Check out the A-level way to solve this: kzitem.info/news/bejne/x2uGyXaBeox9oY4si=ipQkzH32Yfz0C8Bd
@bigbrewer3375
5 ай бұрын
just so everybody knows. This is a question from an AS-level Further maths exam done by 17 yr olds (first year of college) who chose to do a-level further maths (much more difficult than regular maths). On top of that this is an optional module known as further pure mathematics 1. So in other words, the average first year maths student isn't doing stuff this difficult.
@bigbrewer3375
5 ай бұрын
you can also find the paper here: revisionmaths.com/sites/mathsrevision.net/files/imce/8FM0_21_que_20211008.pdf You can find more as level papers here: www.physicsandmathstutor.com/maths-revision/a-level-edexcel/papers-further-as/
@joeythreeclubs
5 ай бұрын
This is easy, Britain needs to step up their game
@overlord3481
5 ай бұрын
@@joeythreeclubs Not everybody is a 300iq genius at the age of 17 like you, and yet, you're not smart enough to know the meaning of "easy".
@overlord3481
5 ай бұрын
That explains a lot
@coatedrak1305
5 ай бұрын
@@joeythreeclubswell believe it or not mate, that’s the first question on the exam.
@GulibleKarma20
4 ай бұрын
answer {x | x(x-1) > (x-1)/x}
@davidbrisbane7206
3 ай бұрын
🤣😂🤣😂
@basstwelve
Ай бұрын
they said use algebra not set theory 😛
@cdkw8254
5 ай бұрын
I choked on my coffee whe I saw Q2 after Q1
@kekw8105
5 ай бұрын
it's actually easy if you have learnt it
@holyshit922
5 ай бұрын
@@kekw8105 If we want exact solution not just approximation this equation is as difficult as first order PDE
@kekw8105
5 ай бұрын
@@holyshit922 yea that's why it's easy
@hi-ld4gg
5 ай бұрын
@@holyshit922 I don't think they'd be getting high schoolers to do exact solutions for non linear odes
@evremn
5 ай бұрын
@@hi-ld4gg i did a level maths and they expect you to solve 2nd order differential equations. Cannot remember if it was for the more advanced module though
@cyrusyeung8096
5 ай бұрын
We can actually cancel out the x - 1, we just need to split into cases. First, note that x cannot be 1, as 1(1 - 1) is not greater than (1 - 1)/1. Hence, we can safely assume x - 1 ≠ 0. Second, we consider x - 1 > 0, i.e. x > 1. Then, x > 1/x. This gives x² > 1 as x > 1 > 0. This means x < - 1 or x > 1. Obviously, x > 1 is the only one that satisfies x - 1 > 0, so one possible interval is x > 1. Third, consider x - 1 < 0, i.e. x < 1. Then, x < 1/x. If x > 0, then x² < 1, which implies 0 < x < -1 (ignore -1 < x ≤ 0 since it contradicts with x > 0). If x < 0, x² > 1, which implies x < -1 or x > 1 (reject as x < 0). Therefore, we have indentified two more possible intervals: 0 < x < 1 or x < -1. In sum, the answer is x < -1 or 0 < x < 1 or x > 1
@GDPlainA
5 ай бұрын
dayum i used the exact same method
@DerGully
5 ай бұрын
Used the same method. The one thing to care is, when you multiply with a negative number, the inequality sign flips. So if you want to cancel out the (x-1) on both sides, split cases x>1 and x
@n8chz
5 ай бұрын
So they could have really made a trick question of it by making it >= rather than >
@DerGully
5 ай бұрын
@@n8chz No, that one is still straightforward. Greater-equal flips to Less-Equal the same way. Just consider the x=1 case separately (in this case true), rest is same.
@browl218
5 ай бұрын
I don't get it. I got x is from -infinity to 0 and from 1 to infinity
@fantasypvp
5 ай бұрын
We were taught to multiply by the denominator squared and then factorise to find the critical values. Then its as simple as drawing a graph and figuring out which ones are asymptotes
@paulgoddard7385
5 ай бұрын
Taking the original inequality: the graph on the lhs is a simple quadratic intersecting the x-axis at 0 and 1; the rhs can be expressed as y = 1 - 1/x. Since the reciprocal graph should be known, the graph of the rhs can be sketched. With a graphical calculators, this process is a whole lot easier. The critical vales can be found algebraically; I'd multiply by x rather than x^2 but that's me. Since there is a double root at x = 1, the curves touch rather than cross at this point. Looking at the graphs, you can see where lhs > rhs.
