No problem
It's in the symmetry of sin(x):
Example: Take any point on the graph here, say 70 degrees. We then see that, due to the symmetry, it is equal to 110. pi  110 = 70; pi  70 = 110.
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MAT Prep Thread  2nd November 2016 watch

somestudent
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 22102016 15:36

KloppOClock
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 22102016 15:38
(Original post by somestudent)
No problem
It's in the symmetry of sin(x):
Example: Take any point on the graph here, say 70 degrees. We then see that, due to the symmetry, it is equal to 110. pi  110 = 70; pi  70 = 110. 
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 22102016 15:39
2015 MAT q4 wtf??

somestudent
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 22102016 15:40
(Original post by Mystery.)
Are vectors a part of mat? I've never seen it apart from section formula at most. Do you need to know like dot product and stuff? 
somestudent
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 22102016 15:42
(Original post by KloppOClock)
you got any other useful trig trivia?
sin(pi/2  x) = cos x

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 22102016 15:47
(Original post by 11234)
2015 MAT q4 wtf?? 
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 22102016 16:23
Help with 2007 paper? It is question 1: F, I and J. Thank you!

KloppOClock
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 22102016 16:31
(Original post by molly221)
Help with 2007 paper? It is question 1: F, I and J. Thank you!
for i, assume log10b=0
for j, set n=1 and add up to arithmetic series together. 
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 22102016 16:40
(Original post by molly221)
Help with 2007 paper? It is question 1: F, I and J. Thank you!
For 1I, you want the greatest possible value of a which means you must have the smallest possible value of b. Setting b = 1 means that it (logb)^2 = 0
So you are left with 4((loga)^2) = 1. Rearranging this leaves you with 10^1/2 which equals the square root of 10/ 
somestudent
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 22102016 16:42
(Original post by molly221)
Help with 2007 paper? It is question 1: F, I and J. Thank you!
2^3x + 4 = 2^2x + 2*2*(2^x)
(2^x)^3 + 4 = (2^x)^2 + 4(2^x)
let u = 2^x:
u^3 + 4 = u^2 + 4u
u^3  u^2  4u + 4
factorising we get
(u  1)(u^2  4)
2^x = 1 (Clearly x = 0)
u^2 = 4 => 2^x = +2
2^x >= 0 and hence 2^x /= 2. We have two real solutions (c).
I: We want the smallest value of (log_{10} b)^2, which is 0, when b = 1. Therefore 4(log_{10 }a)^2 = 1 => log_{10 }a = 1/2 (can only take positive root). a = 10^(1/2) = sqrt(10). Answer is (c)
J: let n = 1. Therefore we have an arithmetic sequence first term 2, difference 1 and length 100.
Sum: 50(4 + 99) = 50(103) = 5150.
Therefore 5150 > k, answer (d). 
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 22102016 17:27
2008 paper help please? It is question 1: C and D. Thanks

