國立臺灣大學一百零九學年度轉學生入學考試試題詳解

※注意：請於試卷上「非選擇題作答區」標明大題及小題題號，並依序作答。

Any device with computer algebra system is prohibited during the exam.

PART 1: Fill in the blanks.

• Only answers will be graded.
• Each answer must be clearly labeled on the answer sheet.
• points are assigned to each blank.

1. 1. $\displaystyle\lim_{x\to\infty}\left(1+\frac3x\right)^{\left[x\right]}=$　(1)　, where $\left[x\right]$ is the greatest integer which is less than or equal to $x$.
   2. $\displaystyle\lim_{x\to0}\left(\cos ax\right)^{\textstyle\frac1{1-\cos bx}}=$　(2)　, where $a,b$ are constants and $ab\neq0$.
   3. $\displaystyle\lim_{n\to\infty}\sum_{i=1}^n\frac1{\displaystyle2n\sqrt{1-\left(\frac i{2n}\right)^2}}=$　(3)　.
訣竅
運用夾擠定理與自然指數的定義處理第一小題；第二小題需先使用換底公式，再搭配 L'Hôpital 法則求解；第三小題可轉換為 Riemann 和後取極限寫為定積分計算之。
解法

1. 注意到 $x-1<\left[x\right]\leq x$，因此有

   $\displaystyle\left[\left(1+\frac3x\right)^{\frac x3}\right]^3\left(1+\frac3x\right)^{-1}\leq\left(1+\frac3x\right)^{\left[x\right]}\leq\left[\left(1+\frac3x\right)^{\frac x3}\right]^3$.

   又由自然指數的定義可注意到

   $\displaystyle\lim_{x\to\infty}\left[\left(1+\frac3x\right)^{\frac x3}\right]^3\left(1+\frac3x\right)^{-1}=e^3=\lim_{x\to\infty}\left[\left(1+\frac3x\right)^{\frac x3}\right]^3$.

   從而由夾擠定理可知所求亦為 $e^3$。
2. 由換底公式與指數函數的連續性並搭配 L'Hôpital 法則可知

   $\displaystyle\lim_{x\to0}\left(\cos ax\right)^{\textstyle\frac1{1-\cos bx}}=\exp\left(\lim_{x\to0}\frac{\ln\left(\cos ax\right)}{1-\cos bx}\right)=\exp\left(\lim_{x\to0}\frac{-a\sin ax/\cos ax}{b\sin bx}\right)=\exp\left(-\frac{a^2}{b^2}\right)=e^{-\frac{a^2}{b^2}}$.

3. 所求可寫為 $\displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^n\frac1{\displaystyle\sqrt{4-\left(\frac{i}n\right)^2}}$。可將此視為函數 $\displaystyle f\left(x\right)=\frac1{\sqrt{4-x^2}}$ 在區間 $\displaystyle\left[0,1\right]$ 上作 $n$ 等分割的 Riemann 和，故其極限值可化為定積分並計算如下

   $\displaystyle\lim_{n\to\infty}\sum_{i=1}^n\frac1{\displaystyle2n\sqrt{1-\left(\frac{i}{2n}\right)^2}}=\int_0^1\frac{dx}{\sqrt{4-x^2}}=\left.\sin^{-1}\left(\frac x2\right)\right|_0^1=\sin^{-1}\frac12=\frac\pi6.$



2. 1. $\displaystyle f\left(x\right)=\left[\ln\left(1+x^2\right)\right]^x$, $\displaystyle\frac d{dx}f\left(x\right)=$　(4)　.
   2. $x^3-y^2+y^3=x$. At $\left(x,y\right)=\left(0,1\right)$. $\displaystyle\frac{d^2y}{dx^2}=$　(5)　.
   3. $\displaystyle f\left(x,y,z\right)=\int_z^{x^y}e^{\sqrt t}dt$. $\nabla f=$　(6)　.
   4. $\displaystyle f\left(x,y\right)=\frac{\sin\left(x^2y\right)}{x^2+y^2}$ for $\left(x,y\right)\neq\left(0,0\right)$ and $f\left(0,0\right)=0$. The directional derivative of $f$ along ${\bf u}=\left(\cos\theta,\sin\theta\right)$ at $\left(0,0\right)$ is 　(7)　.
訣竅
運用換底公式後使用連鎖律等微分法則可求解第一小題；使用隱函數微分求導函數的值；搭配微分積分定理並留意梯度的定義計算；運用方向導數的定義式計算求解，此題由於分段定義的情況故不可逕自化為梯度與方向的內積，需確認函數的可微性等資訊才可直接使用（而本題屬於不可使用的情形）。
解法

