[xayno]

11. a2 = 6
S(infinity) = 25

a2 = ar = 6
a = 6/r

S(infinity) = a/(1-r) = a / (1- 6/a)
25 = a/ (1-6/a)
25(1-6/a) = a
25 - 150/a = a
a - 25 + 150/a = 0
a^2 - 25a + 150 = 0
(a - 10)(a - 15) = 0
a = 10 r = 0.6
a = 15 r = 0.4

Un = ar^n-1
so
Un = 10 * 0.6^n-1
Un = 15 * 0.4^n-1

15 * 0.4^n-1 : 10 * 0.6^n-1
15 * 0.4 * 0.4^n-2 : 10 * 0.6 * 0.6^n-2
6 * 0.4^n-2 : 6 * 0.6^n-2
0.4^n-2 : 0.6^n-2
5^n-2 * 0.4^n-2 : 5^n-2 * 0.6^n-2 /multiplying both sides by 5^n-2
2^n-2 : 3^n-2

[Emily Jane]

(Original post by xayno)
11. a2 = 6
S(infinity) = 25

a2 = ar = 6
a = 6/r

S(infinity) = a/(1-r) = a / (1- 6/a)
25 = a/ (1-6/a)
25(1-6/a) = a
25 - 150/a = a
a - 25 + 150/a = 0
a^2 - 25a + 150 = 0
(a - 10)(a - 15) = 0
a = 10 r = 0.6
a = 15 r = 0.4

Un = ar^n-1
so
Un = 10 * 0.6^n-1
Un = 15 * 0.4^n-1

15 * 0.4^n-1 : 10 * 0.6^n-1
15 * 0.4 * 0.4^n-2 : 10 * 0.6 * 0.6^n-2
6 * 0.4^n-2 : 6 * 0.6^n-2
0.4^n-2 : 0.6^n-2
5^n-2 * 0.4^n-2 : 5^n-2 * 0.6^n-2 /multiplying both sides by 5^n-2
2^n-2 : 3^n-2

Kicking myself I didn't get the first part

[elliannagibbs]

Did anyone understand question's 9 and 10? :/

[genius017]

(Original post by Polioz)
I think you will only lose 1 mark.

you kept the + in there when you should of multiplied. Log(3) + log(2) = log(6)

essentialy the formula was in the form of y=kx^n which is the same as
logy=logk +nlogx
y = c + mx
logk=2
k=100
n=3

so if you simplify logy=logk +nlogx we get .... logy=log(kx^n) ..... get rid of the logs.... y=kx^n ...... therefore y=100x^3




Im not great at explaining maths sorry

i got y=3x+100
i know thats completely wrong but do you think ill still pick up 1 mark for finding out the gradient? :'( 5 marks is alott to lose on one question

[mathsodoku]

(Original post by enlighten25)
did anyone get a=10 and r=0.6 and a=15 and r=0.4 for the first part of the last geometric qu? about 7marks i think, had no idea how to do part 2 about 3 marks????

yeah i got the same thing. That is defo the right answer

for part (i)
you can deduce from the info in the q
2nd term 6=ar
Sum to infinity = 25
25= (a/(1-r))
Rearranging this gives
25(1-r)=a
You can sub this into first equation
So..
6=(a)*r
Now becomes
6=25(1-r)*r
6=25r-25r^2
This rearranges gives you this quadratic
25r^2 -25r +6=0
This solved gives you
X= 3/5 or 0.6
Putting this into first equation gives you
6= 0.6*r
6/0.6= r
r =10

is this how you did ure's

[WizKidd]

FFS Maaaan! January resit it is. Even with a calculator I added the value for the trapezium rule wrong, my life

[mathsodoku]

(Original post by xayno)
That's how i got it

dy/dx = 6x^(1/2) - 5
integrated = 4x^(3/2) - 5x + c passing thru 4,20
20 = 4(4)^(3/2) -5(4) + c
20 = 32 - 20 + c
8 = c

my bad how many marks do you think i would lose if i did that. I think i did i am not 100percent sure

[senS]

Looking at the answers that you guys posted, im not hoping for an A anymore

[riff210]

(Original post by cgpboy)
what did people write for question for, y=3fx and y=f4x?

(i) (6,9)
(ii) (3/2,3)

[xayno]

Can someone tell me where can i find boundaries for past exams for 100/100 ums ?

[bluekipper]

(Original post by xayno)
Can someone tell me where can i find boundaries for past exams for 100/100 ums ?

They only publish the raw unit marks up to A, but you can get the 100 UMS raw mark by doubling the difference between the raw marks for A and B.

eg. Taking June 2011's C2 boundaries, A was 53/72 (=80 UMS) and a B was 46/72 (=70 UMS). Difference = 7, so 90 UMS would be 60/72 and 100 UMS 67/72.

