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    (Original post by ShatnersBassoon)
    I did this paper a few days ago and although I couldn't find a mark scheme, for question 6 I got exactly the same answers as you with exactly the same reasoning. So we would have got the same score... which is hopefully 15/15. If you did the rest of the paper, do you mind checking to see if your answers were the same as mine? Question 5 was a 'show that', but for questions 1, 2 and 3, I'm not sure what, if anything, I got wrong.
    Spoiler:
    Show

    Question 1:
    A: d
    B: c
    C: b
    D: a
    E: a
    F: b
    G: b
    H: c
    I: a
    J: c

    Question 2:
    (i) x = -1, x = 0
    (ii) |y| < (2√3)/3 or in another form, -2/√3 < y < 2/√3
    (iii) y = (2√3)/3 and x = (-√3)/3

    Question 3:
    (i) A graph, which hits the x axis at y = 0, 1, 2 and has turning points at (1-√3, -2/3√3) and (1+√3, 2/3√3)
    (ii) k = -2/3√3 (or k = (-2√3)/9 if you rationalised)
    (iii) X1 = 1; g(1) = 0.25
    (iv) X2 = 2
    I did all the questions (1-7) - I may have made some mistakes but tried to write some full solutions that may be useful

    MAT 2006 Sample Solutions (might not be correct)

    Edit: amended 1H
    Spoiler:
    Show
    1a: d
    1b: c
    1c: b
    1d: i'm in year twelve - we haven't done this yet
    1e: a
    1f: b
    1g: b
    1h: c
    1i: a
    1j: c

    2(i) When y = 1 we have x^2 + x + 1 = 1. This gives us x^2 + x = 0 \Longrightarrow x(x + 1) = 0 and hence x = 0, x = -1.

    2(ii) The equation can be written as x^2 + xy + (y^2 - 1) = 0. For this we need that b^2 - 4ac &gt; 0 and hence y^2 - 4(y^2 - 1) &gt; 0, and so -3y^2 &gt; -4, dividing by -3 giving us that y^2 &lt; \frac{4}{3}. We get that -\frac{2\sqrt{3}}{3} &lt;y &lt; \frac{2\sqrt{3}}{3}.

    2(iii) Simply taking the top of the inequality, y = \frac{2\sqrt{3}}{3}. When this is so, x^2 + \frac{2\sqrt{3}}{3}x + \frac{3}{9} = 0

. Completing the square and solving we get (x + \frac{\sqrt{3}}{3})^2 - \frac{3}{9} + \frac{3}{9} = 0 and hence x + \frac{\sqrt{3}}{3} = 0 \Longrightarrow x = -\frac{\sqrt{3}}{3}

    2(iv) Calculating the terms, we get:
    x^2 = \frac{1}{3}\cos^2\theta + \frac{2}{\sqrt{3}}\cos\theta\sin \theta + \sin^2\theta
    xy = \frac{1}{3}\cos^2\theta - \sin^2\theta
    y^2 = \frac{1}{3}\cos^2\theta - \frac{2}{\sqrt{3}}\cos\theta\sin \theta + \sin^2\theta

    When we add all of these, we get \cos^2\theta + \sin^2\theta = 1, which of course is valid for all \theta

    3(i) f(x) = x(x-1)(x-2) hence intersections at x = 0, x = 1, x=2.

    f'(x) = 3x^2 - 6x + 2 = 3(x^2 - 2x + \frac{2}{3}). Completing the square we get f(x) = 3[(x - 1)^2 - \frac{1}{3}]. Setting to 0 we get f(x) = 3(x-1)^2 - 1 = 0 \Longrightarrow (x-1)^2 = \frac{1}{3}. Therefore our stationary points occur when x = 1\pm\frac{1}{\sqrt{3}}. Then a sketch of the graph.

