114年專科學力鑑定-微積分詳解

教育部114年自學進修專科學校學力鑑定考試試題本

專業科目(一)： 微積分

解答:$$g(x)=x^2-x-2=(x-2)(x+1) \Rightarrow \begin{cases} g(x)\gt 0,& x\gt 2\\ g(x)\lt 0, & -1\lt x\lt 2\\ g(x)\gt 0, &x\lt -1\end{cases} \\ \Rightarrow \lim_{x\to -2}{|g(x)|\over g(x)} =\lim_{x\to -2}{g(x)\over g(x)} =1，故選\bbox[red, 2pt]{(D)}$$

解答:$$\lim_{x\to 0} {f(x) \over g(5x)} =\lim_{x\to 0} {f'(x) \over 5g'(5x)} ={3\over 5\cdot 2}={3\over 10}，故選\bbox[red, 2pt]{(B)}$$

解答:$$萊布尼茲積分公式\text{(Leibniz’ Rule)}: S'(x)= \sin({\pi x^2\over 2})，故選\bbox[red, 2pt]{(C)}$$

解答:$$g(f(x))=\sqrt{(x+1)^2-1} =\sqrt{x(x+2)} \Rightarrow x(x+2)\ge 0 \Rightarrow x\ge 0或x\le -2，故選\bbox[red, 2pt]{(A)}$$

解答:$$f(x)= (x+3)(x^2+1)^{100} \Rightarrow f'(x)=(x^2+1)^{100}+ 100(x+3)(x^2+1)^{99}(2x) \\=(x^2 +1)^{99} \left( x^2+1+100(x +3)(2x) \right) =(x^2+ 1)^{99}(201 x^2+600x+1) \\ \Rightarrow f''(x)=99(x^2 +1)^{98}(2x)(201x^2+600x+1) + (x^2+1)^{99}(402x +600) \\ \Rightarrow f''(0)=0 +600=600，故選\bbox[red, 2pt]{(D)}$$

解答:$$學生乙:u= \sin(x) \Rightarrow du=\cos (x)dx \Rightarrow \int \sin(2x)\,dx = \int 2\sin(x)\cos (x)\,dx \\=\int 2u\,du =u^2+c_1 = \sin^2 (x)+c_1 ={1\over 2}(1-\cos (2x))+c_1=-{1\over 2}\cos(2x)+{1\over 2}+c _1 \\=-{1\over 2}\cos(2x)+ c_2 = 學生甲, 故選\bbox[red, 2pt]{(B)}, 但公布的答案是\bbox[cyan,2pt]{(A)}$$

解答:

$$\iint_D {x\over y}\,dA =\int_1^2 \int_1^{x^2} {x\over y}\,dydx =\int_1^2 x\ln x^2\,dx =\int_1^4 {1\over 2}\ln u\,du \quad (u=x^2) \\= \left. \left[ {1\over 2} \left( u\ln u-u \right) \right] \right|_1^4 = {1\over 2} \left( 4\ln 4-3 \right) =4\ln 2-{3\over 2}，故選\bbox[red, 2pt]{(C)}$$

解答:$$u=2x \Rightarrow {1\over 2}du=dx \Rightarrow \int_{-2}^2 f(2x)\,dx = \int_{-4}^4 f(u)\cdot {1\over 2}du= {1\over 2}\int_{-4}^4 f(u)\,du，故選\bbox[red, 2pt]{(A)}$$

解答:$${1\over 1-x} =1+x+x^2+\cdots+ x^n+ \cdots \Rightarrow \\f(x)=\ln(1-2x) \Rightarrow f'(x)={-2\over 1-2x} =-2(1+2x+(2x)^2+ \cdots+(2x)^n+\cdots) \\ \Rightarrow f'(x)的x^9係數為-2\cdot 2^9 \Rightarrow f^{{(10)}}(x)的常數項為-2\cdot 2^9\cdot 9!\\ \Rightarrow f^{(10)}(0)=-2^{10}\cdot 9!，故選\bbox[red, 2pt]{(B)}$$

解答:$$\lim_{x\to 1}{ax^3+b\over x^2-1} =\lim_{x\to 1}{3ax^2\over 2x} = \lim_{x\to 1}{3ax\over 2} = {3a\over 2}=1 \Rightarrow a={2\over 3}，故選\bbox[red, 2pt]{(A)}$$

