Only 1% could solve this insanely difficult problem
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- Опубликовано: 16 сен 2024
- If this question is easy for you, I see great things for your future. Thanks to Rahul and Battu for the suggestion! This problem appeared on the 2022 JEE Advanced Paper 1 Mathematics section as Question 8. The JEE Advanced is an extremely difficult exam for admission into India’s prestigious IIT schools. I think this is one of the easier questions of the paper. I give credit to the Unacademy Atoms video about the paper which helped me understand how to solve the problem.
2022 JEE Advanced paper 1 answer sheet (see Q.8)
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Unacademy Atoms video solutions to paper 1 (Q8 is around 59:00)
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Damn, who would have guessed 4 years ago that a question from the exam which decided my college would appear in this channel? I scored barely 35% in this Maths paper, and 50% overall, and am now in a good IIT!
im currently training for jee and im somewhat relieved to hear this 😭
In jee advanced, good percentage is almost impossible, rather a good percentile matters
Which college and which branch?
This makes me feel embarrassed. Back when I had attempted it, I had gotten about 70% in maths but borderline failed chemistry.
@Anmol Roy I won't name the IIT for privacy reasons. But I'm in the EE branch.
According to JEE Advanced 2022 report, only 0.95% of the candidates have answered this question correctly.
Is this even true?
@@climense6915 yes it's true
And they call it easy?
@@anil2296 Oh my god!😭
But it's still a good no.
Using coordinate geometry:
A = (0,0), B = (1,0), C = (0,3), M (midpoint of BC) = (1/2, 3/2), BC = √(AB²+AC²) = √(1²+3²) = √10
Larger circle has center (1/2, 3/2) and radius R = BC/2 = √10/2
Smaller circle has center (r, r) and radius r (since it is tangent to both the x-axis and y-axis).
For circles that are internally tangent, distance between centers = R - r
√((1/2 - r)² + (3/2 - r)²) = √10/2 - r
(1/2 - r)² + (3/2 - r)² = (√10/2 - r)²
1/4 - r + r² + 9/4 - 3r + r² = 10/4 - r√10 + r²
r² = 4r - r√10
Since r ≠ 0, divide both sides by r
r = 4 - √10
The calculations are pretty much the same as in the video, but I find the setup a little simpler.
Hello
Can you help me ?
I want solve this but I cann't
x+y+z=15
Xsquare +Ysquare+Zsquare=75
X=?
Y=?
Z=?
@@dadasdadas2044 x=y=z=5
@@dadasdadas2044 You can't and we can't because there aren't enough information
@@dadasdadas2044 bro it's a hit and trial question.
If you put 5 in place of x, y and z, it satisfies both equations.
There's the answer
@@MeowUnderBlanket I don't seem to be able to find an algebraic way of doing it. It's just a plane intersecting a sphere at a single point.
Setting up the diagram is the key to getting the solvable equations. It was tedious enough to be difficult, and oh boy this was the easiest question on that paper...
Very true.
Easiest was the question from permutations are combinations
Easiest was from logarithm and probability and trigonometry and venn diagram
I also gave this year 2022 Jee Advanced and I am happy to tell you @MindYourDecisions that this questions has taken my 3 mins of Precious time and the result was Amazing . LoL 😁
Did u landed into any IIT then?
@@sriprakash7245 Hell Yes I'm studying right now thats why I'm able to answer so late. 😊
@@itsOnlyPIYUSH congratulations
Bhaiya aap kis iit me ho
@@sriprakash7245 Thanks.
I solved this using coordinate geometry. It is less elegant however very little brainpower is needed which can be helpful in nervous exam situations.
Let vertices of the right angled triangle be (3,0),(0,0),(1,0).
The circumcenter simply by its definition is (1/2,3/2). The circumcircle also passes through (0,0). So we get that the equation of the circle is x^2+y^2-x-3y=0
Now let the circle whose radius we want be x^2+y^2+2gx+2fy+c=0
Now it touches the coordinate axes at two points say (x_1,0),(0,y_1). Putting these points we get two quadratic equations in x_1 and y_1 whose discriminant is 0. Therefore we get that g^2=c=f^2 => g=f
Now the circles touch internally. This means the distance between their centres c_1 and c_2 is equal to the difference of the radius of the larger circle and the smaller circle.i.e, C_1C_2= r_1-r where r_1 is the radius of the circumcircle. [(1/2-g)^2+(3/2-f)^2]^(1/2) = √10/2 - √(-c+g^2+f^2).
