已知 $\mathrm{f}(x) = \ln(1 + 2x^2)$。

1. Substitute the given values of $x$ to form a system of linear equations in $a$, $b$, and $c$.

Given $\mathrm{f}(x) = \ln(1 + 2x^2)$ and $p(x) = ax^2 + bx + c$.

For $x = 1$:
$p(1) = \mathrm{f}(1)$
$a(1)^2 + b(1) + c = \ln(1 + 2(1)^2)$
$a + b + c = \ln 3$ --- (1)

For $x = 1.5$:
$p(1.5) = \mathrm{f}(1.5)$
$a(1.5)^2 + b(1.5) + c = \ln(1 + 2(1.5)^2)$
$2.25a + 1.5b + c = \ln 5.5$ --- (2)

For $x = 2$:
$p(2) = \mathrm{f}(2)$
$a(2)^2 + b(2) + c = \ln(1 + 2(2)^2)$
$4a + 2b + c = \ln 9$ --- (3)

2. Solve the system of linear equations using a graphing calculator.

Using a graphing calculator to solve equations (1), (2) and (3) simultaneously:
$a = -0.227318\dots$
$b = 1.78056\dots$
$c = -0.454637\dots$

Rounding to 3 significant figures:
$a = -0.227$
$b = 1.78$
$c = -0.455$

1. Substitute $2x^2$ into the standard series for $\ln(1+x)$.

From the List of Formulae (MF26), the standard series is:
$\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \dots$

Substituting $u = 2x^2$:
$\mathrm{f}(x) = (2x^2) - \frac{(2x^2)^2}{2} + \frac{(2x^2)^3}{3} - \dots$

2. Simplify the terms up to $x^6$.

$= 2x^2 - \frac{4x^4}{2} + \frac{8x^6}{3} - \dots$
$= 2x^2 - 2x^4 + \frac{8}{3}x^6$

1. State the convergence condition for the standard $\ln(1+u)$ expansion

The standard series expansion for $\ln(1+u)$ is valid for $-1 < u \le 1$.
Substituting $u = 2x^2$, the expansion for $\ln(1+2x^2)$ is valid for:
$-1 < 2x^2 \le 1$

2. Solve the inequality to find the range of values for $x$

Since $2x^2 \ge 0$ for all real $x$, the inequality $-1 < 2x^2$ is always satisfied.
Therefore, we only need to solve:
$2x^2 \le 1$
$x^2 \le \frac{1}{2}$
Taking the square root of both sides gives:
$-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$

1. State the Maclaurin series for $\mathrm{f}(x)$ derived in part (b)

$\ln(1 + 2x^2) \approx 2x^2 - 2x^4 + \frac{8}{3}x^6$

2. Substitute $x^2 = \frac{1}{6}$ into both sides of the expansion

$\ln\left(1 + 2\left(\frac{1}{6}\right)\right) \approx 2\left(\frac{1}{6}\right) - 2\left(\frac{1}{6}\right)^2 + \frac{8}{3}\left(\frac{1}{6}\right)^3$
$\ln\left(1 + \frac{1}{3}\right) \approx \frac{1}{3} - 2\left(\frac{1}{36}\right) + \frac{8}{3}\left(\frac{1}{216}\right)$

3. Simplify the RHS to evaluate the approximation and find $m$

$\ln\left(\frac{4}{3}\right) \approx \frac{1}{3} - \frac{1}{18} + \frac{8}{648}$
$\ln\left(\frac{4}{3}\right) \approx \frac{1}{3} - \frac{1}{18} + \frac{1}{81}$
$\ln\left(\frac{4}{3}\right) \approx \frac{54}{162} - \frac{9}{162} + \frac{2}{162}$
$\ln\left(\frac{4}{3}\right) \approx \frac{47}{162}$
$\therefore m = 47$

1. Apply logarithm properties and factorise the denominator to split the expression.

$\ln\left(\frac{1 + 2x^2}{1 + 2x + x^2}\right) = \ln(1 + 2x^2) - \ln(1 + 2x + x^2)$
$= \ln(1 + 2x^2) - \ln((1 + x)^2)$
$= \ln(1 + 2x^2) - 2\ln(1 + x)$

2. Use the standard series expansion for $\ln(1+x)$ and subtract it from the expansion of $\ln(1+2x^2)$ found in part (b).

From part (b), we have $\ln(1 + 2x^2) \approx 2x^2 - 2x^4$.
Using the standard series for $\ln(1+x)$ up to the term in $x^4$:
$2\ln(1 + x) \approx 2\left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\right) = 2x - x^2 + \frac{2}{3}x^3 - \frac{1}{2}x^4$

Subtracting the two series:
$\ln\left(\frac{1 + 2x^2}{1 + 2x + x^2}\right) \approx (2x^2 - 2x^4) - \left(2x - x^2 + \frac{2}{3}x^3 - \frac{1}{2}x^4\right)$
$= -2x + 3x^2 - \frac{2}{3}x^3 - \frac{3}{2}x^4$