A-Level Edexcel Maths Pure Differentiation: Given that $f(x) = ext{ln}(2x -

Question

Given that $f(x) = 	ext{ln}(2x - 5) + 2x^2 - 30$, we need to show that $f(x) = 0$ has a root $\alpha$ in the interval $[3.5, 4]$.

1. Calculate $f(3.5)$:
   $$f(3.5) = 	ext{ln}(2(3.5) - 5) + 2(3.5)^2 - 30 = 	ext{ln}(2) + 24.5 - 30 = 	ext{ln}(2) - 5.5$$
   Approximating $f(3.5)$ gives us a negative value.

2. Calculate $f(4)$:
   $$f(4) = 	ext{ln}(2(4) - 5) + 2(4)^2 - 30 = 	ext{ln}(3) + 32 - 30 = 	ext{ln}(3) + 2$$
   Approximating $f(4)$ gives us a positive value.

Since $f(3.5) < 0$ and $f(4) > 0,$ by the Intermediate Value Theorem, there must be at least one root in the interval $(3.5, 4)$.

(b) Given $f(4) = 3.099$ and $f'(4) = 16.67$ to 4 significant figures,
     we will apply the Newton-Raphson procedure once to find a second approximation for $\alpha$.

The Newton-Raphson formula is given by:
   $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$

Using $x_0 = 4$:
   $$x_1 = 4 - \frac{3.099}{16.67} \approx 4 - 0.185 = 3.815$$
   Rounding to three significant figures, we obtain:
   $$x_1 = 3.81$$

(c) To show that $\alpha$ is the only root of $f(x) = 0$, we must analyze the properties of $f(x)$.

1. Consider $g(x) = 	ext{ln}(2x - 5)$ and $h(x) = 30 - 2x$:
   - The function $g(x)$ is continuous and differentiable, and $h(x)$ is linear.
2. The derivative:
   $$f'(x) = \frac{2}{2x - 5} + 4x$$
   This indicates that $f'(x) > 0$ for $x > 2.5$;
   Thus, $f(x)$ is monotonically increasing in the interval.

Since $f(x)$ goes from negative to positive and is increasing, $\alpha$ can be concluded as the only root.

Given that $f(x) = ext{ln}(2x - 5) + 2x^2 - 30$, we need to show that $f(x) = 0$ has a root $\alpha$ in the interval $[3.5, 4]$ - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1

Worked Solution & Example Answer:Given that $f(x) = ext{ln}(2x - 5) + 2x^2 - 30$, we need to show that $f(x) = 0$ has a root $\alpha$ in the interval $[3.5, 4]$ - 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 $\alpha$, giving your answer to 3 significant figures.

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

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