@artemis_furrson
5 ай бұрын
@@paulgoddard7385 You should never simply multiply both sides by x in an inequality such as this. This is because x can take on negative values, which would flip the inequality sign. Always multiply by x^2.
@paulgoddard7385
5 ай бұрын
@@artemis_furrson I'm not solving an equality; I'm solving an equation to determine the critical values. The inequality is solved by sketching the graphs of both sides.
@artemis_furrson
5 ай бұрын
@@paulgoddard7385 Oh right that makes sense
@a.jz19
5 ай бұрын
That does work since x^2>0, hence it won't flip the sign!
@terrifier1536
5 ай бұрын
It would be nice to see you answer all the questions on a further math a level paper. Would you try do a video on this ?
@genox633
5 ай бұрын
it would be intriguing to see his thought process when answering questions
@benhaw9678
5 ай бұрын
another method (that i believe most A-Level British students are taught) is to multiply by the square of the denominator, since this is known to be positive and we wont have to do case work. In this case, multiplying both sides of the inequality by x² leads to the inequality x³(x-1) > x(x-1) which is slightly nicer to work with imo
@fantasypvp
5 ай бұрын
Yep we were taught this.gotta do this exam in 2 months 💀
@birch8109
5 ай бұрын
@@fantasypvpgood luck 😀
@jakearmstrong3632
5 ай бұрын
Yeah I’m doing gcse but doing a level further maths and thought the method wasn’t the most intuitive
@yallclowns
4 ай бұрын
this doesnt work, as x can be negative which would flip the inequality, but if we multiply by a square, itll always be positive @lachlanross1
@bjornfeuerbacher5514
4 ай бұрын
"he square of the denominator, since this is known to be positive" It can also be exactly equal to zero. One tends to forget these special cases. :/
@burst3k
5 ай бұрын
Q2 next!
@dulot2001
5 ай бұрын
In France, after factorising, we use sign table and students must know how to calculate the sign of affine function.
@vnarayan18
5 ай бұрын
That's the best method!
@BlueLightningSky
4 ай бұрын
If you took the class, you'd know how to solve this because they make you do the same question 10 times and only change the constants. Very rarely will the exam have something you didn't do several versions of, if it did you will never hear the end of it from the students complaining how they couldn't mindlessly copy the method they've been using.
@lordluxembourg8777
5 ай бұрын
As an a-level student doing further maths further pure 1, I've gotta do this test in like 1 month... question 2 we haven't even been taught but this one is pretty doable for 6 marks.
@bprpmathbasics
5 ай бұрын
Best of luck!
@reload2832
5 ай бұрын
Pretty doable? This is in AS level regular maths as well, this whole chapter sticks out of FP1 like a sore thumb, either way we're still lucky it's here BC 6 marks is insanely free. Good luck for the exams btw
@johnathaniel11
5 ай бұрын
FINALLY HE DOES UK STUFF ☺
@thec2359
5 ай бұрын
You should work through an entire A-level maths paper.
@domosautomotive1929
5 ай бұрын
Time to get some new markers.
@attackoramic8361
5 ай бұрын
youtube timing my comments perfectly to the point where this comment shows up at the exact time the red marker runs out of ink.
@Vidrinskas
3 ай бұрын
Multiply by x^2 take terms to the left and factorise. Graph the quartic. Easy.
@amberspark9434
3 ай бұрын
Watching this after taking calculus II, feels like reading a Magic Tree house book after Ulysses. I- there are numbers. There are signs comparing two functions. There’s no trig or integrals. I understood all the words and the explanation.
@diamondnether90
5 ай бұрын
Instead of critical values, we learned to multiply by x^2 then sketch based on the zeroes (cuts at -1, 0, double root at 1)
@Marxism-Leninism137
5 ай бұрын
Wtf is this method to solve disequalities? It's actually a lot easier with the positive/negative study of numerator and denominator.
@fizisistguy
13 күн бұрын
Just assume x is not 1 and cancel out the terms The answer is x>1
@Nobodyman181
5 ай бұрын
999views, 100 likes, 1 hour ago...
