2=r2(x-a)^2 + (y-b)^2 = r^2
(x6)2+(y+1)2=r2(x-6)^2 + (y+1)^2 = r^2

To work out the radius, you had the centre and a point on the circle. The distance between these points is then equal to the radius. Working it out gave r^2 = 26

So the equation was:

(x6)2+(y+1)2=26(x-6)^2 + (y+1)^2 = 26

8) The geometric series question.

a) They said that profit in 2006 was £50,000 and they expected it to multiply by r every year. Find the general term for the profit.

Profit = 50000r^(n-1)

b) They said that they expect the profit to exceed £200,000 in Year n. Show that n>log4logr+1n > \frac{\mathrm{log}4}{\mathrm{log}r} + 1

You had an expression for the profit and you were told it's greater than £200,000. So:

50000rn1>20000050000r^{n-1} > 200000
rn1>4r^{n-1} > 4
n1>log4logrn-1 > \frac{\mathrm{log}4}{\mathrm{log}r}
n>log4logr+1n > \frac{\mathrm{log}4}{\mathrm{log}r} + 1

c) They told you r = 1.09. They asked you to work out when the first profit exceeds £200,000.

n>log4log1.09+1n > \frac{\mathrm{log}4}{\mathrm{log}1.09} + 1
n>17.086...n > 17.086...

n must be GREATER than 17.086, so n must be 18.

d) Work out the total profits by year 10.

Sn=a(1rn)1rS_n = \frac{a(1-r^n)}{1-r}
Sn=50000(11.0910)11.09S_n = \frac{50000(1-1.09^10)}{1-1.09}
Sn=759,646=£760,000S_n = 759,646 = £760,000

I AM ACTUALLY GOING TO KILL MYSELF I THINK. I MISREAD MY CALCULATOR AND WROTE £80,000 (thinking it was £75,964 or something).

9) The question where you had to sketch y = sin(x + pi/6).

I'm going to talk in degrees for a minute here.

a) Well for the sketch, I started my graph at (0,0.5) and sketched it down to (150,0) and then brought it back up to (360,0.5) (not like LITERALLY, but in a sine wavey way).

b) The points of intersection were:

(0,0.5),(ππ/6,0)and(2ππ/6,0)(0,0.5), (\pi - \pi/6, 0) and (2\pi - \pi/6, 0)

c) Solve the equation sin(x + \pi/6) = 0.65

Letx+π/6=ALet x + \pi/6 = A

sin A = 0.65
A = 0.708, 2.434

x+π/6=Ax + \pi/6 = A
x=Aπ/6x = A - \pi/6
x=1.91,0.18(2dcp)x = 1.91, 0.18 (2dcp)

10) The question with the cuboid.

They gave you a cuboid with sides 2x, x and y. They told you the surface area of the cuboid is equal to 600.

a) Show that the V=200x4x33V = 200x - \frac{4x^3}{3} (V being the volume).

First start by working out the area of each of the face of the cuboid. You know they add up to 600. On a cuboid, faces opposite each other have the same area. There were 3 different faces on the cuboid.

Area of the first face = 2x * x = 2x^2
Area of the second face = 2x * y = 2xy
Area of the third face = x * y = xy

Each face appears twice on the cuboid, so the total surface area can be worked out like:

A=(22x2)+(22xy)+(2xy)A = (2* 2x^2) + (2 * 2xy) + (2 * xy)
600=4x2+4xy+2xy600 = 4x^2 + 4xy + 2xy
600=4x2+6xy600 = 4x^2 + 6xy
6004x2=6xy600 - 4x^2 = 6xy
6004x26x=y\frac{600 - 4x^2}{6x} = y

Ok so now we have y in terms of x.

Volume = length * base * height
V=2xxy=2x2yV = 2x * x * y = 2x^2y
V=2x2(6004x26x)V = 2x^2 * (\frac{600 - 4x^2}{6x})

With a bit of algebra, you can show that:

V=200x4x33V = 200x - \frac{4x^3}{3}

b) Find the maximum volume of the cuboid.

ddxV=2004x2\frac{\mathrm{d}}{\mathrm{d}x} V = 200 - 4x^2

The maximum occurs at a turning point, which means dV/dx = 0, so solve that equation:

2004x2=0200 - 4x^2 = 0
x2=50x^2 = 50
x=50(50x = \sqrt{50} (-\sqrt{50} is invalid)

V=2005045033V = 200 * \sqrt{50} - \frac{4*\sqrt{50}^3}{3}
V = 943 cm^3

c) Make sure it's a maximum turning point.

The second differential = -8x
850-8 * \sqrt{50} is negative, so it is a maximum turning point.

*Edit2*

Ok edited to remove the C1 question which was scaring everyone =P.

