Problem solving strategies

Here are strategies for getting unstuck.

Limited number of concepts being tested
As many equations as there are unknowns
Sketches and diagrams
Working backwards
When to compare coefficients?

Limited number of concepts being tested

There are only so many concepts in the syllabus. It is useful to cycle through the concepts, formulas, or theorems and see which one could be applied in this problem. It is more efficient to remember the problems that can be solved by each approach, rather than the approach to a specific problem.

As many equations as there are unknowns

To solve for a variable, you need an equation.

To solve for two variables, you need two equations.

Now, when you have two variables but only one equation, you either have to obtain a second equation, or eliminate one of your variables. When stuck, try using a different set of variables, hence different equations.

Sketches and diagrams

When interpreting diagrams provided, you cannot make any assumption regarding equality of angles or sides, but you can assume intersections roughly appear where they are shown on the diagram.

Using function transformations, most functions appearing on the exam could be quickly sketched. This can help visualizing intersections and areas. As the course revolves around functions, it is a good habit to find connections between the current topic and relevant ideas from functions.

When solving trigonometric equations, it is often useful to draw a unit circle or graphs of sine and cosine. When solving a triangle, if one is not provided, it helps to draw the triangle or triangles.

In combinatorics (HL) and probability, drawing a Venn diagram, tree diagram, or sample space diagram could help visualize the situation. In normal distribution problems, it helps to draw a bell curve.

In complex numbers (HL), it is helpful to draw an Argand diagram. In vectors (HL), it is helpful to draw 2-D schematics of the vectors, lines, and planes.

Working backwards

While the deductive reasoning is focused on a linear train of thought, it is often productive to brainstorm ideas for the entire problem rather than only the next step. Foreseeing the last step or any intermediate step can be just as productive as seeing the first step. Math is just connecting the dots between the steps you think would be involved. It is why there are so many “show that” questions that give you the answer.

It is important though to present the solution from beginning to end, rather in the order each step came to your mind.

When to compare coefficients?

Comparing coefficients is possible when the terms are linearly independent, meaning none of the non-zero terms can be rewritten as sum of (scalar) multiples of the other terms. This concept is good to know, but not explicitly tested on exams.

Definition: Terms $u_1, u_2, \dots, u_n$ are linearly independent, if and only if the only coefficients $a_1, a_2, \dots, a_n$ that satisfy

a_1u_1 + a_2u_2 + \dots + a_nu_n = 0

are $a_1 = a_2 = \dots = a_n = 0$ . Otherwise, the terms are linearly dependent.

For example $x^2$ and $x^3$ are linearly independent, this means that if

\begin{align*} f(x) = Ax^3 + Bx^2 \\ g(x) = Cx^3 + Dx^2 \end{align*}

then

f = g \iff A = C, B = D

With

\begin{align*} f(x) = Ax^2 + Bx + C(3x^2 + 5x) \\ g(x) = Dx^2 + Ex + F(3x^2 + 5x) \end{align*}

and given $f = g$ and values of $A, B, C$ , it is insufficient to determine $D, E, F$ , because $x^2$ , $x$ and $3x^2 + 5x$ are linearly dependent. In this case, comparing coefficients is not possible.

Finally $3$ and $\sqrt7$ are also linearly dependent, as $-{\sqrt 7}(3) + 3\left(\sqrt{7}\right) = 0$ . So you cannot compare coefficients when working with radicals (surds). This means $3A + B\sqrt7 = 3C + D\sqrt7$ does not imply that $A$ is equal to $C$ or that $B$ is equal to $D$ .

Polynomials, coordinates, complex numbers (HL) in Cartesian form, and position vectors (HL) can be compared.

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