@kamo7293
3 ай бұрын
I'm from the UK, and did this exam you show the front of the paper from (Pearson edexcel) in college many years ago. that was a weird kind of nostalgia. seeing a past paper 😅
@bprpmathbasics
5 ай бұрын
Why we have to be careful when solving non-linear inequalities kzitem.info/news/bejne/w6Bq16Wuq3SHnn4si=A94Ua90fYQqat5T9
@ShiroiTensh1
3 ай бұрын
I believe it’s easier to comprehend if we would solve it by dividing into ranges, x>1 1>x>0 x
@ShiroiTensh1
3 ай бұрын
Answers are straightforward, just every time u multiply/divde make assumption to consider if the sign flips or not
@jensraab2902
5 ай бұрын
I took a different approach that got me to the point to test the different intervals much faster but was a little more annoying to compare the terms. So I started by realizing that on both sides we have x-1, one time multiplied and another time divided by x. So the question we need to answer is, how does is x-1 affected multiplying it by x as opposed to the inverse, and does it get greater or not. The two important aspects to consider is sign and whether the factor smaller or greater than 1. For example, in the interval from -1 to 0 (x-1) will always be negative; the factors x and 1/x will also be negative, therefore the product is positive. Comparing two positive numbers, the one with the larger absolute value will be greater. As the absolute value of any number between -1 and 0 is smaller than its inverse, multiplying by x will result in a smaller number than multiplying by 1/x. Therefore, (x-1)/x > (x-1)*x in that interval, and the inequality we are supposed to assess is _not_ true in that interval. The same logic can be applied in the other intervals of interest. It's not always as complicated. For example in the interval x>1 all terms (x, 1/x, and x-1) are positive. And since x > 1/x for x>1, multiplying x-1 by x will always result in a larger numerical value. Therefore the inequality (x-1)*x > (x-1)/x is true in the interval x>1. That method yields the same result but I found the method shown in the video to be more pleasant.
@dneary
4 ай бұрын
I started by dividing across by x-1. If x>1, that doesn't change the inequality direction, and if x
@QuentinStephens
5 ай бұрын
The question does not exclude complex numbers so what are the complex solutions, or can you prove that there are no complex solutions?
@julioantonio8210
5 ай бұрын
Since {x | 0 < x < 1} and {x | 1 < x}, can we write it as {x | 0 < x}?
@pawnstar721
5 ай бұрын
Except 1
@joeythreeclubs
5 ай бұрын
No, bc when x=1, the equation is equal to, not greater than, 0
@julioantonio8210
5 ай бұрын
Forgot about the 1. Thank you
@bprpmathbasics
5 ай бұрын
You could say {x| x>0 but x is not equal to 1}
@lreactor
5 ай бұрын
Wait, I call shenanigans: you didn't use algebra! You simplified the inequality (tbf, with algebra), but then you plugged in values and relied on continuity...
@bjornfeuerbacher5514
4 ай бұрын
At 2:18, I would have noticed that x = 1 obviously is not a solution, and hence we can say that in the following we only look at x not equal to 1. For that case, (x-1)² is always positive, so we can divide the inequality by that without changing anything and only have to solve the much easier inequality (x+1)/x > 0.
@AwakenedNoob
3 ай бұрын
Cant we just do this? x(x-1)>x-1/x (dividing by (x-1) on both sides) x>(x-1)/x(x-1) cancel out the x-1 x>1/x x²>1 x>1
@YuvrajBihani
5 ай бұрын
Why dont u use wavy curve method and put the answer like x belongs to (-infiinity, -1) union (29,infinity)
@johndavidco8501
5 ай бұрын
I prefer constructing a table of signs. Plugging in different values into the expression is too tedious.
@CeRz
5 ай бұрын
table of signs is the best method and there is no debate about it. Easiest way to avoid mistakes.
