Its class 9 question in india i already solved these type question
@Antriksh-nx1bug5 ай бұрын
For olympiad practice 😂
@shafin33655 ай бұрын
Bro, it's class 8 Bangladeshi general question, too 😅@@Antriksh-nx1bug
@Souparno_Biswas4 ай бұрын
@@shafin3365 As a person who is aware of both the Indian and bangladesi syllabus. Definitely it is not a part of the Bangladesi Syllabus. We have step below questions like this in CBSE and half a step below in ICSE of 9th standard.
@Flashy_itachi4 ай бұрын
@@Souparno_BiswasAs a Indian who is in cbse board these types of questions come in class 10th maths chapter 4 quadratic equation
@Hrishi020053 ай бұрын
Use Complex number X=cosa+i sina 1/x=cosa-isinb X+ 1/x = √3 =>2Cosa=√3 Cosa=√3/2 a=π/6 X^100 + 1/x^100= 2 cos 100.π/6 => - 2 cosπ/3 =>-2(1/2) => -1 ##
@tanishdesai76523 ай бұрын
Fastest solution thanks 🙏🙏🙏
@Smoked.X2 ай бұрын
Wow not every aspirant is maths Major.😂
@Hrishi020052 ай бұрын
@@Smoked.X yes bro, But this is the fastest way to solve this problem
@Smoked.X2 ай бұрын
@@Hrishi02005 I regret not studying maths seriously from 10th standard and onwards.
@Hrishi020052 ай бұрын
@@Smoked.X but still you can also study higher mathematics by your own self
@kavyabhojwani27979 ай бұрын
You're INSANE if you think the average score is 50%. That's what the 2-3% people score in JEE Mains.
@nel_tu_4 ай бұрын
@@boredlife 50th percentile is median not mean
@nagasaiprajith23024 ай бұрын
No it's mean here in India 😢
@JEE-oq1me4 ай бұрын
~14% students according to my calculation from my shift, 27 january.
@karthikchowdary41844 ай бұрын
@@JEE-oq1me lol, my fkd up shift😭😭
@hemangachandragiri84 ай бұрын
@@JEE-oq1methat's the easiest shift we had so far
@krishnagovinda-gc8je10 ай бұрын
Just take 2 tests
@tamannarath4 ай бұрын
they do :)
@KRISHNA-jy3dd4 ай бұрын
Yes we do and the combined score of those 2 tests is 50% hardly 😂
@laxmankumavat74294 ай бұрын
seems your score is 0%
@shahanshahpolonium4 ай бұрын
real
@atharvarathod42623 ай бұрын
'A person who thinks'
@Tiqerboy10 ай бұрын
That second problem is confusing. You didn't extend the x > 0 requirement for it as it is obvious that the equation has no real roots for x. Therefore it makes sense to convert it to polar form of a complex number.
@dianamorningstar201010 ай бұрын
The x>0 only applies to the first question
@TheGeronimo210 ай бұрын
In the second one the only requirement would be x != 0
@pog1638410 ай бұрын
Either I screwed up or got a different answer
@tryndamereagiota853910 ай бұрын
@@pog16384wtf lol
@naturefeels-wn8qv10 ай бұрын
you can't compare real numbers with complex no... that's why including x>0 for question 2 doesn't make any sense at all
@TaylorRen10 ай бұрын
The problem is a bit confusing when "x>0" is somehow "carried over" to the second question. However, if x>0 is a constraint, the 2nd problem has no answer at all.
@hippophile10 ай бұрын
Love the complex answer. You could also look at e^(2iπ/3) and e^(-2iπ/3) and look at them on the complex graph: they have real co-ordinates -1/2 from basic trigonometry, and the complex parts cancel (conjugates), so the sum is -1.
@ericherde110 ай бұрын
3rd way to solve problem 2: start with method 1 until you reach x^6=-1, then recognize that x must be an odd twelfth root of unity. A little investigation of the other equations tells you which pair of such roots it could be, then cancel the x^96’s and recognize that x^4 will have real part equal to -1/2, and like all roots of unity, its multiplicative inverse is its complex conjugate, so the answer is -1.
@memorializers10 ай бұрын
Immediately knew we were gonna need complex numbers, since x⁶ is always positive in the set pf real numbers
@Grecks752 ай бұрын
That's basically the way I did it. With "odd" 12th root of unity you probably mean that x is a (primitive) 12th root of unity. Yes, that's the key insight into the problem.
@-A-SaptarshiDeb7 күн бұрын
🎉@@Grecks75
@Bruno_Haible10 ай бұрын
At 4:11 you can save a little computation work by noting that 100 = 6*17-2; that works nicer than 100 = 6*16 + 4.
@jaimeduncan616710 ай бұрын
This is a typical, don't be afraid, just take the direct approach kind of problem. The complex number solution is super elegant.
@toxiceditzzzz10 ай бұрын
Do you know in india these. Questions are given to children at age of 15
@Feng_Q10 ай бұрын
@@toxiceditzzzznah, question 1 is for 12-13 aged kids
@toxiceditzzzz10 ай бұрын
@@Feng_Q my mistake in India too approx at this age only
@VBM3754 ай бұрын
@@toxiceditzzzz Kuch bhi bologe kya jee toh 17 ki age mein hota hai 😂
@susantparida83694 ай бұрын
@@VBM375padhai to 15-16 se shuru na, 11th me hi complex nos.
