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# P1 Revision topic - 1:15pm 9th June - Edexcel watch

1. (Original post by SkylineGTR)
ooops sorry. didn't notice! is someone living in hong kong here?
theres quite a few! and malaysia, i think KOH is one. i came on here before M1 and they had already done it!
2. Trig-solving harder problems eg sin^2(x+pi/6)=1/2 for -pi<x<pi

Coordinate Geometry-how to get coordinates eg when two equations are given

Diffrentiation-max min practical problems,decreasing/increasing functions

i know its quite a bit but if i can get the correct principles i know i can do well in the exam. thanks! ps rep will be given around 30 points [/QUOTE]

Ok, well that's what this topic for!

Completing the square:

This is basically rearranging the expression:

ax² + bx + c

into the form:

(x + p)² + q

But why would you want to do this michael? Aha. You do this for two reasons:

- If you know this expressions equals zero, you can carry the q to the right hand side (by subtracting both sides by q), then taking the square root of both sides, then taking p to the right hand side and have x = .... (this is basically how to solve a quadratic if it doesn't factorise and you hate the quadratic formula).

- If this equations is actually a CURVE, eg. y = ax² + bx + c, then rearranging will give: y = (x + p)² + q (obviously). Not only does this tell you that q is the extreme value (eg. highest piossible value or lowest possible value), it also tells you that -p is the value of x which gives you this value. Remember these two paragraphs. Go back and read them again now, they're quite important.

Trig-solving harder problems eg sin^2(x+pi/6)=1/2 for -pi<x<pi

Taking your equations as an example, you must write: a = (x + pi/6). It's too easy to confuse things if you don't. Now this translates to:

sin² a = 0.5
sin a = √½ (note if the starting value on the RHS was negative, there would be no real values for sin a)
@ = arcsin √½ ~ 0.7854 (@ - alpha angle, the acute angle found by doing the arcsin of |√½|)

Between -5pi/6 and 7pi/6 (the modified values, due to the brackets in the question that we replaced with a), @ has two values (we can tell from looking at the sine curve and the boundaries:

Now modify these to put for x = ...
x = @ - pi/6

x = 0.261, 1.833

Which is wrong. ARGH
3. (Original post by chats)
theres quite a few! and malaysia, i think KOH is one. i came on here before M1 and they had already done it!

And did he tell you what was on the paper?
4. "solve for -Pi<x<Pi" = -180 < x < 180

Sin²(x + Pi/6) = 1/2
Sin(x + Pi/6) = 1/√2
x + 30 = 45
x + 30 = 45, 90+45
5. yeah i suggest you maybe work it out in degrees first or as a check.

then just divide by 180/Pi to get the radian equivalent of the answer.
6. Thank you! I'm off to make sure i know all this
7. Coordinate Geometry-how to get coordinates eg when two equations are given
There is one big formula for this, or you can use two seperate formulas. The big one is just the 2 smaller formulas stuck together. If your given two points, you name them: (x1,y1),(x2,y2) . To find the gradient you use the formula:
y2 - y1
x2 - x1

then you can use the formula: y - y1 = m(x -x1).
to do it all in one go you use this formula:
y - y1 y2 - y1
-----=------
x - x1 x2 - x1

a bit hard to show on the forumas :/
8. sin^2(x+pi/6)=1/2 for -pi<x<pi

sin^2 a = 1/2
sin a = root(1/2) = 1/root2

@ = pi/4
a = pi/4, 3pi/4

in the range -5pi/6 < x + pi/6 < 7pi/6

x = a - pi/6
x = 3pi/12 - 2pi/12, 9pi/12 - 2pi/12
x = pi/12, 7pi/12

9. Anyway, completing what I set out to finish:

how to get coordinates eg when two equations are given

Ok, you have two linear equations. Call them:

y = ax + b
y = cx + d

You want to find where these intersect. Solve as simultaneous equations:

y = ax + b = cx + d
ax - cx = d - b
x (a - c) = (d - b)
x = (d - b)/(a - c)

y = cx + d
y = c(d - b)/(a - c) + d
y = c(d - b)/(a - c) + d(a - c)/(a - c)
y = [cd - cb + da - dc)]/(a - c)

Obviously this will look a lot less complicated with read numbers rather than letters. eg.

y = 3x - 4
y = x + 7

3x - 4 = x + 7
2x = 13
x = 13/2

y = 13/2 + 14/2 = 27/2

So they meet at: (13/2, 27/2)

Diffrentiation-max min practical problems,decreasing/increasing functions
There is a maximum/minimum point where the gradient is zero. So differentiate, which given you the gradient in terms of x, put that equal to zero, and find the value of x that gives this gradient by rearranging. Simple as that. eg.

