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Math Help - A level Maths

  1. #1
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    Talking A level Maths

    i have some problems with these attached file , need help as soon as possible need to make sure i did them correctly
    Attached Thumbnails Attached Thumbnails A level Maths-untitled-1.jpg   A level Maths-untitled-2.jpg   A level Maths-untitled-3.jpg   A level Maths-untitled-4.jpg   A level Maths-untitled-5.jpg  

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  2. #2
    Bar0n janvdl's Avatar
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     \cos 2x = 0.5

     2x = 60^{\circ}

     2x = 60^{\circ} \pm 360 k

     x = 30^{\circ} \pm 180 k

     \mbox{Or } x = 150^{\circ} \pm 180k<br />
    Last edited by CaptainBlack; May 22nd 2007 at 01:32 PM.
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  3. #3
    Bar0n janvdl's Avatar
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    1. cos 2x = 0.5
    2x = 60 degrees
    2x = 60 degrees \pm 360 k
    x = 30 degrees \pm 180 k
    Or x = 150 degrees \pm 180k

    ------------------------------------

    2. Look at the RHS, where you have  cos^2 \theta
    Remember that  cos^2 = 1 - sin^2

    Therefore:

     4 sin^2 \theta - 2 sin \theta = 4 - 4 sin^2 \theta - 1
     8 sin^2 \theta - 2 sin \theta - 3 = 0

    By solving for theta we find:
    sin \theta = 0.75 \ OR \ sin \theta = -0.5
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  4. #4
    Bar0n janvdl's Avatar
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    3.  A = \pi R^2

    But Perimeter  P = 2 \pi R

     20 = 2 \pi R

     \frac{10}{\pi} = R

    Set R into A

     A = \pi (\frac{10}{\pi})^2
     A = \frac{100}{\pi}

    Now my area formula does not seem at all correct. See if you can figure out where i made a mistake.

    -----------

    Now lets ASSUME we did get the right formula.
    To maximise, we must find the derivative and equal it to 0

    So dA/dr = 0

     25 - (r - 5)^2 = 0

     25 - r^2 + 10r - 25 = 0

     r^2 + 10r = 0

     r(r + 10) = 0

    Therefore  r = 0 \ or \ r = -10

    Set both 0 and -10 into the ORIGINAL equation

    If r = 0
    Area = 0

    If r = -10
    Area = -200
    Last edited by janvdl; May 22nd 2007 at 11:02 AM.
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  5. #5
    Bar0n janvdl's Avatar
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    For the Architect Sum:

    a)  4x + 4y + x \pi

    b)  4xy + \frac{1}{2} x^2 \pi

    c)  100 = 4x + 4y + 2x \pi
     y = 25 - (x + \frac{x \pi}{2})

    Set y into the Area formula

    Then we will eventually get:  A = 50x - 2x^2 + \frac{1}{2} x^2 \pi

    ---------------

    Why cant i get these sums right??
    Right lets assume the given values again

     \frac{dA}{dx} = 100 - 4x + x \pi = 0
     100 = 4x - x \pi
     100 = x(4 - \pi)
     x = \frac{100}{(4 - \pi)}

    Now just set  x into the ORIGINAL equation to find  y
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  6. #6
    Bar0n janvdl's Avatar
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    For the Graph sum:

    a) To find where they intersect we must set y = y
     x^2 - 2x + 2 = -x + 4
     x^2 - x - 2 = 0
     x = 2 \ or \ x = -1

    Set these values into an original equation.

    Then the points are: y = -(2) + 4
    And y = -(-1) + 4

    So point P = (-1 ; 5)
    Point Q = (2 ; 2)

    b) For MPQ to be right-angled, the product of the two formulas must equal -1

    I do not get the product = -1, so i am assuming that they are not right angled.
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  7. #7
    Bar0n janvdl's Avatar
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    I will see if i can answer the rest tomorrow. I hope that the things that i could get right myself helped you.

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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
    For the Graph sum:

    a) To find where they intersect we must set y = y
     x^2 - 2x + 2 = -x + 4
     x^2 - x - 2 = 0
     x = 2 \ or \ x = -1

    Set these values into an original equation.

    Then the points are: y = -(2) + 4
    And y = -(-1) + 4

    So point P = (-1 ; 5)
    Point Q = (2 ; 2)

    b) For MPQ to be right-angled, the product of the two formulas must equal -1

    I do not get the product = -1, so i am assuming that they are not right angled.
    we have M(1 , 1), P(-1 , 5) and Q(2 , 2)

    The slope of the line connecting P and Q is given by:
    m_{1} = \frac {2 - 5}{2 + 1} = \frac {-3}{3} = -1

    The slope of the line connecting M and Q is given by:
    m_{2} = \frac {2 - 1}{2 - 1} = 1

    since 1 is the negative reciprocal of -1, the line connecting P and Q is perpendicular to the line connecting M and Q. Thus this forms a right angle at Q. If we draw a line connecting M and P therefore, we form a right-triangle
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
    3.  A = \pi R^2

    But Perimeter  P = 2 \pi R

     20 = 2 \pi R

     \frac{10}{\pi} = R

    Set R into A

     A = \pi (\frac{10}{\pi})^2
     A = \frac{100}{\pi}

    Now my area formula does not seem at all correct. See if you can figure out where i made a mistake.
    Let's find the arc length first of all.

