115年台綜大轉學考-微積分B詳解

臺灣綜合大學系統 115 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:$$\textbf{(a) } \lim_{x \to 1} \frac{\log_2(2x - 1) - x^2}{x + 3} ={0-1\over 1+3} =\bbox[red, 2pt]{-{1\over 4}} \\ \textbf{(b) } \lim_{x \to 0} \frac{\log_3(2x + 1) - x^2}{\sin x} = \lim_{x \to 0} \frac{{d\over dx}(\log_3(2x + 1) - x^2)}{{d\over dx}\sin x} =  \lim_{x \to 0} \frac{ {2\over (2x+1)\ln 3}-2x}{\cos x} = \bbox[red, 2pt]{2\over \ln 3}$$

解答:$$\textbf{(a) } f(x) = \frac{x^2}{x-1}  \Rightarrow f'(x) = \frac{2x(x - 1) - x^2(1)}{(x - 1)^2} ={x(x-2) \over (x-1)^2} \\\qquad \Rightarrow \begin{cases} f'(x)\gt 0,& x\lt 0\\ f'(x)\lt 0, &0\lt x\lt 1\\ f'(x)\lt 0, & 1 \lt x\lt 2\\ f'(x)\gt 0, & x\gt 2. \end{cases} \Rightarrow \text{ $f$ is increasing on the intervals:} \bbox[red, 2pt]{(-\infty, 0)\cup(2, \infty)} \\ \textbf{(b) }f''(x)  = \frac{2(x - 1)(x - 1) - (2x^2 - 4x)}{(x - 1)^3} ={2\over (x-1)^3} \lt 0 \Rightarrow x\lt 1 \\\qquad \Rightarrow f \text{ is concave down on }\bbox[red, 2pt]{(-\infty, 1)}$$

解答:$$ \int 3x^3 + 3^x + \sin(3x) \,dx = \bbox[red, 2pt]{{3\over 4}x^4+{1\over \ln 3}3^x-{1\over 3}\cos(3x)+C}$$

解答:$$\cases{w=\ln u\\ dv=du/u^2} \Rightarrow \cases{dw=du/u\\v=-1/u} \Rightarrow \int_1^3{\ln u\over u^2}\,du = \left. \left[ -{\ln u\over u} \right] \right|_1^3 +\int_1^3 {1\over u^2}\,du = \left. \left[ -{\ln u\over u} \right] \right|_1^3 + \left. \left[ -{1\over u} \right] \right|_1^3 \\= \left. \left[ -{\ln u\over u}-{1\over u} \right] \right|_1^3 = \bbox[red, 2pt]{2-\ln3\over 3}$$

解答:$${d\over dx} (x^2 \cos(\pi y) + xy + 2 )= 2x\cos(\pi y)-\pi y'x^2\sin(\pi y)+y+xy'=0  \\\text{Substitute the point $(1, -1)$ to find the slope $y'$}: 2\cos(-\pi)-\pi y'\sin(-\pi)-1+y'=0 \\ \Rightarrow -2-0-1+y'=0 \Rightarrow y'=3 \Rightarrow \text{ equation of the tangent line: }y-(-1)=3(x-1) \\ \Rightarrow y+1=3x-3 \Rightarrow \bbox[red, 2pt]{y=3x-4}$$

解答:$$ \cos(u) = \sum_{n=0}^{\infty} \frac{(-1)^n u^{2n}}{(2n)!}  \Rightarrow f(x)= \cos(x^3) =  \sum_{n=0}^{\infty} \frac{(-1)^n x^{6n}}{(2n)!} \\ 6n=72 \Rightarrow n=12 \Rightarrow \text{ coefficient of }x^{72}={(-1)^{12} \over (2\cdot 12)!}={1\over 24!} ={f^{(72)}(0) \over 72!} \\ \Rightarrow f^{(72)}(0)= \bbox[red, 2pt]{72!\over 24!}$$

