My god I've been scratching my brains out with this exam and couldn't find the questions solved anywhere. I'm really really thankful for this video. Really, thanks man.
⭐in section 2. (3)For ppl who are wondering why the answer was a right triangle is because. **note:squrt = square root** 1.)we have 216 unique dice rolls 201 are triangles (but some aren't unique) so we categorize them using side lengths 1. Equilateral triangles 2. isosceles 3. scalene triangles, there are only 6 possible equilateral triangles squrt(1)squrt(1)squrt(1) squrt(2)squrt(2)squrt(2) ..., for isosceles triangles draw a graph of an arbitrary isosceles triangle (given) a 60degree angle, in all 3 cases of the 60degree angle being one of the base angles or the top angle all angles must be 60degrees meaning we don't have any "pure" isosceles triangles (Define: Isosceles triangles are triangles with at least 2 congruent sides) so all of our isosceles triangles are equilateral. next, we find scalene triangles using combinations (6choose3) we get 20-1(from section2.(1)we know that 1 squrt(2) squrt(6)is not a possible triangle) unique scalene triangle. 2.)Write down all possible 19 scalene triangles using a tree diagram and canceling out permutations now we are left with 25 triangles (6equalateral from previously and 19scalene)for the equilateral triangle the maximum of bc/a is 6*6/6 = 6 .for the other 19scalene triangles we can categorize them further by 1. Right scalene triangles 2. non-right scalene triangles. by using properties of the special 60-30-90 triangle from the 19 scalene triangles we have 5right triangles that have a 60degree angle in them with maximum bc/a = 12. the other 14 non-right scalene triangles. we only consider ones that possible bc/a >or= to 12 (we need to check if there're any other triangles that bc/a exceeds 12)we have 4 candidates being squrt(1)squrt(3)squrt(5) , squrt(1)squrt(3)squrt(6), squrt(1),squrt(4),squrt(6) and squrt(2)squrt(5)squrt(6) which using the cosine law cannot be triangles with 60degree angle in them. ✨✨Thus right scalene triangle with side length squrt(1) squrt(3) squrt(4) is the unique triangle that is the maximum of bc/a.✨✨
In section 1(1) we can use another property loga(b)*logc(d)=logc(b)*loga(d). And then we move all the bases of the logarithm to the left like that: log3(4)*log4(4)*...*log2(2020). It's just my solution and its more intuitive for me)
idk if you stil read the comments here, but in section 1(5) 12:00 , wouldn't it be easier to apply the tan(theta)= [(m2-m1)/(1+m2m1)] m here is slope, so we can use the numbers in the brackets as slope and [ ] here stand for absolute number i cant remember the formula name unfortunately, but it worked here as well as yours!
Hi, thank you for your comment and for sharing your solution! (and yes, I don't check the comments every day but I do sometimes!) Yes, your method sounds quicker 👍
Hi, I don't have any old EJU questions unfortunately. But I have MEXT exams between 2006 and 2015 (some questions are missing). They are not available on the Internet and I got these files from a generous viewer. I could send them to you if you could send me an email at komei.biz@gmail.com 👍
I can't understand why did you assumed that f(x) is a third degree polinomial function. It is divided by a squared function, but it doesn't mean that it is cubic. It could be a fourth degree or higher
that's true but you have to think about what i would call the test strategy, they would not put an unsolvable problem in the test, there fore if it were a fourth degree or higher there would be 3 variables and it would be unsolvable but this does not mean that every task will go the way you want it to go, but every task will be solvable, (if it is not, that's a whole another problem) in the 2 (3) it was wrong to assume that it was a right triangle, but there is a comment that explains why that is the case, even if it's assumed, basically there are only 19 candidates of which only a few satisfy a 60 degree angle, and the answer happens to be a right triangle, there is probably some geometric math that i don't know that would explain why, but it doesn't matter as long as there is a way to find the answer, even if it 10 pages long, it only matters that it is correct
It really does not have a syllabus, but I use the syllabus for EJU. Both exams have almost the same topics. Therefore if you have EJU you have mext. You can search for it as "EJU sullabus for 2013"
If you took the math stream for your A/Ls the topics are pretty much the same as the ones in our syllabus (assuming you're sri lankan from your username 🙃).. you have to focus on coordinate geometry, functions and integration for math b, bc most questions are based off of those lessons
16:54 Why once u find that c < 4 u say that no triangles can be formed with c=4, c=5, c=6, specifically. So that would mean c>7 is possible, but we just stated c
Hello brother plz provide me all pdf of questions of MEXT Scholarship undergraduate level if you have it. I am unable to download it form the website as they have removed it from there... Plzz brother i will be thankful to you..
