HARDEST A Level Maths Tricky Question #16 

CHILL BROSKI

its over for us bro

Walking into hall rn, will tell how cooked I get 👦🏻👦🏽👦🏾👦🏿 or 🧑🏽‍🍳

how was it guys

@@hound819how was it

Oh nvm u showed it

it's literally that easy.......... it's difficult to find any "tricks" in it (sorry i already tried to be humble..)

good luck

Great job! I’m now a massive fan of trig triangles for parametrics!

I did the exact same thing.

Yes, this is what I got, I think we both did Pythagoras on the max point, on desmos the point is actually a little below i think, the circle is not rlly correct, I was shattered when I got it wrong tho

same

you are findin the r value that has no solutions, which is anything greater than 10/3 if less than 10/3 you would get 4 solutions, the equation does apply as we are saying that there are no solutions when b2 -4ac

I have the same question. Also why wouldn’t parametric diff be valid. Can that not be inferred from the diagram?

I think is because in the quadratic equation he uses ‘u’ which equals “x^2” so essentially one solution of “u” equals 2 solutions of “x”, also if the circle was smaller there’ll be 8 solutions, and ig that’s when the discriminant will be >0, So I guess 4 intersections is the same as 1 and 8 will be the same as 2, as the discriminant is normally to do with tangents to a curve, and here the circle is symmetrical in all the axis and C1 also looks symmetrical in all axis

Old comment I know but no you wouldn't . The quadratic you are setting equal to zero is not the quadratic that draws the graph in the diagram (there isn't a quadratic that draws a graph in the diagram, it has to be defined parametrically). If you go ahead and look at the graph of the quadratic with r substituted into it, you will see that there is one (repeated) real solution. Another way you can understand it is the following (I'm not sure if this is necesarily why or if I'm just guilty of confirmation bias), if you graph the cartesian y = 3x(1-(x/2)^2)^(1/2), you will get one half of the curve C1. You get the parts of C1 occupying the bottom left and top right quadrants. When graphing the cartesian form of a circle (C2), you will only ever get one half of the circle, otherwise you wouldn't be graphing a function (functions can not have one-to-many mappings). As such, you cannot graph the cartesian form of C1 & C2 such that they have more than one intersection. cartesian C1 will always occupy two diagonally opposed quadrants, whether you graph C1 or -C1. C2 will always occupy either the top two quadrants or the bottom two. Just a note as to how I got that function for y: I did the first part differently, I got to y = 3xcost, then to get cost in terms of x, I did x/2 = sint, (x/2)^2 = (sint)^2, (x/2)^2 = 1 - (cost)^2, (cost)^2 = 1 - (x/2)^2, cost = sqrt(1-(x/2)^2), then plugged it into y = 3xcost. As I square rooted (cost)^2, it's actually y= +/- 3x(1-(x/2)^2)^(1/2). If you just square root the funtion y^2 from part a, then you will end up with a graph for modulus y, destroying this information.

Never mind I answered my question makes sense😂

Thanks!

Mock set 2 paper 1 pure I believe

which year pls? @@YashfeeSultana

It’s a Mock Set

p Goat

do you mean to diff. parametrically then set dy/dx=0? the points of intersection btwn C1 and C2 do not occur at a flat gradient

I took that approach as well. Can’t it be inferred from the diagram that the points if intersection are the maximum points of C1?

I understand that its not stated explicitly, but allot of maths is inferring from diagrams/wording

@@AlevelStudent-c1y it's never just inferring there's always something that tells you explicitly

bro what

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You’re so kind! 😄