Edexcel IAL October 2021 P4 (WMA14/01) Q4 Binomial Expansion & Approximation
\[f(x)=\sqrt{1-4x^2} \, |x|<\frac{1}{2}\]
(a) Find, in ascending powers of \( x \), the first four non-zero terms of the binomial expansion of f(x). Give each coefficient in simplest form.
(4)
(b) By substituting \( x=\frac{1}{4} \) into the binomial expansion of f(x), obtain an approximation for \(\sqrt{3}\). Give your answer to 4 decimal places.
(2)
Solution to Part a: Find the first four non-zero terms of \( f(x) \).
Key Concepts Used
- General Binomial Expansion for \( n=1/2 \).
- Substitution: Note that the variable term is \( X=-4x^2 \), leading to powers \( x^0,x^2,x^4,x^6,… \).
Step-by-Step Solution
-
Identify \( n \) and \( X \):
\( n=\frac{1}{2} \) and \( X=-4x^2 \).
-
Apply the Binomial Formula:
– Term in \( x^0 \)
\[ x^0=1 \]
– Term in \( x^2 \)
\[ nX=\frac{1}{2}(-4x^2)=-2x^2 \]
– Term in \( x^4 \)
\[ \frac{n(n-1)}{2!}X^2=\frac{(\frac{1}{2})(-\frac{1}{2})}{2}(-4x^2)^2=-\frac{1}{8}(16x^4)=-2x^4 \]
– Term in \( x^6 \)
\[ \frac{n(n-1)(n-2)}{3!}X^3=\frac{(\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})}{6}(-4x^2)^3=\frac{1}{16}(-64x^6)=-4x^6 \]
Final Answer
\[1-2x^2-2x^4-4x^6+…\]
Solution to Part b: Approximate \(\sqrt{3}\) using \( x=\frac{1}{4} \).
Key Concepts Used
- Approximation: Relate the function value at \( x=\frac{1}{4} \) to \( \sqrt{3} \).
- Algebraic Manipulation: Isolate \( \sqrt{3} \).
Step-by-Step Solution (Part b)
-
Relate function to \(\sqrt{3}\)
Substitute \(x = \frac{1}{4}\) into the original function \(\sqrt{1 – 4x^2}\):
\[f\left(\frac{1}{4}\right) = \sqrt{1 – 4\left(\frac{1}{4}\right)^2} = \sqrt{1 – \frac{4}{16}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\]
Thus, the expansion approximates \(\frac{\sqrt{3}}{2}\). -
Calculate the approximation for \(\frac{\sqrt{3}}{2}\)
Substitute \(x = \frac{1}{4}\) into the expansion:
\[\frac{\sqrt{3}}{2} \approx 1 – 2\left(\frac{1}{4}\right)^2 – 2\left(\frac{1}{4}\right)^4 – 4\left(\frac{1}{4}\right)^6\]
\[= 1 – \frac{2}{16} – \frac{2}{256} – \frac{4}{4096} = 1 – \frac{1}{8} – \frac{1}{128} – \frac{1}{1024}\]
\[= \frac{1024 – 128 – 8 – 1}{1024} = \frac{887}{1024}\] -
Isolate \(\sqrt{3}\)
Multiply the fraction by 2:
\[\sqrt{3} \approx 2 \times \frac{887}{1024} = \frac{887}{512}\] -
Convert to 4 decimal places
\[\frac{887}{512} \approx 1.732421…\]
Final Answer
\[\sqrt{3} \approx 1.7324\]