C2 Revision

This thread can act as a revision thread for C2 for those doing it in January, and other people too I guess :smile:

...

Let's start off with geometric series.

"A mortgage is taken out for £80,000. It is to be paid by annual instalments of £5,000 with the first payment being made at the end of the first year that the mortgage was taken out. Interest of 4% is then charged on any outstanding debt. Find the total time taken to pay off the mortgage."

Not a simple geometric series question, let's hope we get simpler ones than this in the exam ^_^ ! Best to make sure we can do the hard stuff though so that we hopefully find the exam not too hard.

Secondly, how about a differentiation question.

"Given the equation y = 2x + 8x^-2 - 5, x > 0.

Use calculus to show that y is increasing for x > 2."

Have fun :smile:

! Feel free to post more questions here for people to practise on.

Reply 1

siu00as

For all the people doing C2 in January, you may want to try this free paper from http://www.examsolutions.co.uk:

C2 Paper

The paper comes with model solutions to help you over any problems.

Hope it helps your revision!

Reply 2

7589200

18

lol people don't want to be randomly set questions :P

Reply 3

Simba

OP

16

Heh, it's good for revision though...

Ah well, not to worry :smile:

...

Best of luck to everyone who has C2 in January :wink:

...

Reply 4

-G-a-v-

16

There was a question in the heinemann C2 question on logs that I'm not sure how to do:

if a^x = b^y = (ab)^xy, show that x + y = 1

I think thats the question anyway, I don't have the textbook but its in the mixed exercise on logarithms in the C2 book.

Reply 5

evariste_3086

1

distortedgav

There was a question in the heinemann C2 question on logs that I'm not sure how to do:

if a^x = b^y = (ab)^xy, show that x + y = 1

I think thats the question anyway, I don't have the textbook but its in the mixed exercise on logarithms in the C2 book.

a^x=b^y => xlna=ylnb...........(1)
now
b^y=(ab)^xy => ylnb=xylna+xylnb
=> ylnb=y[xlna]+xylnb
=> ylnb=y²lnb+xylnb..........using(1)
=>1=y+x

Reply 6

Simba

OP

16

distortedgav

There was a question in the heinemann C2 question on logs that I'm not sure how to do:

if a^x = b^y = (ab)^xy, show that x + y = 1

I think thats the question anyway, I don't have the textbook but its in the mixed exercise on logarithms in the C2 book.

*Hides* That's a question from the logarithm exercise my teacher set me for homework a bit ago... lol, that must be the book he gets the questions from :p:

...

I struggled with that one for a while... eventually got there though :redface:

.

Reply 7

Christophicus

18

distortedgav

There was a question in the heinemann C2 question on logs that I'm not sure how to do:

if a^x = b^y = (ab)^xy, show that x + y = 1

I think thats the question anyway, I don't have the textbook but its in the mixed exercise on logarithms in the C2 book.

a^x = b^y = (ab)^xy
=> xlna = ylnb = xylnab = xylna + xylnb

=> y²lnb + xylnb = ylnb
=> (y²+xy)lnb = ylnb

Compare coefficients.
y² + xy = y (divide by y)
y + x = 1

I struggled with that one when i did it for the first time. Seems easy now.

Reply 8

-G-a-v-

16

When I did it first time round I just ended up getting lost in a mess of letters. Cheers guys :smile:

Reply 9

7589200

18

hmm ok. Yea it is hard :frown:

You had to use logs? Hmm...I used power rules lol.

if a^x = b^y = (ab)^xy

then b = a^(x/y) (1)

and (a^xy)(b^xy) = a^x
(b^xy) = (a^x) / (a^xy)
b^(xy) = a^(x-xy) (2)

Sub (1) into (2)
a^(x/y)^(xy) = a^(x-xy)

x^2 = x(1 - y)
x + y = 1

No logs!

Reply 10

Dekota

Oh great; 3+ replies to the same question giving same answer.

What a waste of time ... so many questions and so little time.

Reply 11

Christophicus

18

V.P. Keys

Oh great; 3+ replies to the same question giving same answer.