@cocolasticot902710 ай бұрын
For the complex solution, just start with the polar form z=aexp(iθ), then use the exponential form of cosine : 2.cos(x) = exp(ix)+exp(-ix)
@uwelinzbauer397310 ай бұрын
In both cases I multiplied by x and got a quadratic equation, then I solved for x. This x I inserted into the second part of the question. That worked well, I compared my found values to those from the video. In the second question I had to deal with complex numbers, that was a little difficult, but also worked. Nice and interesting video, I admire the elegant alternative solving method, I had no idea of. Best greetings!
@epikherolol818910 ай бұрын
Yeah but u don't really solve it by simply plugging in values. Coz usually these types of questions are asked in our Ntse(National Talent Search Examination) exam which is a national level competitive exam and in this we don't have a lot of time to solve each question
@uwelinzbauer397310 ай бұрын
@@epikherolol8189 I know. I am glad, that I am not a student and that I am not going to have an exam. Sometimes I watch the videos, pause them and try to find the solution. Just for fun, without time pressure. Sometimes I am lucky and I figure it out, but sometimes I fail and then I am lucky, if I understand the explanation, as some questions I find really hard. But it is always interesting for me, and I try to keep up. Best greetings!
@epikherolol818910 ай бұрын
@@uwelinzbauer3973 yea in my free time I also solve without putting time limit tho
@permutating9 ай бұрын
In case of the first question you could square the required value to get x+1/x + 2 which is 7 hence the answer is root 7
@permutating9 ай бұрын
oh wait that's what this guy did too nvm lmaoo
@wernerviehhauser9410 ай бұрын
definitely prefer the complex numbers route. That just feels so much more familiar.
@jacekpliszka5326Ай бұрын
Another solution similar to the 1st one: we set f(n) = x^n+1/x^n f(0)=2 f(1)=sqrt(3), f(a)f(b) =f(a+b) + f(a-b), f(a)^2 = f(2a) + 2 , f(2) = 3-2=1 f(4)=1-2=-1 and from here f(2^k) = -1 for k>1 , then f(a+b) = f(a)f(b)-f(a-b) and now f(100)=f(64+36)=f(64)f(36)-f(28)=-f(36)-f(28)=-f(32)f(4)=-(-1)(-1)=-1
@jacekpliszka5326Ай бұрын
Or using f(3)=0 as in the video: 0=f(3)f(a)=f(a+3)+f(a-3) => f(a+6) = - f(a) and f(a+12) = f(a) and f(100)=f(8*12+4)=f(4)=-1
@josephmw43110 ай бұрын
What I learnt about mathematics is that to solve hard problems you are required to use your critical and creative thinking. You'll find out that in today's job market, people who are critical and creative thinkers are on high demand because they are able to solve hard problems. Such kind of people can earn lots of money. My people, you have to understand that mathematics was designed to train you to have critical and creative thinking.
@solanaceousАй бұрын
Yes but in school you are taught to do opposite. This is why I do olympiads math. It trains your intuition and ability to break problems into sub task.
@LogintoMaths4 ай бұрын
The first problem is very obvious and I solved within 15 seconds 😅. In India these problems are considered to be the giver of free marks😂
@Bv-yl5dg4 ай бұрын
Bruh. Not just india, it's obviously a warmup for those who are unfamiliar with the process so they can better understand the second question
@Physics_scholar693 ай бұрын
Yeah ,these were in 8 th standards
@Mathmaniac-vw9ip3 ай бұрын
Stop being delusional. Every people in every country after standard 8 or atleast maximum standard 9 can do it, unless, they never paid attention to the class.
@Adel69702Ай бұрын
Ok pajeet
@Premium_Youtube_UserАй бұрын
@@Adel69702mujeet trying harder
@FreestyleViewer10 ай бұрын
In India, competitive exam candidates are informed of the following basic concepts. Your case involving √3 is one of them. When dealing with equations having complex numbers as roots, the first step is to check if any of the following three forms are present (where "p" can be positive or negative without making a difference): |Form 1| When the equation has the form x^2 + px + p^2 = 0, it leads to a specific conclusion: x^3 = p^3. |Form 2| When the equation has the form x^2 + √2px + p^2 = 0, it leads to a specific conclusion: x^4 = (-p^4). |Form 3| When the equation has the form x^2 + √3px + p^2 = 0, it leads to a specific conclusion: x^6 = (-p^6). An interesting fact follows: In exams at this level, questions are asked based on the above three forms when complex roots are involved. This is because these are the most straightforward cases. Among these, |Form 1|, |Form 2|, and |Form 3| are associated with taking complex numbers at 60°, 45°, and 30° on the Argand Plane, respectively. These are essentially fractions of 180°, namely one-third, one-fourth, and one-sixth, which is why raising these mixed numbers to the power of 3, 4, and 6 respectively yields real numbers.
@Harsh-wi1gx10 ай бұрын
Thanks😐
@twinkle_pie10 ай бұрын
Another way to solve the first problem is to assume x = y² Then √x + 1/√x = y + 1/y And from the first equation , we have y² + 1/y² = 5 (y + 1/y)² - 2 = 5 ( y + 1/y)² = 7 y + 1/y = √7 And as y² = x y = √x So we get our final answer as √x + 1/√x = √7
@unknownwarrior82694 ай бұрын
No need to insert value as y
@unknownwarrior82694 ай бұрын
We in india solve these problems in 8th grade
@twinkle_pie4 ай бұрын
@@unknownwarrior8269 I didn't ask you where you are from and in which grade you guys solve these kind of problems , I just stated an alternate method to solve the discussed question . It's cool that you guys solve these problems from a young age but isn't India's education system essentially the worst
@unknownwarrior82694 ай бұрын
@@twinkle_pie i agree
@livinginsidegemses4 ай бұрын
@@twinkle_pie wait they actually teach stuff in us schools? I thought you guys only had school sh**tings.... btw africa is not a country and oh yeah europe is a continent and no the earth isn't flat and yes your forehead measures 2 football fields... us education system is the biggest joke in the world. cope.