@sammer-samm9-
8 күн бұрын
Method 1) (- x= 3) equation is given Multiplying both sides by (-1) -1*-x=-1*3 Then x=-3 or Method 2) Let the equation be (- x= 3) If we multiply both sides with "MINUS" sign -(- x)= -(3) Then x= -3. Which one is correct or both methods are correct . Please help
@SuryaBudimansyah
5 ай бұрын
I don't understand how you do the first factoring 1:29
@yoyoezzijr
Ай бұрын
Similar way I did is to check three cases Case 1: x>1, so I can cross out x-1 from both sides, then multiply both sides by x. x² > 1 so x > 1 Case 2: 0 < x < 1, so cross out x-1 but flip the inequality, then multiply both sides by x. x² < 1 so 0 < x < 1 Case 3: x < 0, so cross out x-1 and also flip the inequality, multiply both sides by x and flip the inequality. x² > 1 then x < -1 x ∈ (-∞, -1) U (0, 1) U (1, +∞)
@opdesiahter
5 ай бұрын
Hi, i want to ask what did I do wrong here, I start by multiplying both side with x and gets x^2(x-1) > x-1, which is x^3-x^2 > x-1 => x^3-x^2-x+1 > 0 => x^3-x^2-(x-1) > 0 => x^2(x-1)-(x-1) > 0 => (x-1)((x^2)-1) > 0 => (x-1)(x-1)(x+1) > 0 draw graph, and get the result: 1 > x > -1 or x > 1
@uyangapuujee4508
5 ай бұрын
Not me punching the air because of this video! Ps: i study a level maths lol
@cxld8653
5 ай бұрын
Another thought process can be to take 2 cases, one in which x is positive and cross multiplying doesn't change the inequality and the other in which it is negative and multiplying changes the sign of inequality. say x>0: x^2(x-1) -(x-1)>0 (x^2-1)(x-1)>0 (x-1)^2.(x+1)>0 here x cannot be be equal to 1 so we get x+1>0 but since we assumed x to be positive, we only consider x>0 Similarly in the x
@Nisam_bhai_31
4 ай бұрын
That would take much more time
@cxld8653
3 ай бұрын
@@Nisam_bhai_31 maybe for you, the calculation is quite elementary and should take less than a minute to do the entire thing.
@Nisam_bhai_31
3 ай бұрын
@@cxld8653 for me ehh!?? Fuckin kid
@Ahmed-kg2gf
3 ай бұрын
The solution is so dumb (x-1)² is always positive so it is not a value we care abt (assuming he is talking abt real numbers since he drew a numberline when doing the solution) so that implies that (X+1)/X >0 Then u can quickly discuss (x-1)=0
@BezosAutomaticEye
5 ай бұрын
0:46 nope, I'm gone. Night everybody
@adriansanchez4875
3 ай бұрын
😂😂american curriculum is so trash, I'm just now learning this in a precalculus college course, well better late than never, enjoy your head start foreigners, I'll catch up 😈😈
@dneary
4 ай бұрын
The second question is just algebra too. You are given y(0) and y(1) with h=0.1, and from the equation and the given approximations, you get 100(y(n+1) - 2y(n) + y(n-1)) + 75(y(n+1) - y(n-1)) - 3y(n) = n/5, or 175y(n+1) - 203y(n) + 25y(n-1) = n/5 - setting n=1 you get the approximation of y(0.2), then with n=2 you get the approximation of y(0.3).
@F1r1at
5 ай бұрын
I didn't quite understand that solution. But you can divide everything by (x-1), just need to state that it's not 0. For (x-1) < 0 the inequality will change sign. So you'll get those inequalities: x < 1/x and x < 1 - clearly it's ok on intervals [-inf; -1) and (0; 1) x > 1/x and x > 1 - also clear that it's ok on interval (1; inf] Than you need to test (x-1) = 0 on original inequality, we'll have 0 > 0 which is not true, so we don't include x = 1 in our intervals;
@Ninja20704
5 ай бұрын
The graph can only change sign from positive to negative or vice versa if: 1) the graph passes through 0 (when the numerator is 0) 2) it is discontinuous, usually due to vertical asymptotes (when the denomiator is 0) So we can just factor the numerator and denominator. Setting each factor to 0 will give you the critical numbers because those are the only places where the function could possibly change sign. Thus we can just pick any arbitary test value in the inbetween regions to find our answer.
@joshau2346
3 ай бұрын
this is actually easy, i thought this would be equilvalent to an Olevel additional math question.
@HHUud78TDuj
5 ай бұрын
I don't like the testing on the number line. x(x-1) > (x-1)/x IF x 1 (-inf,-1) IF 1>x>0 x(x-1) > (x-1)/x x^2(x-1) > (x-1) x^2 < 1 falses IF x>0 x(x-1) > (x-1)/x x^2(x-1) > (x-1) x^2 > 1 (1,inf) (-inf,-1) U (1,inf) -1>R>1
@jianzhao4208
5 ай бұрын
Why did I have to watch this video? And why did I force myself to do all 5 questions? Q3 was the easiest for me. Q2 I stopped at finding y2 = 1936/875. Q4 was a lot of algebra work. Q5 I used Heron's formula to find the area of the triangle, but you could also use the cross product.
@vpexmc
5 ай бұрын
x^2 -x > (x-1)/x Divide x x > x-1 All possible answers (Is wrong because of -1, 0, 1, etc.)
