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2005 paper 1 - HSN forum

# 2005 paper 1

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### #1x_Claire_x

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Posted 11 April 2006 - 01:31 PM

Question 7)
a)

- Which i got correct : a = 4 and b = 5.

b)
- I dont understand what the domain is or how to go about doing this part of the question.

And also

Question 8)

a)

- I got this part of the question correct : (x-3) (2x-3) (x+1)

b).

- I got this part of the question correct : (-1,0) (3/2,0) and (3,0)
but i think it was just luck , could someone explain how to do this.

c)

- No idea what this question means

Thanks
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### #2dfx

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Posted 11 April 2006 - 02:57 PM

(b) State the domain of (f):

The domain is the total range of all X values that the function can take. Remember, you cannot have Log(0) or Log of a negative number (for a positive base like in this case b = 5). Therefore the domain should be all numbers STRICTLY GREATER THAN (not greater than or equal to) 4. This is because putting in x = 4 in Log(x-4) will result in Log(0) which cannot be part of the domain. So we have domain is all real numbers greater than 4.

### #3Dave

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Posted 11 April 2006 - 03:01 PM

the domain of a function is the value of x (in this case) which satisfy the equation. For log equations the (x-4) part cannot equal zero or be a negative number from the definition of logs. So, you are being asked what must x be greater than for this equation to work

a equation will cross the x-axis when y=0 and it will cross the y-axis when x = o so you use these facts to sub in the values and get your answer

Findind the greatest/least value is to find the stationary points that range. Finding stationary points is done with differentiation

Sadly we cannot allow people to post the actual questions from SQA papers for copyright reasons(Even though it make life easy for people to help). It will be enough just to state the questions you are having problems with and what you have done so far

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### #4x_Claire_x

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Posted 13 April 2006 - 09:28 AM

Oops sorry, i didnt know about not posting the questions sorry!
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Posted 28 April 2006 - 12:04 PM

Bump.

I was just doing Question 8, and in the answers for part C it says that the least value is -35. Is that wrong?

I'm getting -3.

### #6Nathan

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Posted 28 April 2006 - 05:51 PM

You're right in getting -3 as an answer, but it is not the lowest point...try substituting -2 into the f(x) equation, and you should then get -35

I'm also doing this question at the moment...I can get the minimum, but I can't get the maximum :S any ideas?

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Posted 28 April 2006 - 08:32 PM

Ahh, I see where I went wrong now. As for your problem:

Maximum occurs at Maximum Turning Point, which is also a stationary point.

Stationary points occur when dy/dx = 0

i.e when 6x^2 - 14x = 0

One of the values you get is x = 0.

So when x = 0...

2(0)^3 - 7(0)^2 + 9 = 9

=> Maximum is 9.

### #8Nathan

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Posted 28 April 2006 - 08:37 PM

ah, ok, thanks...the maximum is an awful lot of work considering most nature table questions are 5 marks on their own

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