Home / IBDP Maths AHL 1.10 permutations and combinations AA HL Paper 1- Exam Style Questions

IBDP Maths AHL 1.10 permutations and combinations AA HL Paper 1- Exam Style Questions- New Syllabus

IBDP Maths AA HL - Paper 1 - All Topics

Question

(a) Determine the first four terms of the binomial expansion for \(\sqrt{1+5x}\) in ascending powers of \(x\).
Consider the expression \((1 + px)(1 + qx)^{-1}\), where \(p, q \in \mathbb{Q}\).
(b) Find the expansion of \((1 + px)(1 + qx)^{-1}\) in ascending powers of \(x\), up to and including the term in \(x^2\).
The expansions derived in parts (a) and (b) are identical for the first three terms, for specific values of \(p\) and \(q\).
(c) Show that \(q = \frac{5}{4}\).
The expression \(\frac{1 + px}{1 + qx}\), with \(p = \frac{15}{4}\) and \(q = \frac{5}{4}\), can be used as an approximation for \(\sqrt{1+5x}\) where \(|x| < \frac{1}{5}\).
(d) (i) Hence, by identifying an appropriate value for \(x\), find the approximation for \(\sqrt{1.2}\) in the form \(\frac{m}{n}\), where \(m, n \in \mathbb{Z}\).
(ii) Now consider the approximation for \(\frac{\sqrt{5}}{2}\). Explain why the approximation for \(\frac{\sqrt{5}}{2}\) is not as accurate as the approximation for \(\sqrt{1.2}\).

Syllabus Topic Codes (IB Mathematics AA HL):

• AHL 1.10: Extension of the binomial theorem to fractional and negative indices — parts (a), (b)
• SL 1.1: Operations with numbers; approximation and error — part (d)
• SL 2.7: Solution of linear and quadratic equations — part (c)
▶️ Answer/Explanation

(a)
Using the binomial expansion \((1+X)^n = 1 + nX + \frac{n(n-1)}{2!}X^2 + \frac{n(n-1)(n-2)}{3!}X^3 + \dots\) with \(n = \frac{1}{2}\) and \(X = 5x\):
\(\sqrt{1+5x} = 1 + \left(\frac{1}{2}\right)(5x) + \frac{(\frac{1}{2})(-\frac{1}{2})}{2}(5x)^2 + \frac{(\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})}{6}(5x)^3 + \dots\)
\(= 1 + \frac{5}{2}x – \frac{25}{8}x^2 + \frac{125}{16}x^3 + \dots\)
\(\boxed{1 + \frac{5}{2}x – \frac{25}{8}x^2 + \frac{125}{16}x^3}\)

(b)
First, expand \((1+qx)^{-1}\): \(1 + (-1)(qx) + \frac{(-1)(-2)}{2}(qx)^2 + \dots = 1 – qx + q^2x^2 + \dots\)
Now, find the product \((1+px)(1 – qx + q^2x^2)\):
\(= 1 – qx + q^2x^2 + px – pqx^2 + \dots\)
\(= 1 + (p-q)x + (q^2-pq)x^2 + \dots\)
\(\boxed{1 + (p-q)x + (q^2-pq)x^2}\)

(c)
Equating coefficients of \(x\) and \(x^2\) from (a) and (b):
1) \(p – q = \frac{5}{2} \implies p = q + \frac{5}{2}\)
2) \(q^2 – pq = -\frac{25}{8}\)
Substitute \(p\) into the second equation: \(q^2 – q(q + \frac{5}{2}) = -\frac{25}{8}\)
\(q^2 – q^2 – \frac{5}{2}q = -\frac{25}{8} \implies -\frac{5}{2}q = -\frac{25}{8}\)
\(q = \frac{25}{8} \times \frac{2}{5} = \frac{5}{4}\).
\(\boxed{q = \frac{5}{4}}\)

(d)(i)
To find \(\sqrt{1.2}\), set \(1 + 5x = 1.2 \implies 5x = 0.2 \implies x = \frac{1}{25}\) (or \(0.04\)).
Substitute \(x = \frac{1}{25}\), \(p = \frac{15}{4}\), and \(q = \frac{5}{4}\) into \(\frac{1+px}{1+qx}\):
\(\frac{1 + \frac{15}{4}(\frac{1}{25})}{1 + \frac{5}{4}(\frac{1}{25})} = \frac{1 + \frac{3}{20}}{1 + \frac{1}{20}} = \frac{23/20}{21/20} = \frac{23}{21}\).
\(\boxed{\frac{23}{21}}\)

(d)(ii)
Recognize that \(\frac{\sqrt{5}}{2} = \sqrt{\frac{5}{4}} = \sqrt{1.25}\).
For \(\sqrt{1.25}\), \(1 + 5x = 1.25 \implies 5x = 0.25 \implies x = \frac{1}{20}\).
The value \(x = \frac{1}{20}\) is greater than \(x = \frac{1}{25}\) (the value used for \(\sqrt{1.2}\)).
In binomial approximations, accuracy decreases as \(|x|\) increases (moves further from zero).
Therefore, the approximation for \(\sqrt{1.2}\) is more accurate.

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