Write down your answers in order. You should include all the necessary calculation and reasoning.
- Let $g\left(y\right)$ be the inverse function of $y=f\left(x\right)$ with $g\left(0\right)=1$. Find $g''\left(0\right)$, knowing that $f'\left(x\right)=\sqrt{x^3+1}$.
- Determine the values of $A$, $B$ and $C$ so that $101$ occurs as the maximal value of the function
$f\left(x\right)=Ax^5-x^4+Bx^2+C$,
at a unique point on $\left(-\infty,\infty\right)$. - Evaluate the definite integral $\displaystyle\int_0^\pi e^x\cos xdx$.
- Determine the values of $A$, $B$, $C$, $D$ and $E$, given that
$\displaystyle\lim_{x\to0}\frac{\ln\left(x+1\right)-\left(A+Bx+Cx^2+Dx^3+Ex^4\right)}{x^5}=0$.
- Find the volume of the solid bounded by the surface
$x^2+3y^2+z^2=1$.
- 當 $y=0$ 時有 $\theta=0$；
- 當 $y=\sqrt{1/3}$ 時有 $\displaystyle\theta=\tan^{-1}\left(\sqrt2\right)$；
- 求導可知 $\displaystyle dy=\frac{\sec^2\theta}{\sqrt6}d\theta$。
- Evaluate the integral $\displaystyle\int_0^1\int_y^1x\sqrt{x^3+1}dxdy$.
- Suppose $f\left(x\right)$ is a continuous function defined on $\left[-1,1\right]$ so that $\displaystyle\int_{-1}^1f^2\left(x\right)dx=0$. Give a reason why we can or can not conclude that $f\left(x\right)=0$, for every $x\in\left[-1,1\right]$.
- Suppose $f\left(x\right)$ is a differentiable function defined on $\left(-\infty,\infty\right)$ satisfying $f\left(x+y\right)=f\left(x\right)\cdot f\left(y\right)$, for every $x$ and $y$. Show that $f\left(x\right)=a^x$ for some positive number $a$. (Hint: The function $\ln f\left(x\right)$ must have constant derivative.)
- Evaluate the integral $\displaystyle\iint_Dxydxdy$, where
$D=\left\{\left(x,y\right)\in\mathbb{R}^2\mid\,3\leq x+y\leq6,~0\leq x-2y\leq3\right\}$.
- Determine the maximum of the function $f\left(x,y,z\right)=x-2y+3z$ defined on the surface
$\left\{\left(x,y,z\right)\in\mathbb{R}^3\mid~x^2+y^2+5z^2=7\right\}$.
訣竅
運用反函數的定義與連鎖律求解。解法
根據反函數的定義有 $g\left(f\left(x\right)\right)=x$，求導有$g'\left(f\left(x\right)\right)f'\left(x\right)=1$
因此有 $\displaystyle g'\left(f\left(x\right)\right)=\left(x^3+1\right)^{-1/2}$。進一步求導可得$\displaystyle g''\left(f\left(x\right)\right)f'\left(x\right)=-\frac12\left(x^3+1\right)^{-3/2}\cdot3x^2$
再者，因為 $g\left(0\right)=1$，故 $f\left(1\right)=0$。據此我們取 $x=1$ 代入可得$\displaystyle g''\left(0\right)\cdot\sqrt2=g''\left(f\left(1\right)\right)f'\left(1\right)=-\frac12\cdot\left(1+1\right)^{-3/2}\cdot3$
如此所求為 $\displaystyle g''\left(0\right)=-\frac38$。訣竅
