Oxford University Mathematician TORTURES Physicist with MAT Entrance Exam

Join Tom Rocks Maths Oxford University Mathematician and myself attempt the MAT Entrance Exam for Mathematics at the University of Oxford. This exam is great help for students preparing for mathematics exams such as the MAT, PAT, ESAT, STEP. It can also be fun for Physics and Maths Olympiad Students, JEE Advanced Students and the questions can be used as an extension and challenge for A Level Maths and Physics students.
Please note that these are not official solutions, I have attached the mark scheme from Oxford, check it out.

[Viki]

19:35 my heart dropped seeing you divide by x on both sides hahahaha

[tysoninnes]

Good to see everyone forgets maths they haven't used in a while

[jffrysith]

12:06 It's actually still easy without the simple cases.
Consider a prime p, consider p^2 and p^3.
It is clear that p^2 * p^3 = p^5. As p is prime, this is neither a square nor a cube.
Also, consider again p^3 and p^2. (p^3)^2 * (p^2)^3 = p^12 which is both (p^4)^3 and (p^6)^2, so it's both a square and a cube.
Therefore it can be both a square and a cube, or it can be neither a square nor a cube. It's trivial it can be either or as well.

[amritlohia]

You keep saying "inflection point" for what us mathematicians would call a critical point, turning point, or stationary point. We define an inflection point as a point where the graph changes from concave to convex or vice versa.

[justblitz]

Hey I just wanted to say thank you so much for all the help with the physics revision tips and questions you posted for the May/June batch. I got my results for AS and ended up getting a very high A! Cheers mate and thank you once again!

[danerou]

"To cook or not to cook" - issac newton

[albertrichard]

You certainly solved the first question a lot more elegantly than I did. I had a very convoluted method. First, I factored the cubic as (x - a)(x^2 + bx + c) = 0. Since it is a cubic, and complex solutions come in pairs, this either has only one solution or three solutions, or two solutions with repeated roots. Expanding the cubic and comparing terms to the original one gives a = b, ac = 3000, and 300 = a^2 - c.

Now I want to figure out whether the quadratic x^2 + bx + c has two real solutions - if it does, I've got three roots, and if it doesn't, I don't. The discriminant is b^2 - 4c = a^2 - 4c = 300 - 3c. I want to figure out the sign of the discriminant, which means I need the sign of 100 - c. If c > 100, then the discriminant is negative and I have only one root. Now, c > 100 implies ac > 100a (a must be positive if c > 100 since ac = 3000), which implies 3000 > 100a, and thus a < 30. Hence, if there is a root between 0 and 30, it must be unique. Plugging in x = 0, I get -3000, and plugging in x = 30 will give me something positive since 30*30*30 > 30*10 + 30*100 = 30*110. Therefore there is a sign change, and by the intermediate value theorem, I conclude that a solution exists for x < 30, thereby implying that only one root exists.

[venkatnarayanan]

C is simply tan^2 (x) - its thus easy to choose by inspection.

[ericerpelding]

For question F the problem could be written as sin(x) tan^2(x) = -1, and from the graphs of the sin and tan functions one can see that there two solutions over the interval 0 to 360 degrees.

[SirCumference]

For the quadratic in G it's quicker to multiply by 2 to give 2c^2+3c-2 = 0 and then factorise to give (2c-1)(c+2)=0 and then it's clear that c=1/2 is the only solution.

[blizzyxx]

Just wanna say thanks, got an A* in physics this year thanks to you.

[florianbuerzle]

I'm also a physicist (at least by training... now I'm a maths teacher 😂), and I've never heard a physicist use the term "inflection point" to mean a stationary point, nor have I ever read about it in the physics literature. An inflection point is a point at which the curvature changes sign... also in physics 😇 But of course I don't know all the physicists or all the physics literature 😅 I have a suspicion that he got something mixed up...

[ZantierTasa]

41:48 "There are 2 possible sin solutions for every value." What about sin(x) less than or equal to -1?? Then there are 1 or 0 solutions. So to solve this completely, you should check f(-1).

[DishantYadav-zndp]

I think that u should do jee advance maths section

It is hard

[solipse.]

40:48 I think I made the equation sin^3 x = - cos^2 x and draw each of those graph since its really easy to sketch if there are only one trig function to some power. Turn out there is only 2 solution from the sketches!

[miikemike]

Q2: isn't the general case quite trivial? my first thought was prime factors. given a^2*b^3, if factors of a are a subset of factors of b, always a cube, otherway around always a square? and both, if its the same set? or am I thinking too simplistically? 😳

[gcewing]

"Possibly not an infinite number of sines added together..." There's a Mr. J. Fourier who would like a word with you.

[spanishseeker]

thank you sir I got an A in physics because of you

[jabess.]

(Question)
Here at 1:11:00, could you think about it as:
- If amax = 2k/(N+1), while 2k = a0 + a1 + ... + aN, there are 2 possible cases
All terms are equal, then amax = 2k/(N+1), because It's the arithmetic mean of these factors
OR
There is a term that is greater than every other term and the arithmetic mean must be lesser than this mean
So, in both cases, amax is equal to the mean or greater than the mean?

[nickgriffiths]

Great video- F - would you not also need to show the intercept is -1<x<1 otherwise arcsin(x) would give no solutions (although it's MCQ so no actual need to show it for correct answer)?