@MadaraUchiha..4 ай бұрын
These were one of the easiest questions. You should try JEE Advanced maths.
@hrisavmandal487317 күн бұрын
Yes these are so easy, but in SSC exams you have to do it in 15 seconds just to pass the cutoff..
@tacthib139610 ай бұрын
For the first problem I found it easier to multiply both sides of the equation by x and then get x²-5x+1=0 and then solve for x to solve the problem.
@lupus.andron.exhaustus10 ай бұрын
I took the same path, but somewhere on my way I went into a trap, I think: x² - 5x + 1 = 0 x² - 5x = -1 (x - 2.5)² = - 1 + 6.25 (x - 2.5)² = 5.25 x1 = 2.5 + sqr(5.25) ~ 4.7912 x2 = 2.5 - sqr(5.25) ~ 0.2087 So I get two positive values for x, which both satisfy the condition x>0, but only x1 is correct for the first equation. Where did I get it wrong? 🤔 What was your solution?
@tacthib139610 ай бұрын
@@lupus.andron.exhaustus x2 is also correct for the equation isn't it ?
@Tiqerboy10 ай бұрын
@@lupus.andron.exhaustus both of these are correct. You can see from the original equation, that the solution must be a number and then its reciprocal. 1/4.7912 = 0.2087. Remember the original problem didn't ask you to find x, but if you work out the solution of what they want, you can use either value of x and get the same answer.
@robertwagner901410 ай бұрын
I think you are correct. those both would come to to square root of 7 in given formula.@@lupus.andron.exhaustus
@lupus.andron.exhaustus10 ай бұрын
@@Tiqerboy You're right. Must have been due to a lack of coffee. ;) Thanks!
@fantasypvp10 ай бұрын
In a level maths we're just taught to multiply by x and solve the quadratic then expand out for the solution, we would usually only get real roots though, in further maths we would just do the complex method
@YoungPhysicistsClub172910 ай бұрын
nice
@georgesmelki19 ай бұрын
Problem 2: very good method, for someone who has never heard of complex numbers...Otherwise, solve the quadratic equation x^2 - (root3).x +1=0 which has two complex solutions, cos(pi/6)+_i.sin(pi/6), then apply de Moivre's formula.
@zaikindenis177510 ай бұрын
Thank you for the video. If x+1/x=a and f(k)=x^k+1/x^k, then f(k+1)=f(k)*a-f(k-1). Then I found a pattern in this sequence, from some point it repeats every 9 members. Thank you.
@Gerardo_profe10 ай бұрын
I think the repetition is every 12 terms. If k = 1; f = sqrt(3) * If k = 2, f = 1 ** If k = 3; f = 0 *** If k = 4, f = -1 **** If k = 5; f = -sqrt(3) If k = 6, f = -2 If k = 7; f = -sqrt(3) If k = 8, f = -1 If k = 9; f = 0 If k = 10, f = 1 If k = 11; f = sqrt(3) If k = 12, f = 2 If k = 13; f = sqrt(3) * If k = 14, f = 1 ** If k = 15; f = 0 *** If k = 16, f = -1 **** So f(100) = f (12x8 +4) = f(4) = -1
It’s pretty obvious that z^2 - sqrt(3)z + 1 = 0, and both roots complex. thus both roots take form re^(+-it), but product is 1 by Vietta’s Formula ==> r = 1 ==> z = exp(+-it) = cos(t) +_ i sin(t) hence z + 1/z == z + z* = 2Re(z) = sqrt(3) ==> Re(z) = sqrt(3)/2 = cos(pi/6) ==> t = pi/6. Using De Moivre, ==> z^100 + z^-100 = 2Re(z^100) = 2cos(4pi/6) = -1
@mangakhoon4517go3 ай бұрын
A 12th grade won't know this though
@centmillionaire10 ай бұрын
I solved directly by complex numbers and went through a tedious process of calculating successive degrees of x^n + 1/x^n, only to observe that values repeat over a period of 12 (and "anti-repeat" for a period of 6). I had no clue as to why this happens and your solution is so insightful! Bravo 🎉
@auni40783 ай бұрын
brooooo i did exactly same thing in reverse order first i find x^n + 1/x^n for n =1,2,3,4... 16 on n = 16 i observed it's repeating for every 12th n than after like 5 min that complex number method came in mind T_T
@Grecks752 ай бұрын
It happens so because x is a (primitive) 12th root of unity, so x^12=1. And then, of course, x^(n+12)=x^n, so things get periodic. You can visualize that by looking at the rotations of the unit circle with an angle that is a rational part of 2*pi. That's exactly what complex multiplication does when the multiplier is a root of unity.
@zcubing579210 ай бұрын
The first one is actually very basic and straightforward. The second one was a bit tricky. Here's how I did it, 01.x+(1/x)=√3 02.x^2+(1/x^2)=1 03.x^3+(1/x^3)=0 04. By multiplying equation 02 and 03, x^5+(1/x^5)=-√3 05.x^10+(1/x^10)=1 06.x^20+(1/x^20)=-1 07.x^40+(1/x^40)=-1 08.Multiplying equation 06 & 07, x^60+(1/x^60)=2 09.x^80+(1/x^80)=-1 10.Final step:Multiplying equation 09 & 06, x^100+(1/x^100)=-1 I know it looks ridiculous but it works for me😅.