1. 透過換底可將 $f$ 表達為 $f\left(x\right)=\exp\left[x\ln\left(\ln\left(1+x^2\right)\right)\right]$，如此求導有

   $\begin{aligned}f'\left(x\right)&=\exp\left[x\ln\left(\ln\left(1+x^2\right)\right)\right]\cdot\left[\ln\left(\ln\left(1+x^2\right)\right)+x\cdot\frac1{\ln\left(1+x^2\right)}\cdot\frac1{1+x^2}\cdot2x\right]\\&=\left(\ln\left(1+x^2\right)\right)^x\left(\frac{2x^2}{\left(1+x^2\right)\ln\left(1+x^2\right)}+\ln\left(\ln\left(1+x^2\right)\right)\right).\end{aligned}$

2. 運用隱函數微分求導可得

   $\displaystyle3x^2-2y\frac{dy}{dx}+3y^2\frac{dy}{dx}=1$.

   取 $\left(x,y\right)=\left(0,1\right)$ 代入可得 $\displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(0,1\right)}=1$。又進一步求導有

   $\displaystyle6x-2\left(\frac{dy}{dx}\right)^2-2y\frac{d^2y}{dx^2}+6y\left(\frac{dy}{dx}\right)^2+3y^2\frac{d^2y}{dx^2}=0.$

   因此由 $\left(x,y\right)=\left(0,1\right)$ 與 $\displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(0,1\right)}=1$ 可推得 $\displaystyle\left.\frac{d^2y}{dx^2}\right|_{\left(x,y\right)=\left(0,1\right)}=-4$。
3. 按梯度的定義可知

   $\nabla f\left(x,y,z\right)=\left(f_x\left(x,y,z\right),f_y\left(x,y,z\right),f_z\left(x,y,z\right)\right)=\left(yx^{y-1}e^{x^{y/2}},x^ye^{x^{y/2}}\ln x,-e^{\sqrt z}\right).$

4. 由方向導數的定義可知所求為

   $\displaystyle D_{\bf u}f\left(0,0\right)=\lim_{h\to0}\frac{f\left(h\cos\theta,h\sin\theta\right)-f\left(0,0\right)}h=\lim_{h\to0}\frac{\sin\left(h^3\cos^2\theta\sin\theta\right)}{h^3}=\cos^2\theta\sin\theta$.



3. $f\left(x\right)=\begin{cases}\left|x\right|-1,&\text{for}~-1\leq x\leq3\\\left(x-6\right)^2-4,&\text{for}~3<x\leq10.\end{cases}$. $\displaystyle g\left(x\right)=\int_0^xf\left(t\right)dt$, for $-1\leq x\leq10$.
   1. Choose correct statement(s) about $g\left(x\right)$: 　(8)　
      1. $g\left(x\right)$ is discontinuous at $x=3$.
      2. $g\left(x\right)$ is not differentiable at $x=3$.
      3. $g\left(x\right)$ is differentiable at $x=0$ and $g'\left(0\right)=-1$.
      4. $g\left(x\right)<0$ for $-1\leq x<0$.
   2. The local minimum values of $g\left(x\right)$ occur at $x=$　(9)　.
   3. The $x$-coordinates of points of inflection for $y=g\left(x\right)$ are 　(10)　.
訣竅
由函數圖形與積分的意義的概念可回答關於函數 $g$ 的特性，隨後由球導的狀況回答局部極值的位置，並進一步求其反曲點。
解法
1. 由於 $f$ 在 $x=3$ 處左右極限均存在但不相等，從而不定積分 $g$ 在 $x=3$ 處連續但不可導，而在可導處的求導結果與 $f$ 相同，即有： $g'\left(0\right)=f\left(0\right)=-1$ 以及

   $\displaystyle g\left(x\right)=\int_0^xf\left(t\right)dt=\int_x^0\left(1-\left|t\right|\right)dt=\int_x^0\left(1+t\right)dt=-x-\frac{x^2}2=-\frac12\left(x+1\right)^2+\frac12>0$.

