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    Can sombody please work out these Qs for me?!

    6) Given that dy/dx=3 sqrt x –x^2, and that y=2/3 when x=1, find the value of y when x=4. (7 marks)

    7) The first three terms of an arithmetic series are (12-p), 2p and (4p-5) respectively, where p is a constant.

    a) Find the value of p (2 marks)
    b) Show that the sixth term of the series is 50 (3 marks)
    c) Find the sum of the first 15 terms of the series. (2 marks)
    d) Find how many terms of the series have a value of less than 400 (3 marks)

    8) f(x)=2x^2 +3x –2.
    a) Solve the equation f(x)=0 (2 marks)
    b) Sketch the curve with equation y=f(x), showing the coordinates of any points of intersection with the coordinate axes. (2 marks)
    c) Find the coordinates of the points where the curve with equation y=f(1/2x) crosses the coordinate axes. (3 marks)
    d) When the graph of y=f(x) is translated by 1 unit in the positive x-direction it maps onto the graph with equation y=ax^2 +bx +c, where a, b and c are constants. Find the values of a, b and c. (3 marks)
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    6) Given that dy/dx=3 sqrt x –x^2, and that y=2/3 when x=1, find the value of y when x=4. (7 marks)
    Integrate with respect to x, and add your constant. Plug in the the coordinate (1, 2/3) to find the constant. Then substitute in x=4. Remember that sqrtx = x^0.5

    7) The first three terms of an arithmetic series are (12-p), 2p and (4p-5) respectively, where p is a constant.
    For (a) You know that the difference between two of the terms are the same. Ie: 2p-(12-p) = (4p-5)-(2p), as the same constant is added between the expressions.
    For (b) You find the common difference using the work done in (a). Then use the formulae for the nth term.
    For (c) The method is similar to (b), but with the sum formulae.
    For (d) You find the first term with a value greater than 400. The terms are all those below that.

    8) f(x)=2x^2 +3x –2.
    a) Solve the equation f(x)=0 (2 marks)
    Factorise, use the quadratic equation or complete the square.

    b) Sketch the curve with equation y=f(x), showing the coordinates of any points of intersection with the coordinate axes. (2 marks)
    I'm sure you can do some substitutions to find some coordinates and hence sketch it.

    c) Find the coordinates of the points where the curve with equation y=f(1/2x) crosses the coordinate axes. (3 marks)
    Obtain this equation by replacing x in f(x) by (1/2x). Then substitute for x=0 or y=0 to find the coordinates.

    d) When the graph of y=f(x) is translated by 1 unit in the positive x-direction it maps onto the graph with equation y=ax^2 +bx +c, where a, b and c are constants. Find the values of a, b and c. (3 marks)
    Replace x in f(x) by (x-1) and equate coefficients.
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    6.

    dy/dx = sqrt x - x^2
    dy/dx = x^(1/2) - x^2

    integrate (indefinite integral)

    y = (2/3)x^(3/2) - (1/3)x^3 + C

    to find the constant of integration C, substitute the known paid of values of (x, y) in:

    2/3 = (2/3)*1^(3/2) - (1/3)*1^3 + C
    2/3 = 2/3 - 1/3 + c
    c = 1/3

    Now sub this back into the equation:

    y = (2/3)x^(3/2) - (1/3)x^3 + 1/3

    when x=4,

    y = (2/3)*4^(3/2) - (1/3)*4^3 + 1/3
    y = 16/3 - 64/3 + 1/3
    y = -47/3

    check through this yourself becuase:
    (a) I might have made a mistake
    (b) you will probably learn the method better if you try it too
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    lol... I'm getting slow.

    Oh well.

    (Original post by Gaz031)
    Replace x in f(x) by (x+1) and equate coefficients.
    (x-1)
 
 
 
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Updated: January 22, 2005

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