A-Level Maths: B7-10 Graphs: Solving Modulus Inequalities

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  • Опубліковано 8 лис 2024

КОМЕНТАРІ • 46

  • @nuruladlina7674
    @nuruladlina7674 4 роки тому +32

    I NOW UNDERSTAND THIS STUFF!! AAAA MUCH THANK UUUUU

  • @jordansheridan5325
    @jordansheridan5325 7 років тому +22

    You have boosted my A level grade from a D to almost full marks! It would be amazing if you could make a mechanics 1 ocr mei videos

    • @TLMaths
      @TLMaths  7 років тому +10

      I'm not teaching that module I'm afraid, so I won't be making them - my focus is on the new A-Level spec ready for September. Apologies.

  • @realroshrosh
    @realroshrosh Рік тому +2

    This guy deserves more subscribers

  • @quietworld6870
    @quietworld6870 4 роки тому +17

    My exam is tmrw n now I understand 🤦‍♀️🤦‍♀️

  • @devrohilla7978
    @devrohilla7978 2 роки тому +2

    I am indian but love this content very much

  • @Rachel-xi7be
    @Rachel-xi7be 5 років тому +6

    thano you soooo much for this❤️

  • @afghanclick3811
    @afghanclick3811 7 років тому +5

    hi
    can you make a video about, some tricky exam questions in AQA C1 and C2 please.

    • @TLMaths
      @TLMaths  7 років тому +3

      I don't do any more AQA specific videos I'm afraid - I'm focusing solely on the new specification

    • @afghanclick3811
      @afghanclick3811 7 років тому

      Alright Thank you

  • @muhammadizhan2741
    @muhammadizhan2741 2 роки тому

    Good techar tomorow my papar gud bcz of u sar thank you

  • @danielfontain150
    @danielfontain150 4 роки тому +1

    Hi. I saw in a question that said |6-x|=1/2x + k.
    And Has exactly one soloution. Find the value of K. I dont understand why the intersection has to be in the vortex, so (6,0). Surely it can be anywhere in |6-x|.

    • @TLMaths
      @TLMaths  4 роки тому +1

      It's probably easier to visualise it on desmos: www.desmos.com/calculator/hvoztrllg9
      You can then use the slider for k to see how this changes things. Clearly if there is to be only one intersection point, the line will have to intersect the modulus graph at the vertex.

    • @danielfontain150
      @danielfontain150 4 роки тому

      @@TLMaths Ah ok I see. Thanks for the link. I didnt take the graident into account. Cheers sir!

    • @aryansapkota6796
      @aryansapkota6796 3 роки тому

      Is k= -3? What i did was uno that 6-x= + or - (1/2x+k) . Rearranged one of the equations for K. Substitute into the other equation to solve for x which i got as 6 and then plug back into one of the equations, k = -3 (like simultaneous equations)

  • @keltz-06
    @keltz-06 Рік тому

    ty sir

  • @iuseyoutubealot
    @iuseyoutubealot 4 роки тому +4

    how would i solve |2x+1|+|2x+1|=3 ;do i bring one of |2x+1| to the RHS to start

    • @TLMaths
      @TLMaths  4 роки тому +7

      Surely that will simplify to 2*|2x+1| = 3
      |2x+1| = 3/2
      ...?

    • @aryansapkota6796
      @aryansapkota6796 3 роки тому

      |2x+1| + |2x+1|= 3 or -3 , remove mod signs and solve both equations, x= 1/4 or -5/4

  • @Ash-en4ul
    @Ash-en4ul 4 роки тому +2

    When yoy are calulcating it and like changing the signs to acocomodate for the negative gradient, would you change the graident of the eg make it from 2x+3 to =2x+3 , or do you change the value its equal to eg, 2x+3= 4 ; is it -2x+3= 4 or is it 2x+3= -4? Thanks

    • @TLMaths
      @TLMaths  4 роки тому

      You change the sign of anything that is in the modulus sign, so for |2x+3|, you look at -(2x+3) = -2x-3

    • @Ash-en4ul
      @Ash-en4ul 4 роки тому +1

      @@TLMaths thanks!! Do you know where I can get questions that merge ln concepts and Modulus together?? Thanks
      I know it's oddly specific

  • @sharkm0m0
    @sharkm0m0 Рік тому

    Hello. How can we make the decision which values to take after obtaining x without graphing?

  • @tz6011
    @tz6011 2 роки тому

    Thanks for your videos!

