By sketching a suitable pair of graphs, show that the equation x 3 + 10 x = x + 5 has only one root that lies between 0 and 1 . Use the iterative formula x n + 1 = 5 - x n 3 9 , with a suitable value for x 1 , to find the value of this root correct to 4 decimal places. Give the result of each iteration to 4 decimal places.

Given equation is x 3 + 10 x = x + 5 . . . . i We need to show that the given equation has only one root that lies between 0 , and 1 . Equation i can be written as x 3 = - 9 x + 5 Now, we need to draw the graphs of y = x 3 , and y = - 9 x + 5 . The coordinates of the points that satisfies the equation y = x 3 are evaluated as follows: x -coordinate - 2 - 1 0 1 2 y -coordinate - 8 - 1 0 1 8 x , y - 2 , - 8 - 1 , - 1 0 , 0 1 , 1 2 , 8 Thus, to draw the graph of the function y = x 3 , we first plot the above points in x y -plane, and then join them.Similarly, to draw the graph of the function, y = - 9 x + 5 , we first find the coordinates of the points lying on the line. x = 0 ⇒ y = 5 , and x = 1 ⇒ y = - 9 + 5 = - 4 , x = - 1 ⇒ y = 9 + 5 = 14 .Hence, the graph of both the functions on the same x y -plane is given as follows: Since the graph intersects once only.Hence, the given equation has only one root which is the x -coordinate of the point of intersection of both the graphs, and also the root lies between 0 , and 1 Since the root lies between 0 , and 1 . So, we can chose our initial approximation as x 1 = 0 + 1 2 = 0 . 5 We need to find the value of the root of the given equation correct to 4 decimal places using the iterative formula, x n + 1 = 5 - x n 3 9 , and an initial approximation, x 1 = 0 . 5 x 1 = 0 . 5 Are these same correct to 4 decimal places x 2 = 5 - 0 . 5 3 9 = 0 . 5417 No x 3 = 5 - 0 . 5417 3 9 = 0 . 5460 No x 4 = 5 - 0 . 5460 3 9 = 0 . 5375 No x 5 = 5 - 0 . 5375 3 9 = 0 . 5383 No x 6 = 5 - 0 . 5383 3 9 = 0 . 5382 No x 7 = 5 - 0 . 5382 3 9 = 0 . 5382 Yes Hence, the required value of α correct to 4 decimal places is 0 . 5382 .

The equation cos - 1 3 x = 1 - x has a root α . Show, by calculation, that α is between π 15 and π 12 . Show that the given equation can be rearranged into the form x = 1 3 cos 1 - x .

Given equation is cos - 1 3 x = 1 - x ⇒ cos - 1 3 x + x - 1 = 0 Let f x = cos - 1 3 x + x - 1 We need to show by calculation that the root α of given equation lies between π 15 , and π 12 . f π 15 = cos - 1 3 π 15 + π 15 - 1 = cos - 1 π 5 + π 15 - 1 = 1 . 3598 + 0 . 2094 - 1 = 0 . 5692 > 0 f π 12 = cos - 1 3 π 12 + π 12 - 1 = cos - 1 π 4 + π 12 - 1 = 0 . 6675 + 0 . 2618 - 1 = - 0 . 0707 < 0 Change of sign indicates presence of root. Therefore, the root α lies between π 15 and π 12 . Now, consider cos - 1 3 x = 1 - x Taking cosine function on both sides of the equation, we have cos cos - 1 3 x = cos 1 - x ⇒ 3 x = cos 1 - x ∵ cos cos - 1 x = x , if - 1 ≤ x ≤ 1 ⇒ x = 1 3 cos 1 - x . Hence, Proved.

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Numerical Solutions of Equations, Exercise 3: EXERCISE 6B

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Numerical Solutions of Equations, Exercise 3: EXERCISE 6B

Attempt the practice questions on Chapter 6: Numerical Solutions of Equations, Exercise 3: EXERCISE 6B with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Numerical Solutions of Equations, Exercise 3: EXERCISE 6B with Hints & Solutions

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>EXERCISE 6B>Q 3

The equation $(x - 0.5) e^{3 x} = 1$ has a root $α .$

Use the iterative formula $x_{n + 1} = e^{- 3 x_{n}} + 0.5$ with $x_{1} = 0.5$ to find the value of $α$ correct to $2$ decimal places. Give the value of each of your iterations to $4$ decimal places.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>EXERCISE 6B>Q 4

By sketching a suitable pair of graphs, show that the equation $x^{3} + 10 x = x + 5$ has only one root that lies between $0$ and $\underline{1} .$

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>EXERCISE 6B>Q 4

By sketching a suitable pair of graphs, show that the equation $x^{3} + 10 x = x + 5$ has only one root that lies between $0$ and $1$ . Use the iterative formula $x_{n + 1} = \frac{5 - x_{n}^{3}}{9},$ with a suitable value for $x_{1},$ to find the value of this root correct to $4$ decimal places. Give the result of each iteration to $4$ decimal places.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>EXERCISE 6B>Q 5

The equation $\cos^{- 1} 3 x = 1 - x$ has a root $α .$ Show, by calculation, that $α$ is between $\frac{π}{15}$ and $\frac{π}{12}$ .

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>EXERCISE 6B>Q 5

The equation $\cos^{- 1} 3 x = 1 - x$ has a root $α .$ Show, by calculation, that $α$ is between $\frac{π}{15}$ and $\frac{π}{12}$ . Show that the given equation can be rearranged into the form $x = \frac{1}{3} \cos (1 - x) .$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>EXERCISE 6B>Q 5

The equation $\cos^{- 1} 3 x = 1 - x$ has a root $α .$

Using the iterative formula $x_{n + 1} = \frac{1}{3} \cos (1 - x_{n})$ with a suitable starting value, $x_{1},$ find the value of $α$ correct to $3$ decimal places. Give the result of each iteration to $5$ decimal places.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>EXERCISE 6B>Q 6

The terms of a sequence with first term $x_{1} = 1$ are defined by the iterative formula:
$x_{n + 1} = \sqrt{\frac{5 - 2 x_{n} - x_{n}^{3}}{5}}$
The terms converge to the value $α .$

Use this formula to find the value of $α$ correct to $2$ decimal places. Give the value of each term you calculate to $4$ decimal places.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>EXERCISE 6B>Q 9

The terms of a sequence, defined by the iterative formula $x_{n + 1} = \ln (x_{n}^{2} + 4),$ converge to the value $α .$ The first term of the sequence is $2 .$

Find the value of $α$ correct to $2$ decimal places. Give each term of the sequence you find to $4$ decimal places.

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Numerical Solutions of Equations, Exercise 3: EXERCISE 6B

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Numerical Solutions of Equations, Exercise 3: EXERCISE 6B

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Numerical Solutions of Equations, Exercise 3: EXERCISE 6B with Hints & Solutions

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