[xayno]

(Original post by mathsodoku)
my bad how many marks do you think i would lose if i did that. I think i did i am not 100percent sure

I think you would lose 1 Mark because all your calculations were fine its just the answer that's wrong

[smiley12495]

What did you guys get for Q9? The volume of the trench.

[mathsodoku]

(Original post by smiley12495)
What did you guys get for Q9? The volume of the trench.

q.9
Simple use of trapizium rule

=[h/2 (y(0)+y(n)+2(y1+y2+y3+y4....))]

=[0.2/2 (0+0 +2(0.5+0.7+0.75+0.7+0.5)]

area of cross section=0.63

=0.63*50m(width)

=0.63*50

=31.50m^2


(ii)

(A)y=3.8x^4 -6.8x^3 +7.7x^2 -4.2x

y=3.8(0.2)^4 -6.8(0.2)^3 +7.7(0.2)^2 -4.2(0.2)
y= -0.58032

=0.58032 as +area

0.6-0.58032= 0.01968

(B)y=3.8x^4 -6.8x^3 +7.7x^2 -4.2x

Intergrated gives

=(3.8x^5)/5 -(6.8x^4)/4 + (7.7x^3)/3 -(4.2x^2)/2 between 9 and 0.
=(3.8(0.9)^5)/5 -(6.8(0.9)^4)/4 + (7.7(0.9)^3/3) -(4.2(0.9)^2/2)
=-0.4964976
must *50 to get volume

=-24.82488
=+24.82488 as canoot have negative volume.

[sarah:))]

(Original post by mathsodoku)
q.9
Simple use of trapizium rule

=[h/2 (y(0)+y(n)+2(y1+y2+y3+y4....))]

=[0.2/2 (0+0 +2(0.5+0.7+0.75+0.7+0.5)]

area of cross section=0.63

=0.63*50m(width)

=0.63*50

=31.50m^2



(ii)

(A)y=3.8x^4 -6.8x^3 +7.7x^2 -4.2x

y=3.8(0.2)^4 -6.8(0.2)^3 +7.7(0.2)^2 -4.2(0.2)
y= -0.58032

=0.58032 as +area

0.6-0.58032= 0.01968

(B)y=3.8x^4 -6.8x^3 +7.7x^2 -4.2x

Intergrated gives

=(3.8x^5)/5 -(6.8x^4)/4 + (7.7x^3)/3 -(4.2x^2)/2 between 9 and 0.
=(3.8(0.9)^5)/5 -(6.8(0.9)^4)/4 + (7.7(0.9)^3/3) -(4.2(0.9)^2/2)
=-0.4964976
must *50 to get volume

=-24.82488
=+24.82488 as canoot have negative volume.

hey, you know for the trapezium rule i put the corresponding x-values for my mid y-value, would i get 0/4 for this mistake?

[Atropos]

I'll just write it all out (if im wrong somewhere please correct me)

1. y = sqrt(x) + 3/x = x^1/2 + 3x^-1
dy/dx = 1/2x^-1/2 -3x^-2

2.
u1 = 5 Un+1 = Un + 3
u2 = 5 + 3 = 8
u3 = 8 + 3 = 11

Sn = 1/2n[2a + (n-1)d]
a = 5 d = 3
S50 = 1/2 * 50[2 * 5 + (50-1)3]
s50 = 25[10 + 49 * 3]
s50 = 25(157)
s50 = 3925

3.(i) a^2 = b^2 + c^2 - 2bc Cos A
a^2 = 9.8^2 + 6.4^2 - 2*9.8*6.4* cos 53.4 =
= 62.20955168
a = AD = 7.89 cm

(ii) A = 1/2abSin C
A = 0.5 * 7.3 * 9.8 * Sin(180-53.4)
A = 28.72 cm^2

4.
P(6,3)

(i) y = 3f(x)
P'(6,9)
(ii) y= f(4x)
P'(1.5,3)

5.
A = 1/2Ɵr^2
A = 45
Ɵ = 1.6
2A = Ɵr^2
2A/Ɵ = r^2
sqrt(2A/Ɵ) = r
sqrt(2*45/1.6) = r
sqrt(56.25) = r
r = 7.5 cm

Perimeter = 2r + Ɵr (boths sides of lenght r + arc lenght)
= 2 * 7.5 + 1.6 * 7.5 = 27 cm

6.
m = 17-5/5-1 = 3
log y - 5 = 3(log x - 1)
log y = 3log(x) + 2
y = 10^(3log(x) + 2) (that's where i left it (( )
y = 10^log(x^3) * 10^2
y = x^3 * 100
y = 100x^3