解答:$$x^3+y^3 +z^3+ 6xyz=1 \Rightarrow 3x^2+0+3z^2z'+6yz+ 6xyz'=  (3z^2+6xy)z'+ 3x^2+6yz=0 \\ \Rightarrow z'={\partial z\over \partial x} =-{3x^2+6yz\over 3z^2+6xy} =-{x^2+2yz\over z^2+2xy}，故選\bbox[red, 2pt]{(D)}$$

解答:$$\int{2x+7\over x^2+x-2} \,dx =\int{2x+7\over (x+2)(x-1)} \,dx = \int \left( {3\over x-1}-{1\over x+2} \right) \,dx = 3\ln|x-1|-\ln|x+2|+C\\，故選\bbox[red, 2pt]{(B)}$$

解答:$$f'({\pi\over 2}) = \lim_{h\to 0}{f(h+\pi/2) -f(\pi/2)\over h} = \lim_{h\to 0}{f(h+\pi/2)  \over h} =  \lim_{h\to 0}{|h|\cdot|\cos(h+\pi/2) |\over h} \\= \lim_{h\to 0}{|h|\cdot|-\sin(h)| \over h} =\lim_{h\to 0}{|h|\cdot|\sin(h)| \over h} \\ \lim_{h\to 0^+}{|h|\cdot|\sin(h)| \over h} =\lim_{h\to 0^+}{h\cdot \sin(h) \over h} =\lim_{h\to 0^+} \sin(h)=0\\ \lim_{h\to 0^-}{|h|\cdot|\sin(h)| \over h} =\lim_{h\to 0^-}{-h\cdot \sin(-h) \over h} =\lim_{h\to 0^-} {h\cdot \sin(h) \over h}=0 \\  \Rightarrow f'(\pi/2)=0 \\f'(-{\pi\over 2}) = \lim_{h\to 0}{f(h-\pi/2) -f(-\pi/2)\over h} = \lim_{h\to 0}{f(h-\pi/2)  \over h} =  \lim_{h\to 0}{|h-\pi|\cdot|\cos(h-\pi/2) |\over h} \\= \lim_{h\to 0}{(\pi-h)|\sin(h)| \over h} \Rightarrow \cases{\lim_{h\to 0^+}{(\pi-h)|\sin(h)| \over h} =\pi \\\lim_{h\to 0^-}{(\pi-h)|\sin(h)| \over h} =-\pi } \Rightarrow f'(-{\pi\over 2}) 不存在\\，故選\bbox[red, 2pt]{(C)}$$

解答:

$$x=y^2-1\Rightarrow y=\sqrt{x+1} 代入y=\sqrt 2 x \Rightarrow \sqrt 2 x=\sqrt{x+1} \Rightarrow 2x^2-x-1=0\\ \Rightarrow (2x+1)(x-1)=0 \Rightarrow \cases{x=1\\ x=-1/2} \Rightarrow \cases{y=\sqrt 2\\ y=-\sqrt 2/2} \Rightarrow 兩曲線交點為\cases{A(-1/2, -\sqrt 2/2) \\B(1,\sqrt 2)} \\ \Rightarrow \cases{R= \int_{-1}^{-1/2} \sqrt {x+1}\,dx =1/3\sqrt 2\\ S=\int_{-1/2}^1 (\sqrt{x+1}-\sqrt 2 x)= 19/12\sqrt 2\,dx} \\ \Rightarrow 所圍面積=2R+S={2 \over 3\sqrt 2}+{19\over 12\sqrt 2} ={9\over 8}\sqrt 2，故選\bbox[red, 2pt]{(B)}$$

解答:$$\int_0^{\pi} x\cos^2 x\,dx = {1\over 2}\int_0^{\pi} x(\cos(2x)+1)\,dx = {1\over 2} \left. \left[ {1\over 2}x^2+{1\over 2}x\sin(2x)+ {1\over 4}\cos(2x) \right] \right|_0^{\pi} \\={1\over 2}\cdot {1\over 2}\pi^2 ={1\over 4}\pi^2，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{u=(\ln x)^2\\ dv=x\,dx} \Rightarrow \cases{du= {2 \ln x\over x}\,dx\\ v={1\over 2}x^2} \Rightarrow \int x(\ln x)^2\,dx ={1\over 2}x^2 (\ln x)^2-\int x\ln x\,dx \\={1\over 2}x^2 (\ln x)^2-{1\over 2}x^2 \ln x+{1\over 4}x^2+C \Rightarrow \int_0^1 x(\ln x)^2\,dx = \left. \left[ {1\over 2}x^2 (\ln x)^2-{1\over 2}x^2 \ln x+{1\over 4}x^2 \right] \right|_0^1\\={1\over 4}，故選\bbox[red, 2pt]{(C)}$$

