Indeed a positive gradient here is necessary but not sufficient for a function to tend to positive infinity. Proving that ln(x) tends to infinity as x tends to infinity might take a while, and in this short example of interview questions the aim was to explore the question “For what values of m there are zero, one or two intersections between ln(x) and mx?”. It is often the case that an interviewer might choose not to pursue a line of inquiry and move on with the question. In this video it has happened - in different ways - with both questions.
@@bentwonie8287 They technically could, but statistics and mechanics are smaller focuses in the undergraduate maths degree. The aim of the interview is to test your overall thinking, rather than how well you remember niche topics. If you were taking the masters course in maths (MASt Mathematics), then they'd be more focused on asking questions regarding specific areas (pure, statistics, applied or theoretical physics), depending on your degree focus.
@@Josama0214In case you still haven’t understood maths yet, most A level syllabuses cover three parts within maths; Pure, Mech and Stats. So Mechanics and Stats are part of Maths, and can be tested in an interview, like it was in my case
Yeah you can do y=-1/x then y'=1/x^2 and 1/x^2 is always positive when talking about real numbers. So the function's derivative is always positive when x is positive but the function approaches 0. lim(x->inf)(-1/x) = 0.
Indeed, the interviewer's gradient argument will only work if the limit of the definite integral of the derivative (assuming it's non-negative) as the upper threshold tends to infinity is infinite... which is just equivalent to saying that the limit of the function (assumed differentiable) is infinite.
An argument based on bounds and/or comparing functions would indeed have been good, but it could have taken a while. In this short example of interview questions the aim was to explore the question “For what values of m there are zero, one or two intersections between ln(x) and mx?”. It is often the case that an interviewer might choose not to pursue a line of inquiry and move on with the question. In this video this has happened - in different ways - with both questions.
I’m about 75% sure I spoke to her at an open day and she’s already a student at Cambridge maths! She’s just “posing” as a prospective student for the sale of this video
Number of times y = ax intersects y = ln(x) depends on a as follows: Case 1: 'a' is negative. In such a case, y = ax will intersect ln(x) exactly once between x = 0 and x = 1, where the value ln(x) is equal to ax. Case 2: 'a' is zero. In such a case, y = ax is constant function y = 0 and intersects y = ln(x) at x = 1. Hence, it intersects y = ln(x) only once. Case 3: 'a' is positive. In such a case, the curve y = ax may never intersect y = ln(x) given 'a' is larger than a specific value, and may intersect y = ln(x) twice if 'a' is smaller than that specific value. That specific value of 'a' occurs when y = ax touches y = ln(x) at exactly one point. For the touching case, d/dx(ln(x)) = d/dx(ax) at x = x0 => 1/x0 = a; Since (x0,y0) lies on both y = ax and y = ln(x) y0 = ln(x0) = ax0 = 1 => x0 = e and a = 1/e Hence, if a is greater than 1/e the two curves never intersect; and if a less than 1/e the two curves intersect twice. And if a = 1/e, the two curves intersect only once. Hence, the number of point of intersection is as follows: a 1 0 < a < 1/e => 2 a = 1/e => 1 a > 1/e => 0
If I were marking a student who said that a function tends to infinity because its derivative is always positive, as Dr Spivack suggested, that would be worth 0 marks (in light of all such functions that don't tend to infinity).
Indeed a positive gradient is not sufficient for a function to tend to positive infinity. Proving that ln(x) tends to infinity as x tends to infinity might take a while. In this case, the interviewer chose to make the candidate feel at ease and move on, since in this short example the aim was to explore the question “For what values of m there are zero, one or two intersections between ln(x) and mx?”.
@@samuelmoss2480 Yeah that seems like a much better way, might also be worth mentioning that e^x is a continuous strictly increasing function so if e^y = infinity, then y = infinity
positive gradient depicts increment rather than decrement. for Amy's case since she had the log graph, lets say you start it with x= 1, then you keep on plotting the graph with considerations of gradient. so since already the function had the value of 0 at x= 1, she got that incresing nature. however, in your case, your function even from x= 1 starts from -1 but it does increases to 0. thats the main answer. here in your case, your graph never crosses the x axis so you reach only 0. in her case, it crossed and thats the main point. i believe Euler's approximation method might put some insight here as well.
