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WJEC C3 exam.

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Reply 20
Avatar for kito
kito
Which question was number 4? Pretty angry with myself, lost 8-10 marks due to silly errors! was averaging 71 - 73 in my mocks! :frown:
Reply 21
Avatar for RhysM18
RhysM18
Original post by aaroncsn
I believe that the second question should be 0.3070. Your answer and method are perfectly correct. You basically convert sin^2 x to 1-cos^2 x and then integrate both seperately. I am confident that 0.3070 is the right answer.

Looking through it, i think that those that you labelled red are the only mistakes that you have done. Apart from that,I believe the rest are correct, that is if compared to my answers, which i believe i have not done any errors so far. =)

Edit: Oops i overlooked something. your 10(a) shouldn't it be [3+k, infinity) as they asked for the range in terms of k. And also for 8(b)(i) , the x should be 1/3 of the graph in 8(a) i believe due to the f(3x) which makes the graph much steeper when extrapolated upwards.


What do you mean by 1/3 because f(3x)=f(x) when x=0 (y-axis)
Original post by RhysM18
What do you mean by 1/3 because f(3x)=f(x) when x=0 (y-axis)


well, x has to be a 3rd so that y is the same number, making it 3 times less broader.
Reply 23
Avatar for RhysM18
RhysM18
Original post by TheGrinningSkull
well, x has to be a 3rd so that y is the same number, making it 3 times less broader.


f(3x) compresses horizontally not vertically, I got the intersection at (0,-3)
Reply 24
Avatar for RhysM18
RhysM18
Just realised your not talking about intersection
Original post by RhysM18
Just realised your not talking about intersection


Yeah, we're talking about the x-axis intersection :smile:
Reply 26
Avatar for RhysM18
RhysM18
Original post by TheGrinningSkull
Yeah, we're talking about the x-axis intersection :smile:


0.462 is the answer for that, you don't need to divide by 3
Reply 27
Avatar for aberlad93
aberlad93
Btw guys, what was the other graph that we had to draw except e^x?
Reply 28
Avatar for RhysM18
RhysM18
Original post by aberlad93
Btw guys, what was the other graph that we had to draw except e^x?

e^(3x)-4
Reply 29
Avatar for aberlad93
aberlad93
Original post by RhysM18
e^(3x)-4


Thanks XD, just checked online to see if the graph i drew was correct... cant remember if i stupidly said that the asymptote was at y= -3...
I think i said -4.
Reply 30
Avatar for welshyshu
welshyshu
[QUOTE="TheGrinningSkull;35962807"]Have people still not taken the exam? Tell me if I should take this down!

What I highlight in red is what I believe I got wrong, green is the correction of what it should be after discussion.

1. (a) 0.7402
1. (b) 0.3070 (0.2598) (I used the same method because I thought I was wrong using cos2x+sin2x=1cos^2x + sin^2x = 1 (EDIT: Thanks for the corrections, I thought the integral had something to do with it, I just did the Simpson's rule again cos I knew that was the safest. )

2. (a) prove

2. (b) 41.41, 318.59, 109.47, 250.53

3. (a) (i) t4−2t t^4 - 2t

3. (a) (ii) show it

3. (b) show it, change of sign indicates presence of root alpha, final answer being 1.61007 and proving this by taking 0.000005 either side, change of sign indicates presence of root alpha.

4. 2.2

5. (a) 4/(1+16x2) 4/(1+16x^2)

5. (b)
Unparseable latex formula:

3x^2 e^x^3



5. (c) 5x4ln∣x∣+x4 5x^4 ln|x| + x^4 or x4(5ln∣x∣+1) x^4(5ln|x|+1)

5. (d) 4x/(5−4x2)2 4x/(5-4x^2)^2

6. (a) (i) −4cos(x/4)+c -4cos(x/4) + c

6. (a) (ii) 3/2e(2x/3)+c 3/2 e^(2x/3) + c

6. (a) (iii) 7ln∣8x−2∣/8+c 7ln|8x-2|/8 +c

6. (b) 126.4, 4.8, I took original power to be 0.5 instead of -0.5 and so got the new power to be 1.5 :colondollar:

7. (a) x>=2 x>= 2 or x=<1/2 x=< 1/2

7. (b) x=+21,−21 x = +21,-21

8. (a) sketch graph, point of intersection at (0,1) with y-axis

8. (b) (i) sketch graph, for large negative numbers as x tends to minus infinity, y value approaches the asymptote -4

8. (b) (ii) (0,-3)

8. (b) (iii) 4.519, should be 0.462, which I crossed out :angry:, I thought the graph should be broader but now I understand it, hopefully I only lose 1 mark overall for this question.

