# Oxford PAT 2012, Question 6

## 9 thoughts on “Oxford PAT 2012, Question 6”

1. Peter Cong says:

Hello and thanks for all your help, they are very clear. But I can’t quiet get my head around the reason why the dashed line should be rejected as an answer

1. Hi Peter — it’s just because it says in the question ‘just touches (the curve) at x>0’

2. Jonas says:

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.

1. Beatrice says:

Hi, sorry for bringing this back up — but how can we easily apply y = f'(u)*(x-u)+f(u) in this case, taking into account that we’re not given the coordinates of the point the line is tangent to?

1. Hi Beatrice — yes you’re right, this isn’t actually an especially good way of answering the question because you still need to solve the combined equation to find the point of intersection of the line and the curve. Thanks for pointing that out.

3. Jacob says:

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.

1. Jacob says:

Ahh I see. I completely forgot to look at discriminant. Thank you!