@@robmartin4657 what I mean is try to look for the increasing or decreasing nature with the help of the gradient. I know gradient alone is not enough that's why I suggested an idea from Euler's approximation.
I'm so thankful I never have to do thia kind of maths ever in daily life after being gaslit in secondary school being told 'you won't always be able to use a calculator and you'll need to use proper maths'
Taking two times logarithm of base 2, we are comparing the quantities: 2^n and n*log_2(100) + log_2(log_2(100)). It is obvious that the first grows much faster.
How can you argue that just because the derivative is > 0 the function tends to infinity? Isn't the derivative of the sigmoid always positive as well and it just tends to 1?
Indeed a positive gradient is not sufficient for a function to tend to positive infinity. Proving that ln(x) tends to infinity as x tends to infinity might take a while. In this case, the interviewer chose to make the candidate feel at ease and move on, since in this short example the aim was to explore the question “For what values of m there are zero, one or two intersections between ln(x) and mx?”.
True, but I guess it would be possible to prove it using the fact that the range of ln(x) is the real numbers AND that the function is increasing (which I guess would involve a couple other theorems too), if you were for some reason really set on not proving it the normal way
@ianbennett2443 Since the function is increasing for all X in the domain it’s synonymous with a positive derivative for all x. The range is based off the domain of the inverse function e^x which is all real numbers so it has no bound and would likely be the fastest way to do it.
How is possible to evaluate if a person is capable of Cambridge studies from just one question of this kind? In my opinion if a student replies correctly to this question then I could not be sure to accept him/her. Even if a student fails to answer this question it would be a bad indication but I would not reject at one, I would need to ask more questions.
There are usually two interviews. Additionally, the main filter for admissions are the STEP entrance exams, which is after the interviews, and not the interviews themselves Of course, some completely capable people do get rejected after their interview (due to lack of information), but thats just how it goes
@jackhnatejko8830 I don't know about Cambridge but certainly at Oxford there's also the fact that the general setup of the interview is very similar to much of the day to day teaching. Admittedly in a tutorial you'll have done the work beforehand so it's more about going through a problem than tackling it new, but the sort of skills and approaches needed are quite similar
well... at the last few questions, I paused for way too long, just because I hadn't done this question before, I wanted to find out the solutions to the equation and even though I knew pretty early on I wouldn't be successful because (e^(ax))=x is not made to be solved on simple ways, I kept sticking with it for too long, until eventually doing the ye^(y)=ax. I'm way too stubborn. I found the first question much easier than the other ones, as just using the properties of e and ln-functions will carry you to the solution with barely any effort.
ปีที่แล้ว +4
Amy is the person I would like to sit next to during a math exam 🧐
could i ask; would this be considered a "successful interview" would this applicant be offered a place just off this interview performance? not including anything else?
I'm teaching maths at a university and I think this is definitely a successful interview. Keep in mind that the girl just finished high school. She applied her knowledge quickly and goal-oriented, was fast to correct her mistakes when they were pointed out and was able to come up with her own solutions to the problems posed, which is probably the most important part of mathematics.
Does anyone here know what the mean radio of the earth is? I’m lead to believe it’s 3959 miles. Can anyone here confirm and agree this axiom so we can discuss further this. Thanks
wouldn't 16^n be less than 100^100^n? If so, then yes, it seems so obvious, why the need for all the log stuff? I feel same as you...am I missing something?
@@lastfreegeneration984 thanks, now it makes sense. Put them equal to each other for the sake of argument, with the 2^ terms on the LHS, then take logs to the base 2 of both sides, the log of a number is the power to which the base has to be raised to give the number, so that gets rid of the first 2 of the outer bracket on the LHS, on the RHS you’ve got logbase2 of 100, raised to the power 100, itself raised to n. Bring that exponent term to the front of the RHS x logb2 of 100. Take log to the base 2 of both sides again, that drops another 2 off, now on the LHS you’ve only got 2 to the power n. On the RHS you’ve got logb2 of 100^n + logb2(logb2 of 100). With the first term on RHS you can bring the n exponent to the front multiplying logb2 of 100, now as he prompted you raise 2 to a power less than 7 to give 100. The log 100 terms are 6.64 to 3 significant figures, but infinity as posed it gets there first wherever there is 😂. So it’s bigger. Thanks again for the bracket tip, I’m familiar with (x^s)^t = x^(st) for the laws of indices.
