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Happy revising.
Collaborative Mathematics
Post your problem here and hopefully someone will get back to you quite quickly. State the question that is giving you trouble and what you’ve tried so far.
Happy revising.
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Total stuck on question 2, paper one. I thought it seemed really easy at first but now I have no idea of what to do. Please assist immediately lol
Hi Ross. This question is one of those pains in the backside that could lead you up the garden path if you let it. There is no substitution into circle equations or anything like that needed. As the circles are congruent they must have the same radius. This means that P is the mid point of the line AB. Using the formulae list you can get the coordinates of the centres of each circle then use midpoint formula. I think you’ll be able to attempt part b using the same kind of thinking (ie. use distance formula) Hope your holiday is going well. Great weather for revision isn’t it.
Paper 1 Question #8 part c is annoying the heck out of me - I’ve got the max value correct as 9, but the answers state the min value as -35 and all I can muster is -3. The function doesn’t even seem to go that low at all! Where have I gone wrong? (Parts a and b are correct)
Also, question #9 on the same paper has completely flummoxed me - I can get the right answers, but I’m fairly sure it’s the wrong way. What sort of direction should I go in?
Thanks!
it does go that low, just write it into your graphics calculator.
oops
Hi Sam. in answer to #8 c). When you differentiate you’ll get your turning points at x= 0 and x=7/3. Justify these using a nature table. However your minimum value in this case does not lie on one of the turning points. It is on one of your domain limits. For a better explanation of this look at my reply to James on the past paper help for 2006. I hope this helps.
In reply to Gavin it’s a non-calculator paper so you wouldnt be able to do that in the exam !!
Well spotted Martin. Calculators ,who needs them!
would it be possible to receive some help as to how to go about q9 please?
I’ve expanded to: cos^2x -sin^2x
but no clue as to what next. thanks for all you’re other help as well!!!
Hi Pixie.
Use another expansion ie.
Also remember that
square root something at some point is my hint.
cheers I think I was being a bit thick on that 1 and probably this 1 too seeing as it’s only worth 3 marks, q2 has baffled me and i promise this is the last one as any more after now I probably wouldn’t take in anyway, just have to hope nice questions come up tomorrow and I don’t make my stupid mistakes like 1×0 = 1 or something!
he he meant to say q7
hi, i’m really confused about question 7 on paper one, how on earth do you get a=4 from the grap!!!
I have absolutly no idea how to do paper one question 9
cos2x= 7/25
im guessing expand but i got stumped with the fraction!
please help maths exam is a week away ha
Hi Sith
Try using Trig formulae particually [tex] \cos2x = 2\cos^2 x - 1 [tex]
If your having trouble getting some of the solutions and you can’t get hold of a teacher or anyone else- there are very detailed solutions at http://www.sqa.org.uk/sqa/2470.html .
They can be very helpful and there are full solutions for most other subjects and past papers on the SQA site.
Hope I’ve been of some help
Q7 is a tricky little number,
Think of normal log x graphs you’ve come across before –> all of them go through the point (1.0), if the point at y=0 is 5 then the graph has been moved to the right by 4 hense a=4 where f(x) = log b (x-a)
Hi, ive tried doing q(8c) on paper one nad i have looked at the past advice and on the other post, but still cant find the minimum value of ‘f’. i have found that 9 is from the nature table, but i dont understand how you can get -35.
p.s this has been a very useful blog
Ru
i just looked on the link to the sqa website and i have found that you have to do this f(2)=(2x^3)-(7x^2)+9 why would you do this? would you substitute every number of x until you found the lowest?
Ru
i have also got stuck on q(11b) i have used the discriminant to get to (4t^2)-20([t^2]-4)=0 i dont understand how to get from there to t=(5^1/2)
how do u do question 10? ive been tryin for ages but still i am getting it wrong
hey can someone help me with Q.10 on paper 2?
I’m a bit stuck on question 3a). I can’t remember how to get the coordinates of F from being given the other points and the ratio. Can someone help?
Robbie
You take a weighted average of the coordinates of D and B, to get the coordinates of F, using weights of 1/3 and 2/3 respectively (because F divides DB in the ratio 2:1, and is therefore nearer to B than to D - so gets the bigger weight, 2/3).
