Don't panic. There are two basic strategies to solve questions like this

[@klawtawar]

Thanks!

[@Antriksh-nx1bug]

Its class 9 question in india i already solved these type question

[@Antriksh-nx1bug]

For olympiad practice 😂

[@shafin3365]

Bro, it's class 8 Bangladeshi general question, too 😅​@@Antriksh-nx1bug

[@Souparno_Biswas]

@@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_itachi]

​@@Souparno_BiswasAs a Indian who is in cbse board these types of questions come in class 10th maths chapter 4 quadratic equation

[@Hrishi02005]

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 ##

[@tanishdesai7652]

Fastest solution thanks 🙏🙏🙏

[@Smoked.X]

Wow not every aspirant is maths Major.😂

[@Hrishi02005]

@@Smoked.X yes bro, But this is the fastest way to solve this problem

[@Smoked.X]

@@Hrishi02005 I regret not studying maths seriously from 10th standard and onwards.

[@Hrishi02005]

@@Smoked.X but still you can also study higher mathematics by your own self

[@kavyabhojwani2797]

You're INSANE if you think the average score is 50%. That's what the 2-3% people score in JEE Mains.

[@nel_tu_]

​@@boredlife 50th percentile is median not mean

[@nagasaiprajith2302]

No it's mean here in India 😢

[@JEE-oq1me]

~14% students according to my calculation from my shift, 27 january.

[@karthikchowdary4184]

@@JEE-oq1me lol, my fkd up shift😭😭

[@hemangachandragiri8]

​@@JEE-oq1methat's the easiest shift we had so far

[@krishnagovinda-gc8je]

Just take 2 tests

[@tamannarath]

they do :)

[@KRISHNA-jy3dd]

Yes we do and the combined score of those 2 tests is 50% hardly 😂

[@laxmankumavat7429]

seems your score is 0%

[@shahanshahpolonium]

real

[@atharvarathod4262]

'A person who thinks'

[@Tiqerboy]

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.

[@dianamorningstar2010]

The x>0 only applies to the first question

[@TheGeronimo2]

In the second one the only requirement would be x != 0

[@pog16384]

Either I screwed up or got a different answer

[@tryndamereagiota8539]

​@@pog16384wtf lol

[@naturefeels-wn8qv]

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

[@TaylorRen]

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.

[@hippophile]

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.

[@ericherde1]

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.

[@memorializers]

Immediately knew we were gonna need complex numbers, since x⁶ is always positive in the set pf real numbers

[@Grecks75]

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-SaptarshiDeb]

🎉​@@Grecks75

[@Bruno_Haible]

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.

[@jaimeduncan6167]

This is a typical, don't be afraid, just take the direct approach kind of problem. The complex number solution is super elegant.

[@toxiceditzzzz]

Do you know in india these. Questions are given to children at age of 15

[@Feng_Q]

​@@toxiceditzzzznah, question 1 is for 12-13 aged kids

[@toxiceditzzzz]

@@Feng_Q my mistake in India too approx at this age only

[@VBM375]

​@@toxiceditzzzz Kuch bhi bologe kya jee toh 17 ki age mein hota hai 😂

[@susantparida8369]

​@@VBM375padhai to 15-16 se shuru na, 11th me hi complex nos.

[@cocolasticot9027]

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)

[@uwelinzbauer3973]

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!

[@epikherolol8189]

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

[@uwelinzbauer3973]

​@@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!

[@epikherolol8189]

@@uwelinzbauer3973 yea in my free time I also solve without putting time limit tho

[@permutating]

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

[@permutating]

oh wait that's what this guy did too nvm lmaoo

[@wernerviehhauser94]

definitely prefer the complex numbers route. That just feels so much more familiar.

[@jacekpliszka5326]

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

[@jacekpliszka5326]

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

[@josephmw431]

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.

[@solanaceous]

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.

[@LogintoMaths]

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-yl5dg]

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_scholar69]

Yeah ,these were in 8 th standards

[@Mathmaniac-vw9ip]

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.

[@Adel69702]

Ok pajeet

[@Premium_Youtube_User]

​@@Adel69702mujeet trying harder

[@FreestyleViewer]

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-wi1gx]

Thanks😐

[@twinkle_pie]

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

[@unknownwarrior8269]

No need to insert value as y

[@unknownwarrior8269]

We in india solve these problems in 8th grade

[@twinkle_pie]

@@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

[@unknownwarrior8269]

@@twinkle_pie i agree

[@livinginsidegemses]

@@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..]

These were one of the easiest questions. You should try JEE Advanced maths.

[@hrisavmandal4873]

Yes these are so easy, but in SSC exams you have to do it in 15 seconds just to pass the cutoff..

[@tacthib1396]

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.exhaustus]

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?

[@tacthib1396]

@@lupus.andron.exhaustus x2 is also correct for the equation isn't it ?

[@Tiqerboy]

@@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.