@@dehnsurgeon Do you know the questions asked for teaching positions then? A position at google requires you to be a decent developer but the interview questions are simple algorithms knowledge.
The questions you get asked to be a teaching student in nz are: will you show a commitment to the treaty of waitangi and lots of dei questions and a very basic math test that a kid could do and that’s it!
The trouble with this selection process is that it eliminates all potentially exceptional students who have not yet learned the skills of problem solving. Surely that's your job, Cambridge, to teach such skills! Or are you just upping your university rankings by taking in the already-skilled problem solvers?
They're about right for a BSc I think. The interview is as much a test of how well a student can explain their thought process and respond to menotring than a test of pure mathematical skill. Don't forget that there is also an entrance exam which is geared more towards asking tougher questions.
Once again, it's not about the questions, your academic ability is already proven pre interview when you complete your admissions test. The interview is to see how teachable you are.
I hope this doesn't come across as too critical or offensive, but is attitude/composure/professionalism considered in these interviews? I found Amy's frustrated "eugh" early in the interview came across as very unprofessional, and seemed like a disregard for the importance of the interview (and I'm sure she does understand the importance of it). This also obviously has no bearing on her mathematical ability.
First off, this is a parody video. Second, it definitely would be considered. Amy’s conduct in this video wasn’t just bad, it was borderline inappropriate, especially with her mathematical ability
@@sb_dunkokay well I just think it’s fine like it’s a little awkward but I think the interviews really don’t care much, at least if it was me interviewing I would just be looking for her mathematical ability. I think something like that is pretty minor to me.
Take base 2 logarithm once, we get 2^2^n vs C1*100^n Take logarithm one more time: 2^n vs C2 + C3*n Can stop right here - the exponent on the left always beats the linear function on the right for large enough n. How could this solution can take more than two minutes at the most? “Logarithms with base > 1 are monotonously increasing functions” - that’s all she should’ve answered to justify how proof for logarithms proves the original statement. Which she did not. As to the generalized second question, the “strategy” should’ve been stated as follows: for arbitrary functions f(x) and g(x) finding the point x0 where they both touch each other is a matter of solving two equations: f(x0) = g(x0) f’(x0) = g’(x0) In high school in Russia I finished that would be considered an extremely weak performance. Not Cambridge quality … or perhaps Cambridge quality is not what I imagined?
@@sanjaymajhi4428 normal college se toh lakh Guna accha a Hai.IIT na niklne k baad Mera friend v yahi bol raha tha,shayad mai v yahi bolunga agar nahi nikla toh 🤣
In the first question here, they are actually wrong. This is a trick question as N goes to infinity which is a concept that we cannot conceptually grasp, meaning they are equally big (or sized, as their size is actually undefinable)
who are the mathematics and why are they mocking the interview
lol...
lmao
😅😅😅
Looooool fk off. I dont do maths ot anything else cos im stupid dont know how I got here but glad I did because your comment cracked me up mate
hahaha I feel like only someone in the field of mathematics or physics could make such a joke
nightmare blunt rotation
Whys it always nightmare blunt rotation but not dream blunt rotation
@@sigma9025respectfully, just look at them
Loool
I love this channel reminds me of Olympiads and those beautiful challenging math problems
wrotoeshaw has come a long way
he was going on a hike
Indeed a positive gradient here is necessary but not sufficient for a function to tend to positive infinity. Proving that ln(x) tends to infinity as x tends to infinity might take a while, and in this short example of interview questions the aim was to explore the question “For what values of m there are zero, one or two intersections between ln(x) and mx?”. It is often the case that an interviewer might choose not to pursue a line of inquiry and move on with the question. In this video it has happened - in different ways - with both questions.
This was a really helpful video, thank you. Please could you do an example interview with a stats and mechanics question?
It's a Math Interview they aren't going to ask Statistics or Mechanics Questions💀
@@Josama0214 Just done a stats course at uni, it’s heavily maths based
@@bentwonie8287 They technically could, but statistics and mechanics are smaller focuses in the undergraduate maths degree. The aim of the interview is to test your overall thinking, rather than how well you remember niche topics. If you were taking the masters course in maths (MASt Mathematics), then they'd be more focused on asking questions regarding specific areas (pure, statistics, applied or theoretical physics), depending on your degree focus.
