Year 13 Maths Help Thread

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    (Original post by kiiten)
    does that make the ln part

    ln (2^12 / 2^18)

    ln (2^ -6) ?

    I think i misunderstood what you said - i still get ln16 ( ln(2^6 / 4) )
    Then perhaps you took a different approach than me which gave you the 16.

    \ln(2^{-6}) \Rightarrow -6\ln(2) is correct, all is left is to divide by the 4 and you got the ln part. The 28 is correct already.
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    (Original post by RDKGames)
    Then perhaps you took a different approach than me which gave you the 16.

    \ln(2^{-6}) \Rightarrow -6\ln(2) is correct, all is left is to divide by the 4 and you got the ln part. The 28 is correct already.
    Of courseee - i didnt think to move the -6 also ln 2 is not divided by 4 right?

    Instead i changed ln to ln(1/64) then divided by 4 to get ln16?

    I understand now, just need to remember log rules - thanks
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    3/x = 3lnx
    3/x^2 = ?

    - sorry for lots of ques today :3
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    (Original post by kiiten)
    3/x = 3lnx
    3/x^2 = ?

    - sorry for lots of ques today :3
    What are you doing to the LHS to get \displaystyle \frac{3}{x} \Rightarrow \frac{3}{x^2}?? Do the same to the RHS.
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    (Original post by RDKGames)
    What are you doing to the LHS to get \displaystyle \frac{3}{x} \Rightarrow \frac{3}{x^2}?? Do the same to the RHS.
    3lnx^2 ?
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    (Original post by kiiten)
    3lnx^2 ?
    Not really. 3\ln(x^2) = 6\ln(x)

    Just divide RHS by x. Not sure where this question really comes from or why because it doesn't simplify to anything nice.
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    (Original post by RDKGames)
    Not really. 3\ln(x^2) = 6\ln(x)

    Just divide RHS by x. Not sure where this question really comes from or why because it doesn't simplify to anything nice.
    I wanted to convert that to lnx (3/x^2)

    Ill attach the ques....
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    (Original post by kiiten)
    I wanted to convert that to lnx (3/x^2)

    Ill attach the ques....
    8.a) ii)

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    (Original post by kiiten)
    8.a) ii)

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    Where??
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    (Original post by RDKGames)
    Where??
    Sorry i was having trouble attaching it

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    (Original post by kiiten)
    Sorry i was having trouble attaching it

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    The proof for decreasing function?? I'm not quite sure what you were trying to do up there.

    \displaystyle f'(x)=-\frac{2}{e^{2x}}-\frac{3}{x^2}

    Can you see why this differential will always be negative, hence the function is always decreasing?
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    (Original post by RDKGames)
    The proof for decreasing function?? I'm not quite sure what you were trying to do up there.

    \displaystyle f'(x)=-\frac{2}{e^{2x}}-\frac{3}{x^2}

    Can you see why this differential will always be negative, hence the function is always decreasing?
    Yes - i thought you had to show it by doing dy/dx < 0 (or is it the other way round)
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    (Original post by kiiten)
    Yes - i thought you had to show it by doing dy/dx < 0 (or is it the other way round)
    That is the way to show it, what would be the "other way round"? You do NOT have to solve f'(x)&lt;0 if that's what you're asking, it is a transcendental equation so you wouldn't get any exact answers anyway. You simply have to factor out a -1 then show how the inside of the bracket is always positive therefore f'(x)&lt;0 in the given domain of x \in \mathbb{R},x&gt;0
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    (Original post by RDKGames)
    That is the way to show it, what would be the "other way round"? You do NOT have to solve f'(x)&lt;0 if that's what you're asking, it is a transcendental equation so you wouldn't get any exact answers anyway. You simply have to factor out a -1 then show how the inside of the bracket is always positive therefore f'(x)&lt;0 in the given domain of x \in \mathbb{R},x&gt;0
    How could i show inside the bracket is always +ve - just input some values?
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    (Original post by kiiten)
    How could i show inside the bracket is always +ve - just input some values?
    Not really, that wouldn't prove anything for ALL x&gt;0. Since the numerator's are already positive, consider how e^{2x} and x^2 are always positive for all real x
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    (Original post by kiiten)
    How could i show inside the bracket is always +ve - just input some values?
    x^2>0 so 3/x^2>0
    Similarly e^2x>0 for xER so same follows hence -of them is negative


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    (Original post by physicsmaths)
    x^2>0 so 3/x^2>0
    Similarly e^2x>0 for xER so same follows hence -of them is negative


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    Thanks

    If this was a question in an exam do you know what the mark scheme would say? e.g. certain phrases you must include?
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    (Original post by kiiten)
    Thanks

    If this was a question in an exam do you know what the mark scheme would say? e.g. certain phrases you must include?
    It would say e^{x}&gt;0 \Rightarrow e^{2x}&gt;0 and x^2&gt;0 \Rightarrow \frac{3}{x^2}&gt;0 where x&gt;0 therefore f'(x)&lt;0 hence f is decreasing over the interval but mark schemes never really mention all of it. Would touch up on the fact that the two terms are greater than 0, though.

    More inequalities than 'certain phrases' to be honest. This would be like 2 or 3 marks max.
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    Finally, I have a normal A Level question which I'm stuck on (I probably have annoyed many here by posting stuff which should go in the STEP prep thread):

    Find the volume of revolution of the shape modelled by the equation x=\sin t,y = \sin 2t, 0&lt;t&lt;\frac{\pi}{2}.

    V=\pi \int^{\frac{\pi}{2}}_0 y^2 \frac{dx}{dt} dt so I used the fact that y^2={\sin}^2 2t=4{\sin}^2 t {\cos}^2 t and \frac{dx}{dt}= \cos t to get me to 4\pi \int^{\frac{\pi}{2}}_0 {\sin}^2 t {\cos}^3 t dt= 4\pi \int^{\frac{\pi}{2}}_0 {\cos t}^3- {\cos t}^5 dt. What do I do next?

    Am I bizarrely overcomplicating this?
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    (Original post by Palette)
    Finally, I have a normal A Level question which I'm stuck on (I probably have annoyed many here by posting stuff which should go in the STEP prep thread):

    Find the volume of revolution of the shape modelled by the equation x=\sin t,y = \sin 2t, 0&lt;t&lt;\frac{\pi}{2}.

    V=\pi \int^{\frac{\pi}{2}}_0 y^2 \frac{dx}{dt} dt so I used the fact that y^2={\sin}^2 2t=4{\sin}^2 t {\cos}^2 t and \frac{dx}{dt}= \cos t to get me to 4\pi \int^{\frac{\pi}{2}}_0 {\sin}^2 t {\cos}^3 t dt= 4\pi \int^{\frac{\pi}{2}}_0 {\cos t}^3- {\cos t}^5 dt. What do I do next?

    Am I bizarrely overcomplicating this?
    You can use a reduction formula to integrate  \cos^3 t and  \cos^5 t but there's an easier way of doing this.
    If you notice that  \sin^2 t\cos^3 t \equiv \sin^2 t\cos t (1-\sin^2 t ) , it should be much easier to integrate.
 
 
 
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