111年台綜大轉學考-微積分B詳解

臺灣綜合大學系統111學年度學士班轉學生聯合招生考試

科目名稱：微積分B

解答： $$(a)\lim_{x\to 0}{ 4^x-1 \over x} =\lim_{x\to 0}{ (4^x-1)' \over (x)'} = \lim_{x\to 0}{ \ln 4\cdot 4^x \over 1} = \bbox[red, 2pt]{\ln 4} \\(b)\lim_{x\to -6}{ \sqrt{10-x}-4 \over x+6} = \lim_{x\to -6}{ (\sqrt{10-x}-4)' \over (x+6)'} = \lim_{x\to -6} {-1\over 2\sqrt{10-x}} = \bbox[red, 2pt]{-{1\over 8}}$$

解答： $$f(x)= x^2-3x+2 = (x-2)(x-1) \Rightarrow \begin{cases} f(x)\ge 0, & x\ge 2 \\f(x)\le 0, & 1\le x\le 2\\ f(x)\ge 0, & x\le 1\end{cases} \\ \Rightarrow \int_0^3 |x^2-3x+2|\,dx =\int_0^1 (x^2-3x+2)\,dx -\int_1^2 (x^2-3x+2)\,dx +\int_2^3 (x^2-3x+2) \,dx \\ = \left.\left[ {1\over 3}x^3-{3\over 2}x^2 +2x\right]\right|_0^1- \left.\left[ {1\over 3}x^3-{3\over 2}x^2 +2x\right]\right|_1^2+  \left.\left[ {1\over 3}x^3-{3\over 2}x^2 +2x\right]\right|_2^3 \\={5\over 6} -({2\over 3}-{5\over 6})+({3\over 2}-{2\over 3}) =\bbox[red,2pt]{11\over 6}$$

解答： $$\int_0^\pi e^{-x}\sin(\pi -x)\,dx= \int_0^\pi e^{-x}\sin x\,dx= \left. \left[-e^{-x}\sin x \right]\right|_0^\pi+ \int_0^\pi e^{-x}\cos x\,dx \\ = \left. \left[-e^{-x}\sin x -e^{-x}\cos x\right]\right|_0^\pi- \int_0^\pi e^{-x}\sin x\,dx  \Rightarrow 2\int_0^\pi e^{-x}\sin x\,dx =\left. \left[-e^{-x}\sin x -e^{-x}\cos x\right]\right|_0^\pi \\ \Rightarrow \int_0^\pi e^{-x}\sin x\,dx ={1\over 2}\left. \left[-e^{-x}\sin x -e^{-x}\cos x\right]\right|_0^\pi =\bbox[red, 2pt]{{1\over 2} (1+e^{-\pi})}$$

解答： $$u=\sqrt x+1 \Rightarrow du={dx\over 2\sqrt x} \Rightarrow \int_1^9 {1\over \sqrt x(1+\sqrt x)^2}\,dx = \int_2^4 {2\over u^2}\,du = \left.\left[ -{2\over u}\right]\right|_2^4 =-{1\over 2}+1 =\bbox[red, 2pt]{1\over 2}$$

解答： $$\int_0^2{1\over 6x^2+7x +2}\,dx =\int_0^2 \left({2\over 2x+1}-{3\over 3x+2} \right)\,dx \\ =\left.\left[ \ln(2x+1)-\ln(3x+2) \right]\right|_0^2 = \ln 5-\ln 8+\ln 2 =  \bbox[red,2pt]{\ln{5\over 4}}$$

解答： $$y^2(x^2+ y^2)=x^2 \Rightarrow 2yy'(x^2+y^2) +y^2(2x+2yy')=2x\\ 將\cases{x=\sqrt 2/2\\ y=\sqrt 2/2} 代入上式\Rightarrow \sqrt 2y'+{1\over 2}(\sqrt 2+\sqrt 2y')=\sqrt 2 \Rightarrow y'= \bbox[red, 2pt]{1\over 3}$$

解答： $$(a)\;u(x,y)= y+xe^{xy} \Rightarrow u_x=e^{xy} +xye^{xy} \Rightarrow u_x(0,1)=\bbox[red, 2pt]1\\(b)\;u(x,y)= y+xe^{xy} =(2t+1)+ (2s+t)e^{(2s+t)(2t+1)}\\ \Rightarrow {\partial u\over \partial t} =2+ e^{(2s+t)(2t+1)} +(2s+t) (4s+4t+1)e^{(2s+t)(2t+1)} \Rightarrow {\partial u\over \partial t} (0,0)=\bbox[red, 2pt]3$$