@kajaldey265610 ай бұрын
that's actually genius
@zcubing579210 ай бұрын
@@kajaldey2656 Thanks😊
@user-lr3es8kj9m10 ай бұрын
Ayoo you're a cuber too?
@Grecks752 ай бұрын
Sure, that works, and I bet you had a lot of fun with multiplying and the distributive law, and writing it all down. 😂And it's not a silly approach either. Fun fact: You didn't even need knowledge about complex numbers for that, just basic algebra, great! (Having said that, the problem can be solved with a bit less effort, though, if you DO know something about complex numbers. 😉)
@zcubing57922 ай бұрын
@@Grecks75 I know it can be done in a much easier way with complex numbers but I have my limitations. I had solved the equation before I approached in the way I did and found out x=(√3/2)±(i/2) but I have no device or mean which could now give me the answer of (x^100+1/x^100) by entering the value of x. So I used the long way. I hope you understand.
@Grecks752 ай бұрын
Regarding the second problem: The first method (using algebra) is clearly superior. But whatever your way to the solution is, here are a few key insights into the problem that will help you tackle the problem: 1) Realize that x _cannot_ be a real number. Why? 2) From the fact that x is _not_ real but x + 1/x _is_ , derive that the absolute value of x _must_ be 1. How? 3) From the above and the given equation, derive that x is in fact a primitive 12th root of unity. How? (Hint: You can do it by calculating the arg(x) when you know abs(x)=1 and twice the real part ox x is given as sqrt(3) by the given equation, or you can do it by algebraic manipulations as shown in the video). From here on the problem becomes trivial. (Btw: I find that solving the quadratic equation for the two complex-conjugate roots that are x is rather boring, but, of course, it also gets you to the answer.)
@Shreeraksha-ey8nl9 ай бұрын
For 2nd question solve quadratic equation, root is omega , omega to the power 100 + 1/ omega to the power 100 = omega to the power 100+ omega square to the power 100 , since omega and omega square are multiplicative inverse of each other Omega to the power x = omega to the power x/3 Then equation becomes omega + omega square Now wtk omega + omega square +1 =0 This implies that omega + omega square = -1
@himangshubaruah11404 ай бұрын
I think this question is wrong itself because, the value of x + 1/x is either equal to any value from 2 to + ve infinity (if x is +ve) or equal to any value from -2 to -ve infinity (if x is -ve). So any number between -2 and 2 is not possible for any value of x. So x+1/x = square_root(3) is not possible for any value of x.
@KhushChadha4 ай бұрын
Bingo! Ur right
@ParvGupta-so1fcКүн бұрын
That comes from the AM-GM inequality which is only valid for real numbers. Inequalities are not defined for complex numbers.
@NOONO54 ай бұрын
9th class ch1 questions
@knotwilg3596Ай бұрын
*Simplest: quadratic equation and reducing powers* Solve x + 1/x = V3 by the standard formula for thequadratic equation x² - V3x +1 =0 x = (V3+i)/2 (or (V3-i)/2 which will follow the same logic) x² = (3+2V3 i -1)/4 = (1+V3 i)/2 x³ = (V3 +3i + i -V3)/4 = i hence x^6 = -1 and x^12 = 1 x^100 = x^96 * x^4 = x^4 = ix = (-1+ iV3)/2 and 1/x^100 = 1/x^4 = (-1 - iV3)/2 x^100 + 1/x^100 = -1 *Alternative: reducing powers without solving for x* (x+1/x)² = x²+1/x² +2 => 3 = (x²+1/x²) +2 => x²+1/x² = 1 (x+1/x)³ = x³+1/x³ +3(x+1/x) => 3V3 = x³+1/x³ + 3V3 => x³+1/x³ = 0 => x^6+1=0 Hence x^6 = -1 and x^100 = x^96*x^4 = (-1)^16 * x^4 = x^4 x^100 + 1/x^100 = x^4 + 1/x^4 = (x²+1/x²)² - 2 = 1-2 = -1 *Alternative: reducing powers, without fractions or roots* x+1/x = V3 => x²+1 = V3 x => x^4 + 2x² + 1 = 3 x² => x^4 = x² - 1 => x^6 = x^4 - x² = x² - 1 - x² = -1 => x^100 = x^96 * x^4 = x^4 = x² - 1 and 1/x^100 = 1/x4 = - x² => x^100 + 1/x^100 = -1 *Alternative: polar coordinates* First observe that x² - V3x + 1 = 0 has no real solutions, since 3 - 4 < 0. x = r (cos a+ i sin a), working with complex numbers in their polar representation (r = radius, a = angle) 1/x = (cos a - i sin a)/r x+1/x = (r+1/r) cos a + (r-1) i sin a = V3, which is a real number, hence the imaginary part is 0 => r=1 (since a is not 0 for a non-real number) Hence x = cos a + i sin a and 1/x = cos a - i sin a 2 cos a = V3 => cos a = V3/2 and a = pi/6 or -pi/6 x^100 = cos (100 pi/6) / sin (100 pi/6) ~= cos (4pi/6) + i (sin 4pi/6) = cos (2pi/3) + i sin (2pi/3) = -1/2 + iV3/2 1/x^100 = cos (2pi/3) - i sin (2pi/3) = -1/2 - iV3/2 x^100 +1/x^100 = -1
@Ahaandeeps4 ай бұрын
I put that second equation into desmos and, It seems impossible because the lines never touch (It does not have a solution) or maybe it's irrational. It's either around 2^.5 or 0.5 something. It also suggests that 1^.5 may not be 1. Also, I am talking about the value of x. But ((3+5^(.5))/2)^.5 is super duper close. It was pretty hard to find that.
