Let $f(x) = ext{ln}(2x - 5) + 2x^2 - 30$, \, x > 2.5. (a) Show that $f(x) = 0$ has a root $\alpha$ in the interval $[3.5, 4]$. A student takes 4 as the first approximation to $\alpha$. Given $f(4) = 3.099$ and $f'(4) = 16.67$ to 4 significant figures, (b) apply the Newton-Raphson procedure once to obtain a second approximation for $\alpha$, giving your answer to 3 significant figures. (c) Show that $\alpha$ is the only root of $f(x) = 0$.

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Let $f(x) = ext{ln}(2x - 5) + 2x^2 - 30$, \, x > 2.5 - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1

Question 10

$Let-$f(x)-=--ext{ln}(2x---5)-+-2x^2---30$,-\,-x->-2.5-Edexcel-A-Level Maths Pure-Question 10-2017-Paper 1.png$

Let $f(x) = ext{ln}(2x - 5) + 2x^2 - 30$, \, x > 2.5. (a) Show that $f(x) = 0$ has a root $\alpha$ in the interval $[3.5, 4]$. A student takes 4 as the first appr... show full transcript

Worked Solution & Example Answer:Let $f(x) = ext{ln}(2x - 5) + 2x^2 - 30$, \, x > 2.5 - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1

Step 1

Show that $f(x) = 0$ has a root $\alpha$ in the interval $[3.5, 4]$

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Answer

To show that $f(x) = 0$ has a root in the interval $[3.5, 4]$ , we evaluate the function at both endpoints.

Firstly, we calculate:

$f(3.5) = ext{ln}(2(3.5) - 5) + 2(3.5)^2 - 30$

Calculating further,

$f(3.5) = ext{ln}(2) + 24.5 - 30 = -4.8.$

Then, we evaluate:

$f(4) = ext{ln}(2(4) - 5) + 2(4)^2 - 30$

Continuing,

$f(4) = ext{ln}(3) + 32 - 30 = 3.099.$

Since $f(3.5) < 0$ and $f(4) > 0$ , by the Intermediate Value Theorem, there exists at least one root $\alpha$ in the interval $[3.5, 4]$ .

Step 2

apply the Newton-Raphson procedure once to obtain a second approximation for $\alpha$

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Answer

Using the Newton-Raphson method, we take:

$x_1 = 4$ ,

with:

$f(4) = 3.099$

and

$f'(4) = 16.67.$

Now applying the formula:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ ,

we compute:

$x_2 = 4 - \frac{3.099}{16.67}.$

Calculating this gives:

$x_2 \approx 4 - 0.1852 = 3.8148.$

Rounding to 3 significant figures, we have:

$x_2 \approx 3.81.$

Step 3

Show that $\alpha$ is the only root of $f(x) = 0$

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Answer

To demonstrate that $\alpha$ is the only root, we analyze the derivative:

$f'(x) = \frac{d}{dx}\left(\text{ln}(2x - 5) + 2x^2 - 30\right) = \frac{2}{2x - 5} + 4x.$

Since $2x - 5 > 0$ for all $x > 2.5$ , we see that the term $\frac{2}{2x - 5}$ is always positive. The term $4x$ is also positive for all $x > 0$ .

Thus, $f'(x) > 0$ for $x > 2.5$ , indicating that $f(x)$ is strictly increasing beyond this point. Since we have established that there is a root in the interval $[3.5, 4]$ , it must be that this is the only root, as the function cannot turn back down due to it being strictly increasing.

Let $f(x) = ext{ln}(2x - 5) + 2x^2 - 30$, \, x > 2.5 - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1

Worked Solution & Example Answer:Let $f(x) = ext{ln}(2x - 5) + 2x^2 - 30$, \, x > 2.5 - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1

Show that $f(x) = 0$ has a root $\alpha$ in the interval $[3.5, 4]$

apply the Newton-Raphson procedure once to obtain a second approximation for $\alpha$

Show that $\alpha$ is the only root of $f(x) = 0$

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