What a waste of time ... so many questions and so little time.

2 actually and it isn't a waste of time. With maths, practice makes perfect. :wink:

Reply 12

Michael Scofield

Hey Simba I think this thread is a great idea. I'm resitting C2 in January to get from a B to an A so this will be very useful in the coming days.

Reply 13

AlphaNumeric

10

Perhaps I'm just scatter brained from a rather hedonistic evening last night, but with

a^{x}=b^{y}=(ab)^xy

if y=0 then x=0. This is evident from

a^{x} = 1 = 1

It is also the assumption people have made in

y² + xy = y (divide by y)
y + x = 1

since it assumes y isn't 0 too.

Does the question give the constraint x,y not equal to 0 (since the question is symmetric in x<->y and a<->b)?

*please note, my maths skill have been shot in the foot due to last night, so I could be embarrasingly wrong)

Reply 14

Christophicus

18

AlphaNumeric

Perhaps I'm just scatter brained from a rather hedonistic evening last night, but with

a^{x}=b^{y}=(ab)^xy

if y=0 then x=0. This is evident from

a^{x} = 1 = 1

It is also the assumption people have made in

since it assumes y isn't 0 too.

Does the question give the constraint x,y not equal to 0 (since the question is symmetric in x<->y and a<->b)?

*please note, my maths skill have been shot in the foot due to last night, so I could be embarrasingly wrong)

Unfortunately it doesn't. But surely given the constraint in the question as the equation you're trying to prove:
x+y = 1

Then surely you know x,y ≠0 for this particular question?

x = y = 0 is correct but the heinemann books aren't great. They teach the topic really well but aren't particularly accurate.

Reply 15

Godders

Im retaking C2 in January. And it is just so much easier now after doing C3 than it was last year!

Thank you for the website with a paper on. I need as many papers as i can get.

Reply 16

Dekota

Widowmaker

y² + xy = y (divide by y)
y + x = 1

Call it being pedantic if you must, but here we have;

y² + xy = y, you factorise first:
=> y(y+x) = y, then you divide:
=> y + x = 1

Reply 17

siu00as

Godders

Im retaking C2 in January. And it is just so much easier now after doing C3 than it was last year!

Thank you for the website with a paper on. I need as many papers as i can get.

You might like this, there's a link to a C3 paper on this thread
http://www.thestudentroom.co.uk/t183971.html

Reply 18

Godders

siu00as

You might like this, there's a link to a C3 paper on this thread
http://www.thestudentroom.co.uk/t183971.html

I already downloaded it when i did the C2 paper.Thanks for the thought though.

Reply 19

AlphaNumeric

10

V.P. Keys

Call it being pedantic if you must, but here we have;

y² + xy = y, you factorise first:
=> y(y+x) = y, then you divide:
=> y + x = 1

That's pedantic to the point of not even being required. Since it is obvious all terms are products of y, a cancellation can be done.

If you had something more complicated where it wasn't immediately obvious that each term was multiplied by a particular factor, you'd have to do some factorisation, but even then writing

af(x) + ag(x) = ah(x)
a[ f(x) + g(x) ] = ah(x)
f(x) + g(x) = h(x)

is not required. If you muliply by sides by 1/a (assuming a isn't zero :wink:

) then you get

(1/a)(af(x) + ag(x)) = (1/a) ah(x)
(1/a)af(x) + (1/a)ag(x) = h(x)
f(x)+g(x) = h(x)

Basically, your pedantry is not required.

Widowmaker

Unfortunately it doesn't. But surely given the constraint in the question as the equation you're trying to prove:
x+y = 1

Then surely you know x,y ≠0 for this particular question?

I'd chalk it up to a lack of rigour at A Level I suppose. I'm too used to "What possible cases are there for y and x's values and ...." then leading on to physical interpretations of different things. Come to think of it the point that x,y ≠0 is a distinct case from x,y = 0 when doing such questions I don't remember being told at A Level, though my memory is clouded by the fog of time :p:

C2 Revision

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