@_Rainbooow
3 ай бұрын
I'm guessing x is greater than 1 (that means any number above 1)
@m.h.6470
5 ай бұрын
Solution: Since x is in the denominator of the right side, we know that x ≠ 0. Then we multiply by x on both sides. But if x is negative, we need to switch the inequality, so we need two cases: Case x > 0: x²(x - 1) > x - 1 Case x < 0: x²(x - 1) < x - 1 If x = 1, we get 0 > 0 or 0 < 0, which is a contradiction. Therefore we know that x ≠ 1. With this, we can divide both sides by (x - 1), again creating cases for x > 1 and x < 1 Case 0 < x < 1: x² < 1 → this is true for 0 < x < 1, so any x in this range is part of the solution Case x > 1: x² > 1 → this is true for x > 1, so any x in this range is part of the solution Case x < 0: (0 is already smaller than 1, so no need for additional case) x² > 1 → since x < 0, the only values that lead to x² > 1 are < -1. Therefore x < -1 In total, we have: x ∈ {(-∞, -1), (0, 1), (1, ∞)} or x ∉ {[-1,0], 1}
@holyshit922
5 ай бұрын
If we have d^2y/dx^2 +15dy/dx - 3y = 2x or d^2y/dx^2 +15dy/dx - 3y^2 = 0 It would be easier Now when we have d^2y/dx^2 +15dy/dx - 3y^2 = 2x we can try to change it into system of equations rewite it in symmetric form and try solve the same way as first order PDE Here on youtube is video about systems of differential equations in symmetrical form recorded by some Indians Let z = dy/dx + 15y dz/dx =d^2y/dx^2 + 15dy/dx dz/dx=2x+3y^2 dy/dx = z - 15y Rewriting this system in symmetrical form we get dz/(2x+3y^2) = dy/(z - 15y) = dx/1 and we have to solve this system This system is equivalent to the following PDE z_{x} +(z-15y)z_{y} = (2x+3y^2)
@nubidubi23
3 ай бұрын
i don't get it, can't i just multiply both sides by x and have x^2(x-1)>x-1, then divide by x-1 and have x^2>1?
@rontoolsie
3 ай бұрын
If for x with a 0 than the > sign remains intact.
@mathboy8188
4 ай бұрын
That was the way I would do it as well. (I'd also mention the Intermediate Value Theorem being the justification for the "test point in each interval" procedure.) But another way to do it would be to multiply/divide using absolute values of the things in question. That bypasses the concern that the unknown that you're multiplying by might be negative. (In the derivation I'll explicitly forbid the possibilities where the unknown might equal 0.) I don't think this approach is better, but it can be a useful technique sometimes, so I'll show it. It's basically an algebraic way to do all the multiplications and divisions that the problem "makes" you "want" to do, while keeping track of the consequences of the signs, and not doing anything invalid with the inequality. When you do these algebraic manipulations with absolute values, you end up with a bunch of "sign" functions sgn, where sgn(a) = 1 of a > 0, and sgn(a) = - 1 is a < 0, and sgn(0) = 0. That's because you exploit the following algebraic identities (will explicitly rule out sgn(0) situations in the derivation): a / |a| = |a| / a = sgn(a), and a |a| = sgn(a) a^2. So the point is that using absolute values and sgn functions lets you grind through the algebra easily, without worrying all the signs and their consequences on the inequality as you go, and so putting off until the end working out all those specific cases and their consequences. Find all x s.t. x (x-1) > (x-1) / x. By inspection, x not in { 0, 1 }, so exclude those possibilities in what follows: x (x-1) > (x-1) / x if-and-only-if |x| x (x-1) > |x| (x-1) / x if-and-only-if sgn(x) x^2 (x-1) > sgn(x) (x-1) if-and-only-if sgn(x) x^2 (x-1) / |x-1| > sgn(x) (x-1) / |x-1| if-and-only-if sgn(x) sgn(x-1) x^2 > sgn(x) sgn(x-1). Now there are only two cases for the product sgn(x) sgn(x-1)... either it's 1 or it's -1 (have excluded x = 0 and x = 1 at the start). CASE: sgn(x) sgn(x-1) = 1. Then the equation sgn(x) sgn(x-1) x^2 > sgn(x) sgn(x-1) reads x^2 > 1, and so x > 1 or x < -1. Inspection shows that both x > 1 and x < -1 satisfy sgn(x) sgn(x-1) = 1, so both those intervals are solutions. CASE: sgn(x) sgn(x-1) = -1. Then that equation reads - x^2 > - 1, so x^2 < 1, so -1 < x < 1 (and recall x = 0 is always omitted). When -1 < x < 1, have that -2 < x - 1 < 0, so sgn(x-1) = -1 for all x in -1 < x < 1. Since our case is sgn(x) sgn(x-1) = -1, and our x satisfies -1 < x < 1, have that our x satisfies sgn(x-1) = -1, and so must have that sgn(x) (-1) = -1, and so sgn(x) = 1, and so x must be positive. Thus for x to satisfy sgn(x) sgn(x-1) x^2 > sgn(x) sgn(x-1), and also satisfy sgn(x) sgn(x-1) = -1, must have that -1 < x < 1, and also that x > 0. The x which make both of those simultaneously true are 0 < x < 1. So the solution set is: (-infinity, -1) UNION (0,1) UNION (1, infinity). (That should be written in set notation as per the problem's directions and BPRP's solution. Whatever.)
