My first A-level math exam question

[@bprpmathbasics]

Check out the A-level way to solve this: kzbin.info/www/bejne/nGazlHWBdr9_pLssi=ipQkzH32Yfz0C8Bd

[@bigbrewer3375]

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]

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]

This is easy, Britain needs to step up their game

[@overlord3481]

@@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]

That explains a lot

[@coatedrak1305]

@@joeythreeclubswell believe it or not mate, that’s the first question on the exam.

[@fantasypvp]

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]

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]

​@@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]

@@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]

@@paulgoddard7385 Oh right that makes sense

[@a.jz19]

That does work since x^2>0, hence it won't flip the sign!

[@cdkw8254]

I choked on my coffee whe I saw Q2 after Q1

[@kekw8105]

it's actually easy if you have learnt it

[@holyshit922]

@@kekw8105 If we want exact solution not just approximation this equation is as difficult as first order PDE

[@kekw8105]

@@holyshit922 yea that's why it's easy

[@hi-ld4gg]

​@@holyshit922 I don't think they'd be getting high schoolers to do exact solutions for non linear odes

[@evremn]

​@@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]

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]

dayum i used the exact same method

[@DerGully]

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]

So they could have really made a trick question of it by making it >= rather than >

[@DerGully]

@@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]

I don't get it. I got x is from -infinity to 0 and from 1 to infinity

[@GulibleKarma20]

answer {x | x(x-1) > (x-1)/x}

[@davidbrisbane7206]

🤣😂🤣😂

[@basstwelve]

they said use algebra not set theory 😛

[@MajorlyBlue]

​@@basstwelve you use set notation to write your answer.

[@MajorlyBlue]

reason being, the answer to an inequality is a set of values

[@benhaw9678]

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]

Yep we were taught this.gotta do this exam in 2 months 💀

[@birch8109]

@@fantasypvpgood luck 😀

[@jakearmstrong3632]

Yeah I’m doing gcse but doing a level further maths and thought the method wasn’t the most intuitive

[@yallclowns]

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]

"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. :/

[@diamondnether90]

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)

[@pietergeerkens6324]

Not seen that before. I love it!

[@bprpmathbasics]

Why we have to be careful when solving non-linear inequalities kzbin.info/www/bejne/mJuXoqSup6eJoassi=A94Ua90fYQqat5T9

[@ShiroiTensh1]

I believe it’s easier to comprehend if we would solve it by dividing into ranges, x>1 1>x>0 x

[@ShiroiTensh1]

Answers are straightforward, just every time u multiply/divde make assumption to consider if the sign flips or not

[@terrifier1536]

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]

it would be intriguing to see his thought process when answering questions

[@burst3k]

Q2 next!

[@bjornfeuerbacher5514]

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.

[@LittletbigT]

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]

Best of luck!

[@reload2832]

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

[@BlueLightningSky]

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.

[@dulot2001]

In France, after factorising, we use sign table and students must know how to calculate the sign of affine function.

[@vnarayan18]

That's the best method!

[@dneary]

I started by dividing across by x-1. If x>1, that doesn't change the inequality direction, and if x

[@dmytryk7887]

The way I recommend for the last step is 1) imagine a large value of a and you see that all factors are positive so it is positive for x>1. Step 2) you know the zeroes are at -1, 0, and 1 so these are the values where the sign might change, but only one factor could change at any of these values. You can see there will be a sign change at -1 and 0, but not a 1 so you can just fill in the sign for each interval. In problems like this you're really just looking for even powered factors.

[@amberspark9434]

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.

[@thec2359]

You should work through an entire A-level maths paper.

[@Vidrinskas]

Multiply by x^2 take terms to the left and factorise. Graph the quartic. Easy.

[@cxld8653]

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]

That would take much more time

[@cxld8653]

@@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]

@@cxld8653 for me ehh!?? Fuckin kid

[@yoyoezzijr]

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, +∞)

[@kennethvalbjoern]

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.

[@antonyqueen6512]

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

[@Extermiraptor]

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]

howre u finding decision😪

[@okenough2124]

​@@kausarlolzI do decision aswell. It's crap

[@dneary]

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).

[@wqltr1822]

The method taught in the a level further pure 1 textbook is to miltiply by even factors, so in this case multiply both sides by x^2. Then move everything to one side and sketch the resulting polynomial. Don't immediately expand though, try to keep as many common factors before having to expand. Then at the end discard any values which make the original lhs or rhs invalid, so in this case x=0, which you wont habe to worry about since the inequality is strict.

[@jensraab2902]

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.