[@robertwagner9014]

I think you are correct. those both would come to to square root of 7 in given formula.@@lupus.andron.exhaustus

[@lupus.andron.exhaustus]

@@Tiqerboy You're right. Must have been due to a lack of coffee. ;) Thanks!

[@fantasypvp]

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

[@YoungPhysicistsClub1729]

nice

[@georgesmelki1]

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.

[@zaikindenis1775]

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_profe]

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

[@SidneiMV]

*Problem #1* x + 1/x = 5 x > 0 find √x + 1/√x = A A² = x + 1/x + 2 =7 *A = √7* *Problem #2* x + 1/x = √3 find x¹⁰⁰ + 1/x¹⁰⁰ = S x + 1/x = √3 x² + 1/x² = 1 (x² + 1/x²)(x + 1/x) = √3 (x³ + 1/x³) + (x + 1/x) = √3 x³ + 1/x³ = 0 [ I didn't realize that x⁶ = -1 so I solved the problem as follows ] (x³ + 1/x³)(x² + 1/x²) = 0 (x⁵ + 1/x⁵) + (x + 1/x) = 0 x⁵ + 1/x⁵ = -√3 x¹⁰ + 1/x¹⁰ = 1 x²⁰ + 1/x²⁰ = -1 x⁴⁰ + 1/x⁴⁰ = -1 (x²⁰ + 1/x²⁰)(x⁴⁰ + 1/x⁴⁰) = 1 (x⁶⁰ + 1/x⁶⁰) + (x²⁰ + 1/x²⁰) = 1 x⁶⁰ + 1/x⁶⁰ = 2 x⁸⁰ + 1/x⁸⁰ = -1 (x⁸⁰ + 1/x⁸⁰)(x²⁰ + 1/x²⁰) = 1 (x¹⁰⁰ + 1/x¹⁰⁰) + (x⁶⁰ + 1/x⁶⁰) = 1 *x¹⁰⁰ + 1/x¹⁰⁰ = -1*

[@adw1z]

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

[@mangakhoon4517go]

A 12th grade won't know this though

[@centmillionaire]

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 🎉

[@auni4078]

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

[@Grecks75]

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.

[@zcubing5792]

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😅.

[@kajaldey2656]

that's actually genius

[@zcubing5792]

@@kajaldey2656 Thanks😊

[@user-lr3es8kj9m]

Ayoo you're a cuber too?

[@Grecks75]

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. 😉)

[@zcubing5792]

@@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.

[@Grecks75]

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-ey8nl]

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

[@himangshubaruah1140]

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.

[@KhushChadha]

Bingo! Ur right

[@ParvGupta-so1fc]

That comes from the AM-GM inequality which is only valid for real numbers. Inequalities are not defined for complex numbers.

[@NOONO5]

9th class ch1 questions

[@knotwilg3596]

*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

[@Ahaandeeps]

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.

[@martinpetkoski2547]

The solution is complex, so it doesn't show up on Desmos.

[@jimcameron6803]

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.

[@ibnSafaa]

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

[@eternal456]

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!

[@CRnk153]

For first problem i just made form "x^2-5x+1=0 with x>0", thats would be very easy to solve

[@ishaankumar4587]

Those questions were pretty simple , most students have already done many variants of that question in india thus jee uaually doesnt ask these questions

[@mikeeisler6463]

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

[@nickh5086]

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

[@henrytang2203]

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.

[@haidarjafarsaraf7513]

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

[@725etw7w]

Explain please how x^6=-1, is it positive everytime?

[@Tiqerboy]

x is a complex number in the second example. I think the presenter should have made that clear.

[@725etw7w]

@@Tiqerboy thank you

[@Firdouse.]

As an average Indian, i studied this when I was 14yrs old 👍

[@abaddon6078]

I hope that was sarcasm...

[@--_-Anonymous-_--]

​@@abaddon6078 he is talking facts, no sarcasm at all...

[@akshayaggarwal5148]

Yeah, when we are in 11th, we have to study this. This is actually the easiest question in this chapter-complex numbers

[@vivekvk45vk]

These type of problems in India are generally asked in competitive exams like SSC and other government exams.

[@prasunbagdi6112]

Hmm that's means only intelligent people join the government and not the corrupt ones, right? Right?

[@JenyBhatt-be1ed]

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 😊

[@lafq167]

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-kk9un]

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

[@rdspam]

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.

[@oaneric18]

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).

[@sirajzama8080]

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.

[@shasha6538]

the second solution is more accurate because in first solution , even power of x can never be negative.

[@MrWarlls]

For the solution 2b, you can notice that e^(2i*Pi/3) + e^(-2i*Pi/3) = 2cos(2Pi/3)

[@paulortega5317]

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

[@1TuRneD1]

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)

[@mapsm754]

In part 2A, once u get x2 + 1/x2, u could simply raise both sides to power of 50 and get the answer

[@shugneechk6060]

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)

[@kartikpundir764]

Me being a grade 11th math student from India , solved this using complex numbers method and Euler form 🙃(second method).

