Oxford PAT 2014, Question 3



11 thoughts on “Oxford PAT 2014, Question 3

    1. We have constructed this equation: x/(x²+6x+8) = A/(x+4) + B/(x+2). If we multiply both sides by (x+4)(x+2), we get x = Ax + 2A + Bx + 4B. So if we look at coefficients of x, we get 1 = A + B (and also if we look at constants, we get 0 = 2A + 4B).

  1. I only learned integrals in prep for this test. When I came across this, I could not integrate? What technique is used here? Or is it simply problem solving?

    1. Hi Esther, thanks for your question.

      We have written x/((x+4)(x+2)) = A/(x+4) + B/(x+2). That means that A(x+2) + B(x+4) = x. So if we consider the coefficients of x we get Ax + Bx = x, and if we consider the numbers we get 2A + 4B = 0.

        1. Yes it does look a bit funny if you write it like that, doesn’t it? But it’s completely fine.

          What we are trying to do is to find the values of two numbers A and B, such that x/((x+4)(x+2)) = A/(x+4) + B/(x+2), leading us the the equation A(x+2) + B(x+4) = x.

          Since x is a variable that could range across many values, we need to consider the terms that contain x in them separately from the terms that don’t contain x, and this gives us two equations that A and B have to satisfy: Ax + Bx = x (which means that A + B = 0), and also 2A + 4B = 0. If we solve these two equations for A and B, and plug the answers back into the original equation A(x+2) + B(x+4) = x, it should work out.

    1. Hi Harry, thanks for your question:

      We’re not ignoring cos x. We are saying that if you differentiate (1+sin x) with respect to x, you get cos x. Therefore if you differentiate ln(1+sin x) with respect to x, you get (cos x/(1+sin x)). Therefore the integral with respect to x of (cos x/(1+sin x)) is ln(1+sin x).

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