解答:$$f^{-1}(f(x))=x \Rightarrow \left( f^{-1}(f(x)) \right)'=1 \Rightarrow (f^{-1})'(f(x))\cdot f'(x)=1 \Rightarrow (f^{-1})'(f(x))={1\over f'(x)} \\ f(x)=a= 3 \Rightarrow x^3+4x+4=3^2 \Rightarrow x^2+4x-5= (x+5)(x-1)=0 \Rightarrow x=1 \\若x=-5 \Rightarrow x^3+4x+4\lt 0違反根號內不得為負數\\ 又f'(x)={3x^2+4\over 2\sqrt{x^3+4x+4}} \Rightarrow f'(1) ={7\over 6}\\因此(f^{-1})'(f(1)=3)={1\over f'(1)} ={6\over 7}，故選\bbox[red, 2pt]{(B)}$$

解答:$$2x^2-xy+y^2-x-2y=1 \Rightarrow 4x-y-xy'+2yy'-1-2y'=0 \Rightarrow y'={1+y-4x\over 2y-2-x} \\ (A)\bigcirc:  y'(1,3)= 0 \Rightarrow 有水平切線\\ (B)\bigcirc:y'(1/7,-3/7)=0\Rightarrow 有水平切線 \\(C)\times: 圖形為橢圓，一定有兩條鉛垂切線 \\(D)\bigcirc:圖形為橢圓，一定有兩條鉛垂切線\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$x^2+y^2= 16 \Rightarrow \cases{x=4\cos \theta\\ y=4\sin \theta} \Rightarrow f(4\cos \theta, 4\sin \theta) =g(\theta) =32\cos^2\theta+48\sin^2\theta-16 \sqrt 3\cos \theta \sin \theta-5 \\=16\sin^2\theta-8\sqrt 3\sin 2\theta+27 \Rightarrow g'(\theta)=32\sin \theta \cos \theta-16\sqrt 3 \cos 2\theta =16\sin 2\theta-16\sqrt 3\cos 2\theta \\=32 \left( {1\over 2}\sin 2\theta-{\sqrt 3\over 2}\cos 2\theta \right) =32\sin(2\theta-{\pi\over 3}) \Rightarrow g''(\theta) =64\cos(2\theta-{\pi\over 3}) \\ g'(\theta)=0 \Rightarrow \cases{2\theta-\pi/3=0 \Rightarrow \theta=\pi/6 \Rightarrow g''(\pi/6)=64\gt 0\\ 2\theta-\pi/3= \pi \Rightarrow \theta=2\pi/3 \Rightarrow g''(2\pi/3)=-64\lt 0   \\ 2\theta-\pi/3=2\pi \Rightarrow \theta=7\pi/6 \Rightarrow g''(7\pi/6) =64 \gt 0}  \\ \Rightarrow  \theta=2\pi/3 \Rightarrow (x,y)=(-2, 2\sqrt 3) \Rightarrow f(-2,2\sqrt 3)為極大值，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(x,y)=1+x^2+{1\over 2}y^2 \Rightarrow \cases{f_x=2x\\ f_y=y} \Rightarrow 面積S= \iint_D\sqrt{1+4x^2+ y^2} \,dA\\ 取\cases{u=2x\\ v=y} \Rightarrow {\partial (u,v) \over \partial(x,y)} = \begin{Vmatrix}2&0\\0& 1 \end{Vmatrix} =2 \Rightarrow du\,dv =2dx\,dy \Rightarrow S= \iint_{D' }\sqrt{1+u^2+v^2} \cdot {1\over 2}du\,dv \\ 取\cases{u=r\cos \theta\\ v=r\sin \theta} \Rightarrow D'=\{(u,v) \mid u^2+v^2\le 3\}\Rightarrow S= {1\over 2}\int_0^{2\pi} \int_0^{\sqrt 3} \sqrt{1+ r^2\cos^2 \theta+ r^2\sin^2\theta}\cdot  r\cdot drd\theta \\={1\over 2} \left( \int_0^{2\pi }d\theta \right) \left( \int_0^\sqrt{3} r\sqrt{1+r^2}\,dr \right) =\pi\int_0^\sqrt{3} r\sqrt{1+r^2}\,dr = \pi \int_1^4 {1\over 2}\sqrt{w}\,dw \quad (w=1+r^2) \\=\pi \left. \left[ {1\over 3}w^{3/2} \right] \right|_1^4 ={7\over 3}\pi，故選\bbox[red, 2pt]{(D)}$$

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解題僅供參考，其他學力鑑定試題及詳解