But, g^2-c=0 and g^2 = f^2
This allows us to completely simplify the expression into the quadratic equation
f^2+f(√10-4)=0
From here we obtain that f = 4-√10. since the radius of the circle is √(-c+f^2+g^2) and -c+g^2 =0, we have the radius = 4-√10
👍
Yeah I also solved it the same way
I can confidently say there was no way I could have solved this.
but I have solved this quite quickly.😅
@@patrickng8974 honestly I've never seen the "arc BAC = π/2" before and with hindsight I can see that it means that it's a 90° angle, I still wouldn't have been able to get further than that point in the puzzle. But that's what most of us are here for, to learn and figure these things out.
Yes me too i just ignored this question like every other question in the math section 😅
@@patrickng8974 like how?? Many normal high school student cannot even understant that tangent line between two circle, even more conceptualize the triangle inside the circle. Like, it straight up competition level knowledge
@@arolimarcellinus8541 I’m in middle school and solved it, and yes, I do competition math
I gave jee advanced this year and skipped all maths questions except 4 which i attempted just to qualify 😂
Can't be more same 😂
What rank did you get in jee mains?
@@blue_g29 53440 was my rank in jee mains
@@Aditisoukar i am 2023 aspirant and i am fearing of jee mains too much ..can you help me with this?
How can you just skip it? Is thst even allowed?
I am impressed that you used the module, |r - 1.5|. Very nice.
Well that's a good habit to take to use modules when you want a difference and don't know which term is the greater (when, let's say, one of them is a variable or is unknown).
But it was quite useless here, we know r < 1.5 (because r is the length of the square and is smaller than both sides of the triangle) and we'll square it anyways.
We can use coordinate geometry in this question two sides of triangle as y and x axis and the center of smaller circle will lie on y=x
Result will be 4- √10
How it will lie on y=x
@@soundsinteresting208
Two sides of triangle are perpendicular and the circle (not circumcircle) touch the sides (y and x axis ) in this case center of circle will on y=x axis
Ah yes, me watching JEE videos after JEE
There is also a formula for this problem. It's r = a + b - c. Applying with a=1, b=3, c=√10, then r= 1+3-√10 = 4-√10
From where this formula has come could you tell please
@@Cooososoo just take the side lengths as variables and solve it like in the video
@@Cooososoo MD = c/2 - r, ME = |a/2-r| and DE = |b/2 - r|, Pythagoras theorem
c²/4 + r² - rc = a²/4 + r² - ra + b²/4 + r² - rb
As a²+b² = c² they cancel out
r(r-a-b+c) = 0
r = a+b-c
The formula is 2r=a+b-c and its for an incircle. This isn't an incircle
I scored only 15 marks in math.Physics and chemistry saved my rank
Because of maths i lost my iit seat :/
@@saqlain3470 this year both mains and advanced math paper was hard
bhaiya i know you 🤓
@@sampritdatta1752 😎
I haven't normally encountered such type of qs in pyqs, this does however remind me a lot of ioqm pattern papers. Certainly using those techniques helped in solving the question.
Beautiful solution sir. Thank you. Greetings from México City.
Salutaciones Señor.
i remember solving this question in JEE ADVANDED this year
took a little time... but eventually did it
Hey dude, how can angle BAC be π/2
@@vampire9355 look in radians we say pi/2 is 90 degree
That was pretty simple. I just had to watch the video and got the answer!
Tbh using coordinate geometry is made it pretty easy than thinking about all this stuff
Hi Sir iam Battu , thank for providing the solution ☺️
Thank you for giving us a very interesting problem!
Hi Battu, are you JEE Aspirant ??
I am real battu u r fake
@@Avinash_kumar_IITD ok
@@techmaster6587 No
More generally, one can show the following: Let ABC be a triangle with side lengths a=BC, b=CA, c=AB, and let r denote the radius of the circle tangent to AB, AC, and internally the circumcircle.
Then r=sqrt((s-b)(s-c)/(s(s-a)))bc/s (with semi-perimeter s). Moreover, if BAC is a right angle, then r=2(s-a)=b+c-a.
