Any hard IGCSE Edexcel maths question?

I've been doing a lot of past papers and did a lot of hard questions and found out how to do most of them

Anyone got any really hard questions that they think might come in the exam as a similar type?

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Reply 1

nuttygirl

There are 10 beads in a box in a box. n of the beads are red. Meg takes one bead at random and does not replace it. She takes a second bead at random from the box. The probability that she takes 2 red beads is 1/3.
Show that n^2-n-30=0

Simplify fully
2/(x-1) + (11-x)/(x^2+3x-4)

Can anyone help please? Don't get them at all! :smile:

Reply 2

jelly1000

I'm not very good at Maths so this could be subjective.
This is from the specimen paper [3H] and is rated A*:
Q 20:
Express p 1/2 in the form 3^k where k is an integer.

Q22 from May 2005 3H
Simplify Fully:
2/x-1 + x-11/x^2+3x-4
it might just be the ammount of numbers in that making it look hard.

Reply 3

CocoPop

^ nuttygirl:

For the first one, you do this:

n/10 * (n-1)/9 = 1/3

(n^2-n)/90 = 1/3

n^2 - n = 30

n^2 - n - 30 = 0

And for the second (both nuttygirl and jelly1000 posted this question):

2/(x-1) + (11-x)/(x^2+3x-4)

2/(x-1) + (11-x)/(x+4)(x-1)

(2(x^2+3x+4)+(11-x)(x-1))/(x-1)(x+4)

(x^2+18x-19)/(x-1)(x+4)

(x+19)(x-1)/(x-1)(x+4)

Then you simply cross out the x-1:

(x+19)/(x+4)

Reply 4

nuttygirl

CocoPop

^ nuttygirl:

For the first one, you do this:

n/10 * (n-1)/9 = 1/3

(n^2-n)/90 = 1/3

n^2 - n = 30

n^2 - n - 30 = 0

Thanks!

Reply 5

[m a a r i.]

i dont like differientation :s-smilie:

pwease, can someone explain about the maximum and the minimum points?

Reply 6

Pixie-Goblin

2/x-1 + x-11/(x^2+3x-4)
= 2/x-1 + x-11/(x-1)(x+4)
= [2(x+4)+x-11]/(x-1)(x+4)
=(3x-3)/(x-1)(x+4)
=3(x-1)/(x-1)(x+4)
=3/(x+4)

assuming i read it correctly, works out quite nicely..

themariii

i dont like differientation :s-smilie:

pwease, can someone explain about the maximum and the minimum points?

differentiation allows you to find a second function which gives the gradient of the first function at any point. So, at a maximum or a minimum on the first function, the curve will be horizontal at that point, so the gradient is zero. Hence, to find maximums and minimums, it will be where the derivative is zero.

(I dont know if this bit is part of the IGCSE syllabus, please correct me if it's not and i'll remove from my post to avoid unnecessary confusion)
Now, finding out if they are a maximum or minimum, there are two ways to go about it:
1) look at the gradient just before and just after the point where dy/dx=0. If it is positive just before and negative just after then it is a maximum, if it is negative just before and positive just afterwards then it is a minimum. Think about what's going on with the gradient as it approaches the point where the gradient is zero
2) Take a second derivative. So, what you're doing is finding out the rate of change of the gradient. If the second derivative is positive where the first derivative = 0, then the gradient is becoming more positive at that point, so it will be a minimum because it's going to slope upwards. If the second derivative is negative, then it will start sloping downwards, so it is a maximum

Reply 7

originalname

themariii

i dont like differientation :s-smilie:

pwease, can someone explain about the maximum and the minimum points?

At a maximum point and minimum point, the gradient is 0 (dy/dx = 0).

On the left hand side of a maxima, the gradient is positive, and on the right hand side, negative. / \

The opposite is true for minima. \ /

Also, d2y/dx2 (or the second differential of y) will be >0 for minima, <0 for maxima, and =0 for a point of inflection (doubt thatll come up).

I didn't do IGCSE so I don't know how relevant that is, but it's all I think you'd ever need to know about maxima and minima.

Edit.

My teacher used to explain that minima look like 'u's and maxima look like 'n's, which is a bit weak in my opinion but maybe It'll help...

In short, a turning point on the graph is where the gradient = 0... The gradients on either side of the turning point will be opposite signs. You can tell whether it's a maxima or minima by looking at the graph and seeing whether is looks like an n or a u...

Or do it algebraically.

y = f(x)

dy/dx = f'(x)

when f'(x) = 0, solving for x -----------> x=A or B when dy/dx = 0

then d2y/dx2 = f''(x)

when x=A f''(A)>0 (minima), x=B f''(B)<0 (maxima).

Reply 8

Pixie-Goblin

CocoPop

^ nuttygirl:
And for the second (both nuttygirl and jelly1000 posted this question):

2/(x-1) + (11-x)/(x^2+3x-4)

2/(x-1) + (11-x)/(x+4)(x-1)

(2(x^2+3x+4)+(11-x)(x-1))/(x-1)(x+4)

(x^2+18x-19)/(x-1)(x+4)

(x+19)(x-1)/(x-1)(x+4)

Then you simply cross out the x-1:

(x+19)/(x+4)

Close, but not quite. (small note: Because you already have 2 divided by (x-1), it is unnecessary to do the entire cross multiplication). While your method will still work, what you're doing is going from
(11-x)/(x-1)(x-4) to (11-x)(x-1)/(x-1)(x-4). Should actually be (11-x)(x-1)/[(x-1)^2](x-4)

EDIT: makes the problem rather..sticky..

Reply 9

CocoPop

^ Ah, right! Thanks for pointing that out! :smile:

Reply 10

x10

4 questions were asked... 2 answered and 2 not answered:

For the specimen question jelly1000 asked:

You missed out this part "p = 3^8"

So "Express p^1/2 in the form 3^k, where k is an integer"

so you substituite in p so it will be (3^8)^1/2 = 81
3^k = 81
k = 4

Differenttiation for Mariii:

I'll assume you know how to differentiate (eg. y = x^2 - 2x - 4... dy/dx = 2x - 2)

So you will equate 0 to 2x - 2
2x - 2 = 0
2x = 2
x = 1

Then you substituite 1 into the original equation (y = x^2 - 2x - 4)

1^2 - 2(1) - 4 = -5 = y

So the turning point is (1 , -5)

To know whether it is a minimum or a maximum point you look at the equation

If it starts with +x^2 it will have 1 turning point which will be minimum
If it starts with -x^2 it will have 1 turning point which will be maximum

If it is a cubic equation, I think you have to draw it to find out whether it is maximum or minimum

Ask me anything else, and I'll try to help ASAP :smile:

Btw, sorry for bad spelling^^

Turned out people replied by the time I wrote this hehe... But this might give a better picture

Reply 11

[m a a r i.]

ohhhh..
thanks guysss
i understand.. much better
:smile:

Reply 12

Incognito901

[QUOTE='[m a a r i.];12299979']i dont like differientation :s-smilie:

pwease, can someone explain about the maximum and the minimum points?

I think I am late. How is life?

Reply 13

chooti

2X^31. multiply the power, which is 3 in this case, into the base value, which is two.you get?...... 62. keep that aside. now subtract 1 from the power, in this case 3 minus 1youll get 23. now youve got the differentiated value of the base number, which is 6 and youve got the value of the power. REMEMBER: X remains the only change is to its power4. put em together: 6X^2THATS ITTTT!!!!