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really scared for C4 intergration:/:s-smilie:
Original post by kamisaysjambo
really scared for C4 intergration:/:s-smilie:


How come? What part don't you understand?
Reply 622
Original post by jadecross
I'm not sure how to work out the range and domain for C3 :frown:
All i know is the domain is input value and range is output but i'm not sure how to work them out
can someone please help?
e.g. how do you work out the domain and range for y=ex+12y = e^x+\frac{1}{2} Thank you.


Anyone? :smile:
(edited 10 years ago)
Original post by Robbie242
And that was my main problem, I stuck to 2008 on wards, so I didn't have time to experience those earlier papers and familiarise myself with them. The lift question wasn't hard, I just interpreted it wrong and used the lifts components instead as for suvat instead of reaching an equation and finding a value of t then subtracting 10, I tried to put it into the equation and its just all misinterpreted really.
I would like to say this on par with Jan 2013, if I sat jan 13 I would've got the vectors part c) wrong and the last part of the resolving question on the paper. So I'd expect due to the overwhelming amount of marks for moments 63/75 for an A?


Yh lift question isn't hard, it's just so easy to get wrong if you don't understand the question, like taking the wrong mass or whatever. I felt this paper had all the tricky bits thrown at you, lift, parallel vectors, non-uniform moments, force at pulley etc... the only topic this paper was lacking was Thrust in the rope if they included that as well wow...

Vectors part c) took me a good 15 mins to do in the exam as I got it all wrong and had to cross it out and start again but got there in the end! Don't know how I'd compare it, this one seems slightly harder to me but yes 63 for A is reasonable if everyone else did bad.
Reply 624
Original post by Boy_wonder_95
Yh lift question isn't hard, it's just so easy to get wrong if you don't understand the question, like taking the wrong mass or whatever. I felt this paper had all the tricky bits thrown at you, lift, parallel vectors, non-uniform moments, force at pulley etc... the only topic this paper was lacking was Thrust in the rope if they included that as well wow...

Vectors part c) took me a good 15 mins to do in the exam as I got it all wrong and had to cross it out and start again but got there in the end! Don't know how I'd compare it, this one seems slightly harder to me but yes 63 for A is reasonable if everyone else did bad.



Exactly, I entirely agree, Impulse was extremely easy and lorry suvat was also easy but everything else wasn't as simple as it could be and haha yeah.

I see well I'm expecting a low A from this exam unless there is some sort of boundary miracle, wasn't in the right frame of mind coming straight out of economics
Original post by MathsNerd1
How come? What part don't you understand?


Substitution mostly:/
Original post by jadecross
I'm not sure how to work out the range and domain for C3 :frown:
All i know is the domain is input value and range is output but i'm not sure how to work them out
can someone please help?
e.g. how do you work out the domain and range for y=ex+12y = e^x+\frac{1}{2} Thank you.

Normally it would say f(x)=ex+12f(x) = e^x+\frac{1}{2}, but anyway,

The domain is here is obviously R\mathbb{R}, as you can put in any real number.

Now, we know that:

limx(ex+12)=12\displaystyle\lim_{x\to -\infty} \left( e^x+\frac{1}{2} \right) = \frac{1}{2}

limx(ex+12)=\displaystyle\lim_{x\to \infty} \left( e^x+\frac{1}{2} \right) = \infty

So the range would be [12,][ \frac{1}{2}, \infty]
(edited 10 years ago)
Reply 627
Original post by justinawe
Normally it would say f(x)=ex+12f(x) = e^x+\frac{1}{2}, but anyway,

The domain is here is obviously R\mathbb{R}, as you can put in any real number.

