This is MAT 2019 in ten minutes or less I'm James Munro, let's go.
1A. The left-hand side of this equation is a cubic and its turning points are at x = plus or minus 10.
The value on the right hand side is bigger than the value at
the turning points so there's exactly one solution, and the answer is B.
1B. If you try some small square and cube numbers you'll find, for example, 1 times
1 is 1 but 4 times 8 is 32, which is neither square nor cube, so the answer is
sometimes a square sometimes a cube and the answer is C.
1C. This sum is a geometric series, which we could do, or we could just note that when x=0 the sum
is 0 and when x=90 degrees the sum diverges so the graph is D.
1D. The parabolas intersect at x=0 and x= minus a. We need to consider the cases
a<0 and a>0 separately. If we do the integral for the
difference between the parabolas, we find that the area is plus or minus a^3/3
so a has to be 3 or -3, and the answer is B.
1E. We're looking for solutions to the form y=mx+c. We could take y=x (that works!),
or we could take y=x+360 degrees or +720 degrees... there are
infinitely many lines that work, and the answer is E.
1F. Write cos squared = 1 - sin squared, and this becomes a cubic for sin(x).
We'd like to know how many roots this cubic has got, so find its turning points, and it turns out
there's only one solution to the cubic. One value of sin(x) means two solutions
for x so the answer is C.
1G. Get rid of all the logarithms and try and solve these equations.
If we raise the middle one to the power of c, we get b on the
left and b to a quadratic in c on the right hand side. Because b isn't 1 for
these logarithms to make sense we must have the quadratic equal to 1.
Solve this for positive c to find that c is 1/2, and the other equations become a=b^2
and a = 2 ^ - b. A quick sketch shows that there's a unique
solution to these equations the answer is A.
1H. Call the sides a, ar, and ar^2.
Pythagoras gives 1+r^2=r^4. Solve this
quadratic for r^2, to find r is the square root of ((1 + root 5) over 2).
Now tan BAC might be r or it might be 1/r depending on
which corner is A. Some algebra gives the answer to be C.
1I. Both x and 2^x are strictly increasing, so x2^x is a strictly increasing function of x.
That means there's no solutions x<y to this equation and the answer is A.
1J. let's imagine we had a straight line through O and one of the corners of the triangle,
and then we rotated it by an angle of theta.
That function's periodic with period 60 degrees (every time the
line goes through another corner of the triangle) and it's symmetric in theta
because rotating in the opposite direction does the same thing.
The minimum therefore occurs halfway in between those maximum values (when it
goes through the corners), when the line is parallel to one of the sides of the
triangle. In that case some geometry shows that the length of PQ is 2/3 and
the answer is D.
Question 2, part 1. Taking the highest power of x on the
Left we find that n is k times (k+1)/2. Part two. x equals one on
the Left gives 2 to the K but x equals 1 on the right gives the sum of the
coefficients the sum of the coefficients is at most n plus 1 times the biggest
coefficient part 3 what's K is bigger than I we get from PK of X to PK plus 1
of X by multiplying by 1 plus X to the K where X to the K is some huge power of X
that doesn't affect the coefficient of x to the I part for substitute in 1 over X
multiply through by X to the power of n on the Left we get back to PK of X and
on the right we get the coefficients in reverse order part 5 there are only M
plus 1 coefficients but we showed in part 2 that the biggest coefficient is
bigger than 2 to the K over m plus 1 in order to get all of the numbers in the
coefficients we need n plus 1 squares to be bigger than 2 to the K but that's not
true for large K so the student is wrong question 3 part 1 the integral is C
cubed over 6 the area of s translate the parabola in the question left until the
roots are 0 and B minus a then the area of s is exactly the integral we've just
done with C equal to B minus a pop to set em X equal to the equation of the
parabola to get a quadratic we want a repeated root so the discriminant must
be 0 rearrange this for M part 3 the area of s is B 2 squared minus 1 all
cubed over 6 set this equal to the expression for the area
of our and simplify we're interested in solutions to this equation so look at
the quartic in brackets when beter is 1 the value is minus 3 but for very large
beater it's clearly positive so somewhere in between there must be a
solution question 4 part 1 if ap is inside the circle then the closest point
is a B itself if ap is outside the circle then the closest point is on the
edge at a B over the square root of a square plus B squared part 2 for example
we could have the disc x square plus y squared strictly less than 1 with no
edge then for points outside the disc there's no closest point part 3 this
