Solve exponential equation via iteration

6 marks Standard +0.3

4 Using the substitution $u = 3 ^ { x }$, or otherwise, solve, correct to 3 significant figures, the equation $$3 ^ { x } = 2 + 3 ^ { - x }$$

5 marks Standard +0.3

2 Solve, correct to 3 significant figures, the equation $$\mathrm { e } ^ { x } + \mathrm { e } ^ { 2 x } = \mathrm { e } ^ { 3 x }$$

7 marks Moderate -0.3

5

By sketching a suitable pair of graphs, show that the equation $2 + \mathrm { e } ^ { - 0.2 x } = \ln ( 1 + x )$ has only one root.
Show by calculation that this root lies between 7 and 9 . \includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-08_2716_40_109_2009}
Use the iterative formula $$x _ { n + 1 } = \exp \left( 2 + \mathrm { e } ^ { - 0.2 x _ { n } } \right) - 1$$ to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. $\left[ \exp ( x ) \right.$ is an alternative notation for $\left. \mathrm { e } ^ { x } .\right]$ \includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-10_481_789_262_639} The diagram shows the curve $y = \sin 2 x ( 1 + \sin 2 x )$, for $0 \leqslant x \leqslant \frac { 3 } { 4 } \pi$, and its minimum point $M$. The shaded region bounded by the curve that lies above the $x$-axis and the $x$-axis itself is denoted by $R$.

7 marks Standard +0.3

3. $$f ( x ) = 2 ^ { x - 1 } - 4 + 1.5 x \quad x \in \mathbb { R }$$

Show that the equation $\mathrm { f } ( x ) = 0$ can be written as $$x = \frac { 1 } { 3 } \left( 8 - 2 ^ { x } \right)$$ The equation $\mathrm { f } ( x ) = 0$ has a root $\alpha$, where $\alpha = 1.6$ to one decimal place.
Starting with $x _ { 0 } = 1.6$, use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 3 } \left( 8 - 2 ^ { x _ { n } } \right)$$ to calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 3 decimal places.
By choosing a suitable interval, prove that $\alpha = 1.633$ to 3 decimal places.

8 marks Standard +0.3

2. $$\mathrm { g } ( x ) = \mathrm { e } ^ { x - 1 } + x - 6$$

Show that the equation $\mathrm { g } ( x ) = 0$ can be written as $$x = \ln ( 6 - x ) + 1 , \quad x < 6$$ The root of $\mathrm { g } ( x ) = 0$ is $\alpha$.
The iterative formula $$x _ { n + 1 } = \ln \left( 6 - x _ { n } \right) + 1 , \quad x _ { 0 } = 2$$ is used to find an approximate value for $\alpha$.
Calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ to 4 decimal places.
By choosing a suitable interval, show that $\alpha = 2.307$ correct to 3 decimal places.

5 marks Standard +0.3

6 Find the solution to this equation, correct to 3 significant figures. $$\left( 2 ^ { x } \right) \left( 2 ^ { x + 1 } \right) = 10 .$$

11 marks Standard +0.3

6. $\quad f ( x ) = 2 x ^ { 2 } + 3 \ln ( 2 - x ) , \quad x \in \mathbb { R } , \quad x < 2$.

Show that the equation $\mathrm { f } ( x ) = 0$ can be written in the form $$x = 2 - \mathrm { e } ^ { k x ^ { 2 } }$$ where $k$ is a constant to be found. The root, $\alpha$, of the equation $\mathrm { f } ( x ) = 0$ is 1.9 correct to 1 decimal place.
Use the iterative formula $$x _ { n + 1 } = 2 - \mathrm { e } ^ { k x _ { n } ^ { 2 } }$$ with $x _ { 0 } = 1.9$ and your value of $k$, to find $\alpha$ correct to 3 decimal places.
You should show the result of each iteration.
Solve the equation $\mathrm { f } ^ { \prime } ( x ) = 0$.

12 marks Standard +0.8

4 You are given the equation $( 2 x - 1 ) ^ { 2 } - \mathrm { e } ^ { x } = 0$.

Verify that 0 is a root of the equation. There are also two other roots, $\alpha$ and $\beta$, where $0 < \alpha < \beta$.
The iterative formula $x _ { r + 1 } = \ln \left( 2 x _ { r } - 1 \right) ^ { 2 }$ is to be used to find a root of the equation.
(a) Sketch the line $y = x$ and the curve $y = \ln ( 2 x - 1 ) ^ { 2 }$ on the same axes, showing the roots $0 , \alpha$ and $\beta$.
(b) By drawing a 'staircase' diagram on your sketch, starting with a value of $x$ that is between $\alpha$ and $\beta$, show that this iteration does not converge to $\alpha$.
(c) Using this iterative formula with $x _ { 1 } = 3.75$, find the value of $\beta$ correct to 3 decimal places.
Using the Newton-Raphson method with $x _ { 1 } = 1.6$, find the root $\alpha$ of the equation $( 2 x - 1 ) ^ { 2 } - \mathrm { e } ^ { x } = 0$ correct to 5 significant figures. Show the result of each iteration.

8 marks Standard +0.3

7

By sketching the curves $y = \ln x$ and $y = 8 - 2 x ^ { 2 }$ on a single diagram, show that the equation $$\ln x = 8 - 2 x ^ { 2 }$$ has exactly one real root.
Explain how your diagram shows that the root is between 1 and 2 .
Use the iterative formula $$x _ { n + 1 } = \sqrt { 4 - \frac { 1 } { 2 } \ln x _ { n } } ,$$ with a suitable starting value, to find the root. Show all your working and give the root correct to 3 decimal places.
The curves $y = \ln x$ and $y = 8 - 2 x ^ { 2 }$ are each translated by 2 units in the positive $x$-direction and then stretched by scale factor 4 in the $y$-direction. Find the coordinates of the point where the new curves intersect, giving each coordinate correct to 2 decimal places. \includegraphics[max width=\textwidth, alt={}, center]{b8ff33d4-dfe5-4067-855e-86d5765cc249-3_389_917_1117_557} The diagram shows the curve with equation $$x = ( y + 4 ) \ln ( 2 y + 3 ) .$$ The curve crosses the $x$-axis at $A$ and the $y$-axis at $B$.
Find an expression for $\frac { \mathrm { d } x } { \mathrm {~d} y }$ in terms of $y$.
Find the gradient of the curve at each of the points $A$ and $B$, giving each answer correct to 2 decimal places.