   即應選ii. 與 iii.
2. 由於當 $x\neq3$ 時有 $g'\left(x\right)=f\left(x\right)$，從而解 $g'\left(x\right)=0$ 可知有 $x=\pm1$、$x=4$ 與 $x=8$。又不可導的位置有 $x=3$，而邊界為 $x=-1$ 與 $x=10$。逐步考量斜率的遞增遞減的區域可發現：函數 $g$ 在 $\left(-1,1\right)$ 處遞減，隨後在 $\left(1,3\right)$ 上遞增，又於 $\left(3,4\right)$ 上遞增，接著 $\left(4,8\right)$ 遞減，最終於 $\left(8,10\right)$ 上遞增。綜述可知在 $x=1$、$x=8$ 處有局部極值。
3. 在除了 $x=0$ 與 $x=3$ 處有 $g''\left(x\right)=\begin{cases}-1,&\text{for}~-1<x<0,\\1,&\text{for}~0<x<3,\\2x-12,&\text{for}~3<x<10.\end{cases}$。從而可知在 $x=0$、$x=3$ 與 $x=6$ 等處有凹向性的改變。


4. A vertical fence 2 m high is located 1 m away from a wall. The length of the shortest ladder that can extend from the wall over the fence to a point on the ground is 　(11)　.
訣竅
依據題意設定參數並表達出階梯的長度為該參數的函數，並透過微分求其極值。
解法
設此階梯與籬笆的距離為 $x$，那麼按題設可知與高牆的距離為 $x+1$，從而使用相似三角形可知該階梯的長度為 $L\left(x\right)=\left(1+x^{-1}\right)\sqrt{x^2+4}$，其中 $x>0$。對此求導有
$\displaystyle L'\left(x\right)=-x^{-2}\sqrt{x^2+4}+(1+x)\left(x^2+4\right)^{-1/2}=x^{-2}\left(x^2+4\right)^{-1/2}\left(x^3-4\right)$.
因此解 $L'\left(x\right)=0$ 有 $x=2^{2/3}$，且容易注意到 $x\in\left(0,2^{2/3}\right)$ 有 $L'\left(x\right)<0$ 而 $x\in\left(2^{2/3},\infty\right)$ 有 $L'\left(x\right)>0$。故在 $x=2^{2/3}$ 達到最小值。此時階梯長度的最小值為 $\left(1+2^{2/3}\right)^{3/2}$。


5. Compute the following integrals:
   1. $\displaystyle\int_0^{\infty}\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}\,dx=$　(12)　.
   2. $\displaystyle\int\frac{dx}{\sqrt{x^2-6x}}=$　(13)　.
   3. $\displaystyle\int\sin^{-1}\left(\sqrt x\right)dx=$　(14)　.
訣竅
由有理函數的積分技巧、三角代換法與一般的變數代換與分部積分法答題。
解法

1. 首先注意到被積分函數可改寫如下

   $\displaystyle\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}=-\frac{1/5}{x-1}+\frac{x/5}{x^2+4}+\frac{11/5}{x^2+4}$,

   其中 $x=1$ 處為瑕點。因此將給定的瑕積分分拆為兩段如下：

   $\displaystyle\int_0^\infty\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}dx=\int_0^{1^-}\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}dx+\int_{1^+}^\infty\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}dx$.

   然而這兩者皆發散：

   $\displaystyle\begin{aligned}\int_0^{1^-}\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}dx&=\int_0^{1^-}\left(\frac{1/5}{1-x}+\frac{x/5}{x^2+4}+\frac{11/5}{x^2+4}\right)dx\\&=\left.\left(-\frac15\ln\left(1-x\right)+\frac1{10}\ln\left(x^2+4\right)+\frac{11}{10}\tan^{-1}\frac x2\right)\right|_0^{1^-}=\infty.\end{aligned}$

   類似地也有 $\displaystyle\int_{1^+}^\infty\frac{2x-3}{\left(x-1\right)\left(x^2+4\right)}dx=-\infty$，故給定的瑕積分發散。
2. 配方有 $x^2-6x=\left(x-3\right)^2-9$。令 $x-3=3\sec\theta$，那麼有 $dx=3\sec\theta\tan\theta\,d\theta$，如此所求的不定積分可改寫並計算如下

   $\displaystyle\begin{aligned}\int\frac{dx}{\sqrt{x^2-6x}}&=\int\frac{3\sec\theta\tan\theta \,d\theta}{3\tan\theta}=\int\sec\theta\,d\theta=\ln\left|\sec\theta+\tan\theta\right|+C\\&=\ln\left|\frac{x-3+\sqrt{x^2-6x}}3\right|+C=\ln\left|x-3+\sqrt{x^2-6x}\right|+C'.\end{aligned}$