Hi sorry for the very late reply! Since we don't know the coefficient of x^3, we call it 'a' temporarily, and and when we do the long division, we have to multiply x^2 by ax because the quotient (ax + b + 2a) times the dividend (x^2 - 2x + 1) must become ax^3 + bx^2 + 2x + 3. I hope this makes sense 😅
Hi sorry for the late reply and that's a good question. As you pointed out, the question doesn't state that the triangle is a right triangle. However, when the sides are root1 ~ root6 and when one of the angles is 60 degrees, you can only create either an equilateral triangle or a right triangle. It's a lot of work to prove this so I wrote a code that confirms this is true (you can use free websites like www.online-python.com/ to run this code): import math count = 0 for a in range(1,7): for b in range(1,7): for c in range(1,7): a_root = math.sqrt(a) b_root = math.sqrt(b) c_root = math.sqrt(c) if ((a_root + b_root) > c_root) and ((a_root + c_root) > b_root) and ((b_root + c_root) > a_root): angle1 = round(math.degrees(math.acos((a + b - c)/(2 * a_root * b_root)))) angle2 = round(math.degrees(math.acos((b + c - a)/(2 * b_root * c_root)))) angle3 = round(math.degrees(math.acos((a + c - b)/(2 * a_root * c_root)))) if angle1 == 60 or angle2 == 60 or angle3 == 60: count += 1 print("{0}: a={1}, b={2}, c={3} is a triangle".format(count, a, b, c)) print("(A,B,C) = ({0},{1},{2})".format(angle1,angle2,angle3)) print("The value of bc/a = {0}".format(b*c/a)) print(" ")
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My god I've been scratching my brains out with this exam and couldn't find the questions solved anywhere. I'm really really thankful for this video. Really, thanks man.
Thank you for watching!
⭐in section 2. (3)For ppl who are wondering why the answer was a right triangle is because. **note:squrt = square root**
1.)we have 216 unique dice rolls 201 are triangles (but some aren't unique) so we categorize them using side lengths 1. Equilateral triangles 2. isosceles 3. scalene triangles, there are only 6 possible equilateral triangles squrt(1)squrt(1)squrt(1) squrt(2)squrt(2)squrt(2) ..., for isosceles triangles draw a graph of an arbitrary isosceles triangle (given) a 60degree angle, in all 3 cases of the 60degree angle being one of the base angles or the top angle all angles must be 60degrees meaning we don't have any "pure" isosceles triangles (Define: Isosceles triangles are triangles with at least 2 congruent sides) so all of our isosceles triangles are equilateral. next, we find scalene triangles using combinations (6choose3) we get 20-1(from section2.(1)we know that 1 squrt(2) squrt(6)is not a possible triangle) unique scalene triangle.
2.)Write down all possible 19 scalene triangles using a tree diagram and canceling out permutations now we are left with 25 triangles (6equalateral from previously and 19scalene)for the equilateral triangle the maximum of bc/a is 6*6/6 = 6 .for the other 19scalene triangles we can categorize them further by 1. Right scalene triangles 2. non-right scalene triangles. by using properties of the special 60-30-90 triangle from the 19 scalene triangles we have 5right triangles that have a 60degree angle in them with maximum bc/a = 12. the other 14 non-right scalene triangles. we only consider ones that possible bc/a >or= to 12 (we need to check if there're any other triangles that bc/a exceeds 12)we have 4 candidates being squrt(1)squrt(3)squrt(5) , squrt(1)squrt(3)squrt(6), squrt(1),squrt(4),squrt(6) and squrt(2)squrt(5)squrt(6) which using the cosine law cannot be triangles with 60degree angle in them.