[@kamo7293]

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 😅

[@julioantonio8210]

Since {x | 0 < x < 1} and {x | 1 < x}, can we write it as {x | 0 < x}?

[@pawnstar721]

Except 1

[@joeythreeclubs]

No, bc when x=1, the equation is equal to, not greater than, 0

[@julioantonio8210]

Forgot about the 1. Thank you

[@bprpmathbasics]

You could say {x| x>0 but x is not equal to 1}

[@johndavidco8501]

I prefer constructing a table of signs. Plugging in different values into the expression is too tedious.

[@CeRz]

table of signs is the best method and there is no debate about it. Easiest way to avoid mistakes.

[@johnathaniel11]

FINALLY HE DOES UK STUFF ☺

[@m.h.6470]

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}

[@CalculusIsFun1]

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

[@QuentinStephens]

The question does not exclude complex numbers so what are the complex solutions, or can you prove that there are no complex solutions?

[@domosautomotive1929]

Time to get some new markers.

[@attackoramic8361]

youtube timing my comments perfectly to the point where this comment shows up at the exact time the red marker runs out of ink.

[@seanhunter111]

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.

[@mathboy8188]

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.)

[@sugarfrosted2005]

You could also just nix the squared term in the the last inequality and reduce cases

[@SuperCrAzYfLiPpEr]

On the second problem what does the n mean next to the derivatives and y? Can't find how to google that notation.

[@bprpmathbasics]

It’s just a subscription like x1, y1 and you have y’1.

[@bprpmathbasics]

Subscript*

[@jonathanterry1983]

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

[@BezosAutomaticEye]

0:46 nope, I'm gone. Night everybody

[@bartekburmistrz8679]

if we remember that x != 0 we can just multiply both sides by x^2, draw the polynomial and its solved

[@SidneiMV]

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*

[@bobmarley9905]

Well technically you can do those EITHER of the methods bprp mentions NOT to do, but you just gotta be careful with the logic, casework, and multiplying by negatives on both sides of an inequality (which flips the inequality signs)... I'm trying to figure out how to best phrase the logic, but so far I think I got a good way of explaining it I think... which is why it makes sense to just do this method since it's less likely for error, but it doesn't offer much other than just rote memorization and not real deep understanding, something that's significantly diminished in the modern pedagogy of math and adversely affects people's understanding and logical thinking.

[@YuvrajBihani]

Why dont u use wavy curve method and put the answer like x belongs to (-infiinity, -1) union (29,infinity)

[@spaaaaaaaaaaaaaaaaaaaaaaaace]

how about other countries' o-level/a-level exam questions

[@christianherrera4729]

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

[@xno_elysiumx3744]

Are differential equasions teached in UK high schools? We never had this.

[@SuryaBudimansyah]

I don't understand how you do the first factoring 1:29

[@Heimndallr]

I could explain if you still don't understand

[@SuryaBudimansyah]

@Heimndallr Be my guest

[@Heimndallr]

@@SuryaBudimansyah lmao u have discord or something i could explain better with the paperwork

[@DeakthNote]

sir please take a look at A level Further Pure Mathematics 2 2022 and try Q7 please much appreciated

[@bprpmathbasics]

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]

@@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]

@@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]

​@@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]

@@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. :(

[@tescowifi]

can you not just multiply at the start by x? then get x^3-x^2-x+1?

[@fardinzaman100]

Should we not multiply the inequality with x^2 instead of x, as we do not know if x is confirm positive?

[@kamilkonrad9497]

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.

[@F1r1at]

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]

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.

[@tabris1135]

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?

[@jdogg5639]

I guess I don't correctly remember the rules of algebra on inequalities. I thought you could just cancel out the (x-1) on both sides and move the right x to the left to reduce it all to x^2>1. So clearly wrong. I was thinking you can do normal algebra but just have to flip the inequality sign if multiplying both sides by a negative.

[@Mr0nknown]

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?

[@maxime_weill]

You shall not divide by 0. It's undefined.

[@mohamedzafir8566]

Wavy Curve Method?

[@printhallo]

So just simplify and then sign table? Second question is wild after such a first question.

[@AbouTaim-Lille]

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

[@padla6304]

i love "Futurama"

[@yiowruvern2932]

I remember learning this a long long time ago, it was like riding a bicycle, but the bicycle was on fire and everything else was on fire..... Yes, I took A levels further maths (further mads)

[@Omar_MTH]

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)

[@henry55]

how can u divide by x-1? doesn’t that assume x≠1

[@bogydan4223]

You should try this one, trust me, is much harder then it looks: sqrt(x+7)-1

[@mtaur4113]

There is a shortcut, but don't take the shortcut if there's any doubt in your mind. (x-1)^2 is non-negative, so no sign change at x=1. However we still avoid 1 for the strict inequality (and if it were in the denominator, we would also avoid it when looking for equality)

[@charliebrett7510]

Why can’t you get X^2 - x > 1 - 1/x ?

