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6 (a) Find the first three terms, in ascending powers of x, of the binomial expansion of \( \frac{1}{\sqrt{4+x}} \) - AQA - A-Level Maths Pure - Question 6 - 2018 - Paper 1
Question 6
6 (a) Find the first three terms, in ascending powers of x, of the binomial expansion of \( \frac{1}{\sqrt{4+x}} \). 6 (b) Hence, find the first three terms of the ... show full transcript
Worked Solution & Example Answer:6 (a) Find the first three terms, in ascending powers of x, of the binomial expansion of \( \frac{1}{\sqrt{4+x}} \) - AQA - A-Level Maths Pure - Question 6 - 2018 - Paper 1
Step 1
Find the first three terms, in ascending powers of x, of the binomial expansion of \( \frac{1}{\sqrt{4+x}} \).
Answer
To find the binomial expansion of ( \frac{1}{\sqrt{4+x}} ), we can rewrite it as ( (4+x)^{-\frac{1}{2}} ). Using the binomial expansion formula ( (1+k)^n \approx 1 + nk + \frac{n(n-1)}{2}k^2 ) for small ( k ), we have:
[ (1 + \frac{x}{4})^{-\frac{1}{2}} \approx 1 - \frac{1}{2} \cdot \frac{x}{4} + \frac{-\frac{1}{2} \cdot -\frac{3}{2}}{2} \left( \frac{x}{4} \right)^2 \Rightarrow 1 - \frac{x}{8} + \frac{3x^2}{128}. ]
Hence, the first three terms are ( 1 - \frac{x}{8} + \frac{3x^2}{128} ).
Step 2
Hence, find the first three terms of the binomial expansion of \( \frac{1}{\sqrt{4-x^3}} \).
Answer
Rewriting ( \frac{1}{\sqrt{4-x^3}} ) as ( (4-x^3)^{-\frac{1}{2}} ) gives:
[ (1 - \frac{x^3}{4})^{-\frac{1}{2}} \approx 1 + \frac{1}{2} \cdot \frac{x^3}{4} + \frac{\frac{1}{2} \cdot \frac{3}{2}}{2} \left( \frac{x^3}{4} \right)^2 \Rightarrow 1 + \frac{x^3}{8} + \frac{3x^6}{128}. ]
Thus, the first three terms are ( 1 + \frac{x^3}{8} + \frac{3x^6}{128} ).
Step 3
Using your answer to part (b), find an approximation for \( \int_{0}^{2} \frac{1}{\sqrt{4-x^3}} \, dx \), giving your answer to seven decimal places.
Answer
To approximate the integral, we can substitute the first three terms of the expansion:
[ \int_{0}^{2} \left( 1 + \frac{x^3}{8} + \frac{3x^6}{128} \right) dx. ]
Calculating the integral:
- The integral of ( 1 ) from 0 to 2 is ( 2 ).
- The integral of ( \frac{x^3}{8} ) from 0 to 2 is ( \frac{1}{8} \cdot \frac{x^4}{4} \bigg|_{0}^{2} = \frac{1}{8} \cdot 4 = \frac{1}{2}. )
- The integral of ( \frac{3x^6}{128} ) from 0 to 2 is ( \frac{3}{128} \cdot \frac{x^7}{7} \bigg|_{0}^{2} = \frac{3}{128} \cdot \frac{128}{7} = \frac{3}{7}. )
Adding these results gives:
[ 2 + \frac{1}{2} + \frac{3}{7} = 2.5 + 0.428571 = 2.928571.] ]
Thus, the approximation for the integral is approximately ( 2.9285714 ).
Step 4
Explain clearly whether Edward's approximation will be an overestimate, an underestimate, or if it is impossible to tell.
Answer
Edward's approximation is likely to be an underestimate. Since the series being used for approximation consists of positive terms (all terms in the expansion are positive), adding more terms will increase the value of the approximation gradually. Therefore, as the number of terms increases, the total approximation will converge to the actual value of the integral but will always remain less than or equal to it.
Step 5
Explain why Edward's approximation is invalid.
Answer
Edward's approximation is invalid because the binomial expansion used for ( \frac{1}{\sqrt{4-x^3}} ) is only valid within a specific range. The expansion holds true if (|x| < 4), and at value (x = 2), (4 - x^3 = 0). Thus, the approximation breaks down and is no longer reliable at the upper limit of integration.