@martinpetkoski25474 ай бұрын
The solution is complex, so it doesn't show up on Desmos.
@jimcameron6803Ай бұрын
Very interesting. For the second problem, it turns out that if n is divisible by 4, x^n + 1/x^n is equal to -1 if the binary expansion of n has odd parity and 0 if it has even parity.
@ibnSafaa4 ай бұрын
Just considered x is a complex number so we can write it in the Polar formula x = cos(z)+isin(z) And his numerical companion is x' x'=cos(z)-isin(z) x'=1/(cos(z)+isin(z)) =1/x So we can now that x plus 1/x is 2cos(z) x +1/x = 2cos(z) =sqrt(3) 2cos(z)=sqrt(3) cos(z)=sqrt(3)/2 So z = π/6 +2kπ When k=0 » z=π/6 x¹⁰⁰=cos(100z)+isin(100z) So x¹⁰⁰ +1/x¹⁰⁰ = 2cos(100z) = 2cos(100π/6) 2cos(50π/3)=2cos(2π/3)=-1
@eternal4564 ай бұрын
I have started to realise that in mathematics, patience is the key because you never have obvious answers to problems like these. So the key is patience and simple but effective approach. We need get rid of the quick solving mentality to get better if we are not a master at concepts. Great video!
@CRnk15310 ай бұрын
For first problem i just made form "x^2-5x+1=0 with x>0", thats would be very easy to solve
@ishaankumar45874 ай бұрын
Those questions were pretty simple , most students have already done many variants of that question in india thus jee uaually doesnt ask these questions
@mikeeisler6463Ай бұрын
If x^6 = -1 then the cube root of x^6 is x^2 and the cube root of -1 is -1. If x^2 = -1 then then x = i. i^4 = -1 * -1 = 1, (i^4)^25 = i^100, 1^25 = 1. So x^100 = 1. 1 + 1/1 = 2
@nickh50864 ай бұрын
Just declaring x>0 doesn’t do it. This implies x is real. Complex numbers cannot be compared with real numbers. If this a real test, then the ones proposing the test should get an F
@henrytang22032 ай бұрын
I definitely prefer the complex numbers approach to the last problem. It's simple and to the point, with much less reliance on algebraic manipulations.
@haidarjafarsaraf751310 ай бұрын
For the second problem, the square of x+1/x can be takan 50 times in a row. İn the third one , its understood that -1 will always come
@725etw7w10 ай бұрын
Explain please how x^6=-1, is it positive everytime?
@Tiqerboy10 ай бұрын
x is a complex number in the second example. I think the presenter should have made that clear.
@725etw7w10 ай бұрын
@@Tiqerboy thank you
@Firdouse.4 ай бұрын
As an average Indian, i studied this when I was 14yrs old 👍
@abaddon60783 ай бұрын
I hope that was sarcasm...
@--_-Anonymous-_--3 ай бұрын
@@abaddon6078 he is talking facts, no sarcasm at all...
@akshayaggarwal514818 күн бұрын
Yeah, when we are in 11th, we have to study this. This is actually the easiest question in this chapter-complex numbers
@vivekvk45vk10 ай бұрын
These type of problems in India are generally asked in competitive exams like SSC and other government exams.
@prasunbagdi611210 ай бұрын
Hmm that's means only intelligent people join the government and not the corrupt ones, right? Right?
@JenyBhatt-be1ed4 ай бұрын
Just came up with another way to solve the eqn after 5:21 ...got the eqn (x⁴=x²-1) from 2nd and 3rd eqns,I assumed x²=-(w)² where omega or w = cube root of unity and as w⁴=w, we can easily manipulate it into finding (w)¹⁰⁰... Also w+w²+1=0(where omega and omega² are complex cube root of unity...we got w+w²+1 by sum of roots= -b/a in x³=1) Btw ur approach is also amazing and eular form makes it easy to understand 😊
@lafq167Ай бұрын
I used another method (based on the same calculation as 2a) : let a(p) = x^p + 1/x^p. We prove that a(p+q) = a(p)*a(q) - a(p-q) and in paralllel that a(2^k) = -1 for k >1. Then, as we want to know a(100) = a(64+36) we have a(100) = a(64)a(36) - a(28) = -a(32 + 4) - a(28) = - [a(32)a(4) - a(28) ] - a(28) = -1
@HarshSharma-kk9un10 ай бұрын
In india we learn some results i.e. If x+1/x =2 then x=1 X+1/x =(-2 )then x = -1 x+1/x = 1 then x^3 = -1 x+1/x = -1 then x^3 = 1 x+1/x = underroot 3 then x^6= -1 x-1/x = underroot3 then x^6=1
@rdspam8 күн бұрын
Theres a test in the US where the median score last year was 10 out of 120. In 2022 the median score was 1. A median of 0 or 1 is historically common.
@oaneric188 ай бұрын
I have a similar problem from my maths teacher: If x+1/x is a whole number, prove that x^n+1/x^n is also a whole number (x is real and n is a positive integer).
@sirajzama8080Ай бұрын
As a jee aspirant i could tell you how to do this first square on both sides and make numeric term on one side and variabke term on another you will find that x²+(1÷x²)=1, Now similarly square again and separate terms you will get x⁴+(1÷x⁴)=-1,now square again and separate the terms you will get x⁶+(1÷x⁶)=-1 again this will be true for x^n+(1÷(x^n))=-1 for n times squaring therefore at 50th time it will be -1.
