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AQA A2 Mathematics MPC3 Core 3 - Wednesday 15th June 2016 [Official Thread] Watch

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    (SinX)^2
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    (Original post by SF7)
    How do you write sin(^2)x on a calculator?
    Same as you would (sinx)^2
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    i know the reason a function may not have an inverse is because it has one-to-one mapping, but what does one-to-one mapping mean??
    and how do you work out the range of a function if its domain is all real values of x??
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    One to one just means one value of x maps onto one value of y, and it depends on the function for example y=x^2 can have a domain for all real values but we know that f(x)>or equal to 0 as f(x) can't be negative, so just look at what they give and try a few numbers with your calculator both positive and negative and you will see its range.
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    (Original post by humzaakram1)
    Attachment 549955Part D anyone?
    I haven't done it but I would find the equations of the tangents and form a triangle on either side, doesn't matter which,with lengths calculated from using the tangent equations and solve for theta. I'll give it a try now .
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    Can anyone explain why the value of x you get as a solution for bii (2/3) isn't rejected because it's below the domain? There's a very similar question in one of the later papers (This is JUN12) and you drop marks for not rejecting the values you find which fall outside.

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    (Original post by -jordan-)
    Can anyone explain why the value of x you get as a solution for bii (2/3) isn't rejected because it's below the domain? There's a very similar question in one of the later papers (This is JUN12) and you drop marks for not rejecting the values you find which fall outside.

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    For fg(x), you use the domain of g(x). So that value isn't rejected as it is within the domain of g(x).

    For gf(x), you would look at the domain of f(x) to reject answers.
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    (Original post by MahuduElec)
    For fg(x), you use the domain of g(x). So that value isn't rejected as it is within the domain of g(x).

    For gf(x), you would look at the domain of f(x) to reject answers.
    do the values not have to be satisfied by the ranges of both parent functions?
    (i'm just asking in general not this particular question)
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    I've done all the past papers, and i wanna do a challenging one so if someone can tell me a hard paper i would like to do it thanks
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    (Original post by C0balt)
    do the values not have to be satisfied by the ranges of both parent functions?
    (i'm just asking in general not this particular question)
    i doubt it because the functions are not defined by their ranges but rather their domains. If the functions were defined by their ranges then I'd guess that your case would be true.
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    (Original post by ilovecake123)
    I've done all the past papers, and i wanna do a challenging one so if someone can tell me a hard paper i would like to do it thanks
    Do other exam boards such as Edexcel and OCR (MEI).
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    (Original post by TheLifelessRobot)
    i doubt it because the functions are not defined by their ranges but rather their domains. If the functions were defined by their ranges then I'd guess that your case would be true.
    well so they are defined by their domains which will give the range too so yeah I think they both have to be satisfied right
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    Can anyone please solve this? I got up to sec x - tanx

    prove this trig identity

    (1-sinx)/cox = 1/(sec x + tanx)
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    (Original post by liemluji)
    Can anyone please solve this? I got up to sec x - tanx

    prove this trig identity

    (1-sinx)/cox = 1/(sec x + tanx)
    Should that be cosx or cotx in denominator?
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    (Original post by Ano123)
    Should that be cosx or cotx in denominator?
    oops sorry it's supposed to be cos x
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    (Original post by humzaakram1)
    Attachment 549955Part D anyone?
    Okay I figured it out; basically you have to find the gradients of the tangent lines for both tangents, and then using the angle of inclination=tan^-1(m), by doing tan^-1(1)-tan^-1(1/3) you get the value for tan^-1(1/2).
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    (Original post by liemluji)
    Can anyone please solve this? I got up to sec x - tanx

    prove this trig identity

    (1-sinx)/cox = 1/(sec x + tanx)
    Multiply by sec x + tan x.
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    (Original post by C0balt)
    do the values not have to be satisfied by the ranges of both parent functions?
    (i'm just asking in general not this particular question)
    I remember doing a question where if you considered both domains (edit) , you would incorrectly reject one answer. Ever since then, I only look at the function closest to the (x) and have never incorrectly rejected/accepted a value.

    Think of it like this, for fg(x) if the value of x does not satisfy g(x)'s domain, you won't have a value for g(x). And so you won't be able to input a value into f(x), even if the value satisfies f(x)'s domain. As what fg(x) is essentially doing is inputting the value from g(x), into the function f(x). So if your value from g(x) isn't valid/ doesn't work/ math error, you can't input that into f(x) and expect a valid value.
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    (Original post by ilovecake123)
    I've done all the past papers, and i wanna do a challenging one so if someone can tell me a hard paper i would like to do it thanks
    Try edexcel C3 Gold 4, you can find it on physicsandmathstutor
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    (Original post by ilovecake123)
    I've done all the past papers, and i wanna do a challenging one so if someone can tell me a hard paper i would like to do it thanks
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    Try this integral.
 
 
 
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