[@opdesiahter]

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

[@vpexmc]

x^2 -x > (x-1)/x Divide x x > x-1 All possible answers (Is wrong because of -1, 0, 1, etc.)

[@nubidubi23]

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]

If for x with a 0 than the > sign remains intact.

[@jackychanmaths]

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_]

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

[@pino_t0412]

In italy, we study this math at secindsry school ( scuile medie). This is very simple math

[@rubentandy7654]

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

[@LimCun-de7nw]

Do GCE A level further maths paper please

[@lreactor]

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...

[@aperinich]

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]

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.

[@Pedro-fz3yb]

where can I find Q2?

[@WLT111TTTT]

Your set notation seems omitting x belongs to Real number. I don't know if i am right

[@AwakenedNoob]

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

[@Gitanshu-s6m]

just use wavy curve

[@jianzhao4208]

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.

[@cparks1000000]

Since (x-1)^2 is nonnegative, you can cancel it assuming x is not equal to 1.

[@chandrikakhare4796]

PLS USE WAVY CURVE METHOD

[@colinwilcox4266]

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?

[@johnnyrosenberg9522]

My first thought was to just immediately get rid of x-1: x>1/x

[@AdvaitKoshal]

these questions are taught to grade 11 students in the most basic form here in india

[@phoenixarian8513]

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.

[@viserion_137]

Wtf is this method to solve disequalities? It's actually a lot easier with the positive/negative study of numerator and denominator.

[@joshau2346]

this is actually easy, i thought this would be equilvalent to an Olevel additional math question.

[@SmoMo_]

I don’t get this, my thoughts were just 1) X cannot be 0 2) multiple both sides by X, then x^2*(x-1) > ( x-1) 3) x can’t be 1 then , as both sides would be 0 4) therefore dividing by (x-1) is allowed, so x^2 > 1 5) x < -1, x > 1, x!= 0 I feel like I’m missing something because the explanation says I couldn’t multiply by x or divide by (x-1)

[@robertveith6383]

Your statement in 1) is not correct. Do not change to upper case for the letter. That is not the same variable.

[@epikherolol8189]

If u are multiplying by x then u should make 2 cases when x>0 and x

[@yug1445]

(0,-infinity)???

[@snowman2395]

how are they on the same test

[@tarunium_]

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

[@lucaswilkins9217]

Much trickier than it seems, lots of room for careless mistakes, such as missing that x=1 is not a solution where everything around it is. It's a pretty good test question I reckon.

[@HHUud78TDuj]

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

[@_Rainbooow]

I'm guessing x is greater than 1 (that means any number above 1)

[@vadim64841]

No need to plug in any numbers. Just notice that for large positive x all terms are positive. Then on each root the sign alternates, except for terms (x - a)^n where n is even, in which case the sign does not alternate.

[@badrinath546]

Why don’t we just cancel (x - 1) on both sides?

[@mrprime3466]

1) if x =1 then you are dividing by (1-1) = 0. Dividing by 0 is undefined 2) if x < 1, then x - 1 is negative. Dividing or multiplying by a negative real number will change the direction of the inequality sign. For example, 2 > 0. If we multiply (or divide) both sides by -1 and don’t flip the equality then -2 > 0 which is nonsense. If x > 1 then the inequality sign stays in the same direction. So, you’re losing the solution x = 1 and making the equation more complicated by splitting it into regions where the inequality sign changes depending on the values of x.

[@F1r1at]

​@@mrprime3466 yep, you'll have 2 inequalities that way, but they'll be much easier then what you get with the method in video. You don't even need to do anything further with them, just write the answers. Oh, and also you'll need to check original inequality on (x-1) = 0, but that's also quite easy.

[@holyshit922]

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)

[@uyangapuujee4508]

Not me punching the air because of this video! Ps: i study a level maths lol

[@fizisistguy]

Just assume x is not 1 and cancel out the terms The answer is x>1

[@Xebtria]

oh nice if I multiply by x, I have x²(x-1) > x-1, I then can divide by x-1 (defining at the same time that x must not be = 1), so I have the result of x²>1, meaning all X for which x² > 1 are fulfilling the originial inequality. *starts watching video I tell you what not to do: do not multiply by X on both sides GODDAMMIT