在顶点为 A、B 和 C 的三角形中，角 $BAC$ 是直角，且角 $ABC=\frac{\pi}{3}-x$。 证明 $\dfrac{AB}{AC}=\dfrac{1+\sqrt{3}\tan x}{\sqrt{3}-\tan x}$。 由此证明，当 $x$ 足够小以至于可以忽略 $x^2$ 及更高次幂时，$\dfrac{AB}{AC}\approx a+bx$，其中 $a$ 和 $b$ 是待确定的精确常数。 一条曲线由方程 $(1+x^2)\dfrac{dy}{dx}+xy=\sqrt{1+x^2}$ 定义，且 $(0,1)$ 是该曲线上的一个点。 求出 $y$ 的麦克劳林展开式，直到并包括 $x^2$ 项。 由此求出 $e^y$ 的级数展开式，直到并包括 $x^2$ 项。

1. Express the ratio $AB/AC$ in terms of the tangent of the given angle.

In the right-angled triangle $ABC$ with right angle at $A$ (so angle $BAC = \frac{\pi}{2}$), the side opposite to angle $ABC$ is $AC$ and the adjacent side is $AB$. Therefore:
$\tan(ABC) = \frac{AC}{AB}$
Given that angle $ABC = \frac{\pi}{3} - x$, we can take the reciprocal to find $\frac{AB}{AC}$:
$\frac{AB}{AC} = \frac{1}{\tan(ABC)} = \frac{1}{\tan\left(\frac{\pi}{3} - x\right)}$

2. Apply the compound angle formula for tangent to simplify and show the required result.

Using the compound angle expansion formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:
$\tan\left(\frac{\pi}{3} - x\right) = \frac{\tan \frac{\pi}{3} - \tan x}{1 + \tan \frac{\pi}{3} \tan x}$
Knowing that the exact value of $\tan \frac{\pi}{3}$ is $\sqrt{3}$:
$\tan\left(\frac{\pi}{3} - x\right) = \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x}$
Substituting this back into our expression for $\frac{AB}{AC}$:
$\frac{AB}{AC} = \frac{1}{\frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x}}$
$\frac{AB}{AC} = \frac{1 + \sqrt{3}\tan x}{\sqrt{3} - \tan x}$

1. Apply the small angle approximation for $\tan x$.

Since $x$ is small, we use the small angle approximation $\tan x \approx x$.
Substituting this into the result from (a)(i):
$\dfrac{AB}{AC} \approx \dfrac{1+\sqrt{3}x}{\sqrt{3}-x}$

2. Rewrite the denominator and apply the binomial series expansion up to the term in $x$.

Rewrite the expression to prepare for binomial expansion:
$\dfrac{1+\sqrt{3}x}{\sqrt{3}-x} = (1+\sqrt{3}x)(\sqrt{3}-x)^{-1}$
Factor out $\sqrt{3}$ from the second bracket to match the standard series form $(1+y)^n$:
$(\sqrt{3}-x)^{-1} = \left[ \sqrt{3}\left(1-\frac{x}{\sqrt{3}}\right) \right]^{-1} = \frac{1}{\sqrt{3}}\left(1-\frac{x}{\sqrt{3}}\right)^{-1}$
Expand up to the term in $x$:
$\frac{1}{\sqrt{3}}\left(1-\frac{x}{\sqrt{3}}\right)^{-1} \approx \frac{1}{\sqrt{3}}\left[ 1 + (-1)\left(-\frac{x}{\sqrt{3}}\right) \right] = \frac{1}{\sqrt{3}}\left( 1 + \frac{x}{\sqrt{3}} \right)$

3. Multiply the expanded terms, neglect $x^2$, and determine exact constants $a$ and $b$.

Multiply the expanded denominator with the numerator $(1+\sqrt{3}x)$:
$\dfrac{AB}{AC} \approx (1+\sqrt{3}x) \left[ \frac{1}{\sqrt{3}}\left( 1 + \frac{x}{\sqrt{3}} \right) \right]$
$= \frac{1}{\sqrt{3}} \left( 1 + \frac{x}{\sqrt{3}} + \sqrt{3}x + x^2 \right)$
Since $x^2$ and higher powers can be neglected:
$\dfrac{AB}{AC} \approx \frac{1}{\sqrt{3}} \left( 1 + \left(\frac{1}{\sqrt{3}} + \sqrt{3}\right)x \right)$
$= \frac{1}{\sqrt{3}} \left( 1 + \frac{4}{\sqrt{3}}x \right) = \frac{1}{\sqrt{3}} + \frac{4}{3}x$
Rationalising the constant term gives $\frac{\sqrt{3}}{3}$.
Comparing with $a+bx$, we get exact constants $a = \frac{\sqrt{3}}{3}$ and $b = \frac{4}{3}$.

1. Evaluate the first derivative $\frac{dy}{dx}$ at $x = 0$.

Given $(1+x^2)\frac{dy}{dx}+xy=\sqrt{1+x^2}$, substitute $x = 0$ and $y = 1$:

$(1+0^2)\frac{dy}{dx} + (0)(1) = \sqrt{1+0^2}$
$1 \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = 1$

2. Differentiate the given equation implicitly with respect to $x$.

Differentiating both sides with respect to $x$ using the product rule and chain rule:

$\frac{d}{dx}\left((1+x^2)\frac{dy}{dx}\right) + \frac{d}{dx}(xy) = \frac{d}{dx}\left((1+x^2)^{\frac{1}{2}}\right)$

$\left[ (1+x^2)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} \right] + \left[ x\frac{dy}{dx} + y \right] = \frac{1}{2}(1+x^2)^{-\frac{1}{2}}(2x)$

$(1+x^2)\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} + y = \frac{x}{\sqrt{1+x^2}}$

3. Evaluate the second derivative $\frac{d^2y}{dx^2}$ at $x = 0$.

Substitute $x = 0$, $y = 1$, and $\frac{dy}{dx} = 1$ into the differentiated equation:

$(1+0^2)\frac{d^2y}{dx^2} + 3(0)(1) + 1 = \frac{0}{\sqrt{1+0^2}}$

$1 \cdot \frac{d^2y}{dx^2} + 0 + 1 = 0 \implies \frac{d^2y}{dx^2} = -1$

4. Apply the Maclaurin series formula to form the expansion up to the $x^2$ term.

The Maclaurin series for $y$ is given by:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \dots$

Substitute the known values ($y(0)=1$, $y'(0)=1$, $y''(0)=-1$):
$y = 1 + (1)x + \frac{-1}{2}x^2$
$y = 1 + x - \frac{1}{2}x^2$

1. Determine the Maclaurin expansion of $y$ up to the $x^2$ term (as required from part (b)(i)).

2. Express $e^y$ using the expansion of $y$ and factor out the constant $e$.

3. Apply the standard Maclaurin series for $e^u$ and simplify up to the $x^2$ term.