   此處 $C'=C-\ln3$。
3. 令 $u=\sin^{-1}\left(\sqrt x\right)$，如此有 $x=\sin^2u$ 與 $dx=2\sin u\cos u\,du=\sin2u\,du$。據此所求的不定積分為

   $\displaystyle\begin{aligned}\int\sin^{-1}\left(\sqrt x\right)dx&=\int u\sin2u\,du=-\frac{u\cos2u}2+\frac12\int\cos2u\,du\\&=-\frac{2u\cos2u-\sin2u}4+C=-\frac{u\left(\cos^2u-\sin^2u\right)-\sin u\cos u}2+C\\&=-\frac{\left(1-2x\right)\sin^{-1}\left(\sqrt x\right)-\sqrt{x-x^2}}2+C.\end{aligned}$



6. 1. $\displaystyle\iint_Dy\,dA=$　(15)　, where $D$ is the region in the first quadrant that lies between the circles $x^2+y^2=4$ and $\left(x-1\right)^2+y^2=1$.
   2. The volume of the solid $R$ lying below the plane $z=3-2y$ and above the paraboloid $z=x^2+y^2$ is 　(16)　.
訣竅
對於第一小題可運用極座標變換處理即可；第二小題則先求出其交會處，並利用底乘以高的概念列式求其體積。
解法

1. 令 $x=r\cos\theta$、$y=r\sin\theta$，根據積分區域可得變數範圍為 $2\cos\theta\leq r\leq2$、$0\leq\theta\leq\pi/2$。如此所求之重積分可改寫並計算如下

   $\displaystyle\begin{aligned}\iint_Dy\,dA&=\int_0^{\pi/2}\int_{2\cos\theta}^2r\sin\theta\cdot r\,dr\,d\theta=\int_0^{\pi/2}\left.\frac{r^3\sin\theta}3\right|_{2\cos\theta}^2d\theta\\&=\frac83\int_0^{\pi/2}\left(\sin\theta-\sin\theta\cos^3\theta\right)d\theta=\left.\frac83\left(\frac{\cos^4\theta}4-\cos\theta\right)\right|_0^{\frac\pi2}=2.\end{aligned}$

2. 首先先考慮兩曲面的交線由 $x^2+y^2=3-2y$ 所決定，此即 $x^2+\left(y+1\right)^2=4$，如此我們設定 $D=\left\{\left(x,y\right)\in\mathbb R^2\,:\,x^2+\left(y+1\right)^2\leq4\right\}$。那麼所求的體積可列式如下

   $\displaystyle V=\iint_D\left[\left(3-2y\right)-\left(x^2+y^2\right)\right]dA=\iint_D\left[4-x^2-\left(y+1\right)^2\right]dA$.

   那麼令 $x=r\cos\theta$、$y+1=r\sin\theta$，如此變數範圍為 $0\leq r\leq2$、$0\leq\theta\leq2\pi$，則所求的體積可改寫並計算如下

   $\displaystyle V=\int_0^{2\pi}\int_0^2\left(4-r^2\right)r\,dr\,d\theta=2\pi\left.\left(2r^2-\frac{r^4}4\right)\right|_0^2=8\pi.$



7. 1. $C$ is a smooth curve in the upper half plane going from $\left(1,1\right)$ to $\left(0,\sqrt2\right)$. $\displaystyle{\bf F}\left(x,y\right)=\frac{x-3y}{x^2+y^2}{\bf i}+\frac{3x+y}{x^2+y^2}{\bf j}$. Then, $\displaystyle\int_C{\bf F}\cdot d{\bf r}=$　(17)　.
   2. $S$ is the part of the sphere $x^2+y^2+z^2=9$ that lies above the cone $z=\sqrt{x^2+y^2}$ with upward orientation. $\displaystyle{\bf F}\left(x,y,z\right)=\frac{x{\bf i}+y{\bf j}+z{\bf k}}{\left(x^2+y^2+z^2\right)^2}$. Then, $\displaystyle\iint_S{\bf F}\cdot d{\bf S}=$　(18)　.
訣竅
試圖找出位能函數並使用線積分基本定理即可；將取面參數化後直接計算即可。
解法

1. 若函數 $f$ 滿足 $\nabla f={\bf F}$，那麼由 $\displaystyle f_x\left(x,y\right)=\frac{x-3y}{x^2+y^2}$ 可知

   $\displaystyle f\left(x,y\right)=\frac12\ln\left(x^2+y^2\right)-3\tan^{-1}\frac xy+g\left(y\right)$.

