IB Maths AA HL 1.11 Partial fractions Question Bank

Practise IB Mathematics HL 1.11 by applying partial fractions methods to exam-style questions.

[Maximum number: 5]

The following question explores features of a family of curves. The family is then linked to a homogeneous differential equation.
Consider the curve given by $y=\frac{x\left(x^{2}-16\right)}{x^{2}+16}$ .

Using partial fractions, show that $\int \frac{1}{1-v^{2}} \mathrm{~d} v=\frac{1}{2} \ln \left|\frac{A(1+v)}{1-v}\right|$ , where A is a positive

Note: Award no marks for attempts at integration that do not involve partial fractions.
attempt to write $\frac{1}{1-v^{2}}$ in the form $\frac{B}{1+v}+\frac{C}{1-v}$

\begin{aligned} & B=\frac{1}{2} \text { and } C=\frac{1}{2} \\ & \int \frac{1}{1-v^{2}} \mathrm{~d} v=\int \frac{\frac{1}{2}}{1+v}+\frac{\frac{1}{2}}{1-v} \mathrm{~d} v \end{aligned}

attempt to integrate their partial fractions

EITHER

\frac{1}{2} \ln |1+v|-\frac{1}{2} \ln |1-v|+\frac{1}{2} \ln A

Note: Award A1 for first two terms. The moduli must be present for this mark. Award A1 for $\frac{1}{2} \ln A$ .

OR

\begin{aligned} & \frac{1}{2} \ln |1+v|-\frac{1}{2} \ln |1-v|(+c) \\ & =\frac{1}{2} \ln \left|\frac{\mathrm{e}^{2 c}(1+v)}{1-v}\right| \text { where } \mathrm{e}^{2 c}=A \end{aligned}

Note: The moduli must be present for each of these marks.

THEN

=\frac{1}{2} \ln \left|\frac{A(1+v)}{1-v}\right|

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AHL 1.11 (HL)—Partial fractions question 1