where A and B are constants to be found.(4)
(b) Use proof by contradiction to prove that the curve with equation
does not intersect the curve with equation
where A and B are the constants found in part (a).
(Solutions relying on calculator technology are not acceptable.)(4)
- Algebraic Preparation: Factor out 8.
- General Binomial Expansion Formula for \(n=4/3\).
- Coefficient Matching.
- Prepare the expression
Let’s factor out 8.
\[(8-3x)^{\frac{4}{3}}=\left[8\left(1-\frac{3}{8}x\right)\right]^{\frac{4}{3}}=8^{\frac{4}{3}}\left(1-\frac{3}{8}x\right)^{\frac{4}{3}}\]Where,
\[8^{\frac{4}{3}}=({\sqrt[4]{8}})^{4}=2^{4}=16\]
\[f(x)=16\left(1+\left(-\frac{3}{8}x\right)\right)^{\frac{4}{3}}\] - Identify \(n\) and \(X\)
\(n=\frac{4}{3}\) and \(X=-\frac{3}{8}x\).
- Calculate the terms for \((1+X)^{n}\) using binomial expansion formula:
\[(1+x)^{n} = x^{0} + nx + \frac{n(n-1)}{2!}x^{2} + \frac{n(n-1)(n-2)}{3!}x^{3} + …\]
-Term in \( x^0 \):
\[ 1 \]-Term in \( x^1 \):
\[ nx = \frac{4}{3}\left(-\frac{3}{8}\right) = -\frac{1}{2}x \]-Term in \( x^2 \):
\[\frac{n(n-1)}{2!}X^{2} = \frac{\left( \frac{4}{3} \right) \left( \frac{1}{3} \right)}{2} \left( -\frac{3}{8}x \right)^{2} = \frac{2}{9} \left( \frac{9}{64}x^{2} \right) = \frac{1}{32}x^{2}\]
-Term in \( x^3 \):
\[\frac{n(n-1)(n-2)}{3!}X^{3} = \frac{\left( \frac{4}{3} \right) \left( \frac{1}{3} \right) \left( -\frac{2}{3} \right)}{6} \left( -\frac{3}{8}x \right)^{3} = -\frac{4}{81} \left( \frac{27}{512}x^{3} \right) = \frac{1}{384}x^{3}\]
- Multiply by 16:
\[16 \left( 1 – \frac{1}{2}x + \frac{1}{32}x^{2} + \frac{1}{384}x^{3} + … \right) = 16 – 8x + \frac{1}{2}x^{2} + \frac{1}{24}x^{3} + …\]
- Identify \(A\) and \(B\):
Comparing \(16 – 8x + \frac{1}{2}x^{2} + \frac{1}{24}x^{3} + …\) with \(A – 8x + \frac{x^{2}}{2} + Bx^{3} + …\):
\[A = 16 \otimes B = \frac{1}{24}\]Comparing \(16 – 8x + \frac{1}{2}x^{2} + \frac{1}{24}x^{3} + …\) with \(A – 8x + \frac{x^{2}}{2} + Bx^{3} + …\):
\[A = 16 \otimes B = \frac{1}{24}\]
\[
\text{ where } A = 16 \text{ and } B = \frac{1}{24}
\]
- Proof by Contradiction: Assume the curves intersect, meaning \(y_{1} = y_{2}\).
- Algebraic Manipulation: Rearrange the resulting equation to show a contradiction.
-
Form the Assumption
Assume the two curves intersect for some value of \(x\) in the domain
\(0 < x < \frac{8}{3}\):\[y_{1} = 8 + 8x – \frac{15}{2}x^{2}\]
\[y_{2} = 16 – 8x + \frac{1}{2}x^{2} + \frac{1}{24}x^{3}\] -
Equate the two functions (\(y_{1}=y_{2}\))
\[8+8x-\frac{15}{2}x^{2}=16-8x+\frac{1}{2}x^{2}+\frac{1}{24}x^{3}\]
-
Rearrange to set equal to zero
Gather terms on the right-hand side (or multiply by 24 to clear fractions)
\[0=\frac{1}{24}x^{3}+\left(\frac{1}{2}x^{2}+\frac{15}{2}x^{2}\right)-16x+8\]
\[0=\frac{1}{24}x^{3}+8x^{2}-16x+8\]
\[\frac{1}{24}x^{3}+8(x^{2}-2x+1)=0\]
\[\frac{1}{24}x^{3}+8(x-1)^{2}=0\] -
Identify the Contradiction
We are given that \(x\) is positive (\(0<x<\frac{8}{3}\)).
\[\frac{1}{24}x^{3}>0\]
\[8(x-1)^{2}\geq 0\]Therefore,
\[\frac{1}{24}x^{3}+8(x-1)^{2}>0\]This cannot equal zero, giving a contradiction.
\(\frac{1}{24}x^{3}+8(x-1)^{2}=0\), but for all \(x>0\) the expression is
strictly positive, this gives a contradiction. Therefore, the curves do not
intersect in the given range.