解答:$$ f(x,y) = xy - x^2y - xy^2 \Rightarrow \cases{f_x= y(1-2x-y) \\ f_y=x(1-x-2y)}  \Rightarrow \cases{f_{xx}=-2y\\ f_{xy}=1-2x-2y\\ f_{yy}=-2x} \\ \Rightarrow D= f_{xx}f_{yy} -(f_{xy})^2 =4xy-(1-2x-2y)^2\\ \cases{f_x=0\\ f_y=0} \Rightarrow \cases{\cases{y=0\\ 1-2x-y=0}\\ \cases{x=0\\1-x-2y=0}} \cases{\text{Case I: }\cases{x= 0\\y=0} \Rightarrow D(0,0)=-1\lt 0   \\ \text{Case II: }\cases{y=0\\ 1-x-2y=0} \Rightarrow \cases{x=1\\ y=0} \Rightarrow D(1,0)=-1\lt 0  \\ \text{Case III: }\cases{1-2x-y=0\\ x=0} \Rightarrow \cases{x=0\\ y=1} \Rightarrow D(0,1)=-1\lt 0 \\ \text{Case IV: }\cases{1-2x-y=0\\1-x-2y=0} \Rightarrow \cases{x=1/3\\ y=1/3} \Rightarrow D(1/3,1/3)=1/3\gt 0}  \\ \Rightarrow \text{The saddle points: }\bbox[red, 2pt]{(0,0), (1,0), (0,1)}$$

解答:$$\cases{f(x,y,z)=2x+y-3z\\ g(x,y,z)=x^2+y^2-z} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ f_z=\lambda g_z\\ g=0} \Rightarrow \cases{2=\lambda(2x) \\1=\lambda(2y) \\ -3=\lambda(-1) \\ x^2+y^2=z} \Rightarrow \cases{x=1/3\\ y=1/6\\ \lambda=3} \Rightarrow z={1\over 9}+{1\over 36}={5\over 36} \\ \Rightarrow f \left( {1\over 3},{1\over 6},{5\over 36} \right)={2\over 3}+{1\over 6}-{15\over 36}= \bbox[red, 2pt]{5\over 12}$$

解答:$$ \mathbf{r}(t) = t^2\mathbf{i} + 6t\mathbf{j} + 4t^{3/2}\mathbf{k}, \quad 0 \le t \le 1 \Rightarrow  \mathbf{r}'(t) = 2t\mathbf{i} + 6\mathbf{j} + 6\sqrt{t}\mathbf{k}  \\ \Rightarrow  \Vert{}\mathbf{r}'(t)\Vert{} = \sqrt{(2t)^2 + (6)^2 + (6\sqrt{t})^2}= 2\sqrt{t^2+9t+9} \\ \Rightarrow \text{the Unit Tangent Vector $\mathbf{T}(t)$}={r'(t)\over ||r'(t)||} = \bbox[red, 2pt]{\frac{2t \mathbf{i} + 6\mathbf{j} + 6\sqrt{t}\mathbf{k}}{2\sqrt{t^2 + 9t + 9}} } \\ \text{arc length }L=\int_a^b ||r'(t)|| \,dt  = \int_{0}^{1} 2\sqrt{t^2 + 9t + 9} \,dt = \int_{0}^{1} 2\sqrt{ \left( t+{9\over 2} \right)^2-{45\over 4}} \,dt  \\=  2 \int_{9/2}^{11/2} \sqrt{u^2 - \frac{45}{4}} \,du   = 2 \left[ \frac{u}{2}\sqrt{u^2 - \frac{45}{4}} - \frac{45}{8}\ln\left\vert{}u + \sqrt{u^2 - \frac{45}{4}}\right\vert{} \right]_{9/2}^{11/2}  \\  = \bbox[red, 2pt]{\frac{11\sqrt{19} - 27}{2} - \frac{45}{4}\ln\left(\frac{11 + 2\sqrt{19}}{15}\right) }$$

解答:$$ \int_{0}^{4} \int_{0}^{y} y^2 e^{xy} \,dx \,dy = \int_{0}^{4} (y e^{y^2} - y) \,dy = \left. \left[ {1\over 2}e^{y^2}-{1\over 2}y^2 \right] \right|_0^4 =\left( \frac{1}{2} e^{16} - \frac{1}{2} \right) - 8 \\= \frac{1}{2} e^{16} - \frac{17}{2} = \bbox[red, 2pt]{{1\over 2}(e^{16}-17)}$$

====================== END ==========================
解題僅供參考，其他轉學考試題及詳解