有極大值表示 $f$ 最高次為偶數次並且其領導係數為負值；再者透過配方法找出極值發生所在之位置等。解法
假若 $A>0$，那麼 $\displaystyle\lim_{x\to\infty}f\left(x\right)=\infty$；假若 $A<0$，那麼 $\displaystyle\lim_{x\to-\infty}f\left(x\right)=\infty$，兩者皆無最大值，因此僅能有 $A=0$，此時 $f\left(x\right)=-x^4+Bx^2+C$。
進一步，使用配方法可知
$\displaystyle f\left(x\right)=-\left(x^2-\frac{B}2\right)^2+C+\frac{B^2}4$
假若 $B\neq0$，那麼 $\displaystyle x=\pm\sqrt{\frac{B}2}$ 都可產生最大值，這與唯一性矛盾。故 $B=0$，此時有 $f\left(x\right)=-x^4+C$。容易看出最大值發生在 $x=0$，又最大值為 $C=101$。綜上可知 $A=B=0$，而 $C=101$。
訣竅
連續使用兩次分部積分法即可。解法
使用分部積分法可知$\displaystyle\int_0^{\pi}e^x\cos xdx=e^x\cos x\Big|_0^\pi+\int_0^{\pi}e^x\sin xdx=-e^\pi-1+\left[e^x\sin x\Big|_0^\pi-\int_0^{\pi}e^x\cos xdx\right]$
移項便有 $\displaystyle2\int_0^{\pi}e^x\cos xdx=-1-e^\pi$，至終解得 $\displaystyle\int_0^{\pi}e^x\cos xdx=-\frac{1+e^{\pi}}2$。訣竅
利用極限的四則運算定理以及羅必達法則求解即可；亦可由泰勒展開式的經驗得解。解法一
藉由四則運算定理可以發現$\displaystyle\lim_{x\to0}\left[\ln\left(x+1\right)-\left(A+Bx+Cx^2+Dx^3+Ex^4\right)\right]=\lim_{x\to0}\left[x^5\cdot\frac{\ln\left(x+1\right)-\left(A+Bx+Cx^2+Dx^3+Ex^4\right)}{x^5}\right]=0$
因此 $A=0$。接著類似地使用四則運算並搭配羅必達法則，可得$\displaystyle0=\lim_{x\to0}\frac{\ln\left(x+1\right)-\left(Bx+Cx^2+Dx^3+Ex^4\right)}x=\lim_{x\to0}\left[\frac1{x+1}-\left(B+2Cx+3Dx^2+4Ex^3\right)\right]=1-B$
故 $B=1$。進一步，運用相同的方式可知$\displaystyle\begin{aligned}0&=\lim_{x\to0}\frac{\ln\left(x+1\right)-\left(x+Cx^2+Dx^3+Ex^4\right)}{x^2}=\lim_{x\to0}\frac{\frac1{x+1}-\left(1+2Cx+3Dx^2+4Ex^3\right)}{2x}\\&=\lim_{x\to0}\frac{-\frac1{\left(x+1\right)^2}-\left(2C+6Dx+12Ex^2\right)}2=-\frac12-C\end{aligned}$
即有 $\displaystyle C=-\frac12$。順著相同的步驟可以分別得到$\displaystyle\begin{aligned}&\frac13-D=\lim_{x\to0}\frac{\frac2{\left(x+1\right)^3}-\left(6D+24Ex\right)}{6}=0\\&-\frac14-E=\lim_{x\to0}\frac{-\frac6{\left(x+1\right)^4}-24E}{24}=0\end{aligned}$
故 $\displaystyle D=\frac13$ 且 $\displaystyle E=-\frac14$。解法二
經由泰勒展開式可以注意到$\displaystyle\ln\left(x+1\right)=\int_0^x\frac1{1+t}dt=\int_0^x\sum_{k=0}^{\infty}\left(-t\right)^kdt=\sum_{k=0}^{\infty}\left(-1\right)^k\frac{x^{k+1}}{k+1}=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots$
容易推測知道 $A=0$、$B=1$、$\displaystyle C=-\frac12$、$\displaystyle D=\frac13$ 而 $\displaystyle E=-\frac14$。訣竅
可以運用旋轉體表面積的方法計算。解法
可視給定的曲面為 $x=\sqrt{1-3y^2}$ 繞 $y$ 軸旋轉而得的，故其表面積可列式並計算如下$\displaystyle A=\int_{-\sqrt{1/3}}^{\sqrt{1/3}}2\pi x\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy=4\pi\int_0^{\sqrt{1/3}}\sqrt{1-3y^2}\cdot\sqrt{1+\left(\frac{-3y}{\sqrt{1-3y^2}}\right)^2}dy=4\pi\int_0^{\sqrt{1/3}}\sqrt{1+6y^2}dy$
