This is MAT 2020 in 10 minutes or less, I'm James Munro; let's go!
1A The centre is at (3,4) and there's a corner at (1,5)
which is a displacement of (-2,1) as a vector.
If we rotate that vector by 90 degrees then we get the vector
(-1,-2) and that displacement away from the centre at (3,4)
gives us the point (2,2) which is another corner of the square and the answer is (d).
1B If we multiply out these brackets we get {e to the 2x} minus {x squared},
and we can recognize that as the derivative of one half {e to the 2x} minus {x cubed /3}.
If we put the limits in we get (3e^2-5)/6 which is option (d).
1C we have 1-4 which is -3 and then 9-16 which is -7,
and this pattern continues; it gives us an arithmetic progression
with first term -3, common difference -4, and there are 50 terms in this sequence.
Doing the sum of that arithmetic sequence gives us the final answer -5050 which is option (e).
1D if I use cos squared = one minus sine squared,
then this becomes a quadratic in sin(x). If I complete the square then I can find
the maximum value when sin(x) is 1 3 is 13 over 3 which is option (b).
1E The gradient of x squared at x=a is 2a, and if I send that through the point (a,a^2) then
I get the equation of the tangent as y=2ax-a^2 which crosses the x-axis at a/2. The area in the
question is the integral of x^2 minus the area of the triangle made by the x-axis, the perpendicular
to the x-axis from that point, and the tangent line. So I need to integrate x^2 between 0 and a,
and then subtract the area of that triangle. I get a^3 over 3 for the area under the parabola and a^3
over 4 for the area of the triangle, to give me a final answer of a^3 over 12, which is option (c).
1F If I factorize everything inside the brackets I get 2^8 times 3^4 times 5^2 times 7. If I take
out the log base 10 of 7 then I can spot that log base 10 of 2 times 3 is log base 10 of 6,
and pull out 4 times log base 10 of 6, and then spot that log base 10 of 2 times 5 is 1, and
pull out 2 times log base 10 of 10. That leaves me with 2 log 10 of 2 and so the answer is (c).
1G The turning points are at x equals 1 and x equals 3.
So if I take the derivative of this cubic then it should be zero when x is one or x equals three
plugging in one and three to the derivative of this cube it gives me two equations for a and b
and if I solve these simultaneous equations I find that a is minus six and b is nine now
sending this curve through the point one two gives me the value of c that's minus two. And finally
checking the value at x equals three gives me the value of d. I find d is -2 so the answer is (b).
1H If we sketch the derivative of each of these functions then we find that
the derivative of graph a looks like graph e, and the derivative of graph
b looks like graph c, so d is the odd one out and it must be a plot of h(x).
1I This is the sum of the terms for a geometric progression and the sum to infinity is one over
(tan x minus one). If we set that equal to tan x and solve the quadratic for tan x we find
two possible values of tan x. But in fact we need one over tan x to be between -1 and 1
so that the geometric series converges! If we do that then we get
one remaining value of tan x so the answer is (b).
1J When r is very small, A is zero and B is four minus pi r^2. That rules out option (c). On the
other hand when r is very big, A is pi r^2 minus four and B is zero and that rules out option (a).
If we also notice that this sum can never be zero because we can never have both of these
areas being zero at the same time, that rules out options (b) and (e) as well, so the answer is (d).
Question 2, part one. 1100100.
Part two. g of f of g of f of one is 100.
Part three. In binary f adds 00 to the end of a number and g
adds 1 to the end of the number. 200 ends in 000 in binary which we can't make.
Part four. Write the number in binary. It must be made of
1s and 00s. For every 1 write down f and for every 00 write down g.
Part five. Consider u_(k+2) which is elements between 2^(k+2) and 2^(k+3).
Those have either come from applying f to an element between 2^(k+1) and 2^(k+2) or
by applying g to an element between 2^k and 2^(k+1). If we add together the number of
elements in those ranges we can find the number of elements u_(k+2) as the sum of u_(k+1) plus u_k.
Part six. Notice that s_k is u_0 plus u_1 plus (dot dot dot) plus u_k.
If we sum the previous equation over different values of k and use the fact
that u_0 equals 1 and u_1 equals 1 then we can rearrange for the equation here.
Question 3, part 1. Set y equals zero and we get a is -1 and b is 1.
Part 2. If we complete the square then we get one solution for y
precisely when y is 1/2 so delta is 1/2.
Part 3. Where the curve exists we have y is one half plus or minus the square
root of some function of x. That gives us symmetry about the line y equals one half.