@phoenixarian8513
5 ай бұрын
Damn. I thought of proof x(x-1)>(x-1)/x In a few moments I found it false. Symbolic-graphic combination. The left part is a quadratic formula and the right is a hyperbola it is 1 plus y=-1/x. On the range limit x to negative zero the hyperbola will return positive infinity while the quadratic will return something between 0 and 2. So it is false. Wait it is to solve the range of x? This is a lot simpler.
@SacchalKurse
5 ай бұрын
I just thought x(x-1)>x-1/x at x>1 to infinity
@AdvaitKoshal
3 ай бұрын
these questions are taught to grade 11 students in the most basic form here in india
@Omar_MTH
3 ай бұрын
Just divied both sides by x-1 so the inequality becomes => x>1/x or x^2>1 S= all real numbers except the open intevral (-1,1)
@nitanOmkar
2 ай бұрын
The answer is---> R-{(-1,0)}
@DeakthNote
5 ай бұрын
sir please take a look at A level Further Pure Mathematics 2 2022 and try Q7 please much appreciated
@bprpmathbasics
5 ай бұрын
I am not sure if this is the correct exam because I don’t see number 7. qualifications.pearson.com/content/dam/pdf/A-Level/Mathematics/2017/Exam-materials/8fm0-22-que-20220524.pdf
@DeakthNote
5 ай бұрын
@@bprpmathbasics sorry!! i was talking about the A level paper qualifications.pearson.com/content/dam/pdf/A-Level/Mathematics/2017/Exam-materials/9fm0-4a-que-20220628.pdf sorry i meant this one
@DeakthNote
5 ай бұрын
@@bprpmathbasics sorry i meant this one also thank you for responding :) qualifications.pearson.com/content/dam/pdf/A-Level/Mathematics/2017/Exam-materials/9fm0-4a-que-20220628.pdf
@DeakthNote
5 ай бұрын
@@bprpmathbasics I think you are lookng at the AS level which students do in first year sixth form we have 2 years on the last year of highschool we do A level i was talking abt the A level version thx for responding again. qualifications.pearson.com/content/dam/pdf/A-Level/Mathematics/2017/Exam-materials/9fm0-4a-que-20220628.pdf
@randomperson-v7y
5 ай бұрын
@@bprpmathbasics sir i think he means the A level version on the link u sent change the last part where you see /8fm0-22-que-whatever to /9fm0-4a-que-20220628.pdf sorry its just yt wont let me send links. :(
@kennethvalbjoern
3 ай бұрын
The question starts with "Use algebra...". When you do your interpretation of your sign-analysis, you actually use the intermediate value theorem of calculus without mention, using that the function (x-1)^2(x+1)/x is continuous everywhere execpt at x=0. So your solution is not purely algebraic.
@printhallo
5 ай бұрын
So just simplify and then sign table? Second question is wild after such a first question.
@Mr0nknown
3 ай бұрын
Why can x not be 0? If we divide by 0 on the very first equation top left then we get 0 is bigger than negative infinity. And with the equation at the top right, we get infinity is bigger than 0. So why doesn't that work?
@Joffrerap
Ай бұрын
You shall not divide by 0. It's undefined.
@bartekburmistrz8679
4 ай бұрын
if we remember that x != 0 we can just multiply both sides by x^2, draw the polynomial and its solved
@charliebrett7510
4 ай бұрын
Why can’t you get X^2 - x > 1 - 1/x ?