    3(ii) f(x) = k \Longrightarrow x^3 - 3x^2 + 2x - k = 0. We need a repeated root. Setting up an equation for this we get (x-a)^2(x-b) = x^3 - 3x^2 + 2x - k. Hence k = -a^2b. Expanding the brackets we get (x^2 - 2ax + a^2)(x-b) = x^3 - 3x^2 + 2x - k. Comparing coefficients we get the simultaneous equations -2a - b = -3; 2ab + a^2 = 2. Eliminating b we get 2a(3-2a) + a^2 = 2 \Longrightarrow -3a^2 + 6a - 2 = 0 \Longrightarrow -(3a^2 - 6a + 2) = 0. Solving this equation which has been done above, we take the positive root as 1+\frac{1}{\sqrt{3}}. b = 3-2a = 1-\frac{2}{\sqrt{3}}. We find our answer k = -a^2b = -\frac{2\sqrt{3}}{9}

    3(iii) Using the sketch will help. We need to find the value of X_1 between 0 and which the area under the curve is the greatest. We find that X_1 = 1. Calculating our value now, we get:
    g(1) = \int_{0}^{1}f(t) \, \mathrm{d}t = [\frac{1}{4}x^4 - x^3 + x^2]^1_0 = \frac{1}{4}

    3(iv) The minimum effectively becomes a maximum and hence it is clear to see that X_2 = 2

    4(i) Taking the derivative of the parabola, we get -\frac{1}{2}x. When x = t, the gradient is -\frac{1}{2}t. Our equation is therefore y + \frac{1}{4}t^2 = -\frac{1}{2}t(x-t).

    4(ii) Into our initial tangent equation, we substitute in t = 2\sqrt{3}. This yields y = -\sqrt{3}x + 3. Substituting this into our circle equation and rearranging, this yields x^2 + (-\sqrt{3}x + 2)^2 = 1 \Longrightarrow 4x^2 - 4\sqrt{3}x + 3 = 0. Taking the discriminant, we get (-4\sqrt{3})^2 - (4 \times 4 \times 3) = 48 - 48 - which shows that it is also a tangent to the circle as there is only one intersection

    4(iii) Finding the x-intercept at B will be the same for A apart from the negative sign, due to the symmetry. We know the y-coordinate is 0 as they are intersecting the x-axis. Rearranging the equation, we get x = \frac{y-3}{-\sqrt{3}}. When y = 0, x = \frac{-3}{-\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}. From our tangent equation, the y-intercept is 3. The length CA using pythagoras is hence \sqrt{3^2 + (\sqrt{3})^2} = \sqrt{12} = 2\sqrt{3}. We have now established the triangle to be equilaterial and hence CAB is 60 degrees.

    Due to the triangle being equilateral, all the regions in the corners are equal. Therefore the area of R = (area of triangle - area of circle) / 3. Calculating this we get:

    R = \frac{(\frac{1}{2}\times2\sqrt{3 }\times 3) - \pi \times 1^2}{3} = \frac{3\sqrt{3} - \pi}{3} = \sqrt{3} - \frac{\pi}{3}

    5. ^a_c|^b_d is how I can format it.

    (i) Via rule 1
    ^0_c|^0_d = ^{0 \times a}_c|^{0 \times b}_d = 0 \times ^a_c|^b_d = 0

    (ii) Via property 2
    ^a_c|^0_0 = ^0_c|^0_a = 0 as above

    Via property 3
    ^0_0|^b_d = ^0_b|^0_d = 0 as above

    Via properties 3 and 2 respectively
    ^a_0|^b_0 = ^a_b|^0_0 = ^0_b|^0_a = 0 as above

    (iii)Via properties 2 and 3 as above
    ^a_{s \times c}|^b_{s \times d} = ^a_b|^{s \times c}_{s \times d} = ^{s \times d}_b|^{s \times c}_a

    Now using property 1
    ^{s \times d}_b|^{s \times c}_a = s \times ^d_b|^c_a

    We finish by using properties 2 and 3 again to yield
    s \times ^d_b|^c_a = s \times ^a_b|^c_d = s \times ^a_c|^b_d

    (iv) Using properties 2 and 3 as in previous parts we get

    ^a_{c+x}|^b_{d+y} = ^{d+y}_b|^{c+x}_a

    We use property four to yield

    ^{d+y}_b|^{c+x}_a = ^d_b|^c_a + ^y_b|^x_a

    We then use properties 2 and 3 on each symbol to yield our answer
    ^d_b|^c_a + ^y_b|^x_a = ^a_c|^b_d + ^a_x|^b_y

    6(i) If the message on the silver box is true, then the message on the gold box is false and hence the prize would be in the gold box. If the message on the silver box is false, then the message on the gold box must be false (if it was true, exactly one would be true) and hence either way the prize is in the gold box.