@@henru364 Yea but that doesn't really mean anything.
@@Josama0214In case you still haven’t understood maths yet, most A level syllabuses cover three parts within maths; Pure, Mech and Stats. So Mechanics and Stats are part of Maths, and can be tested in an interview, like it was in my case
I think Amy's argument for why log(x) -> inf was better than the interviewers' gradient argument, which doesn't hold in general
yea the derivative of 1/x is always negative, does not mean 1/x diverges to negative infinity as x approaches infinity so a bit of a fallacy there
Whatever her argument, he would have possibly argued the opposite.
Yeah you can do y=-1/x then y'=1/x^2 and 1/x^2 is always positive when talking about real numbers. So the function's derivative is always positive when x is positive but the function approaches 0. lim(x->inf)(-1/x) = 0.
Indeed, the interviewer's gradient argument will only work if the limit of the definite integral of the derivative (assuming it's non-negative) as the upper threshold tends to infinity is infinite... which is just equivalent to saying that the limit of the function (assumed differentiable) is infinite.
An argument based on bounds and/or comparing functions would indeed have been good, but it could have taken a while. In this short example of interview questions the aim was to explore the question “For what values of m there are zero, one or two intersections between ln(x) and mx?”. It is often the case that an interviewer might choose not to pursue a line of inquiry and move on with the question. In this video this has happened - in different ways - with both questions.
The prospective student is really strong, in my opinion. She will have a fun time studying maths :)
I’m about 75% sure I spoke to her at an open day and she’s already a student at Cambridge maths! She’s just “posing” as a prospective student for the sale of this video
Number of times y = ax intersects y = ln(x) depends on a as follows:
Case 1: 'a' is negative. In such a case, y = ax will intersect ln(x) exactly once between x = 0 and x = 1, where the value ln(x) is equal to ax.
Case 2: 'a' is zero. In such a case, y = ax is constant function y = 0 and intersects y = ln(x) at x = 1. Hence, it intersects y = ln(x) only once.
Case 3: 'a' is positive. In such a case, the curve y = ax may never intersect y = ln(x) given 'a' is larger than a specific value, and may intersect y = ln(x) twice if 'a' is smaller than that specific value. That specific value of 'a' occurs when y = ax touches y = ln(x) at exactly one point.
For the touching case,
d/dx(ln(x)) = d/dx(ax) at x = x0 => 1/x0 = a; Since (x0,y0) lies on both y = ax and y = ln(x) y0 = ln(x0) = ax0 = 1 => x0 = e and a = 1/e
Hence, if a is greater than 1/e the two curves never intersect; and if a less than 1/e the two curves intersect twice. And if a = 1/e, the two curves intersect only once.
Hence, the number of point of intersection is as follows:
a 1
0 < a < 1/e => 2
a = 1/e => 1
a > 1/e => 0
Good interview! I think she has a good chance. As for the whiteboard, zoom offers a much better one with lower latency.
If I were marking a student who said that a function tends to infinity because its derivative is always positive, as Dr Spivack suggested, that would be worth 0 marks (in light of all such functions that don't tend to infinity).
Can you give an example of such a function?
C-exp(-x) for any real constant C for example... it tends to C as x approaches inf... and the first derivative is always positive
Indeed a positive gradient is not sufficient for a function to tend to positive infinity. Proving that ln(x) tends to infinity as x tends to infinity might take a while. In this case, the interviewer chose to make the candidate feel at ease and move on, since in this short example the aim was to explore the question “For what values of m there are zero, one or two intersections between ln(x) and mx?”.
@@MathematicsatCambridge wholesome cambridge uni ❤❤
@@MathematicsatCambridge Everybody makes mistakes, no reason to pretend that there was some deep pedagogic thought process behind it...
14:10 why does having a gradient that's always positive mean it tends to infinity? e.g. y=-1/x has a positive gradient whenever x>0 but it tends to 0
It doesn’t. I think a better way to see it is that lnx = y implies x = e^y and if x goes to infinity y must therefore go to infinity ( as e >1).
@@samuelmoss2480 Yeah that seems like a much better way, might also be worth mentioning that e^x is a continuous strictly increasing function so if e^y = infinity, then y = infinity
positive gradient depicts increment rather than decrement. for Amy's case since she had the log graph, lets say you start it with x= 1, then you keep on plotting the graph with considerations of gradient. so since already the function had the value of 0 at x= 1, she got that incresing nature. however, in your case, your function even from x= 1 starts from -1 but it does increases to 0. thats the main answer. here in your case, your graph never crosses the x axis so you reach only 0. in her case, it crossed and thats the main point. i believe Euler's approximation method might put some insight here as well.