The order has been revised as well.
Jsuper
for Q10 did you forget to put your value of x back into the equation? (I think it asked for the maximum volume didn't it?)
for the year in 8) it was 17.something (I think), does that mean it's in the 18th year?
and for 7) I think you made the same mistake as me, iirc the y's were -1 and -2 which would give 1^2, not 3^2, giving 26 not 34 - can't guarentee I remembered the y's though


you're right or we're both wrong lol-i was scared for a minute when i saw 34

If it was the 17th year it was less that 200,000 so has to be minimum 18
DeathAwaitsU
Ok, I got bored of waiting for the whole Edexcel rules thing so I'm going to post my take on the C2 exam. I'd like to stress that my answers aren't necessarily correct and I'm doing this all from memory after 12 hours so I may have forgotten the questions and so I might be posting complete crap. If you do get something different (and you think you're right) please let me know. I can't remember 1 question in the paper though which is annoying and the questions aren't in the correct order at all:

But here it is:

1) The definite integration question.

You had to find the integral giving your answer in the form a+b2a + b\sqrt{2}

18(1/x)dx=2+42\int^8_1 (1/\sqrt{x}) \, \mathrm{d}x = -2 + 4\sqrt{2}

2) The polynomial question with the factor theorem.

a) They gave you some function and wanted you to find the remainder when you divided it by (x-2) I think. So all you had to do was:

f(2)=16f(2) = -16

b) They told you (x+2) is a factor of the function and wanted you to hence show that the function factorises like (x+2)(3x2)(linear)(x+2)(3x-2)(linear)

I would show how it's done but I don't remember the function well. I'm not entirely sure whether the factor was (x-2) or (x+2).

3) The binomial expansion question.

a) Find the first four terms of (1+kx)6(1+kx)^6.

I used the 1+nx+n(n1)x22!...1 + nx + \frac{n(n-1)x^2}{2!}... formula to give:

1+6kx+15k2x2+20k3x31 + 6kx + 15k^2x^2 + 20k^3x^3

b) They told you the coefficient of x is the same as the coefficient of x^2. So:

6k=15k26k = 15k^2
15k26k=015k^2 - 6k = 0
k(15k6)=0k(15k - 6) = 0
k=6/15(k>0)k = 6/15 (k>0)

I actually really do not remember getting 6/15 at all but it seems right now. I could have the question wrong.

c) Find the coefficient of x^3 which was just plugging in k into 20k320k^3.

4) The question with the triangle + cosine rule.

a) They gave you a triangle with sides 4cm, 5cm, 6cm and an angle A between 5cm and 6cm. They asked you to show that cos A=3/4:

cosA=b2+c2a22bccos A = \frac{b^2 + c^2 - a^2}{2bc}
cosA=62+5242256cos A = \frac{6^2 + 5^2 - 4^2}{2*5*6}
cosA=3/4cos A = 3/4

b) Hence or otherwise, find the exact value of sin A.

In a right angled triangle, cos A = adjacent/hypoteneuse and sin A = opposite/hypoteneuse. So draw a right angled triangle, with the hypoteneuse as 4 and the adjacent as 3. Then the opposite can be worked out as:

a2+b2=c2(Pythagorastheorem)a^2 + b^2 = c^2 (Pythagoras' theorem)
32+b2=423^2 + b^2 = 4^2
b=7b = \sqrt{7}

sinA=opposite/hypoteneusesin A = opposite/hypoteneuse
sinA=7/4sin A = \sqrt{7}/4

5) The geometric series question.

a) They said that profit in 2006 was £50,000 and they expected it to multiply by r every year. Find the general term for the profit.

Profit = 50000r^(n-1)

b) They said that they expect the profit to exceed £200,000 in Year n. Show that n>log4logr+1n > \frac{\mathrm{log}4}{\mathrm{log}r} + 1

You had an expression for the profit and you were told it's greater than £200,000. So:

50000rn1>20000050000r^{n-1} > 200000
rn1>4r^{n-1} > 4
n1>log4logrn-1 > \frac{\mathrm{log}4}{\mathrm{log}r}
n>log4logr+1n > \frac{\mathrm{log}4}{\mathrm{log}r} + 1

c) They told you r = 1.09. They asked you to work out when the first profit exceeds £200,000.

n>log4log1.09+1n > \frac{\mathrm{log}4}{\mathrm{log}1.09} + 1
n>17.086...n > 17.086...

n must be GREATER than 17.086, so n must be 18.


6) The trapezium rule question.
a) Fill out the y values on the table. The x values on the table were 0, 0.5, 1, 1.5 and 2.

b) You were asked to work out an approximation for 02(xx3+1)dx\int^2_0 (x\sqrt{x^3 + 1}) \, \mathrm{d}x

Using the trapezium rule:

A=0.52(0+6+2(0.53+1.414+3.137))=4.04(3s.f)A = \frac{0.5}{2}(0 + 6 + 2(0.53 + 1.414 + 3.137)) = 4.04 (3 s.f)

c) They showed you a graph with a line going through the origin and the point (2,6). They also made it clear that the curve goes through the origin and the point (2,6) as well. Find the area between the line and curve, labelled R.

