Year 13 Maths Help Thread

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    (Original post by SeanFM)
    Brilliant, so you've applied the rule correctly - it's the most tricky part, the simplifying, that you just need to do.

    It's difficult to know what you want the simplified version to look like, but ideally you want no fractional powers in the numerator (after combining the two fractions) so you can multiply by the most appropriate numbers (eg if you had (x+1)^(1/2) and (x+1)^(-1/2) in the numerator then it would make sense to multiply both by (x+1)^(1/2) (this gets rid of both fractional powers) and move the thing you multiplied to the denominator.
    Ok thank you, I'll try that.
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    (Original post by SeanFM)
    Brilliant, so you've applied the rule correctly - it's the most tricky part, the simplifying, that you just need to do.

    It's difficult to know what you want the simplified version to look like, but ideally you want no fractional powers in the numerator (after combining the two fractions) so you can multiply by the most appropriate numbers (eg if you had (x+1)^(1/2) and (x+1)^(-1/2) in the numerator then it would make sense to multiply both by (x+1)^(1/2) (this gets rid of both fractional powers) and move the thing you multiplied to the denominator.
    I've tried to simplify it but I still can't get the right answer. Could you show me the step after my first step please?
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    (Original post by osayukiigbinoba)
    I've tried to simplify it but I still can't get the right answer. Could you show me the step after my first step please?
    After applying the quotient rule you should have:
    \displaystyle \frac{(x+1)^{1/2} - \frac{x}{2}(x+1)^{-1/2}}{(x+1)}

    From here just multiply the numerator and denominator by 2(x+1)^{1/2}
    Can you see why?
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    (Original post by RDKGames)
    After applying the quotient rule you should have:
    \displaystyle \frac{(x+1)^{1/2} - \frac{x}{2}(x+1)^{-1/2}}{(x+1)}

    From here just multiply the numerator and denominator by 2(x+1)^{1/2}
    Can you see why?
    I've got the right answer now, thank you!
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    (Original post by osayukiigbinoba)
    I've got the right answer now, thank you!
    but do you understand why and where the 2(x+1)^{1/2} comes from?
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    (Original post by RDKGames)
    but do you understand why and where the 2(x+1)^{1/2} comes from?
    Sorry I just got back home. Yeah it's to remove the fractional powers and the fraction on the numerator.
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    To what extent does M3 help with Year 13 physics?
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    Palette can I volunteer as a helper? I go back to uni (2nd year maths) in 3 weeks and I need to get my brain started again, so this would be ideal. I've helped plenty of people with maths problems on TSR before, too.

    Also to answer your question, in my experience M3 didn't really do much for A2 level Physics (this was AQA and Edexcel respectively), but it was interesting and I would recommend it to Physicists regardless.
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    (Original post by TimGB)
    Palette can I volunteer as a helper? I go back to uni (2nd year maths) in 3 weeks and I need to get my brain started again, so this would be ideal. I've helped plenty of people with maths problems on TSR before, too.

    Also to answer your question, in my experience M3 didn't really do much for A2 level Physics (this was AQA and Edexcel respectively), but it was interesting and I would recommend it to Physicists regardless.
    Added .

    I decided to take statistics modules for next year but I also study A Level physics which was why I asked. It's a relief to know that my doing S2 instead of M3 won't hinder me in A Level physics.
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    (Original post by Palette)
    This thread is for people who are studying A2 Level Maths and Further Maths and is the successor of the Year 12 Maths Help Thread which was created last year. Although I hope to be relaxing much of the time, I understand that other people want to keep their maths skills refreshed over the summer.

    The rules are the same as the Year 12 Maths Help Thread:

    Rules

    1. Keep the discussion related to maths although I do not mind the occasional physics query.
    2. Please post your full working as this will enable a helper to detect any errors in your method.
    3. If you want to help somebody, don't post full solutions. Your help is meant to act as a springboard, not a featherbed as giving full solutions is unlikely to actually make somebody learn.

    Helpers:
    If you want to be a helper, please state your intention to become one and place a brief reason why.

    List:

    Ano123
    B_9710
    SeanFM
    metrize
    RDKGames
    Zacken
    RickmanAlways
    Sonila1
    TimGB
    Relevant Links:
    Spoiler:
    Show
    I volunteer as a helper. I finished college this year and currently on my Gap year. I did edexcel Maths and Further Maths, getting an A* in both subjects. I will want to keep my mind sharp as I prepare for university next year
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    (Original post by desante08)
    I volunteer as a helper. I finished college this year and currently on my Gap year. I did edexcel Maths and Further Maths, getting an A* in both subjects. I will want to keep my mind sharp as I prepare for university next year
    I've added you to the list.
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    I want to know if I've become a practitioner of voodoo mathematics (i.e. if my solution is nonsensical).

    Let a and  b be two numbers such that b\geq a>0.

    Consider the equation \int^b_a x^2 dx=(\int^b_a x dx)^2.
    i) Which value of b satisfies this equation when a=0?

