6 (a) Find the first three terms, in ascending powers of x, of the binomial expansion of
$$rac{1}{rac{1}{4} + x}$$
6 (b) Hence, find the first three terms of the binomial expansion of
$$rac{1}{rac{1}{4} - x^3}$$
6 (c) Using your answer to part (b), find an approximation for
$$\int_0^1 \frac{1}{\frac{1}{4} - x^3} \,dx$$,
giving your answer to seven decimal places - 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
$$rac{1}{rac{1}{4} + x}$$
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
$$rac{1}{rac{1}{4} + x}$$
6 (b) Hence, find the first three terms of the binomial expansion of
$$rac{1}{rac{1}{4} - x^3}$$
6 (c) Using your answer to part (b), find an approximation for
$$\int_0^1 \frac{1}{\frac{1}{4} - x^3} \,dx$$,
giving your answer to seven decimal places - 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 $rac{1}{rac{1}{4} + x}$
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the binomial expansion, rewrite it in a suitable form:
rac{1}{rac{1}{4} + x} = 4(1 + 4x)^{-1}
Using the binomial expansion formula
(1+kx)n=1+nx+2!n(n−1)x2+… for n = -1:
The first term is: [ 1 ]
The second term is: [ -4x ]
The third term is: [ \frac{(-1)(-2)}{2} \cdot 16x^2 = 8x^2 ]
So the first three terms are:
4−16x+32x2.
Step 2
Hence, find the first three terms of the binomial expansion of $rac{1}{rac{1}{4} - x^3}$
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Rewrite the expression:
rac{1}{rac{1}{4} - x^3} = 4 (1 - 4x^3)^{-1}
Using the same binomial expansion:
The first term is: [ 1 ]
The second term is: [ 4x^3 ]
The third term, since there are no terms higher than x3, is: [ 0 ]
Thus, the first three terms are:
4+16x3+0.
Step 3
Using your answer to part (b), find an approximation for $
\int_0^1 \frac{1}{\frac{1}{4} - x^3} \,dx$
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Use the expansion from part (b) and integrate:
∫01(4+16x3)dx.
Calculating the terms:
For 4dx: [ 4x \bigg|_0^1 = 4 ]
For (16x3)dx: [ 16 \cdot \frac{x^4}{4} \bigg|_0^1 = 4 ]
Total approximation:
4+4=8.
So, the approximation is 8.
Step 4
Explain clearly whether Edward's approximation will be an overestimate, an underestimate, or if it is impossible to tell.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Each term in the expansion is positive. Since increasing the number of terms will provide a more accurate estimate, Edward's approximation will therefore be an underestimate.
Step 5
Explain why Edward's approximation is invalid.
97%
117 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The binomial expansion is valid for ∣x∣<41. However, when evaluating ∫02141−x31dx, it exceeds this limit since 21>41. Thus, the approximation derived is invalid.