f(x) = 2 \, ext{sin}(x^2) + x - 2, \quad 0 \leq x < 2\pi (a) Show that f(x)=0 has a root $ \alpha $ between $ x=0.75 $ and $ x=0.85 $. The equation f(x)=0 can be written as $ x = [\text{arcsin}(1-0.5x)]^{\frac{1}{2}} $. (b) Use the iterative formula $ x_{n+1} = [\text{arcsin}(1-0.5x_n)]^{\frac{1}{2}} $ to find the values of $ x_1, \; x_2, \; \text{and} \; x_3 $, giving your answers to 5 decimal places. (c) Show that $ \alpha = 0.80157 $ is correct to 5 decimal places.

A-Level Edexcel Maths Pure Trigonometric Functions: f(x) = 2 \, ext{sin}(x^2) + x -

f(x) = 2 \, ext{sin}(x^2) + x - 2, \quad 0 \leq x < 2\pi (a) Show that f(x)=0 has a root $ \alpha $ between $ x=0.75 $ and $ x=0.85 $ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

Question 2

$f(x)-=-2-\,--ext{sin}(x^2)-+-x---2,-\quad-0-\leq-x-<-2\pi---(a)-Show-that-f(x)=0-has-a-root-$-\alpha-$-between-$-x=0.75-$-and-$-x=0.85-$-Edexcel-A-Level Maths Pure-Question 2-2011-Paper 3.png$

f(x) = 2 \, ext{sin}(x^2) + x - 2, \quad 0 \leq x < 2\pi (a) Show that f(x)=0 has a root $ \alpha $ between $ x=0.75 $ and $ x=0.85 $. The equation f(x)=0 ... show full transcript

Worked Solution & Example Answer:f(x) = 2 \, ext{sin}(x^2) + x - 2, \quad 0 \leq x < 2\pi (a) Show that f(x)=0 has a root $ \alpha $ between $ x=0.75 $ and $ x=0.85 $ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

Step 1

Show that f(x)=0 has a root $ \alpha $ between $ x=0.75 $ and $ x=0.85 $

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Answer

To find if there is a root between ( 0.75 ) and ( 0.85 ), we can evaluate the function at these points:

Calculate ( f(0.75) ): [ f(0.75) = 2 \times \text{sin}(0.75^2) + 0.75 - 2 = -0.18 \quad (approximately) ]
Calculate ( f(0.85) ): [ f(0.85) = 2 \times \text{sin}(0.85^2) + 0.85 - 2 = 0.17 \quad (approximately) ]

Since ( f(0.75) < 0 ) and ( f(0.85) > 0 ), there is a change of sign indicating that a root exists between ( 0.75 ) and ( 0.85 ).

Step 2

Use the iterative formula $ x_{n+1} = [\text{arcsin}(1-0.5x_n)]^{\frac{1}{2}} $ to find $ x_1 $

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Answer

Starting with ( x_0 = 0.8 ):

Substitute ( x_0 ) into the iterative formula: [ x_1 = [\text{arcsin}(1-0.5 \times 0.8)]^{\frac{1}{2}} ] [ x_1 = [\text{arcsin}(1-0.4)]^{\frac{1}{2}} = [\text{arcsin}(0.6)]^{\frac{1}{2}} \approx 0.80219 ]

Step 3

Use the iterative formula to find $ x_2 $

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Answer

Now use ( x_1 ) to find ( x_2 ):

Substitute ( x_1 ) into the formula: [ x_2 = [\text{arcsin}(1-0.5 \times 0.80219)]^{\frac{1}{2}} \approx 0.80213 ]

Step 4

Use the iterative formula to find $ x_3 $

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Answer

Using ( x_2 ) to find ( x_3 ):

Substitute ( x_2 ) into the formula: [ x_3 = [\text{arcsin}(1-0.5 \times 0.80213)]^{\frac{1}{2}} \approx 0.80167 ]

Step 5

Show that $ \alpha = 0.80157 $ is correct to 5 decimal places

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Answer

To verify that ( \alpha = 0.80157 ) is correct to five decimal places:

Evaluate ( f(0.80165) ): [ f(0.80165) = 2 \times \text{sin}(0.80165^2) + 0.80165 - 2 \approx -2.7 \times 10^{-5} ]
Evaluate ( f(0.80157) ): [ f(0.80157) = 2 \times \text{sin}(0.80157^2) + 0.80157 - 2 \approx 8.6 \times 10^{-6} ]

Since there is a change of sign between ( f(0.80165) ) and ( f(0.80157) ), we can conclude that ( \alpha = 0.80157 ) is indeed correct to 5 decimal places.

f(x) = 2 \, ext{sin}(x^2) + x - 2, \quad 0 \leq x < 2\pi (a) Show that f(x)=0 has a root \( \alpha \) between \( x=0.75 \) and \( x=0.85 \) - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

Worked Solution & Example Answer:f(x) = 2 \, ext{sin}(x^2) + x - 2, \quad 0 \leq x < 2\pi (a) Show that f(x)=0 has a root \( \alpha \) between \( x=0.75 \) and \( x=0.85 \) - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

Show that f(x)=0 has a root \( \alpha \) between \( x=0.75 \) and \( x=0.85 \)

Use the iterative formula \( x_{n+1} = [\text{arcsin}(1-0.5x_n)]^{\frac{1}{2}} \) to find \( x_1 \)

Use the iterative formula to find \( x_2 \)

Use the iterative formula to find \( x_3 \)

Show that \( \alpha = 0.80157 \) is correct to 5 decimal places

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