    \int^b_0 x^2 dx=\frac{b^3}{3}=(\int^b_0 x dx)^2=\frac{b^4}{4}.
    As b\neq0 we can divide both sides by b^3 and multiply both sides by 12 to get b=\frac{4}{3}.

    ii. In the case a=1, derive the equation 3b^3-b^2-7b-7=0 and with the aid of a sketch, show that there is only one solution to the equation and that it lies between 2 and 3.

    \int^b_1 x^2 dx=\frac{b^3-1}{3}=(\int^b_1 x dx)^2=(\frac{b^2-1}{2})^2.

    \frac{b^3-1}{3}=\frac{(b-1)(b^2+b+1)}{3} and
    (\frac{b^2-1}{2})^2=\frac{(b+1)^2(b-1)^2}{4}.

    Now cancel (b-1) from both sides to get \frac{b^2+b+1}{3}=\frac{(b+1)^2(  b-1)}{4}.
    We get  4b^2+4b+4=3b^3+3b^2-3b-3 which leads to the required equation of 3b^3-b^2-7b-7=0.

    The y-intercept is (0,-7) and we can find the x-coordinates of the turning points by differentiating the cubic to give 9b^2-2b-7=(9b+7)(b-1). The x-coordinates of the turning points are thus b=1, b=-\frac{7}{9}. f(1)=-12 and f''(1)=16 which is a minimum so
    there is a maximum at b=-\frac{7}{9} and f)(-\frac{7}{9})&lt;0. This all becomes clearer if I can actually show my sketch here! f(2)=-1<0 and f(3)=44>0; the sign changes so there is a root between b=2 and b=3.

    Let p=b+a and q=b-a. Show that 3p^2+q^2=3p^2q and describe p^2 in terms of q. Deduce that 0&lt;q\leq\frac{4}{3}. \int^b_a x^2 dx=\frac{b^3-a^3}{3}=\frac{(b-a)(a^2+b^2+ab)}{3}=\frac{q(a^2+b  ^2+ab)}{3}.

    (\int^b_a x dx)^2=\frac{b^2-a^2}{4}=\frac{(b+a)^2(b-a)^2}{4}=\frac{p^2q^2}{4}.

    Cancelling q from both sides and multiplying both sides by 12, 4a^2+4b^2+4=3p^2q.
    3(b+a)^2+(b-a)^2=3p^2+q^2=3p^2q.

    Thus subtracting both sides by 3p^2 we obtain q^2=3p^2(q-1) to obtain p^2=\frac{q^2}{3(q-1)}.

    As p^2=(b+a)^2&gt;0, 3(q-1)&gt;0 for the equality to hold as we know that q^2 is positive so 3(q-1) can't be negative (neither can it be zero as it is not possible to divide by zero). So q&gt;1. We also know that (b+a)^2\geq (b-a)^2, with equality achieved when a=0. Thus p^2 \geq q^2 so 3(q-1)\leq 1 so q \leq \frac{4}{3}.
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    (Original post by Palette)
    I want to know if I've become a practitioner of voodoo mathematics (i.e. if my solution is nonsensical).

    Let a and  b be two numbers such that b\geq a&gt;0.

    Consider the equation \int^b_a x^2 dx=(\int^b_a x dx)^2.
    i) Which value of b satisfies this equation when a=0?

    \int^b_0 x^2 dx=\frac{b^3}{3}=(\int^b_0 x dx)^2=\frac{b^4}{4}.
    As b\neq0 we can divide both sides by b^3 and multiply both sides by 12 to get b=\frac{4}{3}.

    ii. In the case a=1, derive the equation 3b^3-b^2-7b-7=0 and with the aid of a sketch, show that there is only one solution to the equation and that it lies between 2 and 3.

    \int^b_1 x^2 dx=\frac{b^3-1}{3}=(\int^b_1 x dx)^2=(\frac{b^2-1}{2})^2.

    \frac{b^3-1}{3}=\frac{(b-1)(b^2+b+1)}{3} and
    (\frac{b^2-1}{2})^2=\frac{(b+1)^2(b-1)^2}{4}.

    Now cancel (b-1) from both sides to get \frac{b^2+b+1}{3}=\frac{(b+1)^2(  b-1)}{4}.
    We get  4b^2+4b+4=3b^3+3b^2-3b-3 which leads to the required equation of 3b^3-b^2-7b-7=0.

    The y-intercept is (0,-7) and we can find the x-coordinates of the turning points by differentiating the cubic to give 9b^2-2b-7=(9b+7)(b-1). The x-coordinates of the turning points are thus b=1, b=-\frac{7}{9}. f(1)=-12 and f''(1)=16 which is a minimum so
    there is a maximum at b=-\frac{7}{9} and f)(-\frac{7}{9})&lt;0. This all becomes clearer if I can actually show my sketch here! f(2)=-1<0 and f(3)=44>0; the sign changes so there is a root between b=2 and b=3.

    Let p=b+a and q=b-a. Show that 3p^2+q^2=3p^2q and describe p^2 in terms of q. Deduce that 0&lt;q\leq\frac{4}{3}. \int^b_a x^2 dx=\frac{b^3-a^3}{3}=\frac{(b-a)(a^2+b^2+ab)}{3}=\frac{q(a^2+b  ^2+ab)}{3}.