So, the X coordinate of F is: (1/3)x6 + (2/3)x12 = 10. The Y and Z coordinates follow similarly.
Hey
i am really stuck on paper 2 questions, 2bi, 2bii and 3.
It seemed easy at first but i cant get he same answer as it says. Can anyone help??
Samantha
From the diagram, cos(p) = 8/17, sin(p) = 15/17, cos(q) = 8/10, and sin(q) = 6/10.
So, in b(i), cos(p+q) = cos(p)cos(q) - sin(p)sin(q) = (8/17)x(8/10) - (15/17)x(6/10) = (8×8 - 15×6)/170 = (64 - 90)/170 = -26/170 = -13/85.
Then in b(ii), tan(p+q) = sin(p+q)/cos(p+q) = (84/85)/(-13/85)= -84/13, using the answers from (a) and b(i).
In 3(a), the point halfway along AB is at (3,2). The slope of AB is +1, so the perpendicular bisector has slope -1 and passes through (3,2). Hence you can find its equation to be x + y = 5.
In 3(b), if the tangent at A has equation x + 3y = 1, then it has slope -1/3. Therefore the radius through A has slope +3. Given that this radius CA has slope + 3 and passes through the point A, you can find its equation to be: y - 0 = 3(x - 1), or y = 3x - 3.
In 3(c)(i), to find the centre C, you know it is at the intersection of the perpendicular bisector found in part (a) and the radius found in (b), so solve x + y = 5 and y = 3x - 3, to give: 5 - x = 3x - 3, or 4x = 8, or x = 2. Then y = 3.
In 3(c)(ii), you now know the centre of the circle is at (2,3), and you also know that the radius r is given by length of CA, so r^2 = (1-2)^2 + (0-3)^2 = 10. Hence the equation of the circle is (x-2)^2 + (y-3)^2 = 10.
Thanks for the help!
But when you expanded tan(p+q) why does it go to sin(p+q)/cos(p+q)??
Samantha
It is not that I expanded tan(p+q) - it is just that tan = sin / cos. Recall that, in a triangle, sin = opposite / hypotenuse, and cos = adjacent / hypotenuse. So, since tan = opposite / adjacent, we can always write that tan = sin / cos.
Hi.
I’ve been looking at all these responses and they are all really helpful! Thankyou!
I was wondering, could someone maybe help with question 4 b)i) on paper one please? I have 4a) as (3x-1)^2 +7, which is right.
For b)i), do you complete the square? If so, could someone help me with that, I seriously do not know what to do
Thankyou!
Your expression is already in the correct form so you don’t need to complete the square. If you think about it when the expression is in this form we can see that its minimum value is 7 because the lowest that (3x-1)^2 can be is zero (squares can’t be negative)
The minimum value (y value )of 7 occurs when 3x-1 = 0 so x = 1/3. This gives us a min t.p of (1/3, 7)
To find the range in ii) think about what this parabola with min value 7 will look like. the range is the possible y values.
hey can someone help with question 12 b? iv tried making the radius a perpendicular line to the equation y=2x and that gave me the gradient but im stuck from there on i dont think what iv done is right so far anyway. any pointers?
Hi
Hmm…you are right, the radius idea isn’t going to get you far.
When you are working on these papers remember part b of a question will quite often follow on from part (a) so think about what you had to find in part (a) as you will most likely have to use it.
You want to involve t in some way which is probably why you went for the radius, but remember your equation from (a) is also dependant on t
How would we show that a line is a tangent to a circle….first we show they meet (solve simultaneously) then we show that they meet at one point (a repeated root)….. you can’t factorise so use the discriminant b^2 -4ac = 0 for a repeated root.
Have fun!
i need help with Q9..
cos2x=7/25
could you help me with that please
Sorry no one picked up on this request Ella. Just in case you are still checking…
7,24,25 is a Pythagorean Triple. That might help. Might not!
Also cos(2x)= … several things! - that might help too.
I was wondering if i could get help with question 10 on paper 2.
i have expanded the scalar product to a.a + a.b + a.c and have found that a.a is 9 and a.b is 0 but how do i find the angle for a.c? it doesnt give it in the question and im not sure how to find it?
Thanks for all the responses..they have really helped us and we were stuck on a lot!
thanks again
x
hi can someone please help with question 10b on paper one 2005?