Higher Maths 2012

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Reply 160

daisyclouds

Original post by JaggySnake95

You can also do completing the square, right?

Yes, I would of done it that way :P

Reply 161

Shwi

Original post by daisyclouds

I'm sure you'll do great! :biggrin:

just keep up the work and if there's any questions you're unsure ask away! :smile:

Thankyou!

Its a lot of simple silly things i have trouble with, im not sure how to work out the graphs of logs, for example in the 2009 past paper Q10 in the multiple choice i couldnt work out which graph had the equation y=log5(x-2)

Reply 162

TheUnbeliever

Original post by Shwi

Thankyou!

A graph of

y = \log_b(x)

will always pass through (b, 1). But in that case, you've got the (x-2) instead of just x so it's like you're looking 2 'backwards' - which shifts your graph two units to the right, so you'd be looking for it to pass through (b+2, 1). You can do a similar trick with other powers (e.g. (b^2, 2)) if they're really trying to catch you out and there are several curves through that same point.

Reply 163

daisyclouds

Original post by Shwi

Thankyou!

Aww logs can be confusing, but that question, you kind of have to know the standard log graph that it usually cuts the x-axis at 1 and that question said y=log5(x-2) so it has moved to the right 2 places so the graph should be cutting the x-axis at 3

but don't get yourself caught up in one multiple choice :smile:

Reply 164

saskat00n

Original post by TheUnbeliever

2x^2 + 4x + 7 = 2(x+p)^2 + q
= 2(x^2 + 2xp + p^2) + q
= 2x^2 + 4xp + 2p^2 + q

Equate coefficients of like terms:
4p = 4
p = 1

2p^2 + q = 7
2 + q = 7
q = 5

ahhhh ok, thanks!

Reply 165

Shwi

Thank you so much!! I just cant seem to get them in to my head :s-smilie:

definitely need to pass because there is nooo way they'd give me an appeal and I really don't fancy resitting it next year :frown:

Reply 166

Esmeee

Original post by SQA

You

haha, somehow, I have to beg to differ :P

Reply 167

ScottishDan

This seems like it should be easy but i'm not getting the right answer. Any help would be appreciated.

"Solve 2cos2x-sinx+1=0" x is between or equal to 0 and 2pie. I think the expansion you need is "1-2sin^A".

Reply 168

member592149

Original post by ScottishDan

Just choose the right expansion (these question you need to make sure if there is a cos and sin value to get rid of 1 so only cos or sin values remain) and then factorise and solve from there.

Reply 169

INT2 Warrior

I just timed myself doing all of 2008 (paper 1 and 2) and got 78% overall. Would that be an A?

Reply 170

daisyclouds

Original post by ScottishDan

2(1-2sin^2x) - sinx + 1 = 0
2 - 4sin^2x - sinx + 1 = 0
-4sin^2x - sinx + 3 = 0
4sin^2x + sinx - 3 = 0
(4sinx - 3)(sinx + 1) = 0
4sinx -3 = 0 sinx + 1 = 0
sinx = 3/4 sinx = - 1
x = 48.6 x = 90

360 - 90 = 270
180 - 48.6 = 131.4

x = 48.6, 131.4, 270

hope that helped :smile:

(edited 11 years ago)

Reply 171

ScottishDan

^ thanks a lot i done something silly with the factorising and put 2sinx at the start of each bracket for some reason. :smile:

Reply 172

daisyclouds

Original post by INT2 Warrior

I just timed myself doing all of 2008 (paper 1 and 2) and got 78% overall. Would that be an A?

Yeah usually an A is around 75% but then again, it depends on how difficult of the paper :smile:

Reply 173

ScottishDan

Another question:

"A circle has the following properties:

-The X-axis and the line y=20 are tangents to the circle
-The circle passes through points (0,2) and (0,18)
-The centre lies in the first quadrant

Find the equation of the circle" (6 marks)

So I know the radius is 10 and the centre will be (x,10) but don't know where to go from here.

Reply 174

__houston

Hi, could someone please help me with this question?

A minimum value of 1 - cos(x - pi/3) occurs when x = t. What is the value of t?

Thanks :smile:

Reply 175

TheUnbeliever

Original post by __houston

A minimum value of 1 - cos(x - pi/3) occurs when x = t. What is the value of t?

Well, we minimize this expression by maximizing the cosine term. Maximizing the cosine term means finding where it equals 1. cos(x) = 1 at x = 0 (the question presumably gives you a domain - I'm assuming [0, 2pi) here) but we're shifted right by pi/3. t = pi/3.

Reply 176

TheUnbeliever

Original post by ScottishDan

The points (0, 2) and (0, 18) are 2 units in from the respective tangent lines. Because these lie on the circumference, we know that the centre lies exactly one radius - 10 units - away. This defines two circles, one centred on each point. The centre of the circle we're asked for lies at the intersection of these two circles. There are exactly two intersections, which is why we're told that it lies in the first quadrant.

x^2 + (y - 2)^2 = x^2 + (y - 18)^2
y^2 - 4y + 4 = y^2 - 36y + 324
-320 = -32y
y = 10

Then substitute back into (one of) the original equation(s)
x^2 + 8^2 = 100
x^2 + 64 = 100
x^2 = 36
x = 6

Reply 177

ScottishDan

Thanks a lot for that! A couple of one mark questions that I'm not too sure about:

"write cos^x in terms of cos2x"

And see when the roots are either unequal or irrational, what does that mean again? I always forget!

Thanks again!

Reply 178

Lewis_Mac