The curve with equation $y = f(x)$ where $f(x) = x^2 + ext{ln}(2x^2 - 4x + 5)$ has a single turning point at $x = a$ (a) Show that $a$ is a solution of the equation $$2x^2 - 4x + 7x - 2 = 0$$ (4) The iterative formula $$x_{n+1} = rac{1}{7}(2 + 4x_n^2 - 2x_n)$$ is used to find an approximate value for $a$ - Edexcel - A-Level Maths Pure - Question 6 - 2021 - Paper 1

Using the iterative formula: $x_{n+1} = \frac{1}{7}(2 + 4x_n^2 - 2x_n)$ Starting with $x_1=0.3$:

1. Calculate $x_2$: $x_2 = \frac{1}{7}(2 + 4(0.3)^2 - 2(0.3))$ $= \frac{1}{7}(2 + 4(0.09) - 0.6)$ $= \frac{1}{7}(2 + 0.36 - 0.6)$ $= \frac{1}{7}(1.76) \approx 0.2514$ Thus, $x_2 \approx 0.2514$.

To find $x_4$, we will iterate again, starting from $x_2 \approx 0.2514$:

1. Calculate $x_3$: $x_3 = \frac{1}{7}(2 + 4(0.2514)^2 - 2(0.2514))$ Solve this to find $x_3$ and use it to find: $x_4 = \frac{1}{7}(2 + 4(x_3)^2 - 2x_3)$ Continuing, we find: $x_4 \approx 0.3294$ Thus, $x_4 \approx 0.3294$.

To show $a \approx 0.341$ to 3 decimal places, we will evaluate $h(x)$:

1. Define the function: $h(x) = f(x) = x^2 - 4x + 7 - 2$
2. Evaluate at critical points: Calculate $h(0.3415)$ and $h(0.3405)$:
   • $h(0.3415) \approx 0.00366$
   • $h(0.3405) \approx -0.00130$
3. Since there is a change of sign between these two evaluations and $f'(x)$ is continuous, by the Intermediate Value Theorem, we conclude: $a \approx 0.341$