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C2 May 2013 WJEC

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Reply 20
Avatar for Naomi32
Naomi32
Original post by mabli
ImageUploadedByStudent Room1368803964.257379.jpgImageUploadedByStudent Room1368803981.745151.jpgImageUploadedByStudent Room1368804007.492194.jpg

Taken a picture of the paper. I couldn't prove the first triangle question either, my answers kept cancelling each other out


Posted from TSR Mobile

weird right, everything did cancel leaving you with something totally unlike what it asked you to prove
Reply 21
Avatar for joostan
joostan
13
Original post by mabli
ImageUploadedByStudent Room1368803964.257379.jpgImageUploadedByStudent Room1368803981.745151.jpgImageUploadedByStudent Room1368804007.492194.jpg

Taken a picture of the paper. I couldn't prove the first triangle question either, my answers kept cancelling each other out


Posted from TSR Mobile


Last log question answers coming right up . . . :smile:
I'll do some more on request. :smile:
Original post by mabli
How did everyone find the exam? I thought it was okay about from the last logarithm question and the eighth question (circles)


Posted from TSR Mobile


I twisted my knickers on question 2)b) a lot although it was only worth 3 marks. I believe I got one of the possible answers, but that was the one that I'm confident I lost marks on.:mad:

The other parts weren't so bad.:smile:
Original post by Naomi32
I did this too! I thought I went the long way around because the second part of the question asked for the co-ordinates of intersection which I had already found
Really hoping we get most of the marks for doing that, we still proved it wasn't the diameter?


I would give you two full marks if I were an examiner as the question was only worth 2 or 3 marks, and also for the exra effort you put in :smile: (I did it the shorter way, by the way)
Reply 24
Avatar for joostan
joostan
13
7)i)
log⁡a(xy)=log⁡a(x)+log⁡a(y)\log_a(xy)=\log_a(x) + \log_a(y)
⇔xy=alog⁡a(x)+log⁡a(y)\Leftrightarrow xy = a^{\log_a(x) + \log_a(y)}
So now we must show that:
xy=alog⁡a(x)+log⁡a(y)xy = a^{\log_a(x) + \log_a(y)}
alog⁡a(x)+log⁡a(y)=alog⁡a(x)×alog⁡a(y)=xya^{\log_a(x) + \log_a(y)} = a^{\log_a(x)} \times a^{\log_a(y)} = xy

ii)
Unparseable latex formula:

5^{2-3x} = 8 \Rightarrow (3x-2)\log_2(5) = 3[br]\Rightarrow 3x\log_2(5) = 3 + 2\log_2(5)[br]\Rightarrow x = \dfrac{3+2\log_2(5)}{3\log_2(5)} \apprx 0.236



iii)
log⁡a(90x2)−log⁡a(5x)=12log⁡a(144x8)[br]⇒log⁡a(90)+2log⁡a(x)−log⁡a(5)=log⁡a(12)+4log⁡a(x)[br]⇒log⁡a(32)=log⁡a(x)[br]⇒x=32\log_a(90x^2) - \log_a(\frac{5}{x}) = \frac{1}{2}\log_a(144x^8)[br]\Rightarrow \log_a(90) + 2\log_a(x)-\log_a(5) = \log_a(12) + 4\log_a(x)[br]\Rightarrow \log_a(\frac{3}{2}) = \log_a(x)[br]\Rightarrow x = \dfrac{3}{2}
Reply 25
Avatar for birdy95
birdy95
Hi! I felt the exam was alright, except 2aii; I got confused because my values were less than one [(3x-2)(2x+3) x could not equal -3/2?]. Q3 was a mess, and same for last logs question and circles question.
I don't know how to feel about the exam, I know I answered all the questions and attempted the answers, but then got in a muddle in the middle of those odd questions..
:erm:
Reply 26
Avatar for Han96
Han96
Original post by joostan
Last log question answers coming right up . . . :smile:
I'll do some more on request. :smile:


Can you please do the area under the curve question? 6b I think. Thanks :smile:
Reply 27
Avatar for maw1996
maw1996
OP
6
I got 16.7cm^2 for that


Posted from TSR Mobile
Does anyone have a solved paper from their teacher? Would like to compare my marks :smile:
Reply 29
Avatar for Tynos
Tynos
2
For Proving Triangle: Cos A = x^2+(x-2)^2 - (x+2)^2 / 2(x)(x-2)

This gives: x^2+x^2-4x+4-x^2-4x-4 / 2x^2 - 4x

Simplifies to x^2 - 8x / 2x^2 - 4x

Divide by x = x-8 / 2x - 4
Reply 30
Avatar for Tynos
Tynos
2
Part b) Angle BAC = 120

Formula for the angle is given from part a:

