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Past Paper Problems - HSN forum

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Past Paper Problems


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#1 sparky

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Posted 15 April 2004 - 08:54 PM

Hi

I wondered if anyone could honestly help with a question from the 2000H paper.

It's q6

For what range of values of k does the equation x² + y² + 4kx -2ky -k -2 = 0 represent a circle?

And any tips on question 7, is there an easy way of working out the co-ordinates you need, I mean it only is worth 3 marks.

Thanks
Mark

#2 Allan

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Posted 15 April 2004 - 09:03 PM

For question 6, using the equation given you can work out the circle's centre and its radius

Centre (-2k, k)

Radius = √(g² + f² - c) working out to √(5k² + k + 2)

For any circle the radius must be greater than 0.

You can complete the square of the radius² giving

5(k + 1/10)² + 1(19/20)

This leads to the answer of all values of k (as k as a square will always be greater than or equal to zero, adding on the 1(19/20) means it will always be greater than 0). A square root of a positive will give a postive answer.

#3 Discogirl17

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Posted 17 April 2004 - 12:52 PM

Ok. You know what. I'm just so going to fail Maths. I didnt get that explanation at all. blink.gif
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#4 sparky

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Posted 17 April 2004 - 01:53 PM

To be quite honest I feel the 2000 Higher Still paper is the hardest of all the higher papers!
You can definately tell this was the first year of higher still, contrast the non-calc of 2000 with that of 2003.
2003 is all wee short questions, whereas in 2000 there was a lot more diagrams and prob solv!

I was worried about maths doing this paper, but they're not all like this one! And I shall definately be asking my teacher about a lot of questions from this paper! And q7 was vectors, and I never got this either, it seemed quite a difficult q for only 3 marks

Mark
Mark

#5 sparky

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Posted 17 April 2004 - 01:55 PM

Also if anyone is a genius at maths, they could maybe help me out with a few q's from paper 2. (2000H Still)

2 (b)
4 (a)
6
9 (b) i got cos THETA to be a negative, which would mean 2 answers, anyone else get this?


Thanks

Edited by sparky, 17 April 2004 - 01:56 PM.

Mark

#6 Allan

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Posted 17 April 2004 - 02:16 PM

For 9b, you would get two answers but if you look at the diagram, the 2nd answer of 267.5º is unrealistic. That's the only explanation for that one I can think of!

#7 Allan

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Posted 17 April 2004 - 02:23 PM

For 2b, the question says the line QC is parallel to the y-axis. That means that since the point Q has x-cordinate 1, the point and centre C will also have the x-cordinate of 1.

Since C lies on the line AB, you can substitute that into the forumla from part a, x + 2y = 9 so 1 + 2y = 9 leading to y = 4

So the centre of the circle is (1,4)

Then to find the radius of the circle you can use the distance formula to find the length of QC, which is the radius.

CQ = √(1-1)² + (9-4)²
CQ = √(0)² + (5)²
CQ = √25
CQ = 5

Put that into the equation of the circle:

( x - a )² + ( y - b )² = r²
( x - 1 )² + ( y - 4 )² = 5²
( x - 1 )² + ( y - 4 )² = 25

You could multiply that out but you don't have to

#8 Allan

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Posted 17 April 2004 - 02:36 PM

Onto 4a now!

Any equation of a parabola is in the form

y = ax² + bx + c

Which can be re-written in the form y = k( x - a )( x - b ) [I'd learn that forumla]

The parabola cuts the axis at (0,0) and (4,0). So put that into the formula

y = k(x - 0)(x - 4)

Then you've got a point on the parabola, here I'm using the turning point (2,4). Put that into the equation:

4 = k(2 - 0)(2 - 4)
4 = k × 2 × (-2)
-4k = 4
4k = -4
k = -1

Now you've got k multiply it out from the equation above:

y = -1(x - 0)(x - 4)
y = -1 × x(x - 4)
y = -1 × (x² - 4x)
y = -x² + 4x
y = 4x - x²

#9 Discogirl17

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Posted 17 April 2004 - 02:47 PM

Ah ah ah! Kill me now! sad.gif Maths is too hard! I got a B but all these Qs are freaking me out! I think I might just go hide in a dark place and watch paint dry cos I think they are probably less scary things to do than Maths! ph34r.gif
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#10 Steve

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Posted 17 April 2004 - 02:47 PM

Here's the working for question 6 biggrin.gif
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#11 Discogirl17

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Posted 17 April 2004 - 02:48 PM

Thanx Steve. Ur an angel. biggrin.gif
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#12 Allan

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Posted 17 April 2004 - 02:52 PM

Cheers Steve, didn't have a clue how you multiplied that one out! smile.gif

#13 Discogirl17

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Posted 17 April 2004 - 05:30 PM

Yep I just genearlly didnt have a clue about Maths but thanx to the site I'm getting better!
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#14 sparky

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Posted 17 April 2004 - 08:08 PM

Thanks steve, had your working right down until the last line mad.gif typical me!

Thanks Allan, the formula for the parabola, I don't think we have ever done! lol. But the circle problem I was getting confused! lol

Thanks anyway

Mark

#15 Allan

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Posted 17 April 2004 - 08:46 PM

that parbola stuff is in the Unit 2 Outcome 1 notes on HSN

as for the circle...i dunno how I can clarify that for you, I could write out the working for you if you like?

#16 sparky

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Posted 17 April 2004 - 08:57 PM

No, it's ok, I actually GET what it is now, but at the time I was stumped!

I'll have a look because I've got them printed off somewhere in a folder! lol

Thanks
Mark

#17 Discogirl17

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Posted 18 April 2004 - 06:58 PM

Yeh I printed off the notes too. Thanx Allan. Ur a genious. biggrin.gif
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#18 Ally

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Posted 18 April 2004 - 08:06 PM

QUOTE (Allan @ Apr 17 2004, 02:36 PM)
Onto 4a now!

Any equation of a parabola is in the form

y = ax² + bx + c

Which can be re-written in the form y = k( x - a )( x - b ) [I'd learn that forumla]




An alternative method to answering that question is to use the 'Complete The Square' method/rule which I personally find much easier.

#19 Ally

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Posted 18 April 2004 - 08:10 PM

QUOTE (sparky @ Apr 17 2004, 01:53 PM)

I was worried about maths doing this paper, but they're not all like this one! And I shall definately be asking my teacher about a lot of questions from this paper!  And q7 was vectors, and I never got this either, it seemed quite a difficult q for only 3 marks

Mark

Sparky for this question all you really need to do is to find an alternative route to VK. I would work it out for you but I can't find the right symbols in the character map. However, once you spot that you have to find an alternative route it should be easier from then on.

biggrin.gif

#20 sparky

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Posted 18 April 2004 - 08:28 PM

Thanks I'll have a look at this tommorow, probably when I am sitting in double maths! blink.gif lol


Mark





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