The functions f and g are defined, for 0 ⩽ x ⩽ π , by f ( x ) = e x - 2 and g ( x ) = 5 - cos x The diagram shows the graph of y = f ( x ) and the graph of y = g ( x ) . The gradients of the curves are equal both when x = p and when x = q Given also that 1 . 5 < q < 2 . 5 , use the iterative formula q n + 1 = 2 + ln sin q n to calculate q correct to 2 decimal places, showing the result of each iteration to 4 decimal places.

Given that: The functions f and g are defined, for 0 ⩽ x ⩽ π , by f ( x ) = e x - 2 and g ( x ) = 5 - cos x We have to calculate q correct to 2 decimal places, showing the result of each iteration to 4 decimal places. The gradients of the curves are equal both when x = p and when x = q That is, f ' ( x ) = e x - 2 and g ' ( x ) = sin x Let F x = e x - 2 - sin x = 0 Take log to base e on both sides x - 2 = ln sin x ⇒ x = ln sin x + 2 ⇒ q = ln sin x + 2 when x = q q 1 = 1 . 5 + 2 . 5 2 = 2 q = ln sin x + 2 Are they the same correct to 4 significant figures q 2 = ln sin 2 + 2 = 1 . 9049 No q 3 = ln sin 1 . 9049 + 2 = 1 . 9431 No q 4 = ln sin 1 . 9431 + 2 = 1 . 9290 No q 5 = ln sin 1 . 9290 + 2 = 1 . 9344 No q 6 = ln sin 1 . 9344 + 2 = 1 . 9324 Yes q seems to be 1 . 93 correct to 2 decimal places.

The equation x = cos x + sin x has a root α that lies between 1 and 1 . 4 . Using the iterative formula x n + 1 = 2 cos x n - π 4 , find the value of α , giving your answer correct to 2 decimal places.

We need to find the value of the root of this equation correct to 2 decimal places using the iterative formula, x n + 1 = 2 cos x n - π 4 , and an initial approximation, x 1 = 1 . 2 x 1 = 1 . 2 Are these same correct to 2 decimal places x 2 = 2 cos x 1 - π 4 = 2 cos 1 . 2 - π 4 = 1 . 29439 No x 3 = 2 cos x 2 - π 4 = 2 cos 1 . 29439 - π 4 = 1 . 23493 No x 4 = 2 cos x 3 - π 4 = 2 cos 1 . 23493 - π 4 = 1 . 27370 No x 5 = 2 cos x 4 - π 4 = 2 cos 1 . 27370 - π 4 = 1 . 24893 No x 6 = 2 cos x 5 - π 4 = 2 cos 1 . 24893 - π 4 = 1 . 26498 No x 7 = 2 cos x 6 - π 4 = 2 cos 1 . 26498 - π 4 = 1 . 25467 No x 8 = 2 cos x 7 - π 4 = 2 cos 1 . 25467 - π 4 = 1 . 26133 No x 9 = 2 cos x 8 - π 4 = 2 cos 1 . 26133 - π 4 = 1 . 25704 No x 10 = 2 cos x 9 - π 4 = 2 cos 1 . 25704 - π 4 = 1 . 25981 No x 11 = 2 cos x 10 - π 4 = 2 cos 1 . 25981 - π 4 = 1 . 25802 Yes Hence, the required value of α correct to 2 decimal places is 1 . 26

By sketching each of the graphs of y = sec x and y = π 2 - x π 4 + x for - π 2 < x ⩽ π 2 on the same diagram, show that the equation sec x = π 2 - x π 4 + x has exactly two real roots in the interval - π 2 < x ⩽ π 2 .

When x y = π 2 - x π 4 + x - 2 - 4 . 3370958 - 1 - 0 . 55169761 0 1 . 2337006 1 1 . 0190987 2 - 1 . 1955031 x sec x - 0 . 4 1 . 0857044 - 0 . 2 1 . 0203388 0 1 0 . 1 1 . 0050209 0 . 2 1 . 0203388 From the above graph we can observe , the given equation has exactly two real roots in the interval - π 2 < x ⩽ π 2

Show that the equation sec x = π 2 - x π 4 + x can be written in the form x = 2 π x + π 2 - 8 sec x 8 .