@charliebrett7510
4 ай бұрын
Why can’t you get X^2 - x > 1 - 1/x ?
@johnnyrosenberg9522
5 ай бұрын
My first thought was to just immediately get rid of x-1: x>1/x
@colinwilcox4266
4 ай бұрын
I did my o level maths in the early 80s and this would have been considered easy. It still is. Has education really been so dumbed down in the uk that this is now considered advanced paper material?
@YeAlexYuan
4 ай бұрын
Hahaha, this method is so dogmatic. I tried to graph these two functions, and it's perfectly easy to get the result.
@tabris1135
5 ай бұрын
When we are at the step ((x-1)^2*(x+1))/x>0, couldn't we just simplify that to (x+1)/x>0 as (x-1)^2 is always a positive factor?
@christianherrera4729
5 ай бұрын
I found another way to solve it which is similar but more intuitive for me. Set them equal to each other, this will give us the intervals. (-inf,-1),(-1,1),(1,inf), idk how to write set notation lol Subtract the right expression from the left expression in each of the intervals. If it is positive, the left expression is greater than the right expression in the interval. I forgot the 1 when solving the cubic equation which robbed me of the satisfaction of getting the right answer 😭 it works perfectly if you solve the cubic correctly though
@AbouTaim-Lille
5 ай бұрын
Off course u can cancel out x-1 from both sides. But only after u suppose the first case which is X>1. And in this case I get X> ¹/X. Which is true in this case. And the second case where 0
@SuperCrAzYfLiPpEr
5 ай бұрын
On the second problem what does the n mean next to the derivatives and y? Can't find how to google that notation.
@bprpmathbasics
5 ай бұрын
It’s just a subscription like x1, y1 and you have y’1.
@bprpmathbasics
5 ай бұрын
Subscript*
@awperbhai8275
3 ай бұрын
is this hard?
@Pedro-fz3yb
3 ай бұрын
where can I find Q2?
@aklodhi4257
3 ай бұрын
I solved it in 10 sec
@jonathanterry1983
5 ай бұрын
I would cancel the (x-1) to get the critical values, but use the original equation to find the answer. x > 1/x indicates our critical values are ±1 (where both sides are equal) and 0 (undefined). -2 --> (-2)(-3) > (-3)/(-2) True --> x (-1/2)(-3/2) > (-3/2)/(-1/2) False 1/2 --> (1/2)(-1/2) > (-1/2)/(1/2) True --> 0 1/2 True --> x>1
@waimanchin690
4 ай бұрын
臨尾做法極差!
@filippomazzoli607
5 ай бұрын
What a waist of time!
@WLT111TTTT
5 ай бұрын
Your set notation seems omitting x belongs to Real number. I don't know if i am right
@SidneiMV
5 ай бұрын
x(x - 1) > (x - 1)/x (x - 1)(x - 1/x) > 0 (x - 1)(x² - 1)/x > 0 (x - 1)²(x + 1)/x > 0 x ≠ -1 x ≠ 0 (x + 1)/x > 0 1st case - both terms positive x > 0 2nd case - both terms negative x < -1 and finally *(x > 0 or x < -1) and x ≠ 1*
@kavin48
5 ай бұрын
1'st view❤❤
@purple-47
4 ай бұрын
this question doesn't look hard but from what I know the easier it looks the harder it is.
@kamilkonrad9497
5 ай бұрын
Can we swap the left (x-1) with the bottom right x in order to achieve x squared >1? It would give us x squared - 1 > 0.
@aperinich
5 ай бұрын
Why the long haul? * Bring denominator x over to LHS ; x^2(x-1) > (x-1) * Divide both sides by (x-1) ; x^2 > 1 * Subtract 1 from from sides ; x^2 - 1> 0 * Factorise ; (x+1)(x-1)>0 x not equal to -1, 0 , 1.
@danielarnold9042
5 ай бұрын
Because x and x-1 can be either positive or negative, and when using inequalities, multiplying / dividing by a negative number switches the sign (i.e. greater than turns to less than). Best thing to do is multiply by x^2 since it is positive for all real x, then subtract, factor etc.
@fardinzaman100
5 ай бұрын
Should we not multiply the inequality with x^2 instead of x, as we do not know if x is confirm positive?