9. (a) 1/(x−3)2+2 1/(x-3)^2 + 2
9. (b) [2.5 , 3)

10. (a) [3+k,infinity) [3+k , infinity)

10. (b) Least value of k must be -5

10. (c) (i) 9x2+6xk+k2−6 9x^2 + 6xk +k^2 -6 or (3x+k)2−6 (3x+k)^2 - 6

10. (c) (ii) k=−3 k = -3 (Cannot be -9 as least value of k must be -5)

Bar some minor errors, these answers look good, i cant remember exactly what i had though! Fingers Crossed for March.
Original post by RhysM18
0.462 is the answer for that, you don't need to divide by 3


Yes you do, you had ln(4), dividing by 3 gives 0.462
Reply 32
Avatar for emma mw
emma mw
Original post by RhysM18
e^(3x)-4


i thought if f(x) was f(e^x) the f(3x)-4 would be 3e^x -4 :frown: dear lord now my answer is totally messed urgh!
[QUOTE="TheGrinningSkull;35962807"]Have people still not taken the exam? Tell me if I should take this down!

What I highlight in red is what I believe I got wrong, green is the correction of what it should be after discussion.

1. (a) 0.7402
1. (b) 0.3070 (0.2598) (I used the same method because I thought I was wrong using cos2x+sin2x=1cos^2x + sin^2x = 1 (EDIT: Thanks for the corrections, I thought the integral had something to do with it, I just did the Simpson's rule again cos I knew that was the safest. )

2. (a) prove

2. (b) 41.41, 318.59, 109.47, 250.53

3. (a) (i) t4−2t t^4 - 2t

3. (a) (ii) show it

3. (b) show it, change of sign indicates presence of root alpha, final answer being 1.61007 and proving this by taking 0.000005 either side, change of sign indicates presence of root alpha.

4. 2.2

5. (a) 4/(1+16x2) 4/(1+16x^2)

5. (b)
Unparseable latex formula:

3x^2 e^x^3



5. (c) 5x4ln∣x∣+x4 5x^4 ln|x| + x^4 or x4(5ln∣x∣+1) x^4(5ln|x|+1)

5. (d) 4x/(5−4x2)2 4x/(5-4x^2)^2

6. (a) (i) −4cos(x/4)+c -4cos(x/4) + c

6. (a) (ii) 3/2e(2x/3)+c 3/2 e^(2x/3) + c

6. (a) (iii) 7ln∣8x−2∣/8+c 7ln|8x-2|/8 +c

6. (b) 126.4, 4.8, I took original power to be 0.5 instead of -0.5 and so got the new power to be 1.5 :colondollar:

7. (a) x>=2 x>= 2 or x=<1/2 x=< 1/2

7. (b) x=+21,−21 x = +21,-21

8. (a) sketch graph, point of intersection at (0,1) with y-axis

8. (b) (i) sketch graph, for large negative numbers as x tends to minus infinity, y value approaches the asymptote -4

8. (b) (ii) (0,-3)

8. (b) (iii) 4.519, should be 0.462, which I crossed out :angry:, I thought the graph should be broader but now I understand it, hopefully I only lose 1 mark overall for this question.

9. (a) 1/(x−3)2+2 1/(x-3)^2 + 2
9. (b) [2.5 , 3)

10. (a) [3+k,infinity) [3+k , infinity)

10. (b) Least value of k must be -5

10. (c) (i) 9x2+6xk+k2−6 9x^2 + 6xk +k^2 -6 or (3x+k)2−6 (3x+k)^2 - 6

10. (c) (ii) k=−3 k = -3 (Cannot be -9 as least value of k must be -5)


Sure you didn't get 3-x for the inverse function rather than x-3?
Reply 34
Avatar for roseroserose.
roseroserose.
OP
I don't suppose anyone can remember the inequality that led to +\- 21?

and is anyone doing S1 or FP1 this week.
Reply 35
Avatar for RhysM18
RhysM18
Original post by roseroserose.
I don't suppose anyone can remember the inequality that led to +\- 21?

and is anyone doing S1 or FP1 this week.


(3lxl+1)^(1/3)=4
i'm doing S1
Reply 36
Avatar for hannahwk
hannahwk
I thought that the exam could have been worse so I was quite relieved :smile: I'm also doing s1 next week. yeah....that isn't going down well..
Hey, just wondering what everyone thinks the boundaries will be? They seem to vary alot between years.
Reply 38
Avatar for hannahwk
hannahwk
Original post by welshyshu
Bar some minor errors, these answers look good, i cant remember exactly what i had though! Fingers Crossed for March.


what was the question for 4 and 9a?
Original post by alfredtehbutler
Sure you didn't get 3-x for the inverse function rather than x-3?


Pretty sure, unlucky? :redface:

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