time we could use a circle x squared plus y squared equals 1 and think about
the point in the center part 4 use Pythagoras to find an expression for the
distance D now d squared is a quadratic in X with a positive leading coefficient
so it's got a unique minimum and that's a minimum of D part 5 for points in this
set X plus 2/5 is half of Y minus 1/5 and that's the equation of a straight
line and in fact all of the points on that straight line are in s part 6
suppose there are two closest points inside the set s think about the line
segment joining those two points along the line the distance squared is a
quadratic with equal values of the endpoints of that line segment so
halfway between it takes a new minimum by the property were given that it's
also in the set s so contradiction and even closer point exists question 5 Part
one each subset has two or more elements so you need at least two K elements
overall to have a successful partition part 2 if N equals 1 and this is 0 by
the previous part if n is bigger than 1 then there's exactly one partition for
all the elements together in one subset part 3 in the first case we've got n
choices for which single element we're going to pair with n plus 1 and in the
second case we've got a choices of which subsets as n plus 1 into so we can put
that together to give F n plus 1 K is equal to n times F n minus 1 K minus 1
plus K times F MK part 4 using this result we can build up from small values
of N in K to find that F 7/3 is 105 part 5 we've got two n items and we'd like to
split them into n pairs there are two n factor
real ways to write down the elements in our set and then when we pair them up
the order inside the past doesn't matter and the order of the past doesn't matter
final answer to n factorial over n factorial 2 to the N questions takes
part one following the rules we find the next flexor decimal numbers are T 1 100
101 101 more 1 1 0 1 see you on 200 and then 201 party think about when we carry
over to a new digit we see that the place value of the kaif digit is K
factorial so multiply each digit by the corresponding factorial and sum them up
to find the value in decimal part 3 to convert the number 2 flex decimal we'll
need to find the biggest factorial that's smaller than the target number
the quotient gives us the first digit of the Flex decimal number and then we
continue the calculation down through smaller factorials for example 255 turns
out to be 202 won 104 we're gonna add the digits together starting from the
right and if we find that the digit in the caithe place is too big we can
subtract k plus 1 and carry 1 over to the next digit up part 5 only the last
two digits matter because all the other factorials are multiples of 3 if the
last two digits are 0 0 or 1 1 then the whole number is a multiple of 3 part 6
the list of arrangements is sorted into 6 blocks by the first letter of the
arrangement our flex decimal number starts with 3 so we're somewhere in the
fourth block the arrangements that start with D inside that block we've got
arrangements of the letters ABC e f and we can look at the next digit of the
Flex decimal number to find out how far through the block we are if we continue
this process through the digits at the flash decimal number we find that we're
at d FB AEC question 7 part 1 the box has only got n digits and there's only
finitely many different things it can display after a while then it has to
repeat itself and come back to something we've seen before
part 2 X I plus s is what we see s bottom presses after X I but that's also
what we'll see s button presses after X J part 3 I is bigger than M so we can
write I as n plus s some s then X I plus K P means X
Plus KP + s now we're going to use the previous part we know that X M is equal
to M plus K P so X M plus s is the same thing as X M plus K P plus s part for
first note that if J minus I is a multiple of P then by part 3 X J is the
same thing as X I now suppose instead that J minus I is are more than a
multiple of P with R less than P so there's a point our presses before we
got to XJ with the same values XJ but P is supposed to be the smallest number of
bottom presses four values to repeat so R has to be 0 and that means that J
minus I is a multiple of P part 5u has to be at least M because we've got into
the loop and R for a further u presses we've done a whole number of loops so
you must be a multiple of P putting it together that means that U is the
smallest multiple of P that is at least M part 6 you could reset the boxes and
give one of them a you press head start now press the buttons together until the
displays agree the harbor's after exactly M presses part seven reset the
boxes and press each button M times they're now ready to loop leave one of
them alone and press the other button until the displays agree that happens
after P presses that was Matt 2019 in ten minutes or less for more Matt's go
to www.hsn.com