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    I'm doing the June 2009 C3 OCR paper and this is the very last question:

    Show that, for all non-zero values of the constant m, the curve



    has exactly two stationary points.
    I'll scan in my teachers working for it, and I understand all the steps except where he's used the fact that the discriminant > 0 to show that the curve has 2 stationary points. I don't get how those 2 facts are related?

    Teacher working:
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    Here is the end zoomed in incase you can't read it:
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    Well at stationary points you have the equation \frac{dy}{dx}=0 so mx^2+x(m^2+2)+m=0 This is a quadratic with coefficients in terms of m. From you work on quadratics you should know that there is one (repeated)solution when the discriminant=0, two distinct solutions when it is more than zero, and no real solutions when it is less than zero. Here the discriminat is always going to be greater than zero and thus there will be two solutions for the equation \frac{dy}{dx}=0.
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    (Original post by namedeprived)
    Well at stationary points you have the equation \frac{dy}{dx}=0 so mx^2+x(m^2+2)+m=0 This is a quadratic with coefficients in terms of m. From you work on quadratics you should know that there is one (repeated)solution when the discriminant=0, two distinct solutions when it is more than zero, and no real solutions when it is less than zero. Here the discriminat is always going to be greater than zero and thus there will be two solutions for the equation \frac{dy}{dx}=0.
    Oh you have two equations/solutions where x=0 so that means theres two places where dy/dx = 0 so 2 stationary points? I think I got it now, am I right?
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    (Original post by Josh-H)
    Oh you have two equations/solutions where x=0 so that means theres two places where dy/dx = 0 so 2 stationary points? I think I got it now, am I right?
    Not necessarily where x=0. Cnosider the equation you have: mx^2+x(m^2+2)+m=0 You have solutions at the points x=\frac{-(m^2+2)\pm\sqrt{(m^2+2)^2 - 4m^2}}{2m}, ie 2 diffferent values therefore two stationary points.

    This comes from the quadratic formula. However we don't know the value of m, all that we're told is that it is non-zero.

    the reason that there are two stationary points when \frac{dy}{dx}=0 is because the discriminant for the quadratic equation \frac{dy}{dx}=0 is greater than zero, and therefore there has to be two solutions. We don't know what they are though, we can only write them in terms of the constant m.
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    (Original post by namedeprived)
    Not necessarily where x=0. Cnosider the equation you have: mx^2+x(m^2+2)+m=0 You have solutions at the points x=\frac{-(m^2+2)\pm\sqrt{(m^2+2)^2 - 4m^2}}{2m}, ie 2 diffferent values therefore two stationary points.

    This comes from the quadratic formula. However we don't know the value of m, all that we're told is that it is non-zero.

    the reason that there are two stationary points when \frac{dy}{dx}=0 is because the discriminant for the quadratic equation \frac{dy}{dx}=0 is greater than zero, and therefore there has to be two solutions. We don't know what they are though, we can only write them in terms of the constant m.
    Right so in this instance, where x is the x-coordinate of a stationary point, we use the discriminant to show there's 2 possible values of x so there must be 2 stationary points, got it.
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    (Original post by Josh-H)
    Right so in this instance, where x is the x-coordinate of a stationary point, we use the discriminant to show there's 2 possible values of x so there must be 2 stationary points, got it.

    Sometimes questions similar to this one are set which require the exercise of some mathematical reasoning about the discriminant and the stationary point(s).

    For example:

    f(x) = x^4 + kx^3 + x^2 + 17. Given that f'(x) has 1 stationary point, show that k^2 < 32/9.

    At first sight, the problem is perplexing because the trap is to think that 1 stationary point = 1 root, so b^2 = 4ac. Yet the question asks you to show that b^2 < 4ac. It looks as though the discriminant rule has suddenly developed a contradiction. But this is not so!

    We are dealing with stationary points, so we need the 1st derivative of f(x).

    f'(x) = 4x^3 + 3kx^2 + 2x. However, x is a common factor so we can write: f'(x)= x (4x^2 + 3kx + 2). Its immediately apparent that x = 0 is a solution of f'(x) = 0. Given that there is only 1 stationary point, the quadratic component of f'(x) has zero roots. In which case, using the rule about the discriminant:

    b^2 - 4ac < 0

    Here, the discriminant is 3k - (4x4x2), so 9k^2 < 32 and therefore
    k^2 < 32.
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    (Original post by BrasenoseAdm)
    Sometimes questions similar to this one are set which require the exercise of some mathematical reasoning about the discriminant and the stationary point(s).

    For example:

    f(x) = x^4 + kx^3 + x^2 + 17. Given that f'(x) has 1 stationary point, show that k^2 < 32/9.

    At first sight, the problem is perplexing because the trap is to think that 1 stationary point = 1 root, so b^2 = 4ac. Yet the question asks you to show that b^2 < 4ac. It looks as though the discriminant rule has suddenly developed a contradiction. But this is not so!

    We are dealing with stationary points, so we need the 1st derivative of f(x).

    f'(x) = 4x^3 + 3kx^2 + 2x. However, x is a common factor so we can write: f'(x)= x (4x^2 + 3kx + 2). Its immediately apparent that x = 0 is a solution of f'(x) = 0. Given that there is only 1 stationary point, the quadratic component of f'(x) has zero roots. In which case, using the rule about the discriminant:

    b^2 - 4ac < 0

    Here, the discriminant is 3k - (4x4x2), so 9k^2 < 32 and therefore
    k^2 < 32.
    I do not think the user cares now!

    He asked the question in 2010
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    (Original post by TeeEm)
    I do not think the user cares now!

    He asked the question in 2010
    I guess not - he went on to read engineering at B'ham. I just thought it was an interesting query and examiners seem to 'throw' candidates quite regularly by couching questions about the discriminant in an unfamiliar way.
 
 
 
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