To see this, denote by M the midpoint of this circle k, E the the center of the excircle of ABC opposite A, g the angle bisector of angle BAC, X the intersection of k and line segment AM, Y the intersection of this excircle and line segment AE, Z the intersection of this excircle and g not lieing on the line segment AE.
Invert at a circle centered at A with radius 1. This sends B to B' on ray AB with AB'=1/c, and C to C' on ray AC with AC'=1/b. This maps the circumcircle of ABC to the line B'C', and the lines AB and AC are mapped to AB', AC' (i.e. to themselves). Consequently, k is mapped to some circle k' tangent to lines AB', AC', B'C' and situated outside(!) triangle AB'C' (since k is tangent internally(!) to the circumcircle of ABC). That is to say, k' is the excircle of triangle AB'C' opposite A.
Now follow up with a reflection across the line g and a homothety centered at A with factor bc. Since AB'=1/c, AC'=1/b, this second map sends B' to C, C' to B and consequently triangle AB'C' to ACB and thus k' to the excircle of ABC opposite A.
Since both of these maps leave the line g invariant, one can see that their composition takes X to Z. Thus, bc/AX=AZ (1).
Let Q denote the tangent point of the excircle opposite A with line AB. By secant theorem, AY*AZ=AQ^2 (2). It is well-known that AQ=s, where s is the semi-perimeter.
Let I denote the center of the incircle of ABC, W the intersection of the incircle and line segment AI, and P the tangent point of the incircle on line AB. Let R denote the radius of the incircle.
Since both k and the incircle are tangent to AB and AC, one has r/R=AX/AW (3).
Since both the in- and excircle are tangent to AB and AC, one has AW/AY=AP/AQ=(s-a)/s (4) (using the well-known AP=s-a, AQ=s).
Combining (1), (2), (3), and (4) one obtains
r=R AX/AW=R (AX/AY) (AY/AW)=Rbc/(AY*AZ) s/(s-a)=Rbc/(s(s-a)).
Let F denote the area of ABC. Then F=sR.
Therefore, r=(F/s)bc/(s(s-a)).
If ABC is right-angled in A, then F=bc/2. Moreover, 4s(s-a)=(-a+b+c)(a+b+c)=-a^2+(b+c)^2=-a^2+b^2+c^2+2bc=2bc (by Pythagoras's theorem), so s(s-a)=F. Thus, r=(F/s)bc/F=bc/s=2F/s=2(s-a)=b+c-a.
If ABC is not assumed to be right-angled in A, one may instead apply Heron's formula for F to conclude
r=sqrt{(s-b)(s-c)/(s(s-a))}bc/s
I got close, through clever reasoning, I determined that r must be between 0.5 and 1 exclusively, but couldn't narrow it past that
You could have did it like assuming a as origin and writing coordinates of both centres then c1c2=r1-r3
Yes but classic geometry is far better than coordinate
I got this correct in jee advanced 2022 😀
I faced this question 🙋this year
Me also
bhay JEE result kitna aya?
Me too
Gees congrats to you guys for applying
The schools here in the US are way easier to get into but also not as prestigious
@@johnscarceforpresident1647 but still majority of us are just employers in your companies!!🙂
Remeber the stuff , this has a another approach by mixtilinear incircles , a general problem was in rajeev manocha and similar came in AIME
I had never heard the term mixtilinear circles before, interesting. mathworld.wolfram.com/MixtilinearIncircles.html
@@MindYourDecisions woah! Loved your approach , one of my friend suggested the length chase and i loved it , for first attempt i did by mixitilinear incircles!
@@piyushkumarjha599 dude, Are you Jee Aspirant ...???
@@techmaster6587 no
@@HimanshuRajOk yes?
I have a question suggestion: how to divide a circle in three parts of equal areas using only 2 lines
The two lines will be secants passing through a common point on the circumference
my solution (assuming the circle is centered at the origin and has radius r) make 2 vertical lines one at x=-0.264932084602777*r and the other at x=0.264932084602777*r
that number is the solution of this equation (arcsin(x) + x*sqrt(1-x²) = pi/6)
I took a different approach here. Actually I got it wrong, but its really was hard to find out the mistake if you do it by your own rough diagram. Try it by not looking at myd's diagram or you gotcha. Extend the sides AC and AB and joined the points B' and C' such that the tangent at the point where the both circles meet coincide B'C' ; then smaller circle works like a incircle and we can work out its extended sides and compute radius of circle by conserving area.