@CalculusIsFun1
5 ай бұрын
I just divided by x - 1 and flipped the inequalities. Then multiplied by x. If you take the square root you get plus x is greater than 1 or it can be less than -1. So right of the bat, i have x less than negative 1 and greater than positive 1. The second case i subtract (x-1)/x on both sides and common denominator. The denominator hits a 0 at x = 0 so that’s my case 2. to have the expression be greater than zero the bottom and top have to both be positive or both be negative. I already established that the negative range of values less than -1 works so the only possible range i could prove is between zero and negative 1. no value on that range will result in a positive number so that cannot be a possible range. Another possibility is the range between 0 and positive 1. Here, the top will always be positive and so will the bottom so this range also always works. So all in all we have {x|x
@Extermiraptor
5 ай бұрын
Honestly as someone who does further statistics and decision, this further pure stuff kind of goes over my head at first glance, but it's neat to see videos like this to learn the method to the madness.
@kausarlolz
5 ай бұрын
howre u finding decision😪
@okenough2124
5 ай бұрын
@@kausarlolzI do decision aswell. It's crap
@cparks1000000
5 ай бұрын
Since (x-1)^2 is nonnegative, you can cancel it assuming x is not equal to 1.
@bogydan4223
5 ай бұрын
You should try this one, trust me, is much harder then it looks: sqrt(x+7)-1
@antonyqueen6512
5 ай бұрын
Differently using simplification/mltiplication, but paying attention not to multiply/simplify by a zero factor AND to reverse the inequality sign when multiplying/simplifying by a negative factor, thus: 1. X must be ≠0, since we have 1/x on the right side. 2. We distinguish the cases x1 to simplify by (x-1), If x=1, => 0>0, thus x=1 is not a solution. If x≠1 => we can simplify by x-1 and reverse the inequality sign when x {x1} We multiply by x and reverse inequality sign when x {x^2>1 and x
@seanhunter111
5 ай бұрын
I would solve this using Descartes table of signs rather than the last bit of picking numbers from the number line. As all the factors are linear, it's easy to see that they are each negative when x is less than their critical value. Then it's easy to figure out from the table of signs what the combined sign is. It's a little bit clearer for me personally than the number line method. Also, this is "Further Maths" A level, not just Maths. "Further Maths" is almost like the high school equivalent of an honours class, so it deals with more advanced topics than just normal maths. Knowing the UK system a bit when they say set notation, they don't mean set builder notation they mean using open and closed intervals (with unions etc as needed). So in this case they are expecting something like "x ∈ (∞,-1) U (0,1)", rather than "x>-1 or 0 < x < 1". I don't think they would mark you down for using set builder notation however.
@LimCun-de7nw
4 ай бұрын
Do GCE A level further maths paper please
@xno_elysiumx3744
5 ай бұрын
Are differential equasions teached in UK high schools? We never had this.
@jackychanmaths
5 ай бұрын
This should be a Grade 10 inequality in somewhat additional or further mathematics x(x-1) > (x-1)/x Case 1: x>0 x^2 (x-1) > x-1 (x-1) (x^2 - 1) > 0 (x-1)^2 (x+1) > 0 When x≠1, x+1 > 0 which is always true whenever x>0 and x≠1 When x=1, (x-1)^2 (x+1) = 0 so it is not a solution Case 2: x
@ManiacalPenguin_
5 ай бұрын
It can come up in year 11/gcse/grade 10. I'm unsure why this easy of a question is on FP1 tbh, could easily pass as a GCSE question if they removed the set notation part
@vfxgenie983
5 ай бұрын
This is not high school this is college level 17-18
@talhahussains1b47
4 ай бұрын
i threw up reading question two
@urbigfan96
5 ай бұрын
still dont know what he did at 1:37 XD just cant see it
@snowman2395
4 ай бұрын
how are they on the same test
@tarunium_
3 ай бұрын
in the uk the questions (atleast fot maths) are incimentel in difficulty so the first few questions would be easier than the rest of the paper
@chandrikakhare4796
5 ай бұрын
PLS USE WAVY CURVE METHOD
@fivenightsatfreddys9369
4 ай бұрын
1+1+1 = 1 according to trinity christianity
@aaronwilliamh9905
4 ай бұрын
Did you try 1 * 1 * 1 = ?
@aaronwilliamh9905
4 ай бұрын
Did you try 1 ÷ 1 ÷ 1 =?
@pino_t0412
5 ай бұрын
In italy, we study this math at secindsry school ( scuile medie). This is very simple math
@rubentandy7654
4 ай бұрын
Most people do a level maths at their secondary school, and I’m sure Italy has a nice easy q1 to warm you up too
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