Find the stationary points of the curve y = x^3 - 2x + 9

dy/dx = 3x^2 - 2
0 = 3x^2 - 2
3x^2 = 2
x^2 = 2/3
x = +/- root(2/3)

Increasing/decreasing functions can be found when differentiating a second time:

d^2y/dx^2 = 6x

This means the change in gradient when x is positive is positive, and the change in gradient when x is negative is negative. So for x > 0, the function is increasing, and when x < 0, the function is decreasing. The change in gradient is 0 when x = 0, so when the curve crosses the y axis there is a point of inflection (meaning the change in gradient is zero, but it is not a maximum or minimum point).
10. This means the change in gradient when x is positive is positive, and the change in gradient when x is negative is negative. So for x > 0, the function is increasing, and when x < 0, the function is decreasing. The change in gradient is 0 when x = 0, so when the curve crosses the y axis there is a point of inflection (meaning the change in gradient is zero, but it is not a maximum or minimum point).
Not quite true, there is a local minium and local maximum around the y - axis. when x = -√(2/3) theres a maximum point and x = √(2/3) theres a min. function is increasing when x > √(2/3) and x < -√(2/3), function is decreasing when -√(2/3) < x < √(2/3). The easiest way to solve increasing/decreasing functions is too attempt at drawing the curve and see what its doing.
Rev
11. (Original post by Revelation)
Not quite true, there is a local minium and local maximum around the y - axis. when x = -√(2/3) theres a minimum point and x = √(2/3) theres a max. function is increasing when x > √(2/3) and x < -√(2/3), function is decreasing when -√(2/3) < x < √(2/3)
Rev
Good point. I saw that there were two turning points after I changes x = root(2/3) to x = +/- root(2/3), then confused myself when I found the 2nd derivative gave 6x...

Ah I get it. The gradient is decreasing when x < 0, so there is a min point there, and the same with +ve, and EXACTLY in between the change in gradient is zero because the curve is "changing direction".

Thanks!
12. that is all excellent, thanks all. i will try and absorb all this.
13. so just tdo get this clear:

d^2y/dx^2 < 0 is a maximum turning point

d^2y/dx^2 > 0 is a minimum turning point

and

d^3y/dx^3 = 0 is a point of inflexion.
14. (Original post by Bebop)
so just tdo get this clear:

d^2y/dx^2 < 0 is a maximum turning point

d^2y/dx^2 > 0 is a minimum turning point

and

d^3y/dx^3 = 0 is a point of inflexion.
im pretty sure we dont need to know thhe inflexion bit
15. I think I've seen a question on it before, knowing it won't really require too much effort anyway.
16. yes except but for your poi, if d^2y/dx^2 = 0 and d^3y / dx^3 =/= 0 then a p.o.i occurs.
17. (Original post by Bebop)
I think I've seen a question on it before, knowing it won't really require too much effort anyway.
what is the point of inflexion again?
18. (Original post by TheWolf)
what is the point of inflexion again?
Attached Images

19. do we need to know algebraic division? i don't remeber needing it any paper i've done recently
20. (Original post by Bebop)
so just tdo get this clear:

d^2y/dx^2 < 0 is a maximum turning point

d^2y/dx^2 > 0 is a minimum turning point

and

d^3y/dx^3 = 0 is a point of inflexion.
na. Don't do d^3y/dx^3. Point of inflection is when the gradient (dy/dx) = 0 and the change in gradient (d²y/dx²) = 0 as well.

d^3y/dx^3 describes how the instantaneous change in gradient varies in terms of x, which I'm pretty sure you don't need to know...

(Original post by SkylineGTR)
do we need to know algebraic division? i don't remeber needing it any paper i've done recently
Err.. yes lol! Just read the Heinemann book and do some exersices... it's hard as hell trying to describe it in words on here.

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