Now, we know that:

limx(ex+12)=12\displaystyle\lim_{x\to -\infty} \left( e^x+\frac{1}{2} \right) = \frac{1}{2}

limx(ex+12)=\displaystyle\lim_{x\to \infty} \left( e^x+\frac{1}{2} \right) = \infty

So the range would be [12,][ \frac{1}{2}, \infty]

sorry i don't understand this bit :frown:
limx(ex+12)=12\displaystyle\lim_{x\to -\infty} \left( e^x+\frac{1}{2} \right) = \frac{1}{2}

limx(ex+12)=\displaystyle\lim_{x\to \infty} \left( e^x+\frac{1}{2} \right) = \infty
Original post by jadecross
sorry i don't understand this bit :frown:
limx(ex+12)=12\displaystyle\lim_{x\to -\infty} \left( e^x+\frac{1}{2} \right) = \frac{1}{2}

limx(ex+12)=\displaystyle\lim_{x\to \infty} \left( e^x+\frac{1}{2} \right) = \infty


Well, if you were to draw a graph of y=ex+12y=e^x + \frac{1}{2}, you'd notice that as x approaches negative infinity, y approaches 1/2.

So the lower limit of y (or f(x) in this case) would be 1/2.

Take note, however, that the y values APPROACHES 1/2, but never quite gets there. So, make sure that you notice that it's y>1/2 and y cannot equal 1/2.

Similarly, as x approaches infinity, y approaches infinity. So y < infinity

So 1/2 < y < infinity
Reply 629
Original post by justinawe
Well, if you were to draw a graph of y=ex+12y=e^x + \frac{1}{2}, you'd notice that as x approaches negative infinity, y approaches 1/2.

So the lower limit of y (or f(x) in this case) would be 1/2.

Take note, however, that the y values APPROACHES 1/2, but never quite gets there. So, make sure that you notice that it's y>1/2 and y cannot equal 1/2.

Similarly, as x approaches infinity, y approaches infinity. So y < infinity

So 1/2 < y < infinity

Yes I get it now, thanks a lot! :smile:
Original post by Robbie242
60+60+60=180 for the C grade in AS level, you will need to bag 107 points (If those marks you speak of are ums).... you will need to resit one of them if it comes back like that.

At least I think that's how the boundaries are worked out. If those are raw marks, your ums may be higher for both


I was referring to raw marks.

I was wondering what people might think the UMS boundaries might be.
Original post by Boy_wonder_95
Thrust in the rope



This is not possible. The rope would simply go slack.

1.

The Earth may be assumed to be an isolated sphere of radius 6.4 × 10^3 km with its massof 6.0×10^24 kg concentrated at its centre. An object is projected vertically from thesurface of the Earth so that it reaches an altitude of 1.3 × 10^4 km.


2.


I know how to calculate it but how do I incorporate the altitude?


Original post by DJMayes
This is not possible. The rope would simply go slack.


Sorry I meant those which are like "a car of mass 500kg is travelling at 20ms^-1 and tows a caravan of 750kg and then immediately applies the brakes stopping in 60m calculate the thrust in the tow-bar"
(edited 10 years ago)
Original post by blueberryyums

1.

The Earth may be assumed to be an isolated sphere of radius 6.4 × 10^3 km with its massof 6.0×10^24 kg concentrated at its centre. An object is projected vertically from thesurface of the Earth so that it reaches an altitude of 1.3 × 10^4 km.


2.


I know how to calculate it but how do I incorporate the altitude?




You haven't actually said what you're trying to calculate. However, you add the altitude on to the radius of the Earth in order to find the distance between the object and the Earth's center of mass, which you then use in Newton's law of gravity.
Original post by DJMayes
...


Anyways DJ how would you compare yesterday's M1 paper to the one we sat?
Original post by Boy_wonder_95
Anyways DJ how would you compare yesterday's M1 paper to the one we sat?


More challenging, but not outlandishly so.
Original post by kamisaysjambo
Substitution mostly:/


Which part don't you understand about it?


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Anyone have a C3/4 past paper question? (not difficult ones,,just a basic past paper) not connected rates of change becausr I don't have that
Original post by DJMayes
More challenging, but not outlandishly so.


For you perhaps :tongue:

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