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# MAT Prep Thread - 2nd November 2016 watch

1. (Original post by KloppOClock)
sin(π-x)=sin(x)?
I didn't know that thanks!
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.
2. (Original post by some-student)
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.
you got any other useful trig trivia?
3. 2015 MAT q4 wtf??
4. (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?
The syllabus says "Co-ordinate geometry and vectors in the plane". Dot product isn't in C1 or C2 so I doubt it
5. (Original post by KloppOClock)
you got any other useful trig trivia?
You probably know most or all of this but:

sin(pi/2 - x) = cos x

6. (Original post by 11234)
2015 MAT q4 wtf??
Ways to approach this?? I'm really hoping no geometry ones come up. Really would love a number theory one tho
7. Help with 2007 paper? It is question 1: F, I and J. Thank you!
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8. (Original post by molly221)
Help with 2007 paper? It is question 1: F, I and J. Thank you!
for f set y=2^x and solve as quadratic

for i, assume log10b=0
for j, set n=1 and add up to arithmetic series together.
9. (Original post by molly221)
Help with 2007 paper? It is question 1: F, I and J. Thank you!
For 1J, you can substitute each value of n with 1 as the equation holds for n is greater or equal to 1. After doing that you find you get an arithmetic sequence 2+3+4+5...+101. Using the formula gets the answer of (d).

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/
10. (Original post by molly221)
Help with 2007 paper? It is question 1: F, I and J. Thank you!
F: Using laws of indices we get:

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 (log10 b)^2, which is 0, when b = 1. Therefore 4(log10 a)^2 = 1 => log10 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).
11. 2008 paper help please? It is question 1: C and D. Thanks
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12. (Original post by molly221)
2008 paper help please? It is question 1: C and D. Thanks
For question 1D put x=1 into the equation and you end up with 1+3+5+7+...99
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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13. (Original post by molly221)
2008 paper help please? It is question 1: C and D. Thanks
C: Eliminate x and y respectively each time to get
(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)
14. 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.
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15. Here's a solution for J.

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16. (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
(iii) took quite a long time. I had to look at the solutions to do (iv) - I still don't get why the triangles and sectors subtract to get the middle par although the areas weren't too bad to calculate once I was given the sectors and triangles - but even getting 11+/15 in this question on a paper where the average successful score was 62 will really help boost your mark.

I used http://www.webmath.com/gcircle.html to draw my circles.

Hints
Spoiler:
Show
(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.
Solutions
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).

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 x-axis 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 right-angled 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.
17. How much time is everyone dedicating to MAT over half-term? Not sure how I should be balancing MAT prep with A Level stuff.
18. I was feeling sort of confident for this MAT exam... then I did the 2015 paper.
19. Help pls with q5 of 2002 paper

http://www.mathshelper.co.uk/Oxford%...est%202002.pdf
20. (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.
D: Assume n is odd. Infront of the term involving n, which evaluates to n, we have n - 1 terms, which can be seen as pairs which always have a value of -1.

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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