Monthly Archives: June 2015

Using simple special cases in problem-solving


Click here to download this page as pdf.

You’ve seen how mathematical problems can be solved with less, or even no, detailed calculation by looking at what the problem tells you about the shape of the solution.

Another useful method, at least for checking answers, is to look at simple special cases of the problem, or simplified versions of the problem. We saw some of that when we talked about how to check your answers in FP1 and M2. Continue reading

FP2 inequalities


Since here we deal only with functions which are continuous except maybe at a few points, the answer to all these inequalities problems will be intervals:

something like 0<x<1 or x<−1

or: x>1 or x<0

The job is to find the ends of the intervals, called critical points, and to know on which side of those critical points the inequality is valid and on which it is invalid.

We do this by drawing graphs, and then doing algebra to find the exact critical points. Sometimes it helps to simplify the inequality before drawing the graph.

To solve the inequality (5x−9)/(x+3) > 2 we can draw the graphs

y=(5x−9)/(x+3) [in other words, y = 5 − 24/(x+3)]



and see where the curve y = 5 − 24/(x+3) is above the straight line y=2

To get the exact critical points, we simplify the inequality by multiplying both sides by (x+3)2. [Why (x+3)2 and not just (x+3)?]

I prefer not to do too much multiplying up by things like (x+3)2 before drawing graphs, because drawing the graphs straight or almost straight from the inequality gives you a more direct check on your algebra.

If you have | | in the critical-point equation, you can solve it by turning it into two equations.

For example, the crossing point equation is |x+1| = 2x.

Then x+1=2x or x+1=−2x ⇒ x=1 or x=−⅓

Then, with these modulus problems, we have to check back our answers with the original equation, the one with | | in it.

In this case, x=1 is a real solution to |x+1| = 2x, but x=−⅓ isn’t.

There can be critical points which aren’t crossing points where either LHS or RHS has a singularity, i.e. it goes weird like y=1/x goes weird at x=0.

Those will be the x-values at which the left-hand-side formula or the right-hand-side formula involves trying to divide by zero. The singularity of y=(5x−9)/(x+3) is when x=−3.

Your sketch should be big. (If a line crosses a parabola at all, then it crosses it twice. But it’s easy to end up drawing so that the second crossing-point is beyond the range of your sketch…) The sketch does not have to be super-detailed: just enough to see which side of each crossing point or singularity the inequality is valid for.

Then you can do the algebraic working. For this you must multiply both sides by positive quantities, e.g for the inequality (5x−9)/(x+3) > 2 multiply both sides by (x+3)2 (not just (x+3)).

A. Try Q. 1, 14, and 8 (which is a little trickier) from Ex.1A in the FP2 book.

B. Then do Q. 2-7 from ex.1A

C. Then try the past-paper questions on pages 1-2 of this pdf. Answers are on pages 3-5.