106年台聯大轉學考-微積分A2詳解

台灣聯合大學系統106學年度學士班轉學生考試

科目：微積分
類組別：A2
甲、填充題：共8題，每題8分，共64分

解答：$$f'(x)=\sqrt{x^2+1} \Rightarrow \lim_{h\to 0} {1\over h}\int_1^{1+h} \sqrt{1+t^2}\,dt = f'(1) =\bbox[red,2pt] {\sqrt {2}}$$

解答：$$顯然當x^2-1=0時，即x=\pm 1，f不可微；又f(x)=x^2-1 \Rightarrow f'(x)=0 \Rightarrow x=0\\ 因此x=\pm 1, 0為臨界點，共\bbox[red, 2pt]3 個$$

解答：$$\lim_{x\to 0}{\sqrt{ax+b}-2 \over x} =1,當x\to 0,分母為0，因此分子也必須為0\Rightarrow \sqrt b-2=0 \Rightarrow b=4\\ 因此\lim_{x\to 0}{\sqrt{ax+b}-2 \over x} \Rightarrow \lim_{x\to 0}{\sqrt{ax+4}-2 \over x} =\lim_{x\to 0}{(\sqrt{ax+4}-2)' \over (x)'} =\lim_{x\to 0}{ a\over 2\sqrt{ax+4}-2 }\\ ={a \over 2} =1 \Rightarrow a=2 \Rightarrow \bbox[red, 2pt]{\cases{a=2\\ b=4}}$$

解答：$$當p(或q)遞增時，x及y同時遞減，因此兩者為\bbox[red,2pt]{complementary(互補)}$$

解答：$$f(x)=(\tan x)^2 +e^x \Rightarrow f'(x)= 2\tan x\sec ^2 x +e^x \Rightarrow f'(0)=\bbox[red, 2pt]1$$

解答：$$\int_0^1 \int_0^x e^{-x^2}\,dydx =\int_0^1  x e^{-x^2}\, dx = \int_0^1 {1\over 2}e^{-u}\,du \;(u=x^2\Rightarrow du =2xdx) =\left. \left[ -{1\over 2}e^{-u} \right]\right|_0^1 \\= \bbox[red, 2pt]{{1\over 2}(1-e^{-1})}$$

解答：$$此題相當於求f(x,y)-150x-250y的最大化，限制條件為150x+250y\le 50000\\ 因此令\cases{P(x,y)=f(x,y)-150x-250y =100x^{3/4} y^{1/4}-150x-250y\\ g(x,y)= 150x+250y-50000} \\ \Rightarrow \cases{P_x =\lambda g_x\\ P_y = \lambda g_y \\ g=0} \Rightarrow \cases{75x^{-1/4}y^{1/4}-150 = 150\lambda \\ 25x^{3/4}y^{-3/4} -250 = 250\lambda} \xrightarrow{兩式相除} {75x^{-1/4}y^{1/4}-150  \over 25x^{3/4}y^{-3/4} -250} ={3\over 5} \\ \Rightarrow 375x^{-1/4} y^{1/4}=75 x^{3/4} y^{-3/4}  \Rightarrow 5y =x 代入g(x,y) \Rightarrow 750y +250y=50000 \Rightarrow y=50  \\\Rightarrow x=250  \Rightarrow 當\bbox[red,2pt]{\cases{x=250\\ y=50}}時，能最大化 P(x,y)$$

解答：$$\text{the average value of }f \text{ on [0,1] is 2} \Rightarrow {\int_0^1 f\,dx \over 1-0} =2 \Rightarrow \int_0^1 f\,dx =2 \\ 因此\int_0^1 x^2f''(x) = \left. \left[x^2f'(x)-2xf(x) \right]\right|_0^1+2 \int_0^1 f(x)\,dx =f'(1)-2f(1)+2 \int_0^1 f(x)\,dx \\=2-2\cdot 2+ 2\cdot 2=\bbox[red, 2pt]2$$

乙、計算、證明題：共3題，每題12分，共36分

解答：$$f(x,y)=e^{-xy/4} \Rightarrow \cases{f_x = -{y\over 4}e^{-xy/4} \\ f_y= -{x\over 4}e^{-xy/4}} \Rightarrow \cases{f_{xx}= {y^2\over 16}e^{-xy/4}\\ f_{xy}= -{1\over 4}e^{-xy/4} +{xy\over 16}e^{-xy/4} \\ f_{yy}= {x^2\over 16}e^{-xy/4}}\\ \Rightarrow d(x,y)=f_{xx}f_{yy}-f_{xy}^2\\因此\cases{f_x=0 \Rightarrow y=0\\ f_y=0 \Rightarrow x=0} \Rightarrow d(0,0)= -{1\over 16} \lt 0 \Rightarrow (0,0,f(0,0))為一鞍點，非極值\Rightarrow \bbox[red, 2pt]{無極值}$$

解答：$$\int_0^6 \int_{x/3}^2 x\sqrt{y^3+1}\,dydx = \int_0^2 \int_0^{3y} x \sqrt{y^3+1}\,dxdy =\int_0^2  \left.\left[ {1\over 2}x^2 \sqrt{y^3+1} \right] \right|_0^{3y} \\ =\int_0^2  {9\over 2}y^2 \sqrt{y^3+1}\,dy =\int_1^{9} {3\over 2}\sqrt u\,du (u=y^3+1) = \left.\left[ u^{3/2}\right] \right|_1^9 =27-1=\bbox[red, 2pt]{26}$$

解答：$$\lim_{n \to \infty} \left|{a_{n+1} \over a_n} \right| =\lim_{n \to \infty} \left|{(n+1)! \over (n+1)^{n+1}} \cdot {n^n\over n!} \right|=\lim_{n \to \infty} {n+1\over n} \cdot ({n\over n+1})^n ={1\over e} \lt 1\\ \Rightarrow \bbox[red,2pt]{絕對收斂}$$

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