Question
(i) By sketching a suitable pair of graphs, show that the equation\( x^{3}=3-x\)has exactly one real
root. [2]
(ii) Show that if a sequence of real values given by the iterative formula \(X_{n+1}=\frac{2x_{n}^{3}+3}{3x_{n}^{2}+1}\)
converges, then it converges to the root of the equation in part (i). [2]
(iii) Use this iterative formula to determine the root correct to 3 decimal places. Give the result of
each iteration to 5 decimal places. [3]
Answer/Explanation
3 (i) Sketch a relevant graph, e.g.y=\(x^{3}\)
Sketch a second relevant graph, e.g. y = 3 –x, and justify the given statement
3(ii) State or imply the equation\( x (2x^{3}++3)/(3x^{2}+1)\)
Rearrange this in the form \(x^{3}=3-x\) or commence work vice versa
3(iii) Use the iterative formula correctly at least once Obtain final answer 1.213
Show sufficient iterations to 5 d.p. or more to justify 1.213 to 3 d.p., or show there is a
sign change in the interval (1.2125, 1.2135)
Question
The sequence of values given by the iterative formula
\(X_{n+1}=\frac{2{x_{n}}^{6}+12x_{n}}{3x_{n}^{5}+8}\)
with initial value\( x_{1}\) = \(2, converges to \alpha\)
(i) Use the formula to calculate ! correct to 4 decimal places. Give the result of each iteration to
6 decimal places.
(ii) State an equation satisfied by ! and hence find the exact value of \(\alpha \)
Answer/Explanation
(i) Use the iterative formula correctly at least once
Obtain answer 1.3195
Show sufficient iterations to 6 d.p. to justify 1.3195 to 4 d.p., or show there is a sign change in (1.31945, 1.31955)
(ii) State x = \(\frac{2x^{6}+12x}{3x^{5}+8}\) , or equivalent
State answer\( \sqrt[5]{4}\)
Question
The curve with equation\( y = =e^{-2x}In (x-1)\) has a stationary point when x = p.
(i) Show that p satisfies the equation x =\( 1 + exp\left ( \frac{1}{2(x-1)} \right )\),\ , where exp(x )denotes \(e^{x}\)
(ii) Verify by calculation that p lies between 2.2 and 2.6.
(iii) Use an iterative formula based on the equation in part (i) to determine p correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
Answer/Explanation
(i) Use correct product rule Obtain correct derivative in any form\( \frac{dy}{dx}= -2e^{-2x}In (x-1)+\frac{e^{-2x}}{x-1}\) Equate derivative to zero and derive\( x=1+e^{\frac{1}{2(x+1)}} or p= 1+\frac{1}{2(p-1)}\)
(ii) Calculate values of a relevant expression or pair of relevant expressions at x = 2.2 and x = 2.6
\(f(x)= In (x-1)-\frac{1}{2(x-1)}\Rightarrow f(2.2)=-0.2334,f(2.6)=0.317\)
\(f(x)=2e^{-2x} In (x-1)+\frac{e^{-2x}}{x-1}\Rightarrow f(2.2)=0.005…ff(2.6)=-0.0017..\) Complete the argument correctly with correct calculated values
(iii) Use the iterative process \( p_{n+1}=1+exp\left ( \frac{1}{2(p_{n-1})} \right )\) correctly at least once Obtain final answer 2.42
Show sufficient iterations to 4 d.p. to justify 2.42 to 2 d.p., or show there is a sign
change in the interval (2.415, 2.425)