    (\int^b_a x dx)^2=\frac{b^2-a^2}{4}=\frac{(b+a)^2(b-a)^2}{4}=\frac{p^2q^2}{4}.

    Cancelling q from both sides and multiplying both sides by 12, 4a^2+4b^2+4=3p^2q.
    3(b+a)^2+(b-a)^2=3p^2+q^2=3p^2q.

    Thus subtracting both sides by 3p^2 we obtain q^2=3p^2(q-1) to obtain p^2=\frac{q^2}{3(q-1)}.

    As p^2=(b+a)^2&gt;0, 3(q-1)&gt;0 for the equality to hold as we know that q^2 is positive so 3(q-1) can't be negative (neither can it be zero as it is not possible to divide by zero). So q&gt;1. We also know that (b+a)^2\geq (b-a)^2, with equality achieved when a=0. Thus p^2 \geq q^2 so 3(q-1)\leq 1 so q \leq \frac{4}{3}.
    1. It's nonsensical if you say a&gt;0 and then ask about a=0. Pretty sure that should be b &gt; a \geq 0. Otherwise that part is fine.

    2. You can't cancel b-1 without showing that b\neq 1. Which follows from b &gt; a = 1 (but you've written b \geq a - 1, so that's wrong...) mention that the cubic is continuous as well instead of just the sign change. Otherwise fine.

    3. Your inequality in (\int^b_a x dx)^2=\frac{b^2-a^2}{4}=\frac{(b+a)^2(b-a)^2}{4}=\frac{p^2q^2}{4} doesn't make sense.

    4. You can't cancel q unless you know that b \neq a which follows from b&gt;a (but you've written b \geq a).

    So all in all, seems fine apart from a few stupid typos
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    (Original post by Zacken)
    1. It's nonsensical if you say a&gt;0 and then ask about a=0. Pretty sure that should be b &gt; a \geq 0. Otherwise that part is fine.

    2. You can't cancel b-1 without showing that b\neq 1. Which follows from b &gt; a = 1 (but you've written b \geq a - 1, so that's wrong...) mention that the cubic is continuous as well instead of just the sign change. Otherwise fine.

    3. Your inequality in (\int^b_a x dx)^2=\frac{b^2-a^2}{4}=\frac{(b+a)^2(b-a)^2}{4}=\frac{p^2q^2}{4} doesn't make sense.

    4. You can't cancel q unless you know that b \neq a which follows from b&gt;a (but you've written b \geq a).

    So all in all, seems fine apart from a few stupid typos
    Thank you for your feedback.
    The typos in 1.,3. and 4. were the result of the lack of proofreading for which I apologize. I think 2. is the only part which would me marks had I photographed the written version of the solution and the question, and hopefully it shouldn't be too costly! The question source is STEP I 2014 Q3.

    Having refreshed C3 and C4 over the summer (with the glaring exception of vectors, half of which I haven't learnt yet) and with six more months of mathematical experience, I decided to take another shot at some STEP questions and this was part of the result.
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    (Original post by Palette)
    The question source is STEP I 2014 Q3.
    Yeah, I know. Good attempt.
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    Is it correct that \int^\infty_a x^{-2} dx=\frac{1}{a}, a&gt;0?

    Also, should I add a link to the STEP 2017 thread on the front page?
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    (Original post by Palette)
    Is it correct that \int^\infty_a x^{-2} dx=\frac{1}{a}, a&gt;0?

    Also, should I add a link to the STEP 2017 thread on the front page?
    That would be the limit, yes. I think it would make more sense to rewrite it as:

    \int^b_a x^{-2} dx=\frac{1}{a}, 0&lt;a&lt;b, b\rightarrow \infty

    but I'm not sure. Something about limits should be in there anyway.
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    I would like to volunteer as a helper. I achieved A* in Maths and Further, and I am soon starting a joint course in Maths and Computer Science.
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    (Original post by NamelessPersona)
    I would like to volunteer as a helper. I achieved A* in Maths and Further, and I am soon starting a joint course in Maths and Computer Science.
    (Original post by RDKGames)
    That would be the limit, yes. I think it would make more sense to rewrite it as:

    \int^b_a x^{-2} dx=\frac{1}{a}, 0&lt;a&lt;b, b\rightarrow \infty

    but I'm not sure. Something about limits should be in there anyway.
    Find the exact value of  \displaystyle \int _0^{\frac{1}{2} \pi } \cos^{100} x \ dx in the form.  \displaystyle \frac{(a!) \pi }{b(c!)^2} where  a, b, and  c are integers.
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    (Original post by RDKGames)
    That would be the limit, yes. I think it would make more sense to rewrite it as:

    \int^b_a x^{-2} dx=\frac{1}{a}, 0&lt;a&lt;b, b\rightarrow \infty

    but I'm not sure. Something about limits should be in there anyway.
    (Original post by NamelessPersona)
    I would like to volunteer as a helper. I achieved A* in Maths and Further, and I am soon starting a joint course in Maths and Computer Science.
    Any ideas?
 
 
 
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