解答： $$(a)\;e^x= 1+x+ {x^2\over 2!}+ {x^3 \over 3!}+\cdots+ {x^n\over n!}+\cdots \\ \Rightarrow e^2 =1+2+ {2^2\over 2!}+{2^3\over 3!} +\cdots+ {2^n\over n!}+\cdots \\ \Rightarrow 2e^2 = 2+2^2+{2^3\over 2!}+ \cdots +{2^n\over (n-1)!} +\cdots  =\sum_{n=1}^\infty {2^n\over (n-1)!}= \sum_{n=1}^\infty na_n \\ \Rightarrow \sum_{n=1}^\infty na_n=\bbox[red, 2pt]{2e^2}\\(b)\;\int_1^\infty {1\over x(\ln x)}\,dx = \left.\left[\ln(\ln x) \right]\right|_1^\infty =\infty  \Rightarrow \sum_{n=1}^\infty {1\over n(\ln n)} \bbox[red,2pt]{發散}$$

解答： $$f(x,y)=3x^2+2y^2-4y \Rightarrow \cases{f_x= 6x\\ f_y=4y-4} \Rightarrow \cases{f_{xx}=6\\ f_{yy}=4\\ f_{xy}=0} \Rightarrow D(x,y)=f_{xx}f_{yy}-f_{xy}^2 \gt 0\\ 因此\cases{f_x=0 \\f_y=0} \Rightarrow (x,y)=A(0,1) 非鞍點(\because D \gt 0)\\ 接著求邊界點:\cases{y=x^2 \Rightarrow g(x)=f(x,x^2) =  2x^4-x^2 \Rightarrow g'(x)=0 \Rightarrow 8x^3-2x=0 \Rightarrow x=0,\pm 1/2\\ y=4 \Rightarrow h(x)=f(x,4)=3x^2+16 \Rightarrow h'(x)=0 \Rightarrow x=0\\ y=x^2=4 \Rightarrow x=\pm 2} \\ \Rightarrow 邊界點\cases{B(0,0)\\ C(1/2,1/4) \\D(-1/2,1/4)\\ E(0,4)\\ F(2,4)\\ G(-2,4)} \Rightarrow \cases{f(A)=-2 \\f(B)=0 \\f(C)=f(D)=-1/8 \\ f(E)=16\\ f(F)=f(G)=28} \Rightarrow \bbox[red, 2pt]{\cases{\text{absolute maxima =}28 \\ \text{absolute minima}=-2} }$$

解答： $$\cases{f(x,y)= {5\over 4}x^2 +{7\over 4}y^2- {\sqrt 3\over 2}xy\\ g(x,y)=x^2+y^2-1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{{5\over 2}x-{\sqrt 3\over 2}y = \lambda (2x) \cdots(1)\\ {7\over 2}y-{\sqrt 3\over 2}x = \lambda(2y) \cdots(2)\\ x^2+y^2=1 \cdots(3)} \\ 因此 \cases{{(1)\over (2)} \Rightarrow xy={\sqrt 3\over 2}(x^2-y^2) \\ (3) \Rightarrow \cases{x=\cos\theta \\ y=\sin \theta}} \Rightarrow \cos\theta \sin\theta ={\sqrt 3\over 2}(\cos^2\theta -\sin^2\theta) \Rightarrow \sin 2\theta =\sqrt 3\cos 2\theta\\ \Rightarrow \tan 2\theta =\sqrt 3 \Rightarrow 2\theta = {\pi\over 3}+k\pi \Rightarrow \theta = {\pi\over 6}+{k\pi\over 2},k\in \mathbb{Z} \\\Rightarrow \theta =\pi/6,2\pi/3, 7\pi/6,5\pi/3 \Rightarrow (x,y)=\cases{A(\sqrt 3/2, 1/2)\\B(-1/2,\sqrt 3/2) \\C(-\sqrt 3/2,-1/2) \\D(1/2,-\sqrt 3/2)} \Rightarrow \cases{f(A)= f(C)=1 極小值\\f(B)=f(D)=2 極大值} \\ \Rightarrow 極值為\bbox[red, 2pt]{1及2}，其中2為極大值、1為極小值;$$

==================== END ===========================

未公告答案，解題僅供參考