Oxford PAT 2012, Question 6


5 thoughts on “Oxford PAT 2012, Question 6

  1. Hi and thanks for these solutions, they have been extremely helpful.
    I was wondering if for this question you would be allowed to use the point-slope-form of a tangent?
    So y= f'(u)*(x-u)+f(u), where u is the point to which the line will be tangent to, f is the function and f'(u) the slope at the point (u,f(u))

    with this equation you can solve questions like these much more quickly and very reliably.

    1. Hi Jonas — I’m glad the site has been useful.

      In response to your question, I think there are absolutely no restrictions on how you are expected to solve the problems, as long as you get them right and the examiner can see that your reasoning makes sense.

      Your way of solving the problem is nice and elegant so I would certainly feel free to use it.

  2. Hello, if I may ask, I can not see how you are getting from x^2-x(4+m)+2=0 to the next line, you say has only one….? Can not read the last part 😛

    1. Hi Jacob. Sorry sometimes my handwriting is really bad. The phrase is “has only one root”. That is, I am saying that if you were to consider the function (x-2)^2 – (mx + 2), then this would equal 0 at point that the the line and the curve meet. So when we rewrite this to x^2-x(4+m)+2=0, which is a quadratic of the form ax^2+bx+c=0. We can then consider the standard solution for quadratics and because we have only one root we can say that b^2 – 4ac = 0, and that is how we get the next line.

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