   如此對 $y$ 求導便有

   $\displaystyle f_y\left(x,y\right)=\frac y{x^2+y^2}+3\cdot\frac x{y^2+x^2}+g'\left(y\right)=\frac{3x+y}{x^2+y^2}$.

   故 $g'\left(y\right)=0$，從而 $g$ 為常數函數。如此取位能函數為

   $\displaystyle f\left(x,y\right)=\frac12\ln\left(x^2+y^2\right)-3\tan^{-1}\frac xy+C$.

   因此所求為

   $\displaystyle\int_C{\bf F}\cdot d{\bf r}=f\left(0,\sqrt2\right)-f\left(1,1\right)=\frac{3\pi}4$.

2. 將曲面參數化為 ${\bf r}\left(\theta,\phi\right)=\left(3\cos\theta\sin\phi,3\sin\theta\sin\phi,3\cos\phi\right)$，其中參數範圍 $0\leq\theta\leq2\pi$、$0\leq\phi\leq\pi/4$，那麼

   $\displaystyle\begin{aligned}\left({\bf r}_\theta\times{\bf r}_\phi\right)d\theta\,d\phi&=\left(\left(-3\sin\theta\sin\phi,3\cos\theta\sin\phi,0\right)\times\left(3\cos\theta\cos\phi,3\sin\theta\cos\phi,-3\sin\phi\right)\right)d\theta\,d\phi\\&=\left(-9\cos\theta\sin^2\phi,-9\sin\theta\sin^2\phi,-9\sin\phi\cos\phi\right)d\theta\,d\phi\\&=-9\sin\phi\left(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi\right)d\theta\,d\phi\end{aligned}$

   表指向球心的有向面元，如此所求的曲面積分為

   $\displaystyle\begin{aligned}\iint_S{\bf F}\cdot d{\bf S}&=\int_0^{\pi/4}\int_0^{2\pi}\frac{\left(3\cos\theta\sin\phi,3\sin\theta\sin\phi,3\cos\phi\right)}{9^2}\cdot9\sin\phi\left(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi\right)d\theta\,d\phi\\&=\frac13\int_0^{\pi/4}\int_0^{2\pi}\sin\phi\,d\theta\,d\phi=\frac13\cdot2\pi\left(1-\frac{\sqrt2}2\right)=\frac{\left(2-\sqrt2\right)\pi}3.\end{aligned}$



8. 1. $\displaystyle\sum_{n=0}^{\infty}\frac{3^n}{\left(2n\right)!}=$　(19)　.
   2. The Maclaurin series (the Taylor series at $a=0$) for $\cos^{-1}x$ is 　(20)　.
訣竅
利用經典的 Taylor 級數求給定的級數值並運用廣義二項式定理求冪級數。
解法

1. 首先觀察知道

   $\displaystyle\frac{e^x+e^{-x}}2=\frac12\left(\sum_{n=0}^\infty\frac{x^n}{n!}+\sum_{n=0}^\infty\frac{\left(-1\right)^nx^n}{n!}\right)=\sum_{n=0}^\infty\frac{x^{2n}}{\left(2n\right)!}$.

   因此取 $x=\sqrt3$ 可得

   $\displaystyle\sum_{n=0}^\infty\frac{3^n}{\left(2n\right)!}=\frac{e^{\sqrt3}+e^{-\sqrt3}}2=\cosh\left(\sqrt3\right)$.

2. 首先回憶起 $\displaystyle\cos^{-1}x=\frac\pi2-\sin^{-1}x$，又

   $\displaystyle\begin{aligned}\sin^{-1}x&=\int_0^x\frac{dt}{\sqrt{1-t^2}}=\int_0^x\left(1-t^2\right)^{-1/2}dt=\int_0^x\sum_{n=0}^{\infty}{-1/2\choose n}\left(-t^2\right)^ndt\\&=\int_0^x\sum_{n=0}^\infty\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^2}t^{2n}dt=\sum_{n=0}^\infty\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^2\left(2n+1\right)}x^{2n+1}.\end{aligned}$

   因此所求的冪級數為

   $\displaystyle\cos^{-1}x=\frac\pi2-\sum_{n=0}^\infty\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^2\left(2n+1\right)}x^{2n+1}.$

PART 2:

• Solve the following problems. You need to write down complete arguments.
• 10 points are assigned to each problem.