    6(ii) If the message on the lead box is false, at least two messages would be true, and then the prize would be in both the silver and gold box - this is a contradiction. Hence the message is true. Therefore the messages on both the gold and silver boxes must be false (as the lead message is true) and hence he should choose the lead box.

    6(iii) If the message on the lead box is true, the dagger is in the silver box. He should choose either gold or lead.

    If the message on the lead box is false we have that at most one message is false, and as the lead message is false - both gold and silver messages are true and hence he should choose lead or silver.

    The only consistent option is lead and hence he should choose the lead box.

    7. All numbers assumed to be integers.
    (i) Start with AB. Use rule 1 to get ABB, then rule 1 again to get ABBBB.
    (ii) We always start with AB, and using rule 1 doubles the number of Bs. As AB is produced with no applications of rule 1, we apply this rule k times to achieve \mathrm{AB}^{2^k} where k \geq 0.
    (iii) Start with AB. Apply rule 1 four times to get \mathrm{AB}^8. Use rule 2 to get \mathrm{CAB}^5 - then apply it again to get \mathrm{CCABB}.
    (iv) \mathrm{C}^n\mathrm{AB}^{2^k-3n}, where 2^k \geq 3n \geq 0. Apply rule 1 k times to get \mathrm{AB}^{2^k} as in (ii). Then apply rule 2 n times to get \mathrm{C}^n\mathrm{AB}^{2^k-3n}.
    (v) Any word of the form \mathrm{C}^n\mathrm{AB}^m, where n \geq 0 and m \geq 0. Apply rule 1 k times such that 2^k-3n \geq m \Leftrightarrow k \geq \log_2(m+3n) to get \mathrm{AB}^{2^k}. Apply rule 2 n times to get \mathrm{C}^n\mathrm{AB}^{2^k-3n}. Finally apply rule 3 (2^k-3n)-m times to get \mathrm{C}^n\mathrm{AB}^m.
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    (Original post by KloppOClock)
    ..

    Doesn't it say that a is less than -1 so a^2 must be greater than 1 so therefore if 7b^2 = a^2 then

    b^2 = (a^2) / 7

    as a^2 must be greater than 1 then (a^2)/7 must be greater than 0 as it will still be positive. So b^2 must be greater than 0
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    (Original post by some-student)
    I did all the questions (1-7) - I may have made some mistakes but tried to write some full solutions that may be useful
    Thank you; that's very helpful. It's reassuring to see we got the same answers for everything but 1 (h). I don't know if you're doing A levels but part (d) required C3 [a year 13 module] knowledge so it wouldn't come up on any of the MAT papers today.

    With part (h), I'm guessing this is also something you haven't studied yet. The right hand side of the equation is a geometric series, for which there's a formula when summing to infinity. Your range of sinx and sin^2x is correct, but there are plenty of infinite sums which equal 2, for instance 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .... In this case, the sum  \frac{2}{3} + (\frac{2}{3})^2 + (\frac{2}{3})^3 + ... = 2 is relevant, and sinx = \frac{2}{3} has two solutions between 0 and 2π, so I think the answer is (c).

    For everything else, it looks like we had the same method, with one exception:
    For 3(ii), rather than using your algebraic method (which is ingenious, but I think my method is simpler), I used my sketch in part (i). You know that y = f(x) should intersect the line y = -k at two points, one with a negative x value and another with a positive x value. You should be able to see from your graph that y = -k must be a tangent to the minimum you found at x = 1 + 1/√3. The y co-ordinate will be f(1 + 1/√3) = (1 + 1/√3)(1/√3)(1/√3 -1) = 2/(3√3). Thus -k = 2/(3√3), so [after rationalising] k = (-2√3)/9.