@@ankursardar4707 do you mean if it crosses the axis and has a positive gradient it has to go to infinity? because that's also not true
@@robmartin4657 what I mean is try to look for the increasing or decreasing nature with the help of the gradient. I know gradient alone is not enough that's why I suggested an idea from Euler's approximation.
I'm so thankful I never have to do thia kind of maths ever in daily life after being gaslit in secondary school being told 'you won't always be able to use a calculator and you'll need to use proper maths'
Surprisingly All questions asked, were from computer science engineering, design and analysis of algorithms, asymptotic functions analysis chapter.
Taking two times logarithm of base 2, we are comparing the quantities: 2^n and n*log_2(100) + log_2(log_2(100)). It is obvious that the first grows much faster.
I was watching people falling down stairs. What am I doing here
that's how people become math students = broken people :)
I haven’t got a clue what they’re talking about. Haven’t done maths for 21 years! 😂
How can you argue that just because the derivative is > 0 the function tends to infinity? Isn't the derivative of the sigmoid always positive as well and it just tends to 1?
Indeed a positive gradient is not sufficient for a function to tend to positive infinity. Proving that ln(x) tends to infinity as x tends to infinity might take a while. In this case, the interviewer chose to make the candidate feel at ease and move on, since in this short example the aim was to explore the question “For what values of m there are zero, one or two intersections between ln(x) and mx?”.
At 15:00 I do not agree that lnx goes to infinity when x goes to infinity just because the derivative is positive.
True, but I guess it would be possible to prove it using the fact that the range of ln(x) is the real numbers AND that the function is increasing (which I guess would involve a couple other theorems too), if you were for some reason really set on not proving it the normal way
or the fact that the function is increasing and has no upper bound would be enough
@ianbennett2443 Since the function is increasing for all X in the domain it’s synonymous with a positive derivative for all x. The range is based off the domain of the inverse function e^x which is all real numbers so it has no bound and would likely be the fastest way to do it.
@@ianbennett2443 -e^-x satisfies all the conditions you gave and does not approach infinity. you've got to use the second derivative too
increasing function can have a converging point in some exceptional cases.
Loved the interview!
How is possible to evaluate if a person is capable of Cambridge studies from just one question of this kind? In my opinion if a student replies correctly to this question then I could not be sure to accept him/her. Even if a student fails to answer this question it would be a bad indication but I would not reject at one, I would need to ask more questions.
There are usually two interviews. Additionally, the main filter for admissions are the STEP entrance exams, which is after the interviews, and not the interviews themselves
Of course, some completely capable people do get rejected after their interview (due to lack of information), but thats just how it goes
It's not really about the question + answer, but about problem solving and process of thought
@jackhnatejko8830 I don't know about Cambridge but certainly at Oxford there's also the fact that the general setup of the interview is very similar to much of the day to day teaching. Admittedly in a tutorial you'll have done the work beforehand so it's more about going through a problem than tackling it new, but the sort of skills and approaches needed are quite similar
Why am i Here?
I know . I can't stand maths
real 😭
well... at the last few questions, I paused for way too long, just because I hadn't done this question before, I wanted to find out the solutions to the equation and even though I knew pretty early on I wouldn't be successful because (e^(ax))=x is not made to be solved on simple ways, I kept sticking with it for too long, until eventually doing the ye^(y)=ax. I'm way too stubborn.
I found the first question much easier than the other ones, as just using the properties of e and ln-functions will carry you to the solution with barely any effort.
Amy is the person I would like to sit next to during a math exam 🧐
could i ask; would this be considered a "successful interview" would this applicant be offered a place just off this interview performance? not including anything else?
I am just a student - but I though Q2 was much stronger than Q1. Looks borderline in my opinion
I think it’s fairly doubtful. After all, a “typical” applicant (even that reaches the interview stage) is not accepted (I think).
I'm teaching maths at a university and I think this is definitely a successful interview. Keep in mind that the girl just finished high school.