✨✨Thus right scalene triangle with side length squrt(1) squrt(3) squrt(4) is the unique triangle that is the maximum of bc/a.✨✨
Finally got it from this channel,appreciate you .best of luck
Thank you for watching!
In section 1(1) we can use another property loga(b)*logc(d)=logc(b)*loga(d). And then we move all the bases of the logarithm to the left like that: log3(4)*log4(4)*...*log2(2020). It's just my solution and its more intuitive for me)
hey! I was wondering what the process you did around 09:00 to divide was called so I can learn how to do it myself, thanks
Hi sorry for the very late reply! The process is called 'polynomial long division'
@@KomeisChannel oh no worries, cheers!
It's always funny how this shit has you sweating hard first time you learn it then you can do it close eyed lol, keep up the good work!!
Thanks!
How can you find the first answer without a calculator???
4:10 i dont really get that part, why is the quadratic equation has only 1 solution? how to know that?
What you did to cancel log5 with log2019?
He didnt; he skipped the process of all the divisions by putting three dots between log5/log4 and the last term
Use the theorem
logb(a) = logc(b)/logc(a) 😃
Thank u
i was waiting ur vd for a long time
Thank you for watching!
idk if you stil read the comments here, but in section 1(5) 12:00 , wouldn't it be easier to apply the tan(theta)= [(m2-m1)/(1+m2m1)]
m here is slope, so we can use the numbers in the brackets as slope
and [ ] here stand for absolute number
i cant remember the formula name unfortunately, but it worked here as well as yours!
Hi, thank you for your comment and for sharing your solution! (and yes, I don't check the comments every day but I do sometimes!)
Yes, your method sounds quicker 👍
In section 1 number 4, how did you infer that the original equation is cubic just from two divisors ? Could you please elaborate on that?
I'd like to know where can i get both MEXT and EJU old questions.Where can I found pdf files?
Hi, I don't have any old EJU questions unfortunately. But I have MEXT exams between 2006 and 2015 (some questions are missing). They are not available on the Internet and I got these files from a generous viewer. I could send them to you if you could send me an email at komei.biz@gmail.com 👍
I've sent ,pls check your email.Thank you.
@@KomeisChannel emailed you :)
Send me mext papers2005 to all
I can't understand why did you assumed that f(x) is a third degree polinomial function. It is divided by a squared function, but it doesn't mean that it is cubic. It could be a fourth degree or higher
that's true but you have to think about what i would call the test strategy, they would not put an unsolvable problem in the test, there fore if it were a fourth degree or higher there would be 3 variables and it would be unsolvable
but this does not mean that every task will go the way you want it to go, but every task will be solvable, (if it is not, that's a whole another problem)
in the 2 (3) it was wrong to assume that it was a right triangle, but there is a comment that explains why that is the case, even if it's assumed, basically there are only 19 candidates of which only a few satisfy a 60 degree angle, and the answer happens to be a right triangle, there is probably some geometric math that i don't know that would explain why, but it doesn't matter as long as there is a way to find the answer, even if it 10 pages long, it only matters that it is correct
for the polynominal question how did you find that f(x) is a cubic function?
Please can you also for Physics?? 2020,2018 and 2019
can you do chemistry too?
Hi, unfortunately I don't teach chemistry 😔
@@KomeisChannel oh, btw thnx for the maths one😘
Did you find any for Chemisty?
I never had chemistry in my life but I have to study for the exam lol
Thanks a lot sensei. Can u make a video of 2019 mathsB. And can tell how to acess all the previous year mext paper detailed answers.