How long would the altitude to the hypotenuse be, on the triangle in your profile picture?
Hey vedant did you answer ioqm this year? If yes what was your score?
@@notananimenerd1333 ah dude I remember you! I didn't gave ioqm, coz i would need to prepare for it separately than jee which is not for me.. and it requires big brains..
@@carultch i/0, which is.. *_*
@@vedants.vispute77 oh I see.. you ought to check the paper man.. it is considered as one of the toughest ioqms (prmos ) of all times 😬
can also do by coordinate easily as we know circumcentre coordinates and internally tangent circle condition is c1c2= r1-r2
it's just r = x + y - sqrt(x^2+y^2)
I gave this exam this year
Got 102 marks which fetches around 7500 rank out of 2,25,000 candidates who wrote it
The general solution is interesting r=a+b-c, where c is the hypotenuse!
Damn I couldn't solve it during the exam 🙂
The vertical distance of MD is not r-1.5 but 1.5-r. Since it gets squared later, this results in the same thing but still
He took the mod
WE KNOW IT BETTER FROM THE PERSPECTIVE OF COORDINATE GEOMETRY... WE ASSUME A TO BE ORIGIN AND AB AND AC BE THE COORDINATE AXIS THEN THE QUESTION TAKES ONLY 30 SECONDS...
explain
If r=0 WAS allowed, would that mean that the second "circle" was point A?
Yes
I love your humbleness
Some simple applied integral: divide the area into immeasurably thin rings parallel to the perimeter. Go progressively inside from there, ending up at the center of the inner circle. Then the formula of the relationship between the perimeter, area and inner circle will become clear.
Didn't solv this in the exam hall but now I m happy to be in iit Guwahati
i was able to pretty much follow everything until r = 4 - Sq10. It seemed like a big unexplained jump :(
Starting from r(r-4+sq10) = 0. As r > 0, both sides can be divided by r, resulting in: r-4+sq10 = 0 or r = 4-sq10
@@ThePaliwalie Oh my days. If I'd spent 10 seconds thinking about it I would have got that... it was early in the morning for me - that's my excuse. Thanks!
Just remember! You have to think and solve the question AND you should know the value of rt.10 correctly upto 2nd decimal, all within 3 minutes.
didnt need to know root 10. Writting 4 - root(10) was enough.
you don't have to do it within 3 mins, not only do u not have to solve all the questions in the exam, it's almost never a good idea to do so, even doing half the questions is considered to be really good+ some questions obviously take more time than others and maths generally has the most lengthy quesions so....3 mins is very over exagerrated
No need to solve all problems
But they give this is 2place decimal which which made many people do incorrect
At the end instead of focusing on the small right triangle, you can also use the intersecting chords theorem to get the answer. That's what I did.
Damn.... nostalgia's hitting me hard now ......i literally attempted this question in jee adv22.....don't remember if I got the correct answer though😂
Kitna bana?
@@aviralmishraofficial1626 92 bro.... currently In my drop year
. . . which happens to be twice the radius of the inscribed circle!
The problem doesn't say the angle was measured in radians.
It’s expressed in terms of pi with no unit. It doesn’t get much more explicit than that. You don’t say that a right angle is ninety, with no units, do you?
Following you since 2020
Brother, can you solve the complete paper of JEE Advanced in just 1 hour??? Please
No
Pretty sure this is a grade 9 difficulty problem. It's not easy for an normal grade 9 students but considered medium difficulty for any good high school students.
After all you just set the center of the circle's coordinates as (x, y) and pretty mechanically writing down the 3 equations to get r = BotSide + LeftSide - Hypo
man u can use the circumcenter coordinate geometry principle where the center of the circucircle lies on intersection of perpendicular bisectors of sides or even better jus use the distance formula take a point (h,k) and equate its distances from each vertex(cuz radius of circle is equal) its a wayyy more practical way to solve this qs
How can I find out ( h,k)?
in 5:00 shouldnt ME be 1.5 - r since 1.5 is the bigger value than r?
I get it that after squaring it it won't matter but still it's wrong
In the video, it is in modulus so it doesn't matter
So that's why they specified the radius should be greater than 0...
ME = 1.5 - r
But this is not changing the outcome because the power of 2
It's unfair asking a question in this way. Really needs a diagram.
