Write down your answers in order. You should include all the necessary calculation and reasoning.
- Let g(y) be the inverse function of y=f(x) with g(0)=1. Find g″, 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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