Part 4. These points have y equals one half so x^3-x+1/4 is zero.
Part 5. Suppose this cubic factorized as (x minus alpha)(x minus beta)(x minus gamma).
Then if we multiply that out we'd get an x squared coefficient of minus
(alpha plus beta plus gamma) but our cubic has no x squared coefficients, so that sum must be zero.
Part 6. We know the centre of the circle is at ((alpha plus beta)/2 , 1/2). If we write down
the equation for the circle and substitute for (y-1/2)^2 we get a cubic equation for x.
Now this cubic has an x squared coefficient of one, so the sum of its roots must be minus one.
We already know two of the roots - they're alpha and beta - so the other root must be
minus one minus alpha minus beta which is equal to gamma minus one.
OK, pause the clock. This question has a typo in it and I'm really sorry about that. The definition
of an odd function has too many minus signs in it, and if that misled you and meant that you
couldn't do this question then we're going to try and mark very carefully and fairly,
so that you can still get credit for the work that you wrote down. On with the question.
Question 4, 1a. An even function has reflectional symmetry in the y-axis
and an odd function has rotational symmetry of order two about the origin.
Part 1b. Think about a tangent line. If you reflect that line
in the y-axis then you'll get a line with the opposite gradient,
but if you rotate it by 180 degrees you'll get a line with the same gradient.
Part 2a. The area A(minus theta) is congruent to the unshaded triangle in the diagram, so adding
them together we get the isosceles right angle triangle with short side one and area one half.
Part 2b. Write that area as the sum of two triangles,
add their areas together to get one plus tan theta, all over four.
Part c. My sketch looks like this.
Part d. This has rotational symmetry of order two about the point (0,1/4).
Part e. This symmetry means that the derivative is an even function
so the second derivative is an odd function. By the first part of the question,
putting zero into an odd function always gives you zero.
Question 5, part 1. If the 15 sunny days come first then Miriam eats the most possible sweets.
A quick arithmetic progression sum gives 345 sweets. On the other hand if the 15 sunny days
come as the last 15 days then she eats the fewest sweets possible and that's 120 sweets
Part 2. We have to sum a couple more
arithmetic progressions and we get the same numbers as before.
Part 3. The effect of this swap is that Miriam eats an extra sweet because there's a sunny day
earlier but Adam also eats an extra sweet because the rainy day comes one later.
Part four. Yes. Start with the sunny days at the end and use the swaps to move them back in time.
Adam and Miriam's totals start the same and increase by one for each
swap so the totals stay the same as each other.
Question 6, part 1. There's only one tennis ball and k stockings so that's k
choices of which stocking ball Santa puts the tennis ball in.
Part 2. There are n tennis balls and only one stocking
so Santa must put all the balls into that stocking
Part 3. This splits the possibilities for g(n,k) into two cases. In the first case Santa has to
distribute n-1 balls between k children whereas in the second case Santa distributes n balls among k
minus one children giving nothing to the first child. These are all the possibilities for g(n,k)
so we should add together these two expressions to get g(n,k).
Part 4. Use the previous expression to express g(7,5) in terms of g of smaller
numbers. Apply the facts that g(1,k) is k and g(n,1) is 1 and we find that g(7,5) is 330.
Part 5. Let's make Santa give one tennis ball
to each child. Now Santa's got two tennis balls left and five stockings.
Any possibility from g(2,5) is a possibility for splitting up these remaining tennis balls
in a way that satisfies the requirements of part 5, so the answer is g(2,5) is 15.
Question 7 part 1. Lots of possibilities! We could join the first row to itself and get CCACCBCCCC
or we could join the first row to the second row and get CCC A CC B {five C's}.
Part two. You can't get to AB because you've always got at least one C
and you can't get to things like CBCAC because B is always on the right of A.
Part three. Yes. For example join the first row to itself five times
and the second row to itself four times then join those together.
Part four. No because the first block of C's can only be at most twice as long
as the middle block of C's. The original rows satisfy this and any join satisfies this too.
Part five. Write down the basic rows then join the second one to itself. Do it again
then do it again. Join the two basic rows together then join that to the third row
join that to the fourth row and then join that to the fifth row, for the goal row.
Part six. If you're not joining a basic row each time then you can do a longer sequence by joining
the things that your non-basic row is made of. If you join a basic row each time then the number of
C's in the middle goes up by one each time so you need 15 joins in the slowest possible sequence.
That was MAT 2020 in 10 minutes or less. There's more information on that
on www.maths.ox.ac.uk/r/mat. Thanks for watching.