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    Hi, I am really stuck on this. Please can someone show me how to do it!

    f(x) = a/x
    g(x) = ax + k

    The line y = g(x) is a tangent to the curve y = |f(x)| at the point P, and cuts the curve y = |f(x)| at the point Q.

    (b) Show that k = 2a.

    Thank you!!
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    (Original post by 1a2b3c4d)
    Hi, I am really stuck on this. Please can someone show me how to do it!

    f(x) = a/x
    g(x) = ax + k

    The line y = g(x) is a tangent to the curve y = |f(x)| at the point P, and cuts the curve y = |f(x)| at the point Q.

    (b) Show that k = 2a.

    Thank you!!
    Can you show us the full question, please?
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    (Original post by Zacken)
    Can you show us the full question, please?
    That is the full question. Part (a) asks you to sketch the curve 'with equation y = |f(x)|.
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    (Original post by Zacken)
    Can you show us the full question, please?
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    (Original post by 1a2b3c4d)
    That is the full question. Part (a) asks you to sketch the curve 'with equation y = |f(x)|.
    *shrugs* |f(x)| = g(x) \iff ax + k = \pm \frac{a}{x}

    Now consider the two equations:

    ax + k = \frac{a}{x} \Rightarrow ax^2 + kx - a = 0 or ax + k = -\frac{a}{x} \Rightarrow ax^2 + kx + a = 0

    For |f(x)| and g(x) to be tangent, one of these has to have a zero discriminant.
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    Sorry could you take it one step further, I still can't quite get it..
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    (Original post by 1a2b3c4d)
    Sorry could you take it one step further, I still can't quite get it..
    What is the discriminant of those two quadratic equations?
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    (Original post by 1a2b3c4d)
    Hi, I am really stuck on this. Please can someone show me how to do it!

    f(x) = a/x
    g(x) = ax + k

    The line y = g(x) is a tangent to the curve y = |f(x)| at the point P, and cuts the curve y = |f(x)| at the point Q.

    (b) Show that k = 2a.

    Thank you!!
    Following on from Zacken's post, the discriminants are:
    (1) k^2-4a(-a) = k^2+4a^2, and (2) k^2-4a(a)=k^2-4a^2.
    So in case 1, k^2=-4a^2, which is impossible as the LHS is positive and the RHS is negative.
    Thus we go to case 2: k^2=4a^2 -> k=+/-2a, and since k and a are positive, k=2a, as required.
 
 
 
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