The curve with equation $y = f(x)$ where $f(x) = x^2 + ext{ln}(2x^2 - 4x + 5)$ has a single turning point at $x = \alpha$. (a) Show that $\alpha$ is a solution of the equation $$2\alpha^2 - 4\alpha + 7\alpha - 2 = 0$$ (4) The iterative formula $$x_{n+1} = \frac{1}{7}(2 + 4x_n^2 - 2x_n)$$ is used to find an approximate value for $\alpha$. Starting with $x_1 = 0.3$. (b) (i) calculate, giving each answer to 4 decimal places, (i) the value of $x_2$ (ii) the value of $x_4$ (3) Using a suitable interval and a suitable function that should be stated, (c) show that $\alpha$ is 0.341 to 3 decimal places.

A-Level Edexcel Maths Pure Vectors in 2 Dimensions: The curve with equation $y = f(x

The curve with equation $y = f(x)$ where $f(x) = x^2 + ext{ln}(2x^2 - 4x + 5)$ has a single turning point at $x = \alpha$ - Edexcel - A-Level Maths Pure - Question 6 - 2021 - Paper 1

Question 6

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The curve with equation $y = f(x)$ where $f(x) = x^2 + ext{ln}(2x^2 - 4x + 5)$ has a single turning point at $x = \alpha$. (a) Show that $\alpha$ is a solution of... show full transcript

Worked Solution & Example Answer:The curve with equation $y = f(x)$ where $f(x) = x^2 + ext{ln}(2x^2 - 4x + 5)$ has a single turning point at $x = \alpha$ - Edexcel - A-Level Maths Pure - Question 6 - 2021 - Paper 1

Step 1

Show that $\alpha$ is a solution of the equation

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Answer

To show that $\alpha$ is a solution of the equation, we start by differentiating the function:

Differentiate $f(x)$ : $f'(x) = 2x + \frac{4 - 4x}{2x^2 - 4x + 5}$
Set the derivative equal to zero for the single turning point: $2\alpha + \frac{4 - 4\alpha}{2\alpha^2 - 4\alpha + 5} = 0$ Rearranging gives: $2\alpha (2\alpha^2 - 4\alpha + 5) + 4 - 4\alpha = 0$ Which simplifies to: $2\alpha^2 - 4\alpha + 7\alpha - 2 = 0$ .

Step 2

calculate, giving each answer to 4 decimal places, (i) the value of $x_2$

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Answer

Using the iterative formula with $x_1 = 0.3$ :

Calculate $x_2$ : $x_2 = \frac{1}{7}(2 + 4(0.3)^2 - 2(0.3)) = \frac{1}{7}(2 + 0.36 - 0.6)$ $x_2 = \frac{1}{7}(2.12) \approx 0.3030$

Step 3

calculate, giving each answer to 4 decimal places, (ii) the value of $x_4$

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Answer

Using the iterative process:

Starting with $x_2 \approx 0.3030$ , calculate $x_3$ : $x_3 = \frac{1}{7}(2 + 4(0.3030)^2 - 2(0.3030))$ Compute $x_3$ : $x_3 \approx 0.3294$

Next, calculate $x_4$ using $x_3$ : $x_4 = \frac{1}{7}(2 + 4(0.3294)^2 - 2(0.3294)) \approx 0.3398$

Step 4

Using a suitable interval and a suitable function that should be stated, show that $\alpha$ is 0.341 to 3 decimal places.

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Answer

Let the function $h(x) = x^2 - 4x + 7x - 2$ . We analyze:

Evaluate $h(0.3415)$ and $h(0.3405)$ : $h(0.3415) \approx 0.00636$ $h(0.3405) \approx -0.00130$
As $h(0.3415) > 0$ and $h(0.3405) < 0$ , there is a change of sign, indicating that a root exists in the interval $(0.3405, 0.3415)$ .

Thus: $\alpha \approx 0.341$ to 3 decimal places.

The curve with equation $y = f(x)$ where $f(x) = x^2 + ext{ln}(2x^2 - 4x + 5)$ has a single turning point at $x = \alpha$ - Edexcel - A-Level Maths Pure - Question 6 - 2021 - Paper 1

Worked Solution & Example Answer:The curve with equation $y = f(x)$ where $f(x) = x^2 + ext{ln}(2x^2 - 4x + 5)$ has a single turning point at $x = \alpha$ - Edexcel - A-Level Maths Pure - Question 6 - 2021 - Paper 1

Show that $\alpha$ is a solution of the equation

calculate, giving each answer to 4 decimal places, (i) the value of $x_2$

calculate, giving each answer to 4 decimal places, (ii) the value of $x_4$

Using a suitable interval and a suitable function that should be stated, show that $\alpha$ is 0.341 to 3 decimal places.

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