She applied her knowledge quickly and goal-oriented, was fast to correct her mistakes when they were pointed out and was able to come up with her own solutions to the problems posed, which is probably the most important part of mathematics.
thanks so much for the response, appreciate the insight
@@oooBASTIooo
thats wrong, dont spread misinformation.@@alexandersanchez9138
Why did this video summon some of the most annoying people on the planet in the comments 😭
Is the mathematics in the room with us right now
Does anyone here know what the mean radio of the earth is? I’m lead to believe it’s 3959 miles.
Can anyone here confirm and agree this axiom so we can discuss further this.
Thanks
why would they interview students for mathematics? is is for prefential admission to a specific program?
cambridge has interviews for all of its courses
If this is the entrance interview, what's the exit interview? What is 2+2 and why should you switch to a state public school?
jokeman
why is she saying only 2 to the 2 to the n while its 2 to the 2 to the 2 to the n?
16^n > 100^100^n obviously...am i missing something?
@@bentwonie8287 never knew that brackets make a different sense...thanks
wouldn't 16^n be less than 100^100^n? If so, then yes, it seems so obvious, why the need for all the log stuff? I feel same as you...am I missing something?
oh I see, 2^(2^(2^n)) is not the same as (((2)^2)^2)^n , is that it?
@@lastfreegeneration984 thanks, now it makes sense. Put them equal to each other for the sake of argument, with the 2^ terms on the LHS, then take logs to the base 2 of both sides, the log of a number is the power to which the base has to be raised to give the number, so that gets rid of the first 2 of the outer bracket on the LHS, on the RHS you’ve got logbase2 of 100, raised to the power 100, itself raised to n. Bring that exponent term to the front of the RHS x logb2 of 100. Take log to the base 2 of both sides again, that drops another 2 off, now on the LHS you’ve only got 2 to the power n. On the RHS you’ve got logb2 of 100^n + logb2(logb2 of 100). With the first term on RHS you can bring the n exponent to the front multiplying logb2 of 100, now as he prompted you raise 2 to a power less than 7 to give 100. The log 100 terms are 6.64 to 3 significant figures, but infinity as posed it gets there first wherever there is 😂. So it’s bigger. Thanks again for the bracket tip, I’m familiar with (x^s)^t = x^(st) for the laws of indices.
How can I apply for Phd in Cambridge
Was she a student coming out of hs and applying to an undergraduate program or a higher level like graduate?
To answer your question, she would be a HS student in the British equivalent of US 12th grade
She was an undergraduate in her 2nd year (of a 3/4 year course) during this
Is this done for teaching positions?
this kind of interview is just for student applicants I'd assume
lol imagine if these were the questions to get teaching positions
@@dehnsurgeon Do you know the questions asked for teaching positions then?
A position at google requires you to be a decent developer but the interview questions are simple algorithms knowledge.
The questions you get asked to be a teaching student in nz are: will you show a commitment to the treaty of waitangi and lots of dei questions and a very basic math test that a kid could do and that’s it!
What is this? Do i have to do this when i apply for cambridge chemical Engineering and biotechnology?
It’s for math major …
would love to see one for economics!
They don’t really do these kinds of things for economics.
economics isnt a real subject
that is crazy@@7T7777
@@7T7777nigga what
😂
That was quite less challenging than one would think for Cambridge, It was fun regardless
This is a faked/parody video
tf ru talkin abt@@shadowkille8r99
its not about the difficulty, its how u think and approach unseen problems.
@@宻 wrong
ur wrong lol@@shadowkille8r99
The trouble with this selection process is that it eliminates all potentially exceptional students who have not yet learned the skills of problem solving. Surely that's your job, Cambridge, to teach such skills! Or are you just upping your university rankings by taking in the already-skilled problem solvers?
You have to be extremely competent in problem solving and adaptable in your learning to even be able to survive a Cambridge maths degree
is this for undergraduate or graduate courses
under
undergrad
Apart from the croaky voice she nailed the Qs.
Obviously f(n)=2^2^2^n grows faster than g(n)=100^100^n ... log(log(log(f(n)))) / log(log(log(g(n)))) ~ n / log n -> infinity.
yeah Thats what was doing tooo, but I did l2^n/n goes to infinity ,so
why does the top left look like harry from the sidemen
This logs not leaving me alone even now😅😅
Why does the guy on the left kinda look like Harry from Sidmen 😭😭😭
w2s moving nuts
w2s?
genius try not to be chopped challenge impossible
arent these questions too easy???
they are not when u are given so much room to express your answers.
