Last video I showed you China's toughest exam, the gaokao, and today, I'd like to take a
look at how it compares to another notoriously difficult exam, India's JEE Advanced. I'll
show you some of the toughest problems from each exam. But first, let's think about how
they compare overall. Both exams are taken to gain entrance into university. China's
gaokao is sat by around 10 million students each year. And to be eligible a student needs
to have finished high school, India's JEE Advanced is sat by only around 200,000 students.
But to get there you need to have already passed through one round of exams, the JEE
Mains and made the cutoff so only the best of the best students are able to sit the exam.
The JEE is an engineering exam and assesses physics, chemistry and math, allowing entry
into the Indian Institutes of Technology, such as IIT Bombay with a world ranking of
172 and IIT Delhi, with a world ranking of 193. The gaokao has a broader scope, covering
the science subjects as electives, as well as languages and essays. Doing well could
allow entry into some of the world's best universities, including Tsinghua with a world
ranking of 15. And Peking at 23.
Preparation for both exams is intense, and starts years before the exam, with both countries
having a lot of private exam coaching. And China has Maotanchang, a town almost entirely
dedicated to helping students score highly.
I would say the stakes seem a little higher in China, just because if you fail the exam,
your only options are to either give up or wait an entire year to try again. Whereas
in India, there are a few other pathways and other exams, such as UPSC or NEET.
So let's see how the actual questions compare. In my last gaokao video, I showed a math question
that many of you said was a bit on the easy side. So today, I've picked a harder one.
This is from the 2016 Jiangsu exam.
As shown below, there is a straight line l given by this equation, and a parabola C given
by this equation. Part one asks, If line l passes the focus of the parabola, then what
is the equation of the parabola? So they're asking you really to find out what p is. And
that determines the shape, somewhere in here would be the focus of the parabola. And you
can see in the solutions that I've written out that this first part would be handy to
know how to find the focus of the parabola. And if you did, you'd be able to find the
focus for our one to be p over two, zero, then put that point into the line since we
know the line passes it and you will find that p is equal to four. So you get your parabola
equation.
Part two is more difficult. It says there are two points P and Q on our parabola, such
that P and Q are symmetric with respect to line l. a) prove that the midpoint of the
line segment PQ is this.
Pause the video here and try it yourself to really see how tricky this question can be.
I interpret points P and Q to look something like this. And a bit of information we know
is that a line between P and Q must be perpendicular to l. Since the points are symmetric. From
the equation of the line l we can work out its gradient is one. So the gradient of line
PQ must be equal to minus one. That's the first brainwave that you need to get going.
You can write the gradient of PQ from the equation you know about the parabola and simplify
it down into this form where y one plus y two over two is equal to minus p. Now that
in fact will be the y component of the midpoint.
Now we just need to find the x component. To do that you'll need the second brainwave,
which is to remember that the midpoint of PQ must lie on the line l since P and Q are
an equal distance from l. Use the equation of line l, put in the coordinates of the midpoint
and then you will find the x component of the midpoint to be two minus p. Giving the
final answer to that part of the question.
Part B I stared at for a long time before I was able to figure out what you meant to
do, they're asking what is the range of possible values of p, p appears in this midpoint of
PQ and so there must only be a range of values for p for which this is possible. The way
that I ended up doing this was to write an equation of the line PQ, which was something
we hadn't done yet. Although we do already know the gradient, and we know it passes through
the midpoint, so we're able to write this equation as y is minus x plus two minus two
p, then combine this linear equation with the equation of the parabola that we were
given y squared is to p x, that will ultimately give you this quadratic equation here, which
is a relationship between y and the parameter p.
To find the range of values of p, we're going to be using the discriminant. The discriminant
of a quadratic in this form a x squared plus bx plus c would be b squared minus four ac.
And if that discriminant is bigger than zero, they will be two real and unequal routes or
solutions. And that is the situation that we want. So we work out our discriminant,
set it as being larger than zero, and find that p would have to be between zero and four
thirds.
I found that was really challenging, not so much that any of the individual steps were
too hard to do. But it took a lot of forethought to think of a method you would use to get
to the answer.