    Recall:

    s = r \theta
    where s is the arc length and \theta is the angle subtending the arc in radians

    We are told that the perimeter is 20 cm this means that:
    P = r + r + s = 2r + r \theta = r(2 + \theta), where P is the perimeter

     \Rightarrow r(2 + \theta) = 20 \Rightarrow 2 + \theta = \frac {20}{r} \Rightarrow \theta = \frac {20}{r} - 2

    Now, the area A of the sector will be given by:

    A = \frac {1}{2} r^2 \theta
    = \frac {1}{2} r^2 ( \frac {20}{r} - 2)
    = r^2 ( \frac {10}{r} - 1)
    = 10r - r^2

    By completing the square, you will get the desired result




    Now lets ASSUME we did get the right formula.
    To maximise, we must find the derivative and equal it to 0

    So dA/dr = 0

     25 - (r - 5)^2 = 0

     25 - r^2 + 10r - 25 = 0

     r^2 + 10r = 0

     r(r + 10) = 0

    Therefore  r = 0 \ or \ r = -10

    Set both 0 and -10 into the ORIGINAL equation

    If r = 0
    Area = 0

    If r = -10
    Area = -200
    Umm, you forgot to find dA/dr janvdl, you did the manipulations on A, which is why you had weird values. (The negative valus should have hinted to you that you did something wrong)

    We wish to maximize A, so we must find A' and set it equal to zero, a value for r that causes this to happen is the value we are looking for.

    From above,  A = 25 - (r - 5)^2
     \Rightarrow A = 10r - r^2
     \Rightarrow A' = 10 - 2r
    Set A' = 0 we get:
     10 - 2r = 0
     \Rightarrow r = 5

    Note, the formula for a is a downward opening parabola, all we needed to do was find the value of r that gives the vertex. we didn't even need calculus here, we could have found this value by using:
    r = \frac {-b}{2a} , where a is the coefficient of r^2 and b is the coefficient of r.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
     \cos 2x = 0.5

     2x = 60^{\circ}

     2x = 60^{\circ} \pm 360 k

     x = 30^{\circ} \pm 180 k

     \mbox{Or } x = 150^{\circ} \pm 180k<br />
    Remember to only choose the values such that  - \pi < x < \pi , you should list them, it'll only be like four or so, also, use radians instead of degrees, since the limits were given in radians
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    I assume you actually know how to draw the graphs, draw them both on the same pair of axis. see the graphs below.

    a)

    Note that we have the roots of x^2 + e^x - 4 = 0 where the graphs intersect (do you see why?). from the graphs we see that the intersections occur such that one is for a positive x-value and the other is for a negative x-value, thus we will have a positive and negative root.


    b)
    I can't really see the power of e very well, i think it is x_{n} so let's run with that.

    x_{n+1} = - \sqrt {4 - e^{x_{n}}}
     \Rightarrow x_{1} = - \sqrt {4 - e^{x_{0}}} = - \sqrt {4 - e^{-2}} = -1.96588
    \Rightarrow x_{2} = - \sqrt {4 - e^{x_{1}}} = - \sqrt {4 - e^{-1.96588}} = -1.96468
    \Rightarrow x_{3} = - \sqrt {4 - e^{x_{2}}} = - \sqrt {4 - e^{-1.96468}} = -1.96463
    so the approx for the negative root to three decimal places is -1.965


    c)
    the approximation will go towards the negative root, since within any reasonable range \sqrt {4 - e^{x_{n}} will give a positive result. so with the minus on the outside, we will get a negative result

    EDIT: So i believe i have taken care of all the problems janvdl didn't do or the ones he expressed doubt about, for the others, i think you're in good hands with him on the job, so i won't even check them
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  12. #12
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Jhevon View Post
    Remember to only choose the values such that  - \pi < x < \pi , you should list them, it'll only be like four or so, also, use radians instead of degrees, since the limits were given in radians
    What are radians?
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by janvdl View Post
    What are radians?
    An angle in radians is defined to be
    \theta = \frac{s}{r}
    where s is the length of the arc the angle subtends in a cirlce of radius r.

    For conversion purposes, you can use this formula to determine that there are \pi radians in 180 degrees.

    -Dan
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