@shasha653821 күн бұрын
the second solution is more accurate because in first solution , even power of x can never be negative.
@MrWarlls10 ай бұрын
For the solution 2b, you can notice that e^(2i*Pi/3) + e^(-2i*Pi/3) = 2cos(2Pi/3)
@paulortega5317Ай бұрын
Let f(n)=x^n+1/x^n. Note f(0)=2 Use formula f(u+v)=f(u)*f(v)-f(u-v) For first problem let y=sqrt(x). f(2)=5. Find f(1). f(2)=f(1)*f(1)-f(0)=f(1)^2-2=5 f(1)^2=7 f(1)=sqrt(7) For second problem, note if f(n)=2 then f(kn)=2 for k=0,1,2,3,... the sequence repeats after n terms Given f(1)=sqrt(3) f(2)=f(1)*f(1)-f(0)=3-2=1 f(4)=f(2)*f(2)-f(0)=1-2=-1 f(8)=f(4)*f(4)-f(0)=1-2=-1 f(12)=f(4)*f(8)-f(4)=1+1=2 Series repeats after 12 terms 100 = 4 mod 12 so f(100)=f(4)=-1
@1TuRneD14 ай бұрын
I doubt the ans for 2nd question is -1 because by using AM GM inequality you can prove that "t+1/t" will never have solution for t in range (-2,2)
@mapsm75429 күн бұрын
In part 2A, once u get x2 + 1/x2, u could simply raise both sides to power of 50 and get the answer
@shugneechk606010 ай бұрын
It took me around 10 to 20secs to solve these questions and im in grade 11 They are nowhere near to actual jee advanced questions(except for the easy one)
@kartikpundir7649 ай бұрын
Me being a grade 11th math student from India , solved this using complex numbers method and Euler form 🙃(second method).
@user-bl4zj6wt9v10 ай бұрын
There is another way that is easier to think of though it's a bit more process. FIrst step, try to get an answer for x^10+1/x^10, then you need to calculate (x^8+1/x^8)*(x^2+1/x^2). In the process you need to calculate (x^4+1/x^4)*(x^2+1/x^2). Once you get x^10+1/x^10=1, then you can get the final answer is -1.
@gibbogle3 ай бұрын
x^10 + 1/x^10 is my f(5). My method is clearly like yours, but I don't see how knowing x^10 + 1/x^10 gives you x^100 + 1/x^100.
@user-bl4zj6wt9v3 ай бұрын
@@gibbogle Sorry, my bad. I made a mistake (x¹⁰+1/x¹⁰)²=x¹⁰⁰+1/x¹⁰⁰+2...though it leads to x¹⁰⁰+1/x¹⁰⁰=-1, but it is wrong. I failed in this question though it can do (x¹⁰+1/x¹⁰)²=x²⁰+1/x²⁰+2, x²⁰+1/x²⁰=-1, (x²⁰+1/x²⁰)², (x⁴⁰+1/x⁴⁰)², (x⁸⁰+1/x⁸⁰)(x²⁰+1/x²⁰)=x¹⁰⁰+1/x¹⁰⁰+x⁶⁰+1/x⁶⁰...finally you may get the right answer but the process is too complex.
The good thing is, you don’t need to know this in the real world.
@Naitik-15093 ай бұрын
In SSC cgl exam we need to solve both questions under 45 second.😊
@dipanjansarkhel40872 ай бұрын
No... You can solve problem 1 with the trick.. but it will not go for problem 2. It's a problem for JEE advanced and can't be solved with shortcut
@Naitik-15092 ай бұрын
@@dipanjansarkhel4087 but problem 2 is a PYQ in ssc exams asked many times.
@dipanjansarkhel4087Ай бұрын
@@Naitik-1509 ek bar dekh lena.. because. Jab tak mujhe lag raha hay problem 2 to complex number ke bina nahi hoga... Main UPSC aur other competitive ka preparation karta tha ek time par Mila kuch nahi😺😺😺 anyway agar ab CGL mein complex number add hua to ho sakta hay ek bar check kar lena
@Naitik-1509Ай бұрын
@@dipanjansarkhel4087 bhaiya,complex number to add nahi hua but this 2nd problem PYQ hai and teachers ise direct learn karwa dete hai. LIKE X+1/X= root2 hai. then , Xsquare +1/Xsquare=? cube root ka bhi karate hai
@jaadugar591426 күн бұрын
Bhai second wala trick se solve nhi hoga.. Apko bar bar n square - 2 krna hoga tb jake wo x k power 100 pe pahuch payenge 😊😊
@gibbogle3 ай бұрын
Interesting. I did this by a completely different method. First I squared x + 1/x and showed that x^2 + 1/x^2 = 1 Then by repeatedly multiplying by x^2 + 1/x^2, (e.g. (x^2 + 1/x^2)(x^2 + 1/x^2) = 1*1 = 1 = x^4 + 1/x^4 + 2, therefore x^4 + 1/x^4 = 1 - 2 = -1 then (x^4 + 1/x^4)(x^2 + 1/x^2) = -1*1 = -1 = x^6 + 1/x^6 + x^2 + 1/x^2 = x^6 + 1/x^6 + 1, therefore x^6 + 1/x^6 = -1 - 1 = -2 etc.) and setting f(n) = x^2n + 1/x^2n I got f(1) = 1 f(2) = -1 f(3) = -2 f(4) = -1 f(5) = 1 f(6) = 2 f(7) = 1 then it is clear that the cycle repeats, i.e. f(n+6) = f(n) x^100 + 1/x^100 = f(50) = f(44) = f(38) = f(32) = f(26) = f(20) = f(14) = f(8) = f(2) = -1 or, more directly, since 50 = 8*6 + 2, f(50) = f(2). Once again, I got the right answer by the wrong method. It was fun though.
