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Official WJEC C1 Discussion Thread 14/01/2013

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Reply 60
Original post by SamHedges
Know where I can get my hands on a C1 Summer 12 paper? I've done all the others and want more practice!


I've the paper somewhere here but no mark scheme :frown:
I'm not too bothered about the mark scheme, I just want more practice.. Is it on your computer?
Reply 62
Original post by SamHedges
I'm not too bothered about the mark scheme, I just want more practice.. Is it on your computer?


On my phone, can't seem to upload so perhaps email?
Can I PM you my email?
Reply 64
Original post by SamHedges
Can I PM you my email?


Sure:smile:
Reply 65
Hi,
Can somebody help me with a C1 question please?

I don't understand what to do with questions such as:

Given that the equation:
x^3 - 6x^2 + 20 = k
has three distinct real roots, find the range of possible values for k.

We have previously worked out that there is a maximum at (0,20) and minimum at (4,-12)
(Answer is -12<k<20 )

Thanks!
Tom
Reply 66
Original post by tomfishh
Hi,
Can somebody help me with a C1 question please?

I don't understand what to do with questions such as:

Given that the equation:
x^3 - 6x^2 + 20 = k
has three distinct real roots, find the range of possible values for k.

We have previously worked out that there is a maximum at (0,20) and minimum at (4,-12)
(Answer is -12<k<20 )

Thanks!
Tom


well it has to be within the "turning"points, to intersect :tongue: and have real roots... So between -12 and 20:smile:
Reply 67
Original post by L'Evil Fish
well it has to be within the "turning"points, to intersect :tongue: and have real roots... So between -12 and 20:smile:


ohh! It's obvious now you've said it :L

Cheers :biggrin:
Reply 68
Original post by tomfishh
ohh! It's obvious now you've said it :L

Cheers :biggrin:


:yes: tbh I wouldn't have known what to think unless I had the m/s.
Reply 69
Original post by L'Evil Fish
:yes: tbh I wouldn't have known what to think unless I had the m/s.


What about this one:

x^3 - 12x + 11 = k
Has only one real root, find the range of possible values for k.
Max at (-2, 27)
Min (2, -5)

I would've said 27>k>-5, but the m/s wants:
k > 27
k < –5
Reply 70
Original post by tomfishh
What about this one:

x^3 - 12x + 11 = k
Has only one real root, find the range of possible values for k.
Max at (-2, 27)
Min (2, -5)

I would've said 27>k>-5, but the m/s wants:
k > 27
k < –5


it only has one also your min y value seems wrong.
Reply 71
That's what I though originally, but the m/s shows this (

2013-01-13_14-33_Maths_A,AS_markschemes_Jan.jpg

Unless there's a mistake...
Reply 72
Original post by tomfishh
That's what I though originally, but the m/s shows this (

2013-01-13_14-33_Maths_A,AS_markschemes_Jan.jpg

Unless there's a mistake...


Hmmm... But the value of k would have to be known...

I get the x co ordinates to be -2 and 2.

But don't know why the y co ordinates are so...

What paper is this? I don't recall seeing it
Reply 73
Original post by L'Evil Fish
Hmmm... But the value of k would have to be known...

I get the x co ordinates to be -2 and 2.

But don't know why the y co ordinates are so...

What paper is this? I don't recall seeing it


Jan '08, Q10 :/
Reply 74
Original post by tomfishh
Jan '08, Q10 :/


x^3-12x+11

(2,-5)
(-2,27)

They're right :facepalm:

If there is one real root it has yo be outside those barriers because otherwise it'll cross over loads
Reply 75
Original post by L'Evil Fish
x^3-12x+11

(2,-5)
(-2,27)

They're right :facepalm:

If there is one real root it has yo be outside those barriers because otherwise it'll cross over loads


But it doesnt cross the x-axis outside of that range?
>Side question, what is a real root on a graph? :P
Reply 76
Original post by tomfishh
But it doesnt cross the x-axis outside of that range?
>Side question, what is a real root on a graph? :P


A real root is where the graph would cross the x axis. Y is 0 and therefore it's a solution to the equation. It's real because it's a real number. Cubic graphs either have 3 real roots, or 1 and 2 imaginary I think.
Reply 77
Original post by L'Evil Fish
A real root is where the graph would cross the x axis. Y is 0 and therefore it's a solution to the equation. It's real because it's a real number. Cubic graphs either have 3 real roots, or 1 and 2 imaginary I think.


Sorry, being stupid here :/

A y value greater than 27 cant cross the x axis?
And neither can a y value of less than -5?

Or am I thinking about this the wrong way?
Reply 78
Original post by tomfishh
Sorry, being stupid here :/

A y value greater than 27 cant cross the x axis?
And neither can a y value of less than -5?

Or am I thinking about this the wrong way?


:facepalm:

I'm being an idiot sorry! K is a value of y.
Reply 79
Original post by L'Evil Fish
:facepalm:

I'm being an idiot sorry! K is a value of y.


Well and truly confused now! Is the m/s right?

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