@@akihuanakamori2578 wtf does this even mean can you please write something coherent
They're about right for a BSc I think. The interview is as much a test of how well a student can explain their thought process and respond to menotring than a test of pure mathematical skill. Don't forget that there is also an entrance exam which is geared more towards asking tougher questions.
@@Reojokerdo you mean for enrolling in BSc?
I think im lost, I was looking for the cute cats channel
how did mange to get here lol
damn jee aspirants could do this in 5 mins and leave
Once again, it's not about the questions, your academic ability is already proven pre interview when you complete your admissions test. The interview is to see how teachable you are.
Yeah 90% of you arrogant “I’m JEE everything else is so easy” lot won’t survive one term of Cambridge maths
I thought It was a coding Interview hehe DSA
Ah, this reminds me of my mathematics interview at imperial college London 22 years ago, which I aced, might I say. 😁
They lost me when they asked the question.
That first question was a horrible question to start with.
She reminds me of Natalie from Peep Show
Impressive young lady
that was easy bro ong i couldve made it to cambridge
Doubt your interview questions won’t be anywhere near as easy as this
I hope this doesn't come across as too critical or offensive, but is attitude/composure/professionalism considered in these interviews?
I found Amy's frustrated "eugh" early in the interview came across as very unprofessional, and seemed like a disregard for the importance of the interview (and I'm sure she does understand the importance of it). This also obviously has no bearing on her mathematical ability.
First off, this is a parody video. Second, it definitely would be considered. Amy’s conduct in this video wasn’t just bad, it was borderline inappropriate, especially with her mathematical ability
you are a rube
Why was her attitude unprofessional I thought it was polite and respectful???
@@jayrun43412:33 was my main example, I think that saying "aargh" during an interview should be avoided personally.
@@sb_dunkokay well I just think it’s fine like it’s a little awkward but I think the interviews really don’t care much, at least if it was me interviewing I would just be looking for her mathematical ability. I think something like that is pretty minor to me.
Im living turkey and ı admit turkey math lessons are more dificult then cambridge lessons
Take base 2 logarithm once, we get 2^2^n vs C1*100^n
Take logarithm one more time: 2^n vs C2 + C3*n
Can stop right here - the exponent on the left always beats the linear function on the right for large enough n.
How could this solution can take more than two minutes at the most?
“Logarithms with base > 1 are monotonously increasing functions” - that’s all she should’ve answered to justify how proof for logarithms proves the original statement. Which she did not.
As to the generalized second question, the “strategy” should’ve been stated as follows: for arbitrary functions f(x) and g(x) finding the point x0 where they both touch each other is a matter of solving two equations:
f(x0) = g(x0)
f’(x0) = g’(x0)
In high school in Russia I finished that would be considered an extremely weak performance. Not Cambridge quality … or perhaps Cambridge quality is not what I imagined?
shut up
the magic number is e^(1/e)
This is elementary school maths
Wtf is this I'm preparing for IIT (India).we do solve this type of problems in basic maths level 1😮
yes it’s easy,why not try for cam😏🤣
IITs overrated
@@sanjaymajhi4428 normal college se toh lakh Guna accha a
Hai.IIT na niklne k baad Mera friend v yahi bol raha tha,shayad mai v yahi bolunga agar nahi nikla toh 🤣
@@skullman424 high five. My boy
ur so cool mate
tf is w2s doing
Mocking mathematics???? Are you insane??
God, I hate maths.
Oh this is intermediate-level maths in college
In India these are taught in 11th standard
W
In the first question here, they are actually wrong. This is a trick question as N goes to infinity which is a concept that we cannot conceptually grasp, meaning they are equally big (or sized, as their size is actually undefinable)
she was noobie
Future woke interviews
As an indian, i already know the bigger number is 2^2^2^n...
Just math intuition
Intuition is good but you also have to be able to make a sound argument why that’s the case.
doesnt matter, if u cant explain it u wont be offered a place to study Mathematics. And why do u need to relate everything to where ur from?
Nobody cares you're Indian bro
Not diverse enough to get a place.
Maths is nonsense and unnatural but she seems nice so i hope she gets a place
w2s if he didnt do hardcore drugs:
😂