111年高師大轉學考-微積分詳解

國立高雄師範大學 111 學年度學士班轉學生招生考試試題

系所別:電機工程學系 二年級
科 目:微積分(全一頁)

◆計算題 (需詳列計算過程)每大題 10 分

解答： $$\mathbf{(a)}\;y=\sqrt[4] x-2x \Rightarrow y'={1\over 4}x^{-3/4}-2 \Rightarrow y'(1)= -{7\over 4} \Rightarrow 切線方程式: y=-{7\over 4}(x-1)-1 \\ \quad\Rightarrow \bbox[red, 2pt]{7x+ 4y=3}\\ \mathbf{(b)}\; h(x)=|x-1|+|x+2| =\begin{cases} 2x+1 & x\ge 1\\ 3& -2\le x\le 1\\ -2x-1 & x\le -2 \end{cases} \Rightarrow h'(x)=\begin{cases} 2& x\ge 1\\ 0& -2\le x\le 1\\ -2 & x\le -2\end{cases}\\\quad 圖形如下:$$

解答： $$\mathbf{(a)}\;\int_0^\pi \cos^2 \theta\,d\theta = {1\over 2}\int_0^\pi \cos 2\theta +1\,d\theta = {1\over 2}\left. \left[ {1\over 2}\sin 2\theta +\theta \right]\right|_0^\pi = \bbox[red, 2pt]{\pi \over 2}\\ \mathbf{(b)}\; \int_{-\pi/2}^0 \cos(x)\sin (x)\,dx ={1\over 2} \int_{-\pi/2}^0 \sin(2x)\,dx ={1\over 2}\left. \left[ -{1\over 2}\cos (2x) \right]\right|_{\pi/2}^0 =\bbox[red,2pt]{-1\over 2}$$

解答： $${x^2\over 9} +{y^2\over 4}=1 \Rightarrow y^2=4-{4x^2\over 9} \Rightarrow y=\sqrt{4-{4x^2\over 9}}\;(只考慮第一象限)\\ \Rightarrow 第一象限面積=\int_0^3 \sqrt{4-{4x^2\over 9}}\,dx = 2\int_0^3 \sqrt{1-({ x\over 3})^2}\,dx  \\取x=3\sin u,則dx=3\cos u\,du，因此上式變為 2\int_0^{\pi/2} \cos u\cdot (3\cos u)\,du =6\int_0^{\pi/2} \cos^2 u\,du \\=3\int_0^{\pi/2} \cos 2u +1 \,du =3\left.\left[ {1\over 2}\sin 2u+u \right]\right|_0^{\pi/2} ={3\over 2}\pi \Rightarrow 橢圓面積={3\over 2}\pi \times 4 =\bbox[red, 2pt]{6\pi}$$

解答： $$\mathbf{(a)}\;f(x)={3(x^2+2x-1) \over 2x^3+x^2-x} \\={A\over x} +{ B\over 2x-1} +{C\over x+1} \Rightarrow 3(x^2+2x-1)=A(2x-1)(x+1) +Bx(x+1) +Cx(2x-1)\\ \Rightarrow \cases{2A+ B+2C=3\\ A+B-C= 6\\ -A=-3} \Rightarrow \cases{B+2C=-3\\ B-C=3\\ A=3} \Rightarrow \bbox[red, 2pt]{\cases{A=3 \\B=1\\ C=-2}} \\ \mathbf{(b)}\; \int f(x)\,dx = \int {3\over x}+{1\over 2x-1}-{2\over x+1} \,dx = 3\ln |x| +{1\over 2}\ln |2x-1|-2\ln|x+1| +C\\ =\bbox[red, 2pt]{\ln\left|{x^3\sqrt{2x-1} \over (x+1)^2} \right| +C}$$

解答： $$\mathbf{(a)}\;-|f(x)| \le f(x)\le |f(x)| \Rightarrow \int_a^b -|f(x)|\,dx \le \int_a^b f(x)\,dx \le \int_a^b|f(x)|\,dx \\ \Rightarrow \left| \int_a^b f(x)\,dx \right| \le \int_a^b|f(x)|\,dx，\bbox[red,2pt]{故得證} \\ \mathbf{(b)}\; \left| \int_a^b f(x)\cos x\,dx \right| \le \int_a^b |f(x)\cos x|\,dx \le \int_a^b |f(x)||\cos x|\,dx  \le \int_a^b |f(x)|\cdot 1\,dx \\ \Rightarrow \left| \int_a^b f(x)\cos x\,dx \right| \le  \int_a^b |f(x)|\,dx，\bbox[red,2pt]{故得證}$$

解答： $$\lim_{x\to 1^+} \left({1\over \ln x}-{1\over x-1}\right) =\lim_{x\to 1^+}  {x-1-\ln x\over \ln x(x-1)}  =\lim_{x\to 1^+}  {(x-1-\ln x)'\over (\ln x(x-1))'}  \\=\lim_{x\to 1^+}  { 1-1/x\over \ln x+1-1/x}  = \lim_{x\to 1^+}  {( 1-1/x)' \over (\ln x+1-1/x)'} =\lim_{x\to 1^+}  { 1/x^2\over 1/x+ 1/x^2} =\bbox[red, 2pt]{1\over 2}$$

解答： $$P\in \Gamma: y^2=2x \Rightarrow P(t^2/2,t),t\in \mathbb{R} \Rightarrow \overline{PQ} =\sqrt{(t^2/2-1)^2 +(t-4)^2},其中Q(1,4)\\ 令f(t)=(t^2/2-1)^2 +(t-4)^2 ={1\over 4}t^4-8t+17，因此f'(t)=0 \Rightarrow t^3-8=0 \Rightarrow t=2\\ \Rightarrow 當t=2時，f(t)有最小值，此時P=(2^2/2,2)=\bbox[red, 2pt]{(2,2)}$$

解答： $$\pi\int_0^1 \left(x^2-x^4\right)\,dx =\pi \left.\left[{1\over 3}x^3- {1\over 5}x^5\right]\right|_0^1 = \bbox[red, 2pt]{{2\over 15}\pi}$$

解答：

$$y=\sqrt{4-x^2} \Rightarrow {dy\over dx}={-x\over \sqrt{4-x^2}} \Rightarrow dS =\sqrt{1+({dy\over dx})^2} ={2 \over \sqrt{ 4-x^2}} \\ \Rightarrow 繞x軸旋轉表面積=\int 2\pi \sqrt{4-x^2}\,dS = \int_{-1}^1 4\pi\,dx=\bbox[red, 2pt]{8\pi}$$

解答： $$\cases{P(1,3,2)\\ Q(3,-1,6) \\ R(5,2,0)}  \Rightarrow \cases{\vec u =\overrightarrow{PQ} =(2,-4,4)\\ \vec v= \overrightarrow{PR} =(4,-1,-2)} \Rightarrow \vec n= \vec u\times \vec v=(12,20,14)=2(6,10,7) \\ \Rightarrow 欲求之平面為過P且法向量為\vec n，即6(x-1)+10(y-3)+ 7(z-2)=0 \Rightarrow \bbox[red, 2pt]{6x+10y+7z = 50}$$

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