    So if you're in year 12, why have you been doing MAT papers? Are you planning on applying to Oxford next year? Is it for fun? Or maybe a bit of both?
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    (Original post by ShatnersBassoon)
    Thank you; that's very helpful. It's reassuring to see we got the same answers for everything but 1 (h). I don't know if you're doing A levels but part (d) required C3 [a year 13 module] knowledge so it wouldn't come up on any of the MAT papers today.

    With part (h), I'm guessing this is also something you haven't studied yet. The right hand side of the equation is a geometric series, for which there's a formula when summing to infinity. Your range of sinx and sin^2x is correct, but there are plenty of infinite sums which equal 2, for instance 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .... In this case, the sum  \frac{2}{3} + (\frac{2}{3})^2 + (\frac{2}{3})^3 + ... = 2 is relevant, and sinx = \frac{2}{3} has two solutions between 0 and 2π, so I think the answer is (c).

    For everything else, it looks like we had the same method, with one exception:
    For 3(ii), rather than using your algebraic method (which is ingenious, but I think my method is simpler), I used my sketch in part (i). You know that y = f(x) should intersect the line y = -k at two points, one with a negative x value and another with a positive x value. You should be able to see from your graph that y = -k must be a tangent to the minimum you found at x = 1 + 1/√3. The y co-ordinate will be f(1 + 1/√3) = (1 + 1/√3)(1/√3)(1/√3 -1) = 2/(3√3). Thus -k = 2/(3√3), so [after rationalising] k = (-2√3)/9.

    So if you're in year 12, why have you been doing MAT papers? Are you planning on applying to Oxford next year? Is it for fun? Or maybe a bit of both?
    I'm applying next year but I'm doing MAT papers for fun really, but I feel an idiot now because I will have literally none to revise for next year when I apply. However I have told myself to not look at the 2016 paper until I start year 13. We've only started C2 now but I am looking forward to the harder modules.

    Your method for 1H makes sense. I found the formula for an infinite sum in a geometric series and tried out 1H so that I could learn it (as I haven't got there in C2 yet):

    We have the common ratio r=\sin x, first term a = \sin x

    \sum_{k=0}^\infty \sin^kx = \frac{\sin x}{1 - \sin x}

    \frac{\sin x}{1 - \sin x} = 2 \Longrightarrow \sin x = 2 - 2\sin x

    3\sin x = 2 \Longrightarrow \sin x = \frac{2}{3}

    I understand now - thanks

    I did think my method for 3(ii) was a bit overblown but I hadn't done the sketch so I spent ages doing it that way .

    What are you applying for? I think I'll go for CS or Maths & CS. Good luck to you and everyone else!
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    Part ii
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    (Original post by Zacken)
    Integrate both sides of the given identity from -1 to 1, remember that the existing integral in that identity is just a number, you might find it helpful to call it u = \int_{-1}^1 f(t) \, \mathrm{d}t and then integrate both sides like so:

    \int_{-1}^1 6 \, \mathrm{d}x + \int_{-1}^{1} f(x) \, \mathrm{d}x = 2\int_{-1}^1 f(-x) \, \mathrm{d}x + u\int_{-1}^1 3x^2 \, \mathrm{d}x

    which then simplifies down to 12 + u = 2u + 2u where u is the answer you want.
    I feel dumb, but could you explain how
      

\[\int_{1}^{-1}f(t)dt = u = \int_{1}^{-1}f(x)dx\]

    or am I missing something in the simplification?
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    (Original post by theaverage)
    Part ii
     \displaystyle \frac{(x+1)^2}{2} is perfectly valid as well since it's derivative is clearly x + 1. If you expand this you'll notice it differs from what you have only by a constant term, and constant terms don't matter since they disappear when you're taking the definite integral.

    edit: had this page open for a while, by your edit I assume you figured it out
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    Could anyone explain how you would work out 2013 Section 1 H, I and J? I've read through the solutions but I just can't figure out how I'd go about doing them if they came up in a different form next week!
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    (Original post by alexhazmat)
    I feel dumb, but could you explain how
      

\[\int_{1}^{-1}f(t)dt = u = \int_{1}^{-1}f(x)dx\]

    or am I missing something in the simplification?
    They're definite integrals so you don't care what the variable inside is called.
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    (Original post by 13 1 20 8 42)
    They're definite integrals so you don't care what the variable inside is called.
    I'm confused :confused:

    I don't understand how we know that the integral of t w/respect to t is the same as the integral of x w/respect to x? Same limits, but aren't they completely different functions?
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    (Original post by alexhazmat)
    I feel dumb, but could you explain how
      

\[\int_{1}^{-1}f(t)dt = u = \int_{1}^{-1}f(x)dx\]

    or am I missing something in the simplification?
    (Original post by Zacken)
    okay, integrate x dx between -1 and 1, what number do you get?Integrate t dt between -1 and 1, what number do you get?

    The t and x inside the integral don't matter, they're called dummy variables as long as the integral is a fefinite one with limits, then the letter doesn't matter since the integral just represents a number anyway.

    It's a bit (actually, almost exactly) why in summation notation, it doesn't matter what letter you choose. The sum from k=1 to 100 of k is the same as the sum from m=1 to 100 of m. The letters are just placeholders.

    This has tripped up many a student in the past.
    ...remember that if you have a definite integral, that's just a really long way of writing a NUMBER. It's a CONSTANT. No matter what variable you use inside the integral, the entire thing eventually evaluates to some number. This is in contrast to indefinite integrals where it evaluates to a FUNCTION.
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    (Original post by alexhazmat)
    I'm confused :confused:

    I don't understand how we know that the integral of t w/respect to t is the same as the integral of x w/respect to x? Same limits, but aren't they completely different functions?
    No, the function is f in either case..
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    (Original post by alexhazmat)
    I'm confused :confused:

    I don't understand how we know that the integral of t w/respect to t is the same as the integral of x w/respect to x? Same limits, but aren't they completely different functions?
    Say that f(x) = x^3 (just for an example); then f(t) = t^3... t is simply a different variable; the function f is simply performing operations on whatever variable is input. If you were asked to sketch both, you would get the same curve... so integrating between the same limits would yield the same result

    Posted from TSR Mobile
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    (Original post by Zacken)
    ...remember that if you have a definite integral, that's just a really long way of writing a NUMBER. It's a CONSTANT. No matter what variable you use inside the integral, the entire thing eventually evaluates to some number. This is in contrast to indefinite integrals where it evaluates to a FUNCTION.
    (Original post by 13 1 20 8 42)
    No, the function is f in either case..
    (Original post by some-student)
    Say that f(x) = x^3 (just for an example); then f(t) = t^3... t is simply a different variable; the function f is simply performing operations on whatever variable is input. If you were asked to sketch both, you would get the same curve... so integrating between the same limits would yield the same result
    It's worrying that this has tripped me up and the exam is in 6 days lol.

    Thanks guys!
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    (Original post by alexhazmat)
    It's worrying that this has tripped me up and the exam is in 6 days lol.

    Thanks guys!
    It trips everybody up for some reason, probably because you aren't taught the distinction beteeen free/bound/dummy variables at school level in the A-Level sham of mathematics.
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    Good luck to everyone taking mat.
    Multiple choice is probably the hardest part of the test and why marks are low take ur time and it will be decent. not the hardest mathematically but people tend to rush it and it is worth a lot.
    dont be scared of q5 as sometimes it is literally trivial,in other instances it isn't.
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    can someone do 2004 Q1F and tell me what they get?
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    (Original post by 11234)
    i found that the old ones are much easier than the new ones
    the average marks decrease through the year, u should subtract how much from ur scores depends on much the previous average differs from the current average.
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    (Original post by physicsmaths)
    Good luck to everyone taking mat.
    Multiple choice is probably the hardest part of the test and why marks are low take ur time and it will be decent. not the hardest mathematically but people tend to rush it and it is worth a lot.
    dont be scared of q5 as sometimes it is literally trivial,in other instances it isn't.
    how long should i spend on the MCs? I can solve about 3~3.5 questions entirely out of the 4 free response ones. Even though I am not exactly sure how much time I take as I haven't do the free response by it self.
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    (Original post by Mystery.)
    can someone do 2004 Q1F and tell me what they get?
    If you can post the link for 2004 I can solve it for you.
 
 
 
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