I think the challenge is to come up with the diagram. The only part of the question I could see that really needs clarification, is that the word "touches" is vague. Really should say that the two circles are tangent to each other, since tangent has a more precise meaning than "touches". I'm aware that the namesake of the word tangent, is "to touch", so you could infer that they meant to specify it that way.
I would like it if you reformulated the question because "touches" isn't that accurate of a word , good video tho.
What was your initial thought when it said "touches"? I'm just curious to see how else it translates in this context....
@@rakshithbs1642 touches meant to me : they share at least one point, so them being tangente isn't the only option but the first option
@@EliasSipsTea7030 Fair enough, but the definition of tangent is that it "touches" a curve at a point.
Wiki definition: "In geometry, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that "just touches" the curve at that point."
You can also use coordinate geometry. Centers are (0.5,1.5) and (r,r). Now use CC1 = MT-DT result. Same result. We don't even need to draw topologically correct diagram. Just saves time.
What does CC1 MT DT mean?
How is center of the 2nd circle (R,R) and not ( R1, R2)?
@@leif1075 the distance between two circle's center, CC1 = subtraction of the two radius ( MT - DT ) ....as the circles touched internally
@@kongkonasahadola2949 It touches both the axes. Hence radius are same.
@@kongkonasahadola2949 oh you never said what DT and MT were that's why so.dont think anyone would understand...ty for clarifying
The hardest part was to find value of 4-√10 without calculator in examination hall and correctly upto 2 digits
Why don't you try other questions 🤪
that is basic
NV SIR ON TOP ⚛️ ⚛️ ATOMS OP 🔥🔥
Pela construção geométrica é irrelevante escrever em módulo a diferença entre os lados já que é visível qual lado é maior que o outro. Inclusive foi utilizado a visualização em relação aos raios da circunferência.
Abraço
loved it man
problem was only knowing the value of root 10 as it was not provided in the question
It would've been much easier with coordinate geometry
i found that the radius length is 5/2 or 2.5 units
I solved it in the same way
that 2.5 unit is radius of the bigger circle
How?
I solved it during the exam using coordinate geometry
Excellent. Now can you tackle Twitter?
😂 It's a bit much to ask for I guess
Lmao 😂
It would have been instructive to see the circle of r not >0 in the solution.
r=0 is a solution. It touches all the required things at A so needed to be excluded.
I gave JEE Advanced 2022, did not solve this question, still got selected and now sitting in IIT Kharagpur.
Which branch?
Great solution.
This question is just so easy, i solved it using basic coordinate geometry
Good to see indian unacadamy being featured
It was easier with coordinate geometry
The problem is hard to understand.
2r=a+b-c, c=\sqrt{a^2+b^2}
That is incircle radius formula
Incircle touches all the sides but in the question the circle touches only two sides
Hence we can't apply this formula
It's exactly twice the inscribed circle radius. I wonder if that's true for all right triangles?
More generally, r = AB + AC - BC.
If this question is easy,
What is the definition of hard?
Something like this and more :
If f(x) is a differentiable function and g(x) is a double differentiable function such that |f(x)|≤1 and f'(x)=g(x) . If f2(0)+g2(0)=9 . Prove that there exists some c∈(-3,3) such that g(c).g''(c)
solved it the same way, although it took me like 7 minutes whereas in the competition i think u would have like 3 minutes or something to do this problem
I’ve been in construction so long I’ve forgotten to default to radians. 😅
I didn't know that "touches" was the same as "tangent". sigh
Touching is the namesake of the word tangent. It is a little unclear to specify the tangency of the circles with the word "touches", since two circles that intersect at two points, would also technically touch.
Good one👍
Bruh i literally gave this exam and got 18000 rank
Nice sir I am preparing for jee exam this year
2:55 I m in 9 grade and I was able to solve upto here
Here's one that people got into a heated debacle over:
Tim can cut through a piece of wood in 5 minutes of time. If Tim were to continue cutting through pieces of wood at this pace, how long would it take him to cut a new wooden piece into three pieces. Some people believe that 10 minutes is the correct answer where as others believe 15 minutes is the correct answer. (The issue here mainly comes down to the wording of the problem.)
