IBDP Maths SL 1.9 The binomial theorem AA HL Paper 2- Exam Style Questions- New Syllabus

Step 1 — Rewrite the expression
We express the function in a form suitable for binomial expansion: \( \frac{1}{\sqrt{q – x^2}} = (q – x^2)^{-\frac{1}{2}} = q^{-\frac{1}{2}} \left( 1 – \frac{x^2}{q} \right)^{-\frac{1}{2}} \).

Step 2 — Apply the binomial theorem
Using the general binomial expansion for \( (1 + y)^n \) with \( n = -\frac{1}{2} \) and \( y = -\frac{x^2}{q} \): \( (1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots \) Here, \( y = -\frac{x^2}{q} \). We need the term containing \( x^6 \), which comes from the \( y^3 \) term: \( \frac{n(n-1)(n-2)}{3!} y^3, \quad \text{where } n = -\frac{1}{2}. \)

Step 3 — Calculate the coefficient
\( \frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2} – 1\right)\left(-\frac{1}{2} – 2\right)}{3!} = \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6} = \frac{-\frac{15}{8}}{6} = -\frac{15}{48} = -\frac{5}{16}. \) The \( y^3 \) term is \( \left( -\frac{x^2}{q} \right)^3 = -\frac{x^6}{q^3} \). Thus the term in \( x^6 \) is: \( q^{-\frac{1}{2}} \times \left( -\frac{5}{16} \right) \times \left( -\frac{x^6}{q^3} \right) = q^{-\frac{1}{2}} \times \frac{5}{16} \times \frac{x^6}{q^3} = \frac{5}{16} q^{-\frac{7}{2}} x^6. \)

Step 4 — Equate to given coefficient
The coefficient of \( x^6 \) is given as 5120: \( \frac{5}{16} q^{-\frac{7}{2}} = 5120 \) \( q^{-\frac{7}{2}} = 5120 \times \frac{16}{5} = 16384 \) \( q^{\frac{7}{2}} = \frac{1}{16384} \) \( q^{\frac{7}{2}} = 16384^{-1} = (2^{14})^{-1} = 2^{-14} \) \( q = (2^{-14})^{\frac{2}{7}} = 2^{-4} = \frac{1}{16}. \)

\( q = \boxed{\frac{1}{16}} \)