C1 Maths extra practise

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cant solve it atm but multiply the whole thing by the denominator with a flipped sign so - to + then simplify it? is that correct?

Reply 21

Zacken

Original post by not_lucas1

cant solve it atm but multiply the whole thing by the denominator with a flipped sign so - to + then simplify it? is that correct?

Not quite. This is an example where you need to rationalise twice.

BTW: Did you see my earlier link?

Reply 22

C-rated

Original post by SeanFM

Here is a question for you to try :h:

A sequence of numbers

a_0, a_1, a_2, ...

is given by

a_{n+1} = 2a_{n} + 6 - 4n

.

Given that

a_2 = 6

, find

a_1

and

a_3

.

Hence find the value of

\sum_{i=1}^{10} a_{i}

.

Hence, or otherwise, find the value of

\sum_{i=0}^{10} a_{i}

Jeeeeeeeesus, you have to do that in core1 in your exam board?! That's core2 for me
I'm so glad to have MEI :burnout:

Reply 23

Kevin De Bruyne

Original post by C-rated

Jeeeeeeeesus, you have to do that in core1 in your exam board?! That's core2 for me
I'm so glad to have MEI :burnout:

I do not think anything like that would be asked in an exam, it's just an extension question to test people on arithmetic series, recurrence relations and sum notation :tongue:

I think MEI is more of a pain from what I've heard :tongue:

not in AS but with a comprehension paper in A2 + coursework or something like that.

Reply 24

C-rated

Original post by SeanFM

I do not think anything like that would be asked in an exam, it's just an extension question to test people on arithmetic series, recurrence relations and sum notation :tongue:

I think MEI is more of a pain from what I've heard :tongue:

not in AS but with a comprehension paper in A2 + coursework or something like that.

The coursework is total bs, but the comprehension paper is alright. Both are only worth 6.6666% of the total grade put together though :tongue:

Reply 25

Someboady

Original post by Zacken

On the mean side, but whatever - show that:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{\sqrt{3} + \sqrt{2} - 1}{\sqrt{3} - \sqrt{6} + 1} = 2 + a\sqrt{2} + b\sqrt{3} + c\sqrt{6}\end{equation*}

where

a,b,c

are integers to be determined.

Er are these the correct answers:
I probably messed up somewhere :/

Spoiler

Reply 26

Zacken

Original post by Someboady

Er are these the correct answers:
I probably messed up somewhere :/

Spoiler

Not quite, but very close.

Reply 27

Kevin De Bruyne

Original post by C-rated

The coursework is total bs, but the comprehension paper is alright. Both are only worth 6.6666% of the total grade put together though :tongue:

Eeek, I see. I don't know how difficult they are (it looks pretty difficult :s-smilie:

). But I guess that's some good news for people trying to get an A*.

Reply 28

Kevin De Bruyne

Oops - my question is slightly incorrect, hang on a second :colondollar:

Rationalise

\frac{5 + \sqrt{3}}{5 - \sqrt{3}}

Hence, or otherwise, find the equation of the tangent at the point (22,10) on the line given by the equation

y= \frac{2.5 + 0.5\sqrt{3}}{5 - \sqrt{3}}x^2 - 34x + 384 - 110\sqrt3

Original post by not_lucas1

cant solve it atm but multiply the whole thing by the denominator with a flipped sign so - to + then simplify it? is that correct?

Original post by Someboady

Er are these the correct answers:
I probably messed up somewhere :/

Spoiler

Original post by imran_

Hi, im resitting C1 this year and aiming for at least 95%
ive finished ALL past papers from edexcel and ALL solomon papers
coming close to the exam does anyone have any good resources to revise c1 from?
ive tried MADASMATHS papers and some of the questions being asked are ridiculously weird

Try that if you want, but no pressure - it looks scarier than it is, especially the equation :tongue:

(edited 7 years ago)

Reply 29

odera_dim

a=1
b=1
c=1

This was a nasty question and I'm not too sure about the answer but this is what I got 😩

Reply 30

techfan42

Original post by odera_dim

a=1
b=1
c=1

This was a nasty question and I'm not too sure about the answer but this is what I got 😩

Same, I got that too

Reply 31

JN17

^ I also got a,b,c all equal to 1

Reply 32

imran_

Original post by odera_dim

a=1
b=1
c=1

This was a nasty question and I'm not too sure about the answer but this is what I got 😩

how did you do it?

Reply 33

techfan42

it becomes 16+8(root2 +root3 +root6)/8 which equals 2 + root2 +root3 +root6

Reply 34

saffronlord

Original post by SeanFM

Here is a question for you to try :h:

A sequence of numbers

a_0, a_1, a_2, ...

is given by

a_{n+1} = 2a_{n} + 6 - 4n

.

Given that

a_2 = 6

, find

a_1

and

a_3

.

Hence find the value of

\sum_{i=1}^{10} a_{i}

.

Hence, or otherwise, find the value of

\sum_{i=0}^{10} a_{i}

isnt this c2?

Reply 35

techfan42

Original post by imran_

how did you do it?

As Zacken said you have to rationalise it twice First by root3 +root6 +1 and then by 2root3+2

(edited 7 years ago)

Reply 36

odera_dim

Original post by imran_

how did you do it?

rationalise the denominator first by flipping the signs, you should come out with the denominator still being in surd form so rationalise again and simplify

I'm terrible at explaining btw 😭

Reply 37

JN17

Original post by imran_

how did you do it?

Rationalize the denominator the way you normally would, collect all like terms, then do it again, and simplify to be in terms of root 6, root 3 and root 2.