Where can we get the syllabus for the MEXT scholarship exam? I'm really confused about what topics to prepare for. Help!!!
It really does not have a syllabus, but I use the syllabus for EJU. Both exams have almost the same topics. Therefore if you have EJU you have mext. You can search for it as "EJU sullabus for 2013"
If you took the math stream for your A/Ls the topics are pretty much the same as the ones in our syllabus (assuming you're sri lankan from your username 🙃).. you have to focus on coordinate geometry, functions and integration for math b, bc most questions are based off of those lessons
@@Warenya with due respect,m a Nigerian and I would love to partake in the scholarship exam..is there any syllabus or topics for Maths
Thank you SO MUCH for this video, it was very instructive and sooo helpful 🙏
16:54 Why once u find that c < 4 u say that no triangles can be formed with c=4, c=5, c=6, specifically. So that would mean c>7 is possible, but we just stated c
Because the question is asking on a dice. which has 6 faces.
Hello brother plz provide me all pdf of questions of MEXT Scholarship undergraduate level if you have it. I am unable to download it form the website as they have removed it from there... Plzz brother i will be thankful to you..
Broo which book is best for mext
Where did you learn probability? Did you use any book?
Great vídeo👍
I've got one question. On the section 1 (4) why do you multiply the resulting expression by ax?
Hi sorry for the very late reply!
Since we don't know the coefficient of x^3, we call it 'a' temporarily, and and when we do the long division, we have to multiply x^2 by ax because the quotient (ax + b + 2a) times the dividend (x^2 - 2x + 1) must become ax^3 + bx^2 + 2x + 3.
I hope this makes sense 😅
Please do chemistry too 🥺🥺🥺
Can you solve the MEXT math exam 2018 for specialized training? Your videos really help me, thanks a lot!!!
Hi, just uploaded the Special Training College maths exam (2018) on my channel 👍
@@KomeisChannel Thanks dude, you are great
In section 2.(3) how did you know that the triangle is a right triangle?
Hi sorry for the late reply and that's a good question.
As you pointed out, the question doesn't state that the triangle is a right triangle. However, when the sides are root1 ~ root6 and when one of the angles is 60 degrees, you can only create either an equilateral triangle or a right triangle.
It's a lot of work to prove this so I wrote a code that confirms this is true (you can use free websites like www.online-python.com/ to run this code):
import math
count = 0
for a in range(1,7):
for b in range(1,7):
for c in range(1,7):
a_root = math.sqrt(a)
b_root = math.sqrt(b)
c_root = math.sqrt(c)
if ((a_root + b_root) > c_root) and ((a_root + c_root) > b_root) and ((b_root + c_root) > a_root):
angle1 = round(math.degrees(math.acos((a + b - c)/(2 * a_root * b_root))))
angle2 = round(math.degrees(math.acos((b + c - a)/(2 * b_root * c_root))))
angle3 = round(math.degrees(math.acos((a + c - b)/(2 * a_root * c_root))))
if angle1 == 60 or angle2 == 60 or angle3 == 60:
count += 1
print("{0}: a={1}, b={2}, c={3} is a triangle".format(count, a, b, c))
print("(A,B,C) = ({0},{1},{2})".format(angle1,angle2,angle3))
print("The value of bc/a = {0}".format(b*c/a))
print("
")
@@KomeisChannel oh wow tysm
and 1 more question is there a specific theorem covering this topic? because I'm confused as to how I'm supposed to know this in the exam room.
which level or grade of math is this ?
High school from Geo to Calc
Thanks a lot. Really appreciate it.
Thank you for watching!
I wanted to apply next year but seeing this I realized I have no chance of approving that exam ;'/
Komeiiii thank you a lot (new sub :)), LOVE your videos!!! Could you solve MEXT Undergraduate Maths B 2018 exam?
Hey, también estoy estudiando para la beca. Quieres ser mi compañero de estudio?
I hope I can soon!
Sir Komei, please can we talk on gmail, please 🥺🤲🏽