Edexcel Mathematics: Core C1 6663 17th May 2017 [Exam Discussion] Watch

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junayd1998
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(Original post by Cryptokyo)
Yes, exactly! Two parallel lines have the same gradient. So the line parallel to our tangent has the same gradient as our tangent.
Okay i gotcha! Thanks
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Cryptokyo
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Chittesh14 Here are some problems I have dreamed up over the course of the day regarding discriminants. Question 1 is Core 2 I believe but it is a good bit of fun.

I may have gone a bit overkill...

Error on Q4. Please read spoiler.
Spoiler:
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Q4 part c. The answer should be c=-\frac{31}{8}
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Chittesh14
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(Original post by Cryptokyo)
Chittesh14 Here are some problems I have dreamed up over the course of the day regarding discriminants. Question 1 is Core 2 I believe but it is a good bit of fun.

I may have gone a bit overkill...
Thanks, excellent questions. I'll add these in tomorrow. It'd be great if you can add some hints too (or tell me them and I'll add them in the layout on TSR using spoilers) :P.
Btw, for Q2 extension - can you use the discriminant from the point after you have reduced the equation of the intersection to 2adx - ad^2 = ax^2. Because, rearranging this gives: ax^2 - 2adx + ad^2, where a = a, b = -2ad, c = ad^2.

Using the discriminant for a quadratic equation, b^2 - 4ac = 0 (equal roots) or b^2 = 4ac.
b^2 = (-2ad)^2 = 4a^2d^2 and 4ac = 4(a)(ad^2) = 4a^2d^2. As b^2 = 4ac, the quadratic and the tangent intersect at two identical points - so there is only one point of intersection.

Also, shouldn't c = -31/8 for last part 4(c) ? maybe typo
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junayd1998
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http://prntscr.com/e46f3n

Need help one with this one have no idea where to go

Cryptokyo
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RDKGames
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(Original post by junayd1998)
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Cryptokyo
It says they do not cross or touch, therefore the discriminant is less than 0 for the quadratic you end up after you take 3x-7 away from both sides.
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Cryptokyo
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(Original post by junayd1998)
http://prntscr.com/e46f3n

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Cryptokyo
As RDKGames already said. All your working is correct you just need to rearrange 3x-7=2px^{2}-6px+4p into a quadratic that is equal to 0. Then as there are no solutions (no intersections between curve and line) to this quadratic, the discriminant (b^{2}-4ac) is less than 0.
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Chittesh14
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(Original post by junayd1998)
http://prntscr.com/e46f3n

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Cryptokyo
lol that was in our mock. Lovely question I was clueless too at start.


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JezDayy
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Resitting C1 and C2 this year while doing C3, C4 and M1 (fun, lol)
For me I thought 2016's C1 paper was awful, anytime I saw any fractions I panicked and just skipped the question. I thought C2 was a lot easier but then the last question I think I got all wrong.

Hopefully going to pull up my 78 and 73 UMS to 95+, I've been doing lots of practice and I'm getting a lot more confident now!

Only thing I'm worried about is arithmetic sequences in C1, when there's context I always slip-up somewhere or get 'n' wrong.

Good luck and hope 2017 isn't too nasty!
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Cryptokyo
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(Original post by Chittesh14)
Thanks, excellent questions. I'll add these in tomorrow. It'd be great if you can add some hints too (or tell me them and I'll add them in the layout on TSR using spoilers) :P.
Btw, for Q2 extension - can you use the discriminant from the point after you have reduced the equation of the intersection to 2adx - ad^2 = ax^2. Because, rearranging this gives: ax^2 - 2adx + ad^2, where a = a, b = -2ad, c = ad^2.

Using the discriminant for a quadratic equation, b^2 - 4ac = 0 (equal roots) or b^2 = 4ac.
b^2 = (-2ad)^2 = 4a^2d^2 and 4ac = 4(a)(ad^2) = 4a^2d^2. As b^2 = 4ac, the quadratic and the tangent intersect at two identical points - so there is only one point of intersection.

Also, shouldn't c = -31/8 for last part 4(c) ? maybe typo
Yes you can use the discriminant. It does the same thing essentially. I just like the a(x-d)^{2}=0 form as it just seems a bit nicer.

Cheers for the error spot. Have put in a spoiler. I will work on some hints tomorrow.
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Ph8
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(Original post by JezDayy)
Resitting C1 and C2 this year while doing C3, C4 and M1 (fun, lol)
For me I thought 2016's C1 paper was awful, anytime I saw any fractions I panicked and just skipped the question. I thought C2 was a lot easier but then the last question I think I got all wrong.

Hopefully going to pull up my 78 and 73 UMS to 95+, I've been doing lots of practice and I'm getting a lot more confident now!

Only thing I'm worried about is arithmetic sequences in C1, when there's context I always slip-up somewhere or get 'n' wrong.

Good luck and hope 2017 isn't too nasty!
OMG IM IN THE SAME SITUATION AS YOU ahahahha
But im only resitting C1
Wasn't the exam just AWFULLLL!!! Yeah there was SO MANY iffy questions i doubted everything

Mate 78 is good I got 66 LOOOOOOOL it was abysmal but I got 92 in C2 so hopefully I can get an A this year </3 C3 and C4 are a JOKE. C1 is so easy compared to those omg
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JezDayy
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(Original post by Ph8)
OMG IM IN THE SAME SITUATION AS YOU ahahahha
But im only resitting C1
Wasn't the exam just AWFULLLL!!! Yeah there was SO MANY iffy questions i doubted everything

Mate 78 is good I got 66 LOOOOOOOL it was abysmal but I got 92 in C2 so hopefully I can get an A this year </3 C3 and C4 are a JOKE. C1 is so easy compared to those omg
Yeah I agree, I doubted everything and I during the exam I didn't realise you could split the area question into a trapezium and triangle so I just skipped it!