I'd say the answer is 10 minutes, and we'd need more information to conclude it is 15 minutes instead. Tim starts with a new piece of wood, and he has to make 2 cuts through it, to end up with 3 pieces of wood. Thus it takes twice the time to make 1 cut, in order to make the 2 cuts and end up with three pieces.
It didn't specify that whether or not the three pieces add up to the full length (minus the 2 saw kerfs). But since it also didn't specify that he had three pieces plus a scrap piece.
If this was easy then what about hard?
Something like this and more :
If f(x) is a differentiable function and g(x) is a double differentiable function such that |f(x)|≤1 and f'(x)=g(x) . If f2(0)+g2(0)=9 . Prove that there exists some c∈(-3,3) such that g(c).g''(c)
I will be attending this exam next year.. scary
yeea mann... :''(
I'm confused, I thought the formula for this question is r=(abc)/(√ ((a+b+c) (b+c−a) (c+a−b) (a+b−c))) since we know each side length of the triangle? If this correct the answer r ≈ 1.58 and Prech mentions at 2.25 that BC is the diameter of the circumcircle so therefore r= (√10)/2 ≈ 1.58 so I' don't understand why tangent circles are used regardless if the circumcircle touches the triangle of ABC or am I misinterpreting the question? Imo the radius never changes and the question didn't ask for the different between the radius of both circumcircles or am I wrong and if so can someone please explain why?
I also don't understand what the question means by "touch". Perhaps schools in India use math terms different from the USA?
The question is not asking for the radius of the circumcircle, it's asking for the radius of the circle that is tangent to side AB, side AC and the circumcircle.
@@tedr.5978 touch aka tangent
@@tedr.5978 Touch implies the line is a tangent. How is that difficult to understand?
@@redeyexxx1841 I thought touch meant either tangent or went through AB,AC, and the circumcircle
I didn't understand because my geometry is very weak
i gave this paper live.....couldn't solve in time so had to leave it
Thank you so much! My manager at McDonalds would not move me up to fry chef until I came up with the correct answer.
So r is twice the radius of inscribed circle.
Some enlightenment please. Angle BAC is given as pi/2. In other words 3,14 divided by 2 = 1,57 degrees. How did you get to 90 degrees?
cdn-academy.pressidium.com/academy/wp-content/uploads/2021/03/Unit-circle.png
pi/2 radians
Its in radians, 360 degree is 2 Pi Radians
Bro you are in which class 💀💀
I drew this to scale and am having a hard time finding a circle that meets the three points of tangency. In the diagram used, AB is drawn much longer than 1/3 of AC. Something is wrong here, beyond describing segment ME as (r-1.5) rather than (1.5-r) in the diagram at 4:48 (which becomes irrelevant when you expand the terms). I get that the math works, but what is the x,y coordinate of the two-circle point of tangency? Presh can you show this to scale and show where is point D actually in relation to M, and the point of circle tangency? Edit: I guess I can see it now, point T is quite close to point B, and D is indeed below and a little to the right of point M. No follow up needed, I suppose.
0:06 r = 1/4 pi
Damn it, not the closest answer
R stands for Racecar
This is a very easy ques
Idk why so many people weren't able to solve it
I solved it in 2 min
Actually Presh! I solved it in a much easier way than yours , we could solve it more easily by considering perpendicular sides as x and y axes.
bruhhhhhh this is ioqm level
Hardest paper from a university that ain't even in world top 100.
Then don't give this exam go to your top 100 University,ooh u can't even solve this 😂
IIT Bombay and IIT Delhi are in top 100 engineering Universities.
@@ishansharma3944 NO ITS NOT
@@himanshumech133 it is, check the qs university rankings of top 100 engineering universities.
@@wakeawake2950 He might not even be Indian. As a middle class Indian, IIT's are a good to go option and hence its our helplessness that leads us to give this exam (atleast most of us!), many high class Indians literally go to those top 100 universities.
That is hard. I tried using coordinates and its still hard.
If this is an easy problem I wonder what the hard would be
This is why jee advanced is considered the hardest exam for admission to undergraduate engineering. You can find other questions on Google.
Something like this and more :
If f(x) is a differentiable function and g(x) is a double differentiable function such that |f(x)|≤1 and f'(x)=g(x) . If f2(0)+g2(0)=9 . Prove that there exists some c∈(-3,3) such that g(c).g''(c)
try to say "cicrumcircle" 10 times quickly
and we are supposed to do it in 2 minutes