  • @Mostpeoplecant
    @Mostpeoplecant Рік тому

    cheers

  • @davidhill8163
    @davidhill8163 2 роки тому

    Many thanks I find this topic kinda tricky

  • @augustinezulu6143
    @augustinezulu6143 2 роки тому +1

    So the signs changes if the line is going to the left side only?

    • @augustinezulu6143
      @augustinezulu6143 2 роки тому

      I'm confused on the last question

    • @TLMaths
      @TLMaths  2 роки тому +1

      The left hand part of the graph has negative gradient, so will be y = -3x + 7, and the right hand part of the graph has positive gradient so will be y = 3x - 7

  • @darosaleh830
    @darosaleh830 7 років тому +1

    Are you gonna make mechanics videos by the end of may?

    • @TLMaths
      @TLMaths  7 років тому

      No, I'm afraid not - I won't be that far along

  • @مرتضیافغان-ك3ض
    @مرتضیافغان-ك3ض 7 років тому +1

    can you make a video about, how to study or get an A in AQA a level maths. because I am revising a lot but can't do good enough in the exam
    please please

    • @TLMaths
      @TLMaths  7 років тому

      What sort of revision are you doing?

    • @مرتضیافغان-ك3ض
      @مرتضیافغان-ك3ض 7 років тому

      I doing post exam papers, and watching UA-cam video, I did 2016 c2 Peper today and I thought i will get a good result but I think I did bad

    • @مرتضیافغان-ك3ض
      @مرتضیافغان-ك3ض 7 років тому

      sorry about my English writing, I am not good in English

    • @TLMaths
      @TLMaths  7 років тому +21

      Use www.physicsandmathstutor.com/a-level-maths-papers/ to get your papers - and don';t be afraid to dip into other exam boards for C1 and C2 - they're all useful. You need to work through all the papers of the board you are sitting - make sure you record which ones you have done and the score you got each time (you want to see progress and improvement). You will need to mark yourself using the mark scheme, and be harsh - this will get you used to what the examiner is looking for. Time yourself also - you want to train yourself to work at pace so that you don't run out of time in the exam. Write lists of the topics you continue to struggle on and add to this after every paper you complete - these topics will require more emphasis in your study. Watch videos on these and ask your teacher for help. Seek help from somebody else studying the exam or who is in their second year, or even a tutor - the one-to-one opportunity to ask for help is invaluable.
      Ultimately, you need to know what you don't know. If you sit down to do a paper and you think it went well, when in fact it didn't, then you'll not be in a good position. You need to constantly give yourself short low-stakes tests so that you can determine what you do and don't know. Break your revision up into chunks and interleave topics (don't do two hours of polynomials), do half an hour of one topic, then have a break, then half an hour of another topic, then a break, then half an hour of another topic. Try writing out a crib sheet - can you fit everything you need to remember about Core 1 onto a single side of A4 for example? This test of memory forces you to remember topics and concepts and is good for brain-training.

    • @مرتضیافغان-ك3ض
      @مرتضیافغان-ك3ض 7 років тому +3

      Thanks

  • @korypolexa3714
    @korypolexa3714 3 роки тому +2

    I don't get how for number two the y intersect is positive 5 if it was -5

    • @TLMaths
      @TLMaths  3 роки тому +1

      |2x-5|
      Sub in x=0
      |2*0-5| = |-5| = 5

  • @JK-vm7yo
    @JK-vm7yo 2 роки тому

    How come for the first question we put a minus sign on the 1 and not the 2x-7 ?

    • @TLMaths
      @TLMaths  2 роки тому

      When solving |2x-7| = 1, you can consider that either 2x-7 = 1 or 2x-7 = -1 because |1| = 1 and |-1| = 1

  • @Phoenix-nh9kt
    @Phoenix-nh9kt 2 роки тому

    But how do i solve questions wherein y is inside the modulus along with x

    • @TLMaths
      @TLMaths  2 роки тому

      Can you give an example and where you’ve seen it in A-Level Maths?

  • @ronedavin7573
    @ronedavin7573 4 роки тому +1

    I don't understand why for the second question the line touch +ve 5 instead of -5. I'm don't understand this part...Please Help me i'm really confuse

    • @TLMaths
      @TLMaths  4 роки тому +5

      The outputs of the modulus function are always greater than or equal to zero, so |f(x)| is always above or on the x-axis. The graph of y = |2x - 5| doesn't go below the x-axis for this very reason. Think of it like the line y = 2x -5 but everything that has gone below the x-axis has been reflected up in the x-axis. See: www.desmos.com/calculator/utgpjs5blz