@michallesz26 ай бұрын
z + 1/z = V3 ERROR To understand what values the expression x + 1/x takes, we need to perform a proof. The given equation shows that x cannot be zero (0) because it is divided by zero. However, for the value x=1, the result of the equation will be the number 2. For the remaining values of x, we can assume that x = a/b, so 1/x will be b/a. this way we will obtain the equation that x + 1/x = a/b + b/a. Now let's solve the equation a/b + b/a. It looks like: a/b + b/a = ( a^2 + b^2 ) / ab As you can easily see, from the given expression we can use a right triangle for analysis, where a^2 + b^2 give us c^2, which also means the area a square with side c, while "ab" is the result of the surface area of the rectangle which we will denote as S=ab. Therefore, we will obtain the formula that x + 1/x = c^2 / S. This means that we divide a square with side "c" by the area of a rectangle with sides "a, b". For these calculations, let's assume the individual values of these sides, using the Pythagorean theorem, i.e. a=3, b=4, c=5. Using this data, we obtain that c^2 = 25 and ab = 12. Now let's calculate c^2/ab = x + 1/x = 25/12 = 2+1/12. And this is proof that x + 1/x is equal to or greater than 2. However, for negative numbers it will be equal to or less than -2.
@dabest877710 ай бұрын
For the 2nd question: I actually did a very long method where I found x^4 + 1/x^4 = -1 and then continuously squared to get to x^64 + 1/x^64 and then I found x^36 + 1/x^36 = 2 so then, I multiplied those 2 equations to get x^100 + 1/x^100 + x^28 + 1/x^28 = -2 and then I found x^28 + 1/x^28 (it’s -1) and plugged it in to get x^100 + 1/x^100 -1 = -2. Therefore: x^100 + 1/x^100 = -1. A much less elegant solution than the given one, but hey! If it works, then it works
@Ayuwokibaby8 ай бұрын
Well you probably lost a lot of time
@F.M.Dostoyevsky16 күн бұрын
In quest 1, we can even take simpler approach by taking values ( sure it will take a while but its easy) if x>0 and x+1/x=5 the the value of x lies between 4< x>5 since its not given integral value of we will also consider decimals take for eg. x=4.5 which does not equal 5 and x as 5 which brings the value as 6 not 5 so take value as 4.8 which brings the value as 5 ( what we need)
@TheIntrovertYoutuber10 ай бұрын
Second question was easily doable by Complex Variables (I did it in my head with correct answer)
@shrijan1062 ай бұрын
In problem 2 if X is real then 1st condition x+1/x =root 3 is impossible.....
@ManishaMahadeoMundeАй бұрын
Yes
@lowlife_nolife604710 ай бұрын
JEE mains is the Qualifier here. Try JEE advenced.
@thechessplayer83284 ай бұрын
Pretty easy compared to the international olympiad
@Mojo.Jojo.10 ай бұрын
this is simple maths question . we usually solve these in school
@rishikumarchourasia4 ай бұрын
Average score is 10%. 50% is scored by top 3% people.
@-hc__3 ай бұрын
bro getting 300/300 in JEE Mains is not uncommon these days, 50% is a joke
@alien32003 ай бұрын
Average NEET user be like:
@devondevon436610 ай бұрын
For the see the first one 0:06 let x = (sqrt x)^2 (eg 4= (sqrt 4)^2 = 2^2 =4) Hence 1/ (sqrt x)^2 + (sqrt x)^2 =5 equation 1 Let 1/sqrt x + sqrt x = n (1/sqrt x + sqrt x)^2 = n^2 square both side 1/(sqrtx)^2 + (sqrt x)^2 +2 = n^2 5 + 2 = n^2 (substitute the value for equation 1) 7 = n^2 n= sqrt 7 easy problem
@hrisavmandal487317 күн бұрын
Now, the fact is one have to do these solutions in just 12-15 seconds just to pass the cutoff of the exam..
@premkadam260610 ай бұрын
Hey guys. The second one could have been solved much more easily using just another complex no. method. First, we know that one of the complex roots of unity (3rd root). i.e. omega, = minus1/2 + (root3)/2. When we observe the RHS. We see root 3. That is, two times (root3)/2. So after multiplying omega by (-i) (-root(-1)). We get (-i omega). We see that this is the root of the above equation (x + 1/x = root3). Hence again substituting (- i omega) in the req. value. we get omega^100 + omega^200. Using the concept of complex roots of unity, we can say that it is equal to omega + omega^2. And it's value is -1. And hence the answer. It may seem complicated when read here. But if done personally, it will be a question with a solution of just 3 lines. /
@Grecks752 ай бұрын
That's a nice and sound solution, I like it! I validated it, and it works. Please note, though, that you have only worked with one of the two possible x solutions to the given equation x + 1/x = sqrt(3), namely x=-i*omega. For completeness, you should also check that the other solution is x=+i*omega^(-1), and that this other x also gives the same result (which it does, of course).
@premkadam26062 ай бұрын
@@Grecks75 Yes, this is a quadratic equation, so if one root is (say) z, then another will be z bar. or known as z conjugate. When observed, i (omega)^(-1) is the conjugate of -i (omega). So, they will give same results. You are correct.
@averageboulderer4 ай бұрын
I actually feel this is quite easy for jee mains.
@AryanSMH3 ай бұрын
Abhas Saini student here, and I can already say all these questions you have put here are v easy.