Cheers haha, but honestly I was hoping for an A at least at the time I took the exam. Nice one on 92 in C2! Friend got the same as you and found out he got 70/75, that's amazing! I got 59 lol which is really disappointing for me because I really like C2.
But I actually think me resitting is better since C1 and 2 are bloody easy after doing C3 and 4 hahaha
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Ph8
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(Original post by JezDayy)
Yeah I agree, I doubted everything and I during the exam I didn't realise you could split the area question into a trapezium and triangle so I just skipped it!

Cheers haha, but honestly I was hoping for an A at least at the time I took the exam. Nice one on 92 in C2! Friend got the same as you and found out he got 70/75, that's amazing! I got 59 lol which is really disappointing for me because I really like C2.
But I actually think me resitting is better since C1 and 2 are bloody easy after doing C3 and 4 hahaha
Woahhh was it 70/75??? damnnn
The 2016 was a very iffy exam paper. Lets try and 100 UMS this year cause C3 and C4 are so much harder
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JezDayy
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(Original post by Ph8)
Woahhh was it 70/75??? damnnn
The 2016 was a very iffy exam paper. Lets try and 100 UMS this year cause C3 and C4 are so much harder
Yep, that's what I'm really hoping for! I just hope exam nerves don't get the better of me. I feel a lot more prepared this year!
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B_9710
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Now this is probably an extension question since it uses limits that isn't really discussed anyway at AS. This also used C2 as well - binomial theorem. I think this is a decent question because it actually tells us what the derivative is by its definition.

 \text{Given that the definition of the derivative } f' \text{of a function } f \text{ is given as}
 \displaystyle \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} ,
 \text{show that the derivative of } f  \text{ when } f(x)=x^n \text{ is } nx^{n-1} .
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crashMATHS
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(Original post by B_9710)
Now this is probably an extension question since it uses limits that isn't really discussed anyway at AS. This also used C2 as well - binomial theorem. I think this is a decent question because it actually tells us what the derivative is by its definition.

 \text{Given that the definition of the derivative } f' \text{of a function } f \text{ is given as}
 \displaystyle \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} ,
 \text{show that the derivative of } f  \text{ when } f(x)=x^n \text{ is } nx^{n-1} .
A good exercise, but not a great proof: it only works for natural numbers unless you introduce chain/product rules.

It's quicker and easier to prove it for all non-zero integers by induction and use the exponential definition of powers otherwise.

Either way, for C1/2, the vast majority will only be able to prove this for natural numbers (or integers if they are aware of induction).
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DFranklin
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(Original post by crashMATHS)
A good exercise, but not a great proof: it only works for natural numbers unless you introduce chain/product rules.
Well, I think proving it only for \mathbb{N} was implicit.

I think it's a good thing to be have some experience of proving these things from "first principles", even if there are other ways using more machinery. And it's not like you're not going to use very similar concepts time-and-time again at university level (i.e. the idea that (x+a)^n behaves like x^n + anx^{n-1} x (1 + something we can ignore) is a very important one at degree level).

(I must admit my enthusiasm for looking for 'elementary' analysis proofs is perhaps a bit unusual).

Also,university proofs tend to be all very glib about "use the exponential series and define a^x as exp(x log a)" without ever showing that this definition of a^x really does what you'd expect. It's more "here's some rigourous stuff about series and radius of convergence and defining exp(x) and differentiting it to prove things. And then <10 minutes saying "we can define a^x = exp(x log a) and everything will be fine, honest. Kthxbye!".
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crashMATHS
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(Original post by DFranklin)
Well, I think proving it only for \mathbb{N} was implicit.

I think it's a good thing to be have some experience of proving these things from "first principles", even if there are other ways using more machinery. And it's not like you're not going to use very similar concepts time-and-time again at university level (i.e. the idea that (x+a)^n behaves like x^n + anx^{n-1} x (1 + something we can ignore) is a very important one at degree level).

(I must admit my enthusiasm for looking for 'elementary' analysis proofs is perhaps a bit unusual).

Also,university proofs tend to be all very glib about "use the exponential series and define a^x as exp(x log a)" without ever showing that this definition of a^x really does what you'd expect. It's more "here's some rigourous stuff about series and radius of convergence and defining exp(x) and differentiting it to prove things. And then <10 minutes saying "we can define a^x = exp(x log a) and everything will be fine, honest. Kthxbye!".
Agreed. But I do love this exercise and I think it is a very good one for students to do.

Edit: we are fortunate in that our analysis lecturer places strong focus on showing (and proving) that these definitions really do have the defining properties we want (as in picture): https://www.dropbox.com/s/mdw8e9x7za...34.25.png?dl=0
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B_9710
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(Original post by crashMATHS)
A good exercise, but not a great proof: it only works for natural numbers unless you introduce chain/product rules.

It's quicker and easier to prove it for all non-zero integers by induction and use the exponential definition of powers otherwise.

Either way, for C1/2, the vast majority will only be able to prove this for natural numbers (or integers if they are aware of induction).
Yeah it's only meant for natural numbers as its C1/C2 but the reason I think people should have a go is because the result comes directly from the definition of the derivative which I think many A level students probably lose sight of or it has never been mentioned to them or only briefly so.
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crashMATHS
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(Original post by B_9710)
Yeah it's only meant for natural numbers as its C1/C2 but the reason I think people should have a go is because the result comes directly from the definition of the derivative which I think many A level students probably lose sight of or it has never been mentioned to them or only briefly so.
Yeah it's great
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Chittesh14
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I've updated the thread guys, bloody hell - that Latex took me over 1 hours lmao :'(.
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