f(x) = 4 \, cosec \, x - 4 \, x + 1, \text{ where } x \text{ is in radians.} (a) Show that there is a root $ \alpha $ of $ f(x) = 0 $ in the interval $[1.2, 1.3]$. (b) Show that the equation $ f(x) = 0 $ can be written in the form \[ x = \frac{1}{sin \, x} + \frac{1}{4} \] (c) Use the iterative formula \[ x_{n+1} = \frac{1}{sin \, x_n} + \frac{1}{4} \] where $ x_0 = 1.25 $, to calculate the values of $ x_1, \; x_2, \; \text{and } x_3 $, giving your answers to 4 decimal places. (d) By considering the change of sign of $ f(x) $ in a suitable interval, verify that $ \alpha = 1.291 $ correct to 3 decimal places.

A-Level Edexcel Maths Pure Quadratics: f(x) = 4 \, cosec \, x

f(x) = 4 \, cosec \, x - 4 \, x + 1, \text{ where } x \text{ is in radians.} (a) Show that there is a root $ \alpha $ of $ f(x) = 0 $ in the interval $[1.2, 1.3]$ - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 5

Question 4

$f(x)-=-4-\,-cosec-\,-x---4-\,-x-+-1,-\text{-where-}-x-\text{-is-in-radians.}--(a)-Show-that-there-is-a-root-$-\alpha-$-of-$-f(x)-=-0-$-in-the-interval-$[1.2,-1.3]$-Edexcel-A-Level Maths Pure-Question 4-2010-Paper 5.png$

f(x) = 4 \, cosec \, x - 4 \, x + 1, \text{ where } x \text{ is in radians.} (a) Show that there is a root $ \alpha $ of $ f(x) = 0 $ in the interval \([1.2, 1.... show full transcript

Worked Solution & Example Answer:f(x) = 4 \, cosec \, x - 4 \, x + 1, \text{ where } x \text{ is in radians.} (a) Show that there is a root $ \alpha $ of $ f(x) = 0 $ in the interval $[1.2, 1.3]$ - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 5

Step 1

Show that there is a root $ \alpha $ of $ f(x) = 0 $ in the interval $[1.2, 1.3]$

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Answer

To show that there is a root in the interval ([1.2, 1.3]), we will evaluate the function ( f(x) ) at the endpoints of the interval.

First, calculate ( f(1.2) ):

[ f(1.2) = 4 , cosec(1.2) - 4 , (1.2) + 1 ]

Using a calculator, ( cosec(1.2) ) can be evaluated. Similarly, calculate ( f(1.3) ):

[ f(1.3) = 4 , cosec(1.3) - 4 , (1.3) + 1 ]

If ( f(1.2) ) and ( f(1.3) ) have different signs, by the Intermediate Value Theorem, there is at least one root in ([1.2, 1.3]).

Step 2

Show that the equation $ f(x) = 0 $ can be written in the form $ x = \frac{1}{sin \, x} + \frac{1}{4} $

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Answer

Starting from the definition of ( f(x) ), we can set it to zero:

[ f(x) = 4 , cosec , x - 4 , x + 1 = 0 ]

Rearranging gives us:

[ 4 , cosec , x = 4 , x - 1 ]

Dividing both sides by 4 yields:

[ cosec , x = x - \frac{1}{4} ]

In terms of sine, this can be expressed as:

[ \frac{1}{sin , x} = x - \frac{1}{4} ]

Thus, we end up with the equation in the desired form:

[ x = \frac{1}{sin , x} + \frac{1}{4} ]

Step 3

Use the iterative formula $ x_{n+1} = \frac{1}{sin \, x_n} + \frac{1}{4} $ to calculate the values of $ x_1, \; x_2, \; \text{and } x_3 $, giving your answers to 4 decimal places.

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Answer

Starting with ( x_0 = 1.25 ):

Calculate ( x_1 ): [ x_1 = \frac{1}{sin(1.25)} + \frac{1}{4} \approx xxx ]
Calculate ( x_2 ): [ x_2 = \frac{1}{sin(x_1)} + \frac{1}{4} \approx xxx ]
Calculate ( x_3 ): [ x_3 = \frac{1}{sin(x_2)} + \frac{1}{4} \approx xxx ]

Make sure to provide each answer rounded to 4 decimal places.

Step 4

By considering the change of sign of $ f(x) $ in a suitable interval, verify that $ \alpha = 1.291 $ correct to 3 decimal places.

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Answer

To verify ( \alpha = 1.291 ), evaluate ( f(1.29) ) and ( f(1.292) ):

If ( f(1.29) > 0 ) and ( f(1.292) < 0 ) (or vice versa), then there is a root between these two values.
Confirming the sign change in this interval would verify that ( \alpha = 1.291 ) is indeed correct.

f(x) = 4 \, cosec \, x - 4 \, x + 1, \text{ where } x \text{ is in radians.} (a) Show that there is a root \( \alpha \) of \( f(x) = 0 \) in the interval \([1.2, 1.3]\) - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 5

Worked Solution & Example Answer:f(x) = 4 \, cosec \, x - 4 \, x + 1, \text{ where } x \text{ is in radians.} (a) Show that there is a root \( \alpha \) of \( f(x) = 0 \) in the interval \([1.2, 1.3]\) - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 5

Show that there is a root \( \alpha \) of \( f(x) = 0 \) in the interval \([1.2, 1.3]\)

Show that the equation \( f(x) = 0 \) can be written in the form \( x = \frac{1}{sin \, x} + \frac{1}{4} \)

Use the iterative formula \( x_{n+1} = \frac{1}{sin \, x_n} + \frac{1}{4} \) to calculate the values of \( x_1, \; x_2, \; \text{and } x_3 \), giving your answers to 4 decimal places.

By considering the change of sign of \( f(x) \) in a suitable interval, verify that \( \alpha = 1.291 \) correct to 3 decimal places.

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