# PC2 Maths, AQA 24/05/10Watch

#181
Yes i suspect you will get the follow on marks, unless the examiner doesn't realise what you've done of course!
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#182
(Original post by Weeman5872)
I think in bi, they say an arithmetic series has 4th term = 1st term of the geometric, and 8th term = 2cnd term of gp.

Then bii, they say that Un is the nth term in the arithmetic series. So I guess since its all part b, you would assume the arithmetic series is the same as above. Using this, I got 70 for the summaton [a = 11.5, d = -0.5]

Can anyone confirm -1.7 in the integral, and 5.5 in the trapezium rule?

Yes they're both correct.

And also you're right about bi & bii; the point is that if def' askes for the sum of the arithmetic like i did. Thanks a lot!
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8 years ago
#183
(Original post by AlexHutson1)
Oh yeah, quick question, how do we get A* on this course? My teacher has been really vague about it!
90% average of pure core 3 and pure core 4 and 80% total in the whole A level
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8 years ago
#184
(Original post by AlexHutson1)
Oh yeah, quick question, how do we get A* on this course? My teacher has been really vague about it!
you need 90% average across C3/C4..

for the ordinates in trap rule.. were they 0.. 1/6.. 2/6 .... 1?

I always get trapezium rule wrong...
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8 years ago
#185
(Original post by Jfranny)
you need 90% average across C3/C4..

for the ordinates in trap rule.. were they 0.. 1/6.. 2/6 .... 1?

I always get trapezium rule wrong...
clearly not jfranny you ******* retard, that's 7 ordinates.. wow - I'm an idiot xD..

how many marks will I get for getting h correct and stating the formula correct?..

I suspect not even one?
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8 years ago
#186
90% average of pure core 3 and pure core 4 and 80% total in the whole A level
Thanks so much , makes me feel better since I think this exam went really poorly.
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8 years ago
#187
(Original post by Jfranny)
clearly not jfranny you ******* retard, that's 7 ordinates.. wow - I'm an idiot xD..

how many marks will I get for getting h correct and stating the formula correct?..

I suspect not even one?
2, one for h, one for summing the formula, and another if you put values in?

I got like 5.55 for that one.
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8 years ago
#188
ah.. was it 7 ordinates or what? I don't remember the Q...
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8 years ago
#189
What did people do for the question asking to deduce the curve had no max point.
I ended up finding the x coordinate of the stationary point, then put that into d/dx(dy/dx), which came out as 5. Then I said as this was positive, it was a minimum, so no maximum [from the graph I saw it only had a minimum]. But after the exam, I think I graphed something else, as when I checked again, it had no stationary points at all. :s
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8 years ago
#190
(Original post by Jfranny)
ah.. was it 7 ordinates or what? I don't remember the Q...
I'm pretty sure it was 5 ordinates.
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8 years ago
#191
who remembers the trapezium question here?
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8 years ago
#192
(Original post by Jfranny)
ah.. was it 7 ordinates or what? I don't remember the Q...
6 ordinates (5 strips). I remember getting h = 0.2, and the integral was from 0 to 1.
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8 years ago
#193
(Original post by AlexHutson1)
I'm pretty sure it was 5 ordinates.
you're probably right, as I said - I always get trap rule wrong even though it's the easiest..
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8 years ago
#194
2, one for h, one for summing the formula, and another if you put values in?

I got like 5.55 for that one.
Same.
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8 years ago
#195
(Original post by AlexHutson1)
I'm pretty sure it was 5 ordinates.
6 ordinates, 5 strips.
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8 years ago
#196
(Original post by sp92)
6 ordinates, 5 strips.
That's the one :P

Honestly, I am still angry about the translation questions, they really were cruel. We hadn't been taught to use our calculators for those questions.
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8 years ago
#197
I'll only loose one mark so it's fine.. only need 85%+
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#198
The graph was only defined for values of x greater than 0 (specified in quesiton), so when you put any positive value into 2 + "1/5"x^-5/2, the value was always positive which implies any stationary points are minimums.
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8 years ago
#199
6 ordinates (5 strips). I remember getting h = 0.2, and the integral was from 0 to 1.
Correct

The graph was only defined for values of x greater than 0 (specified in quesiton), so when you put any positive value into 2 + "1/5"x^-5/2, the value was always positive which implies any stationary points are minimums.
yep you didn't have to actually work it out.

You just had to put (Differention worked out) > 0 hence cannot be a maximum point.
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8 years ago
#200
(Original post by Jfranny)
yes as it's wrong..

I got pi/6 and 2pi/6 or 3pi/5 can't remember which it was... may be niether.. I was right though

pi/6 is the samething as1/6*pi
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