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WJEC C3/C4 8th/17th June 2016

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Reply 80
Avatar for Sznsnsn
Sznsnsn
1
Original post by FusionNetworks
If I remember correctly: -1/8cos(3t)


I had -sec^3(3t)/8 I think?
Some answers:

1. 0.74379 to 5dp
Multiply this by e: 2.02183 to 5dp.

2. 30, 150, 196.6, 343.4 degrees

146.3,326.3 degrees

3. 23\dfrac{2}{3}

4. −12tan⁡3t-\frac{1}{2}\tan 3t
−18sec⁡33t-\frac{1}{8}\sec^3 3t
−1y3-\dfrac{1}{y^3}

5. Perimeter = Arc Length + Chord Length, arc length is easy, chord length can be found with a right-angled triangle with angle θ2\dfrac{\theta}{2}.
Usual stuff with iterations etc; for function use f(θ)=θ+2sin⁡θ2−4.5f(\theta)=\theta +2\sin\frac{\theta}{2}-4.5

6. An easy counterexample is a=b=1,c=2,d=3a=b=1, c=2, d=3

7. x=−13x=-\dfrac{1}{3} (x=2x=2 doesn't work)

8. a=−3,5a=-3,5
b=−23b=-\frac{2}{3}

9. f−1(x)=12−3ln⁡(x−8)f^{-1}(x)=12-3\ln(x-8)
Domain is [9,+∞)[9,+\infty)

10. To show hh(x)=xhh(x)=x simply expand out the left.
So it is clear that h is its own inverse and so h−1(−1)=19h^{-1}(-1)=\frac{1}{9}

That's all I can remember, please correct anything that's wrong :smile:.

EDIT: Just realised I left out diff/int questions.

Differentiation: ln⁡(cos⁡x)\ln(\cos x) gives −tan⁡x-\tan x,

tan⁡−1(x3)\tan^{-1}(\frac{x}{3}) gives 39+x2\dfrac{3}{9+x^2}

e6x(3x−2)4e^{6x}(3x-2)^4 gives 18xe6x(3x−2)318xe^{6x}(3x-2)^3

Integration: a=2.3a=2.3
(edited 7 years ago)
Reply 82
Avatar for Mango88
Mango88
2
Original post by Sznsnsn
I had -sec^3(3t)/8 I think?


i got that!!
Reply 83
Avatar for Sznsnsn
Sznsnsn
1
Original post by IrrationalRoot
Some answers:

1. 0.74379 to 5dp
Multiply this by e: 2.02183 to 5dp.

2. 30, 150, 196.6, 343.4 degrees

146.3,326.3 degrees

3. 23\dfrac{2}{3}

4. −12tan⁡3t-\frac{1}{2}\tan 3t
−18sec⁡33t-\frac{1}{8}\sec^3 3t
−1y3-\dfrac{1}{y^3}

5. Perimeter = Arc Length + Chord Length, arc length is easy, chord length can be found with a right-angled triangle with angle θ2\dfrac{\theta}{2}.
Usual stuff with iterations etc; for function use f(θ)=θ+2sin⁡θ2−4.5f(\theta)=\theta +2\sin\frac{\theta}{2}-4.5

6. An easy counterexample is a=b=1,c=2,d=3a=b=1, c=2, d=3

7. x=−13x=-\dfrac{1}{3} (x=2x=2 doesn't work)

8. a=−3,5a=-3,5
b=−23b=-\frac{2}{3}

9. f−1(x)=12−3ln⁡(x−8)f^{-1}(x)=12-3\ln(x-8)
Domain is [9,+∞)[9,+\infty)

10. To show hh(x)=xhh(x)=x simply expand out the left.
So it is clear that h is its own inverse and so h−1(−1)=19h^{-1}(-1)=\frac{1}{9}

That's all I can remember, please correct anything that's wrong :smile:.


I had different ans for q2 I and 210 and 330 for 2 of them
Reply 84
Avatar for 98matt
98matt
13
Does anyone have the paper yet? Also who usually makes unofficial mark schemes / when will one be up?

Posted from TSR Mobile
Original post by 98matt
Does anyone have the paper yet? Also who usually makes unofficial mark schemes / when will one be up?

Posted from TSR Mobile


I posted one above :smile:.
For Q7) x=2 didnt work but I believe x=-2 did, so I had x=-1/3 and x=-2
Original post by SamDavies98
For Q7) x=2 didnt work but I believe x=-2 did, so I had x=-1/3 and x=-2


Wasn't the equation ∣5x+4∣=−7x|5x+4|=-7x? (I might be mistaken.)
Yeah either that or 5X-4, I'm hoping it was 5X-4 now!
Original post by SamDavies98
Yeah either that 5X-4, I'm hoping it was 5X-4 now!


But then x=−13x=-\frac{1}{3} is not a solution so it can't be 5x−45x-4.
I'm slightly surprised that people found that easier than last year's paper... I'm anticipating getting 60/61 on that, which definitely won't be 90 UMS... I'll have to compensate on C4. :sigh:

Original post by IrrationalRoot
Wasn't the equation ∣5x+4∣=−7x|5x+4|=-7x? (I might be mistaken.)


It was.
Original post by Hydeman
I'm slightly surprised that people found that easier than last year's paper... I'm anticipating getting 60/61 on that, which definitely won't be 90 UMS... I'll have to compensate on C4. :sigh:



It was.


Did you check my unofficial ms, maybe you picked up more marks than you thought? :smile: (Although I can't guarantee they're all correct.)
Yep you're right, damn
Original post by Hydeman
I'm slightly surprised that people found that easier than last year's paper... I'm anticipating getting 60/61 on that, which definitely won't be 90 UMS... I'll have to compensate on C4. :sigh:



It was.


Is it 90 UMS in BOTH c3 and c4, or 180 over both papers for an A*?

I'm not even aiming for an A* but I'm not sure what the boundaries are for it
Can someone run through how question 2a was supposed to be done? I only managed to get two solutions, so I'm guessing I mucked up the algebra. Hopefully I won't lose all 6 marks...
Does anyone remember how integration and differentiation question looked like? I remember only ln(cosx)
Original post by Salamon16
Does anyone remember how integration and differentiation question looked like? I remember only ln(cosx)


e^6x (4x-5)^4?

I think I've misremembered the inside of the bracket though but the rest is right.

and tan^-1 (X/3)
Original post by Jack1066
Is it 90 UMS in BOTH c3 and c4, or 180 over both papers for an A*?

I'm not even aiming for an A* but I'm not sure what the boundaries are for it


180 over both.

Original post by IrrationalRoot
Did you check my unofficial ms, maybe you picked up more marks than you thought? :smile: (Although I can't guarantee they're all correct.)


I did, and it doesn't seem that I did, I'm afraid. :colondollar:
I also got a much more complicated expression for the second derivative in terms of y in question 3b)...
WJEC and its weird papers. :facepalm:
what is 54/75 UMS wise?
also what did everyone get for the value of a in 7b integration question, i got 0.1

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