You had to realise that the area under the line was a triangle whose area you could work out doing:

Area=1/2baseheight=1/226=6Area = 1/2 * base * height = 1/2 * 2 * 6 = 6

This is because between the origin and the point (2,6) the x distance is 2, so the base is 2. And between the origin and the point (2,6), the y distance is 6, so the height was 6.

The area under the curve you worked out in part b as 4.04.

So the area of R was:

R = Area of line - area of curve = 6 - 4.04 = 1.96 (3 s.f)

The question where you had to sketch y = sin(x + \pi/6).

I'm going to talk in degrees for a minute here.

a) Well for the sketch, I started my graph at (0,0.5) and sketched it down to (150,0) and then brought it back up to (360,0.5).

b) The points of intersection were:

(0,0.5),(ππ/6,0)and(2ππ/6,0)(0,0.5), (\pi - \pi/6, 0) and (2\pi - \pi/6, 0)

c) Solve the equation sin(x + \pi/6) = 0.65

Let x + \pi/6 = A

sin A = 0.65
A = 0.708, 2.434

x+π/6=Ax + \pi/6 = A
x=Aπ/6x = A - \pi/6
x=1.91,0.18(2dcp)x = 1.91, 0.18 (2dcp)


8) The question where you had to sketch the cubic graph.

a) They told you a f'(x) equals something and the curve f(x) passes through (5,65)

Integrating f'(x) gave some polynomial equation + c. You know the curve passes through (5,65). So you could work out c by solving f(5) = 65. I got c = 0.

b) You had to factorise it or something and show what they asked.

c) Sketch it. I remember it passed through (-3/2, 0), the origin and (4,0) or something.

10) The question with the cuboid.

They gave you a cuboid with sides 2x, x and y. They told you the surface area of the cuboid is equal to 600.

a) Show that the V=200x4x33V = 200x - \frac{4x^3}{3} (V being the volume).

First start by working out the area of each of the face of the cuboid. You know they add up to 600. On a cuboid, faces opposite each other have the same area. There were 3 different faces on the cuboid.

Area of the first face = 2x * x = 2x^2
Area of the second face = 2x * y = 2xy
Area of the third face = x * y = xy

Each face appears twice on the cuboid, so the total surface area can be worked out like:

A=(22x2)+(22xy)+(2xy)A = (2* 2x^2) + (2 * 2xy) + (2 * xy)
600=4x2+4xy+2xy600 = 4x^2 + 4xy + 2xy
600=4x2+6xy600 = 4x^2 + 6xy
6004x2=6xy600 - 4x^2 = 6xy
6004x26x=y\frac{600 - 4x^2}{6x} = y

Ok so now we have y in terms of x.

Volume = length * base * height
V=2xxy=2x2yV = 2x * x * y = 2x^2y
V=2x2(6004x26x)V = 2x^2 * (\frac{600 - 4x^2}{6x})

With a bit of algebra, you can show that:

V=200x4x33V = 200x - \frac{4x^3}{3}

b) Find the maximum volume of the cuboid.

ddxV=2004x2\frac{\mathrm{d}}{\mathrm{d}x} V = 200 - 4x^2

The maximum occurs at a turning point, which means dV/dx = 0, so solve that equation:

2004x2=0200 - 4x^2 = 0
x2=50x^2 = 50
x=50(50x = \sqrt{50} (-\sqrt{50} is invalid)

V=2005045033V = 200 * \sqrt{50} - \frac{4*\sqrt{50}^3}{3}
V = 942 cm^3

c) Make sure it's a maximum turning point.

The second differential = -8x
850-8 * \sqrt{50} is negative, so it is a maximum turning point.


Woot Woot...i got exactly the same answers...100% for both of us maybe! :biggrin:. Except q10b is 943cm^3 as it was to the nearest whole no.
Reply 15
Not sure I entirely remember your question 8 tbh. :s-smilie:

That is worrying...

However apart from that it looks ok :smile:
I just pray M1,C3,C4 go just aswell.
Reply 17
Are you sure Q8 is C2 mate? I seriously don't remember that question - nothing from it rings a bell :s-smilie:

Cheers
yay, deathwaitsyou woot i got practically the same as you expect i forgot to plug the formula for q11!!!
Reply 19
Dark and Lovely
Woot Woot...i got exactly the same answers...100% for both of us maybe! . Except q10b is 943cm^3 as it was to the nearest whole no.


Yay :biggrin:. I think you're right about the 943 as I remember writing it down, I must have not been concentrating/tired when typing that :p:.

Guitarman5
Not sure I entirely remember your question 8 tbh. :s-smilie:

That is worrying...

However apart from that it looks ok :smile:


Oh that must explain why I've got 11 questions instead of 10. I think the cubic one may have been in C1.

Glad you agree with the answers though.