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 22102016 18:01
(Original post by molly221)
2008 paper help please? It is question 1: C and D. Thanks
You can then use the sum formulae (can't remember the exact name of he equation) to get 50/2(1+99). It's fifty because it you go from x^0 to x^49 which is 50 numbers. This gives you 2500
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somestudent
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 22102016 18:18
(Original post by molly221)
2008 paper help please? It is question 1: C and D. Thanks
(sin^2(θ) + cos^2(θ))y = y
(sin^2(θ) + cos^2(θ))x = x
Both of which are true for all values of θ as (sin^2(θ) + cos^2(θ)) = 1. Answer (a)
D: Dividing by x  1 and getting the remainder is the same as substituting in x = 1. This gives us an arithmetic sequence length 50, difference 2, first term 1.
We sum this to get
25(2 + 98) = 25 * 100 = 2500 (b) 
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 22102016 21:13
Thank you to everyone who helped me with my previous question. Very much appreciated and helped a lot.
These are the questions I was unsure how to do question 1: D, H, I, J in the 2009 paper and I was wondering if anyone could explain how to do these questions?
Thank you so much in advance. 
alfmeister
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 22102016 21:24
Here's a solution for J.
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somestudent
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 22102016 21:55
(Original post by 11234)
Ways to approach this?? I'm really hoping no geometry ones come up. Really would love a number theory one tho
I used http://www.webmath.com/gcircle.html to draw my circles.
HintsSpoiler:SolutionsShow(i) Use the circle equation and substitute in separately both coordinates. Expand brackets and rearrange.
(ii) Use information from part (i)  substitute in (x0, y0) and rearrange. The circle will be unique if you can eliminate h and m.
(iii) Use a diagram  and consider the equation you worked out in (ii). Work out the areas and radii, and h (y of center) of A, B, and C. Now use the diagram to find out the largest section of A and work this out.
(iv) Draw another diagram. Work out the centers of A and C. What way of dividing circle B will minimise the lopsidedness? (Think about how changing the size of one area will affect another area and hence the lopsidedness). Can you equally split any of your regions into pi/6? What would this region be composed of? Calculate areas.Spoiler:Show(i) [3] Circle equation: (x  m)^2 + (y  h)^2 = r^2.
Substituting in the coordinates separately and expanding/rearranging we get:
m^2  2m + 1 = r^2
m^2 + 2m + 1 = r^2
Setting these equal we get 4m = 0 and hence m = 0.
Using m = 0, we get r^2 = 1 + h^2 and hence r = sqrt(1 + h^2)
(ii) [3] Subbing in (x0, y0), knowing that m = 0, we get:
(x0)^2 + (y0)^2  2hy0 + h^2 = r^2
r^2 = 1 + h^2 and hence:
(x0)^2 + (y0)^2  2hy0 = 1
(x0)^2 + (y0)^2 = 1 + 2hy0
Working out h, we get:
r^2 = 1 + h^2
h^2 = r^2  1
h = sqrt(r^2  1)
Eliminating h we get
(x0)^2 + (y0)^2 = 1 + 2y0 * sqrt(r^2  1)
This is a unique circle as it does not depend on (m, h).
(iii) [5] For each circle we can work out the area by using the given coordinates as (x0, y0) and substituting them into the above equation. For A this eventually rearranges to r = sqrt(2), the center (0, 1), and hence the area is 2π. We find the radius of B to be 1, the center (0, 0), and hence the area is π. We find the radius of C to be 17/8, the center (0, 15/8) and area to be 289π/64.
Using a diagram we can work out A's lopsidedness (A: red, B: grey, C: blue)
Attachment 588618
We can visibly see the area of A we want. To get this, we must subtract a half of B and the segment.
Half of B is simply π/2 and the segment is (1/4 * 2π)  1 = 1/2π  1 (triangle base 2, height 1, sector angle pi/2).
We add these to get
π/2 + π/2  1 = π  1
We subtract this from 2π to get π + 1 and divide this by 2π to get the fraction  our lopsidedness is
(π + 1) / 2π
(iv) [4] Drawing a diagram will help.
B is still like in part (iii), but A and C are not. We work out what we can. This leads to us find for A the center is (0, p) and for C it is (0, p). To get the smallest lopsidedness of B, it must be split up into three equal sections  otherwise, we will have at least one section bigger than this value  which is π / 3. The middle region is split in two by the xaxis and hence each part has an area of π / 6.
The middle region can be composed of two sectors minus two triangles. The triangles have vertices (1, 0), (1, 0) and the centers of the circles. The sectors are of the circles A and C, formed by the triangles.
Therefore the area of a sector minus a triangle is π / 6.
The area of a sector:
the radius is found by using pythagoras on the triangle:
r^2 = p^2 + 1
The area of A or C is therefore (p^2 + 1)π
The angle of the sector is found by using rightangled trigonometry, which yields (1/2x as the trig. will get half of the angle)
tan(1/2x) = 1/p
therefore 1/2x = tan^1(1/p).
x = 2tan^1(1/p)
Therefore the area of the sector is found by multiplying the area of the circle by the angle, then divide by 2π
((p^2 + 1)π * 2tan^1(1/p)) / 2π
= (p^2 + 1)tan^1(1/p)
The area of a triangle:
base length: 2
height: p
area = 1/2bh = 1/2 * 2p = p
Therefore
(p^2+1)tan^1(1/p)  p = π / 6. 
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 22102016 22:14
How much time is everyone dedicating to MAT over halfterm? Not sure how I should be balancing MAT prep with A Level stuff.

KloppOClock
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 22102016 22:20
I was feeling sort of confident for this MAT exam... then I did the 2015 paper.

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 22102016 23:00
Help pls with q5 of 2002 paper
http://www.mathshelper.co.uk/Oxford%...est%202002.pdf 
somestudent
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 22102016 23:25
(Original post by molly221)
Thank you to everyone who helped me with my previous question. Very much appreciated and helped a lot.
These are the questions I was unsure how to do question 1: D, H, I, J in the 2009 paper and I was wondering if anyone could explain how to do these questions?
Thank you so much in advance.
Therefore n  1/2(n  1) >= 100
2n  (n  1) >= 200
n + 1 >= 200
n >= 199
The smallest value of n is 199 (c)
I: we want x^2  1 to be a factor. Therefore f(1) = 0 and f(1) = 0.
f(1) = 0
n^2  25n + 150 = 0
(n  10)(n  15) = 0
n = 10, n = 15
f(1) = 0
We need to consider when n is even or odd, as (1)^x will be 1 if x is even and 1 if x is odd
When n is even
n^2 + 25n  150 = 0
(n^2  25n + 150) = 0
(n  10)(n  15) = 0
n = 10, n = 15
n can only be 10 here as 15 is not even
When n is odd
n^2  25n  150 = 0
(n^2 + 25n + 150) = 0
(n + 10)(n + 15) = 0
n = 10, n = 15
n can only be 15 here as 10 is not odd
The only consistent value of n is 10 and hence the answer is (B)
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