Quick Inequality Q

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  1. member910132's Avatar
    1 11-05-2012  09:42
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    Quick Inequality Q
    Q1:

    \displaystyle (1+2x)^{0.5} \ \text{is valid for} \ |2x|<1 \ \text{and so after rearranging it is valid for} \ |x|<0.5= -0.5<x<0.5, \ \text{but why has the MS got is as} -0.5<x\leq 0.5 ?

    Q2:

    \displaystyle \ln (1-3x) \ \text{ is valid for} \ -1<-3x\leq 1 \Rightarrow \dfrac{-1}{3} \leq x < \dfrac{1}{3} \ \text{ But why has the MS got} \dfrac{-1}{3} < x \leq \dfrac{1}{3} ?

    Thnx
    Last edited by member910132; 11-05-2012 at 09:44.
  2. Intriguing Alias's Avatar
    2 11-05-2012  09:54
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    • Location: Yorkshire
    Re: Quick Inequality Q
    (Original post by member910132)
    Q1:

    \displaystyle (1+2x)^{0.5} \ \text{is valid for} \ |2x|<1 \ \text{and so after rearranging it is valid for} \ |x|<0.5= -0.5<x<0.5, \ \text{but why has the MS got is as} -0.5<x\leq 0.5 ?

    Q2:

    \displaystyle \ln (1-3x) \ \text{ is valid for} \ -1<-3x\leq 1 \Rightarrow \dfrac{-1}{3} \leq x < \dfrac{1}{3} \ \text{ But why has the MS got} \dfrac{-1}{3} < x \leq \dfrac{1}{3} ?

    Thnx
    Now I'm not 100% sure but I don't think there should be any equality sign. Especially clear in the second answer. The mark scheme implies the second one is valid for ln(0).

    Is this from a past paper or textbook?
  3. raheem94's Avatar
    3 11-05-2012  10:39
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    • Posts: 5,512
    Re: Quick Inequality Q
    (Original post by member910132)
    Q1:

    \displaystyle (1+2x)^{0.5} \ \text{is valid for} \ |2x|<1 \ \text{and so after rearranging it is valid for} \ |x|<0.5= -0.5<x<0.5, \ \text{but why has the MS got is as} -0.5<x\leq 0.5 ?

    Q2:

    \displaystyle \ln (1-3x) \ \text{ is valid for} \ -1<-3x\leq 1 \Rightarrow \dfrac{-1}{3} \leq x < \dfrac{1}{3} \ \text{ But why has the MS got} \dfrac{-1}{3} < x \leq \dfrac{1}{3} ?

    Thnx
    I also don't see any reason to put the \le , it should be < , according to me.
  4. member910132's Avatar
    4 11-05-2012  16:01
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    Re: Quick Inequality Q
    (Original post by hassi94)
    Now I'm not 100% sure but I don't think there should be any equality sign. Especially clear in the second answer. The mark scheme implies the second one is valid for ln(0).

    Is this from a past paper or textbook?

    (Original post by raheem94)
    I also don't see any reason to put the \le , it should be < , according to me.
    Q 4 and 5 on page 33 and the answer is on page 136
    http://store.aqa.org.uk/qual/pdf/AQA-MFP3-TEXTBOOK.PDF

    For the second one my AQA formula book says the Maclaurin's series for ln (1+x) is valid for -1<x\leq 1. Assuming that is correct, as it's in the formula book, would my answer to the second one be correct or their ? Are my re-arrangements correct ?

    For the first one it's just the simple (1+x)^n is valid for |x|<1 and am I right in thinking that has no equal sings ?
  5. member910132's Avatar
    5 11-05-2012  23:43
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    Re: Quick Inequality Q
    Bump
  6. Intriguing Alias's Avatar
    6 11-05-2012  23:46
    • TSR Idol
    • Location: Yorkshire
    Re: Quick Inequality Q
    (Original post by member910132)
    Q 4 and 5 on page 33 and the answer is on page 136
    http://store.aqa.org.uk/qual/pdf/AQA-MFP3-TEXTBOOK.PDF

    For the second one my AQA formula book says the Maclaurin's series for ln (1+x) is valid for -1<x\leq 1. Assuming that is correct, as it's in the formula book, would my answer to the second one be correct or their ? Are my re-arrangements correct ?

    For the first one it's just the simple (1+x)^n is valid for |x|<1 and am I right in thinking that has no equal sings ?
    Sounds to me like you were right on both occasions.
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