A tough JEE advanced question that I found on a similar topic is this one here from the
2018 paper 2.
Let T be the line passing through the points P and Q, let F1 be the set of all pairs of
circles, S1 and S2, such that T is tangent to S1 at P and tangent to S2 at Q and also
such that S1 and S2 touch each other at a point say M. Let E1 be the set representing
the locus of M as the pair S1, S2 varies in F1. Let the set of all straight line segments
joining a pair of distinct points of E1 and passing through the point (1,1) be F2. let
E2 be the set of the midpoints of the line segments in the set F2, then, which of the
following statements are true?
Now this question is completely out of control, and seems to get worse with every single sentence.
Again, pause here to have a little try yourself.
It sounds impossible to understand when you read it. But I think something like this is
going on. We have our points P and Q, and circles which are tangent at those points.
These circles can change in size, but they will always touch each other at this point
M.
Here, it's essential to know what a locus is. And it is the set of all points whose
locations satisfies a given condition. So we're asked to find E1, which is the locus
of M. And that's sort of all the places that M could possibly be, you first need to notice
that the angle between P M and Q will be 90 degrees. And that gives you the condition
you need to work out all the places M could be because with that condition, we know that
the gradient of PM times the gradient of QM would be equal to minus one. These lines are
perpendicular. Plug in your points and your workout this equation, which is E1, all the
possible spots where M could be.
Now E1 itself is actually a circle. And around about here we could eliminate A as an answer.
A is actually claiming that point P lies in E1. But if M was over here, where P is then
the size of this circle would be radius zero, and this one would have to be a straight line.
So we can say that both P and Q are not actually possible locations for M
and therefore not included in E1. So the circle I guess is open at those points.
The next thing that happens is that we start to draw chords in this circle of E1, and these
chords pass through the point, one, one. Now the midpoints of these chords form their own
locus called E2, E2 will be all the possible midpoints, to get E2 notice that the chord
will be perpendicular to a line from the centre to the midpoint. And so if we call our midpoint
(h, k) then the condition for E2 will be that those two lines were perpendicular, their
gradients will multiply to be minus one, we arrange that and you have the equation for
E2.
You'd think that finding E1 and E2 would be all the hard work done, because that was pretty
tricky. But then the exam gets downright nasty by trying to trick you with some of the multi
choice answers. They say the point four fifths and seven fifths does not lie in E2. Now those
coordinates will satisfy the equation for E2. So it looks like that point is in E2.
But you're supposed to notice that that point they gave you is actually the midpoint of
a chord that would have passed through P. And since P isn't actually in E1, then neither
is the chord that would originate from it. So B is an answer, because that point will
not be an E2.
This was my working to figure that out. I found the equation of a line through the point
they gave you and (1,1), that was the equation there. And then I found the equation of a
line that would pass through P and (1,1). And it was the same equation.
For C you can find that it is not an answer, because this point does not satisfy the equation
of E2. And for D, you will find that it is also an answer, because of this point does
not satisfy the equation of E1.
So that's been a look at a question from each paper. In this case, I did find the JEE question
harder, since the geometry was more difficult to understand. But of course, you can't judge
a whole exam on one question. There will be questions that are harder and easier than
these ones. So the battle is still on. In fact, it was more difficult for me to comb
through the gaokao questions, since it's difficult to get translations. So if you know of a particularly
tricky one, then send in a link to it. And same with any JEE questions. I do enjoy taking
a look at which ones you found the hardest. I've also seen comments from other countries,
including Turkey, Korea and Brazil, and I hope to cover your exams in the future. It's
worth remembering here that the gaokao and the JEE. are measuring different groups of
students for different reasons on different topics. And the difficulty of questions within
an exam I find is often more determined by the kind of preparation you've done, and perhaps
how similar the questions are to those of previous years. Clearly, a lot of hard work
goes into preparing for either of these exams. Thanks to my Patreon supporters for enabling
me to make these videos and a special shout out to today's Patreon Cat of the Day Glacier.
Thanks for watching to the end of the video. I hope to see you next time.