1. Find the equation of the curve on the $xy$-plane that passes through $\left(3,1\right)$ such that for any point $P$ on the curve, the midpoint of the tangent line at $P$ that lies in the first quadrant is $P$ itself.
訣竅
設定參數曲線並按題意考慮切線在兩軸交點的中點列出方程式。
解法
設所求的曲線參數式為 $\left(x\left(t\right),y\left(t\right)\right)$，那麼若 $P=\left(x_0,y_0\right)=\left(x\left(t_0\right),y\left(t_0\right)\right)$，則在該點的切線方程式為
$\displaystyle y-y_0=\frac{y'\left(t_0\right)}{x'\left(t_0\right)}\left(x-x_0\right)$.
此時兩軸上的交點為
$\displaystyle\left(0,y_0-x_0\frac{y'\left(t_0\right)}{x'\left(t_0\right)}\right),\qquad\left(x_0-y_0\frac{x'\left(t_0\right)}{y'\left(t_0\right)},0\right)$.
由中點關係得關係式
$\displaystyle x_0-y_0\frac{x'\left(t_0\right)}{y'\left(t_0\right)}=2x_0,\quad y_0-x_0\frac{y'\left(t_0\right)}{x'\left(t_0\right)}=2y_0$.
由於 $t_0$ 的任意性可知 $\displaystyle\frac{x'\left(t\right)}{y'\left(t\right)}=-\frac{x\left(t\right)}{y\left(t\right)}$，或寫為 $\displaystyle\frac{dx}x+\frac{dy}y=0$，因此積分得 $\ln\left|xy\right|=C$，從而有 $xy=e^C=C'$。再者又因通過 $\left(3,1\right)$，故此曲線為 $xy=3$。


2. $f\left(x,y\right)$ is a differentiable function. On the curve $\displaystyle\tan^{-1}\left(xy\right)+y=2+\frac\pi4$, $f$ obtains local maximum at $\displaystyle\left(x,y\right)=\left(\frac12,2\right)$. Suppose that $\displaystyle f_x\left(\frac12,2\right)=2$.
   1. Find $\displaystyle\nabla f\left(\frac12,2\right)$.
   2. Assume that on another curve $\displaystyle\tan^{-1}\left(xy\right)+y=1.9+\frac\pi4$, $f$ obtains local maximum at $\left(x_0,y_0\right)$ which is near $\displaystyle\left(\frac12,2\right)$. Use linear approximation to estimate $\displaystyle f\left(x_0,y_0\right)-f\left(\frac12,2\right)$.
訣竅
由 Lagrange 乘子法的概念推理。
解法

1. 設函數 $\displaystyle g\left(x,y\right)=\tan^{-1}\left(xy\right)+y-2-\frac\pi4$，則限制條件為 $g\left(x,y\right)=0$。如此由 Lagrange 乘子法的概念可知極值發生的位置滿足條件

   $\displaystyle\left(2,f_y\left(\frac12,2\right)\right)=\left(f_x\left(\frac12,2\right),f_y\left(\frac12,2\right)\right)=\nabla f\left(\frac12,2\right)\parallel\nabla g\left(\frac12,2\right)=\left(g_x\left(\frac12,2\right),g_y\left(\frac12,2\right)\right)$.

   又計算函數 $g$ 的梯度有

   $\displaystyle\nabla g\left(x,y\right)=\left(\frac y{1+\left(xy\right)^2},\frac x{1+\left(xy\right)^2}+1\right)$.

   取 $\displaystyle \left(x,y\right)=\left(\frac12,2\right)$ 代入便有 $\displaystyle\nabla g\left(\frac12,2\right)=\left(1,\frac54\right)$，故知 $\displaystyle f_y\left(\frac12,2\right)=\frac52$，從而 $\displaystyle\nabla f\left(\frac12,2\right)=\left(2,\frac52\right)$。
2. 運用一階逼近可推知

   $\displaystyle\begin{aligned}f\left(x_0,y_0\right)-f\left(\frac12,2\right)&\approx\nabla f\left(\frac12,2\right)\cdot\left(x_0-\frac12,y_0-2\right)=2\nabla g\left(\frac12,2\right)\cdot\left(x_0-\frac12,y_0-2\right)\\&\approx2\left[g\left(x_0,y_0\right)-g\left(\frac12,2\right)\right]=2\cdot\left(-0.1\right)=-0.2.\end{aligned}$