[@user-bl4zj6wt9v]

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.

[@gibbogle]

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-bl4zj6wt9v]

@@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.

[@satrajitghosh8162]

x^2 + 1/ x^2 + 2 x /.x = 3 x^2 + 1/x^2 = 1 x^4 = x^2 - 1 x^6 = x ^4 - x^2 = - 1 Hereby x ^ 100 = x ^ 102 / x^2 = (-1)^17/x^2 = - 1/ x^2 This implies x^100 + 1/.x^100 = - (1/x ^ 2 + x^2) = -1

[@NEWVA805]

Problem 2: special cases + increasing power

[@AndySaenz924]

The good thing is, you don’t need to know this in the real world.

[@Naitik-1509]

In SSC cgl exam we need to solve both questions under 45 second.😊

[@dipanjansarkhel4087]

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-1509]

@@dipanjansarkhel4087 but problem 2 is a PYQ in ssc exams asked many times.

[@dipanjansarkhel4087]

@@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-1509]

@@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

[@jaadugar5914]

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 😊😊

[@gibbogle]

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.

[@michallesz2]

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.

[@dabest8777]

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

[@Ayuwokibaby]

Well you probably lost a lot of time

[@F.M.Dostoyevsky]

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)

[@TheIntrovertYoutuber]

Second question was easily doable by Complex Variables (I did it in my head with correct answer)

[@shrijan106]

In problem 2 if X is real then 1st condition x+1/x =root 3 is impossible.....

[@ManishaMahadeoMunde]

Yes

[@lowlife_nolife6047]

JEE mains is the Qualifier here. Try JEE advenced.

[@thechessplayer8328]

Pretty easy compared to the international olympiad

[@Mojo.Jojo.]

this is simple maths question . we usually solve these in school

[@rishikumarchourasia]

Average score is 10%. 50% is scored by top 3% people.

[@-hc__]

bro getting 300/300 in JEE Mains is not uncommon these days, 50% is a joke

[@alien3200]

Average NEET user be like:

[@devondevon4366]

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

[@hrisavmandal4873]

Now, the fact is one have to do these solutions in just 12-15 seconds just to pass the cutoff of the exam..

[@premkadam2606]

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. /

[@Grecks75]

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).

[@premkadam2606]

@@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.

[@averageboulderer]

I actually feel this is quite easy for jee mains.

[@AryanSMH]

Abhas Saini student here, and I can already say all these questions you have put here are v easy.

[@r8dra]

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..

[@mangakhoon4517go]

That's the problem just fking remember

[@gitarthabordoloi5913]

No I studied in Allen they deeply explain the concepts hardly any coaching institute makes their student remember this values.

[@r8dra]

@@gitarthabordoloi5913 i mean it's all good if we know how they got derived in the first place

[@gitarthabordoloi5913]

@@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.

[@Torch11]

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

[@restinpeace1916]

Second solution is like: just find x from equation x+1/x=sqrt(3) and put the number in expression x^100+x^(-100)

[@janmesh2332]

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.

[@KejriwalBhakt]

Solved the second one with complex numbers. It was obvious way forward

[@PrajwalNayak-so5uv]

7:20 How did you convert the euler form in rectangular form??

[@MUJAHID96414]

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)

[@shreya1159]

Wow!

[@trueriver1950]

There goes my hope of discovering a new trig function...

[@shubh7864]

Without even starting the video, is the answer -1? Took me like 2 mins without even using complex numbers or anything.

[@Shreeraksha-ey8nl]

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 )

[@zerocat888]

It's not omega

[@AbhijitMondal-th2eh]

You can solve it easily using the concept of cube root of unity. x is one of the 3 cube roots of unity.

[@Grecks75]

No, it's not. The cube of x is +/- i. Instead, x is a 12th root of unity.

[@RussellSubedi]

I went the 5 * 5 * 2 * 2 route. Needless to say, yours was quicker.

[@Grecks75]

Wow, I guess that was a lot of writing. 😀

[@IITJEEAspirant2025-fj6hj]

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

[@phungcanhngo]

Awesome problems and solutions.Thank so much, professor.

[@vishalmishra3046]

*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.o7o5]

Solving the question by just looking at the thumbnail is easier 😅

[@HelloIAmAnExist]

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

[@mkrsinfo2859]

these questions are pretty easy expanding the powered binomial is one the things we learnt when we were small

[@kripanshukhandelwal5868]

I Loved your use of the word "Simplify"... My biology brain really doubted the meaning of "simple" after that explanation !!! 😂

[@akultechz2342]

This question is toddler in front of actual ones tbh

[@HARSHJHA-ln5sg]

U have done it very long , it is not too hard , just use concept of cube root of unity

[@trueriver1950]

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-kz9pb]

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.

[@RevisionWithDev]

EASY question for SSC exam students ❤