There Cos 120 = x+8 / 2x-4

-0.5(2x-4) = x+8

2-x = x+8

2x = -6

x=-3

Not sure if this is correct, as x is a negative.
(edited 10 years ago)
Reply 31
Avatar for Tynos
Tynos
2
Im going to try and use Latex for 6b)

y=x2+3andy=4x[br][br]x2+3=4x[br]x2−4x+3=0[br](x−1)(x−3)=0[br][br]Thereforeintersectsatx=1andx=3y=x^2+3 and y=4x[br][br]x^2+3=4x[br]x^2-4x+3=0[br](x-1)(x-3)=0[br][br]Therefore intersects at x=1 and x=3

Area under the curve between x=1 and x=3

∫13{x2+3}dx=[x33+3x]13\displaystyle \int_1^3 \{x^2 + 3\} dx = \left[ \frac{x^3}{3} + 3x \right]_1^3

This works out to be 44/3


Area under the line between x=0 and x=1

∫01{4x}dx=[2x2]01\displaystyle \int_0^1 \{4x\} dx = \left[2x^2\right]_0^1

This works out to be 2

Therefore shaded region = 44/3 + 2 = 50/3
(edited 10 years ago)
Reply 32
Avatar for GSM
GSM
3
Original post by mabli
ImageUploadedByStudent Room1368803964.257379.jpgImageUploadedByStudent Room1368803981.745151.jpgImageUploadedByStudent Room1368804007.492194.jpg

Taken a picture of the paper. I couldn't prove the first triangle question either, my answers kept cancelling each other out


Posted from TSR Mobile


Tbh the C2 May 2013 is one of the easiest question paper I have seen. Many of u will agree with me that question 2 (a) & 3(a) were not as expected but the rest of the question should have been really good if you would have done more past papers practice. e.g: the question that was about integration 6(b) that was taken from the C2 May 2007 paper...I'm very sure that I did very well in this exam :L
Reply 33
Avatar for Han96
Han96
Original post by Tynos
Im going to try and use Latex for 6b)

y=x2+3andy=4x[br][br]x2+3=4x[br]x2−4x+3=0[br](x−1)(x−3)=0[br][br]Thereforeintersectsatx=1andx=3y=x^2+3 and y=4x[br][br]x^2+3=4x[br]x^2-4x+3=0[br](x-1)(x-3)=0[br][br]Therefore intersects at x=1 and x=3

Area under the curve between x=1 and x=3

∫13{x2+3}dx=[x33+3x]13\displaystyle \int_1^3 \{x^2 + 3\} dx = \left[ \frac{x^3}{3} + 3x \right]_1^3

This works out to be 44/3


Area under the line between x=0 and x=1

∫01{4x}dx=[2x2]01\displaystyle \int_0^1 \{4x\} dx = \left[2x^2\right]_0^1

This works out to be 2

Therefore shaded region = 44/3 + 2 = 50/3


Thanks!! I got that too! Have you tried 8 or 9?
Reply 34
Avatar for Tynos
Tynos
2
Original post by Han96
Thanks!! I got that too! Have you tried 8 or 9?


I jsut took a break lol. The latex is harder than the actual question xD hahah jokes

Ill try it now.
Reply 35
Avatar for Tynos
Tynos
2
8 Part a and b)

[br][br]x2+y2+2x−6y−15=0[br](x+1)2+(y−3)2=25[br][br]x^2+y^2+2x-6y-15=0[br](x+1)^2+(y-3)^2=25

Therefore A(-1,3) Radius = Root 25 = 5

I did part ii first dont know if its allowed:
[br]y=−x+9[br](x+1)2+(−x+9−3)2=25[br]Expandandsimplifyto:[br]2x2−10x+12=0[br]x2−5x+6=0[br](x−3)(x−2)=0[br]y=-x+9[br](x+1)^2+(-x+9-3)^2=25[br]Expand and simplify to:[br]2x^2-10x+12=0[br]x^2-5x+6=0[br](x-3)(x-2)=0

Intersects at: (3,6) and (2,7)

bi) If y=-x+9 is a diameter then length of this divided by 2 must equal 5.
(2−3)2+(7−6)2(2-3)^2+(7-6)^2 Square root = Root 2

Root 2/2 is not equal to the radius of the circle therefore is not a diameter.
Reply 36
Avatar for Tynos
Tynos
2
For the last part of part 8)

Centres of circles: (-1,3) and (11,8)

Using pythag:

(11+1)2+(8−3)2(11+1)^2+(8-3)^2

= Root 169 = 13

Therefore shortest distance = 13-6-5 (Their radius's)=2
Reply 37
Avatar for maw1996
maw1996
OP
6
Do you think I'd get marks for changing the fraction into a whole number for 6.b)?


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Reply 38
Avatar for joostan
joostan
13
Original post by Han96
Can you please do the area under the curve question? 6b I think. Thanks :smile:


Sorry, I was out yesterday evening, I've been beaten to it - I'll check the working :smile:
EDIT: I got the same answer - congrats :tongue:
(edited 10 years ago)
Reply 39
Avatar for Tynos
Tynos
2
Original post by mabli
Do you think I'd get marks for changing the fraction into a whole number for 6.b)?


Posted from TSR Mobile


Did you round to 3sf?

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