@r8dra4 ай бұрын
Here in india we are taught to remember the values of cube roots of unity for these types of problems... A typical JEE aspirant can do this without even lifting a pen..
@mangakhoon4517go3 ай бұрын
That's the problem just fking remember
@gitarthabordoloi59133 ай бұрын
No I studied in Allen they deeply explain the concepts hardly any coaching institute makes their student remember this values.
@r8dra3 ай бұрын
@@gitarthabordoloi5913 i mean it's all good if we know how they got derived in the first place
@gitarthabordoloi59133 ай бұрын
@@r8dra in my coaching the derivation is given how it happened,why it happened etc.But I agree school students are mostly made to remember this value's without the logic.
@Torch114 ай бұрын
By solving the complex numbers in the second method if we multiply divide by iota it would become w/i (w - cube root of unity ) by using the and then we just substitute use a property of cube root of unity w² + w = -1 and it gets cancelled out and we get - 1 as the answer
@restinpeace191610 ай бұрын
Second solution is like: just find x from equation x+1/x=sqrt(3) and put the number in expression x^100+x^(-100)
@janmesh23324 ай бұрын
to be honest, the first one wasn't that hard, pretty basic. The second one took some time, but, also solvable if the thought process is correct.
@KejriwalBhakt4 ай бұрын
Solved the second one with complex numbers. It was obvious way forward
@PrajwalNayak-so5uv3 ай бұрын
7:20 How did you convert the euler form in rectangular form??
@MUJAHID9641410 ай бұрын
Mixing two different concepts of math is a nice way to explore.like in a right angle triangle there are 3 lenths base, height and perpendicular, if we take two in a fraction then we can rearrange them into 3p2 ways that is 3×2=6 ways and that's why there are 6 trigonometry functions are exist.(sorry for my bad English)
@shreya115910 ай бұрын
Wow!
@trueriver195010 ай бұрын
There goes my hope of discovering a new trig function...
@shubh786415 күн бұрын
Without even starting the video, is the answer -1? Took me like 2 mins without even using complex numbers or anything.
@Shreeraksha-ey8nl9 ай бұрын
That's too easy for a jee aspirant.... the root of 2nd equation is omega and using its properties one can easily solve all problems ( nth root of unity )
@zerocat8883 ай бұрын
It's not omega
@AbhijitMondal-th2eh9 ай бұрын
You can solve it easily using the concept of cube root of unity. x is one of the 3 cube roots of unity.
@Grecks752 ай бұрын
No, it's not. The cube of x is +/- i. Instead, x is a 12th root of unity.
@RussellSubedi10 ай бұрын
I went the 5 * 5 * 2 * 2 route. Needless to say, yours was quicker.
@Grecks752 ай бұрын
Wow, I guess that was a lot of writing. 😀
@IITJEEAspirant2025-fj6hj27 күн бұрын
My approach For problem 1. x+(1/x)=5 =>x½+(1/x)½+2.x½.(1/x)½=5+2 =>(x½+(1/x)½)²=7 =>x½+(1/x)²=7½ ans. Problem 2. x¹⁰⁰+(1/x)¹⁰⁰+2.x⁵⁰.(1/x)⁵⁰=y+2 =>(x⁵⁰+(1/x)⁵⁰)²=y+2 =>(x⁵⁰+(1/x)⁵⁰+2.(1/x)²⁵.x²⁵)²=y+2+4 =>(x²⁵+(1/x)²⁵)⁴=y+6 For x+(1/x)=3½ =>(x+(1/x))²=3 😢I am stuck
@phungcanhngo6 ай бұрын
Awesome problems and solutions.Thank so much, professor.
@vishalmishra304610 ай бұрын
*General Solution* IF A = x + 1/x and B = x^n + 1/x^n, THEN B is computable from A using the following - IF x = r e^ iT then 1/x = r e^-iT, so x + 1/x = 2r cosT = A x^n + 1/x^n = r e^iTn + r e^-iTn = 2r cos nT = B IF A = 2r cos T and B = 2r cos nT, then B = 2r cos[ n acos(A/2r) ] Here A^2 = 3 and r = 1, so T = 30 deg, so 100 T = pi/6 x 100 = 2pi x 8 + 2pi/3, so 2 cos (100T) = 2 x cos(2 pi/3) = 2 x -1/2 = -1 = x^100 + 1/x^100
@Aditya.o7o54 ай бұрын
Solving the question by just looking at the thumbnail is easier 😅
@HelloIAmAnExist2 ай бұрын
Just solve for x in the first one. It's not even that complicated/long, there's nothing complex, and the results work out. Your solution is pretty, but there's no reason to look for funny business when this just works
@mkrsinfo285912 күн бұрын
these questions are pretty easy expanding the powered binomial is one the things we learnt when we were small
@kripanshukhandelwal58684 ай бұрын
I Loved your use of the word "Simplify"... My biology brain really doubted the meaning of "simple" after that explanation !!! 😂
@akultechz23427 ай бұрын
This question is toddler in front of actual ones tbh
@HARSHJHA-ln5sg10 ай бұрын
U have done it very long , it is not too hard , just use concept of cube root of unity
@trueriver195010 ай бұрын
Yeah, especially if you happen to remember that the cube root of unity is somehow related to root3, as Indian maths student will be taught in their prep for this test. This is more of a struggle for people who haven't been prepared for this specific exam
@tom-kz9pb10 ай бұрын
I could quickly work out problem 1 in my head, but sighed and quickly gave up on problem 2, without trying much. I hope that they grade on the curve, in India.