如此運用三角代換，令 $\displaystyle y=\frac1{\sqrt6}\tan\theta$，那麼$\displaystyle\begin{aligned}A&=4\pi\int_0^{\tan^{-1}\left(\sqrt2\right)}\sec\theta\cdot\frac1{\sqrt6}\sec^2\theta d\theta=\frac{2\sqrt6\pi}3\int_0^{\tan^{-1}\left(\sqrt2\right)}\sec^3\theta d\theta\\&=\left.\frac{\sqrt6\pi}3\left[\sec\theta\tan\theta+\ln\left|\sec\theta+\tan\theta\right|\right]\right|_0^{\tan^{-1}\left(\sqrt2\right)}=\frac{\sqrt6\pi}3\left(\sqrt6+\ln\left(\sqrt3+\sqrt2\right)\right)\\&=2\pi+\frac{\sqrt6}3\ln\left(\sqrt3+\sqrt2\right)=2\pi+\frac\pi6\ln\left(5+2\sqrt6\right)\end{aligned}$
其中 $\displaystyle\int\sec^3\theta d\theta$ 可計算如下$\displaystyle\begin{aligned}\int\sec^3\theta d\theta&=\sec\theta\tan\theta-\int\tan^2\theta\sec\theta d\theta=\sec\theta\tan\theta-\int\left(\sec^2\theta-1\right)\sec\theta d\theta\\&=\sec\theta\tan\theta-\int\sec^3\theta d\theta+\int\sec\theta d\theta\end{aligned}$
移項便有$\displaystyle\int\sec^3\theta d\theta=\frac{\sec\theta\tan\theta+\int\sec\theta d\theta}2=\frac{\sec\theta\tan\theta+\ln\left|\sec\theta+\tan\theta\right|}2+C$
訣竅
交換積分次序求解。解法
可將原積分範圍 $\left\{\begin{aligned}&y\leq x\leq1\\&0\leq y\leq1\end{aligned}\right.$ 改寫為 $\left\{\begin{aligned}&0\leq x\leq1\\&0\leq y\leq x\end{aligned}\right.$，如此所求的重積分可改寫並計算如下$\displaystyle\int_0^1\int_y^1x\sqrt{x^3+1}dxdy=\int_0^1\int_0^xx\sqrt{x^3+1}dydx=\int_0^1x^2\sqrt{x^3+1}dx=\left.\frac13\cdot\frac23\left(x^3+1\right)^{3/2}\right|_0^1=\frac{4\sqrt2-2}9$
訣竅
運用反證法加以證明。解法
假設有某一點 $x_0\in\left[-1,1\right]$ 使 $f\left(x_0\right)\neq0$，那麼由 $f$ 的連續性可知存在 $\delta>0$ 使 $x\in\left[-1,1\right]\cap\left(x_0-\delta,x_0+\delta\right)$ 蘊含 $\displaystyle\frac{\left|f\left(x_0\right)\right|}2<\left|f\left(x\right)\right|\leq\left|f\left(x_0\right)\right|$。如此可以發現$\displaystyle\int_{-1}^1f^2\left(x\right)dx\geq\int_{\left[-1,1\right]\cap\left(x_0-\delta,x_0+\delta\right)}f^2\left(x\right)dx\geq\frac{\left|f\left(x_0\right)\right|^2}4\int_{\left[-1,1\right]\cap\left(x_0-\delta,x_0+\delta\right)}dx>0$
這與題意矛盾，故函數 $f$ 應處處為零。訣竅
運用提示對該形式的函數求導以證明之。解法
設 $g\left(x\right)=\ln f\left(x\right)$，可以注意到 $g\left(x+y\right)=g\left(x\right)+g\left(y\right)$，從而有 $g\left(0\right)=0$。由此按定義求導可知$\displaystyle g'\left(x\right)=\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}h=\lim_{h\to0}\frac{g\left(h\right)}h=g'\left(0\right)$
故 $g\left(x\right)=g'\left(0\right)x$。如此記 $g'\left(0\right)=\ln a$，那麼 $\ln f\left(x\right)=x\ln a$，故 $f\left(x\right)=a^x$。訣竅