7.dy/dx = 6x^1/2 -5 thru point (4,20)
integrated = 4x^3/2 - 5x + c

y = 4x^3/2 - 5x + c
y = 20 x =4
20 = 4(4)^3/2 - 5(4) + c
20 = 32 - 20 + c
20 - 12 = c
8 = c

y = 4x^3/2 - 5x + 8

8.
sin2Ɵ = 0.7 between 0 and 2pi
sin^-1 (0.7) = 2Ɵ
2Ɵ = 0.775 -> Ɵ = 0.388
pi - 0.775 = 2.366 -> Ɵ = 1.18
0.775 + 2pi = 7.0586 -> Ɵ = 3.53
2.366 + 2pi = 8.6494 - > Ɵ = 4.32

9.
(i) A = h/2(y0 + yn + 2(y1 + ....+yn-1))
A = 0.2/2 (2(0.5+0.7+0.75+0.7+0.5))
= 0.2(0.5+0.7+0.75+0.7+0.5)
= 0.63 m^2
V = 0.63 * 50 = 31.5 m^3

(ii) (A) y= 3.8(0.2)^4 - 6.8(0.2)^3 + 7.7(0.2)^2 -4.x(0.2) = -0.58032
0.6 - 0.58032 = 0.01968

(B) Integrated = (3.8x^5)/5 - (6.8x^4)/4 + (7.7x^3)/3 - 2.1x^2
integral from 0 to 0.9 = (3.8*0.9^5)/5 - (6.8*0.9^4)/4 + (7.7*0.9^3)/3 - 2.1(0.9^2) =
= -0.496

A = 0.496

V = 0.496 = 24.8 (rounded sry)

10.(i)y = x^3 -5x
dy/dx = 3x^2 -5
3x^2 - 5 = 0
x^2 - 5/3 = 0
(x-sqrt(5/3))(x+(sqrt(5/3)) = 0
x = sqrt(5/3) = 1.3
x = -sqrt(5/3) = -1.3

(ii) y = x^3 - 5x = x(x^2 - 5) = x(x-sqrt(5))(x+sqrt(5))
x = 0
x = 2.24
x = -2.24



(iii) dy/dx = 3x^2 - 5
x = 1
dy/dx = -2
y + 4 = -2(x -1)
y = -2x +2 -4
y = -2x -2
y = x^3 - 5x
x^3 - 5x = -2x -2
x^3 - 5x + 2x + 2 = 0
x^3 -3x + 2 = 0

(we know they meet at x = 1 so (x-1) is one of the brackets)

long division

x^2 + x - 2
(x-1) / (x^3 -3x +2)
x^3 -x^2
x^2 - 3x
x^2 - x
- 2x + 2
- 2x + 2
======
(x-1)(x^2 + x - 2) = x^3 -3x + 2 = 0
(x+2)(x-1)^2 = 0
meets curve again at x = -2

11. a2 = 6
S(infinity) = 25

(i)a2 = ar = 6
a = 6/r

S(infinity) = a/(1-r) = a / (1- 6/a)
25 = a/ (1-6/a)
25(1-6/a) = a
25 - 150/a = a
a - 25 + 150/a = 0
a^2 - 25a + 150 = 0
(a - 10)(a - 15) = 0
a = 10 r = 0.6
a = 15 r = 0.4

(ii)Un = ar^n-1
so
Un = 10 * 0.6^n-1
Un = 15 * 0.4^n-1

15 * 0.4^n-1 : 10 * 0.6^n-1
15 * 0.4 * 0.4^n-2 : 10 * 0.6 * 0.6^n-2
6 * 0.4^n-2 : 6 * 0.6^n-2
0.4^n-2 : 0.6^n-2
5^n-2 * 0.4^n-2 : 5^n-2 * 0.6^n-2
2^n-2 : 3^n-2

[qwerty187]

Made so many stupid mistakes on this paper, worked out the area of the wrong triangle because i didnt read the question properly, just redid the first part of the last question again and di it quite easily, rearranged it incorrectly in the exam, think i have dropped about 20 marks easily, but have been getting 65-70 on all the past papers! gutted.

[Polioz]

(Original post by genius017)
i got y=3x+100
i know thats completely wrong but do you think ill still pick up 1 mark for finding out the gradient? :'( 5 marks is alott to lose on one question

You wont lose 5 marks . you worked out the gradient and logk . but you didnt put it into equation properly. If you are lucky you will only lose 1 mark, but most likely you got 3/5.

[sarah:))]

on the trapezium rule question... i realised i used the x-values as my mid y-values e.g for (2x(y2+y3+y4+y5...) would i get 0 marks out of 4 for this mistake?

[Atropos]

(Original post by sarah:)))
on the trapezium rule question... i realised i used the x-values as my mid y-values e.g for (2x(y2+y3+y4+y5...) would i get 0 marks out of 4 for this mistake?

you will definitely lose a mark for wrong answer
+ lose mark for wrong use of formula

u will probably get mark for use of trapezium rule

so I would say you will get about 1/2 marks for it