Given sec x = π 2 - x π 4 + x ⇒ sec x = π 2 2 + π 2 x - π 4 x - x 2 ⇒ x 2 = π 4 x + π 2 8 - sec x ⇒ x 2 = 2 π x + π 2 - 8 sec x 8 ∴ x = 2 π x + π 2 - 8 sec x 8

The two real roots of the equation sec x = π 2 - x π 4 + x in the interval - π 2 < x ⩽ π 2 are denoted α and β . Verify by calculation that the smaller root, α , is - 0 . 21 correct to 2 decimal places.

Let f ( x ) = sec x - π 2 - x π 4 + x f ( - 0 . 215 ) = sec ( - 0 . 215 ) - π 2 + 0 . 215 π 4 - 0 . 215 = 0 . 0049 > 0 f ( - 0 . 205 ) = sec ( - 0 . 205 ) - π 2 + 0 . 205 π 4 - 0 . 205 = - 0 . 00928 < 0 Change of sign indicates presence of root. Therefore, α = - 0 . 21 is root of f x

HARD

AS and A Level

IMPORTANT

Earn 100

The functions $f$ and $g$ are defined, for $0 ⩽ x ⩽ π,$ by $f (x) = e^{x - 2}$ and $g (x) = 5 - \cos x$
The diagram shows the graph of $y = f (x)$ and the graph of $y = g (x) .$
The gradients of the curves are equal both when $x = p$ and when $x = q$

Given that $p < q,$ verify by calculation that $p$ is $0.16$ correct to $2$ decimal places.

Important Questions on Numerical Solutions of Equations

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 10

The functions $f$ and $g$ are defined, for $0 ⩽ x ⩽ π,$ by $f (x) = e^{x - 2}$ and $g (x) = 5 - \cos x$
The diagram shows the graph of $y = f (x)$ and the graph of $y = g (x) .$
The gradients of the curves are equal both when $x = p$ and when $x = q$

Show that $q$ satisfies the equation $q = 2 + \ln (\sin q) .$

Question Image

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 10

Given also that $1.5 < q < 2.5,$ use the iterative formula $q_{n + 1} = 2 + \ln (\sin q_{n})$ to calculate $q$ correct to $2$ decimal places, showing the result of each iteration to $4$ decimal places.

Question Image

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 11

The equation $x = \cos x + \sin x$ has a root $α$ that lies between $1$ and $1.4 .$

Show that $α$ is also a root of the equation $x = \sqrt{2} \cos (x - \frac{π}{4}) .$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 11

The equation $x = \cos x + \sin x$ has a root $α$ that lies between $1$ and $1.4 .$

Using the iterative formula $x_{n + 1} = \sqrt{2} \cos (x_{n} - \frac{π}{4}),$ find the value of $α,$ giving your answer correct to $2$ 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>END-OF-CHAPTER REVIEW EXERCISE 6>Q 12

By sketching each of the graphs of

y = \sec x

and

y = (\frac{π}{2} - x) (\frac{π}{4} + x)

for

- \frac{π}{2} < x ⩽ \frac{π}{2}

on the same diagram, show that the equation

\sec x = (\frac{π}{2} - x) (\frac{π}{4} + x)

has exactly two real roots in the interval

- \frac{π}{2} < x ⩽ \frac{π}{2} .

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 12

Show that the equation sec

x = (\frac{π}{2} - x) (\frac{π}{4} + x)

can be written in the form

x = \sqrt{\frac{2 π x + π^{2} - 8 \sec x}{8}} .

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 12

The two real roots of the equation

\sec x = (\frac{π}{2} - x) (\frac{π}{4} + x)

in the interval

- \frac{π}{2} < x ⩽ \frac{π}{2}

are denoted

α

and

β .

Verify by calculation that the smaller root,

α,

- 0.21

correct to

2

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>END-OF-CHAPTER REVIEW EXERCISE 6>Q 12

The two real roots of the equation

\sec x = (\frac{π}{2} - x) (\frac{π}{4} + x)

in the interval

- \frac{π}{2} < x ⩽ \frac{π}{2}

are denoted

α

and

β .

Using an iterative formula based on the equation given with an initial value of

1,

find the value of

β

correct to

2

decimal places. Give the result of each iteration to

4

decimal places.

The functions f and g are defined, for 0⩽x⩽π, by f(x)=ex-2 and g(x)=5-cosx The diagram shows the graph of y=f(x) and the graph of y=g(x). The gradients of the curves are equal both when x=p and when x=q Given that p<q, verify by calculation that p is 0.16 correct to 2 decimal places.

Important Questions on Numerical Solutions of Equations

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