由積分範圍可推測應運用變數變換改寫重積分，其中應留意 Jacobian 行列式的計算。解法
令 $u=x+y$、$v=x-2y$，那麼由積分範圍知道 $\left\{\begin{aligned}&3\leq u\leq6\\&0\leq v\leq3\end{aligned}\right.$。再者，其 Jacobian 行列式可計算如下$\displaystyle\Big|\left|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\right|\Big|=\Big|\left|\frac{\partial\left(u,v\right)}{\partial\left(x,y\right)}\right|\Big|^{-1}=\Big|\begin{vmatrix}\displaystyle\frac{\partial u}{\partial x}&\displaystyle\frac{\partial u}{\partial y}\\\displaystyle\frac{\partial v}{\partial x}&\displaystyle\frac{\partial v}{\partial y}\end{vmatrix}\Big|^{-1}=\Big|\begin{vmatrix}1&1\\1&-2\end{vmatrix}\Big|^{-1}=\frac13$
因此所求的重積分可改寫並計算如下$\displaystyle\begin{aligned}\iint_Dxydxdy&=\int_3^6\int_0^3\frac{2u+v}3\cdot\frac{u-v}3\cdot\frac13dvdu=\frac1{27}\int_3^6\int_0^3\left(2u^2-uv-v^2\right)dvdu\\&=\frac1{27}\int_3^6\left.\left(2u^2v-\frac{uv^2}2-\frac{v^3}3\right)\right|_0^3du=\frac19\int_3^6\left(2u^2-\frac{3u}2-3\right)du\\&=\left.\frac19\left(\frac{2u^3}3-\frac{3u^2}4-3u\right)\right|_3^6=\frac19\left[\left(144-27-18\right)-\left(18-\frac{27}4-9\right)\right]=\frac{43}4\end{aligned}$
訣竅
運用柯西不等式即可；亦可使用拉格朗日乘子法。解法一
使用柯西不等式可知$\displaystyle\frac{238}5=7\cdot\frac{34}5=\left(x^2+y^2+\left(\sqrt5z\right)^2\right)\left(1^2+\left(-2\right)^2+\left(\frac3{\sqrt5}\right)^2\right)\geq\left(x-2y+3z\right)^2$
如此有 $\displaystyle x-2y+3z\leq\frac{\sqrt{1190}}5$，此時等號成立條件為 $\displaystyle x=\frac{y}{-2}=\frac{5z}3$，即 $\displaystyle\left(x,y,z\right)=\left(\frac{\sqrt{1190}}{34},-\frac{\sqrt{1190}}{17},\frac{3\sqrt{1190}}{170}\right)$。解法二
設拉格朗日乘子函數如下$\displaystyle F\left(x,y,z,\lambda\right)=x-2y+3z+\lambda\left(x^2+y^2+5z^2-7\right)$
據此解聯立方程組$\left\{\begin{aligned}&F_x\left(x,y,z,\lambda\right)=1+\lambda\cdot2x=0\\&F_y\left(x,y,z,\lambda\right)=-2+\lambda\cdot2y=0\\&F_z\left(x,y,z,\lambda\right)=3+\lambda\cdot10z=0\\&F_{\lambda}\left(x,y,z,\lambda\right)=x^2+y^2+5z^2-7=0\end{aligned}\right.$
各式移項平方有 $4x^2\lambda^2=1$、$4y^2\lambda^2=4$、$\displaystyle20z^2\lambda^2=\frac95$，三式相加並運用第四式可知 $\displaystyle\frac{34}5=28\lambda^2$，即得 $\displaystyle\lambda=\pm\sqrt{\frac{17}{70}}=\pm\frac{\sqrt{1190}}{70}$。如此得座標$\displaystyle\left(x,y,z\right)=\mp\left(\frac{\sqrt{1190}}{34},-\frac{\sqrt{1190}}{17},\frac{3\sqrt{1190}}{170}\right)$
如此代入可知 $x-2y+3z$ 的最大值與最小值分別為 $\displaystyle\pm\frac{\sqrt{1190}}5$。
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