114年政大科技管理碩士班-微積分詳解

國立政治大學114學年度碩士班暨碩士在職專班招生考試試題

考試科目 微積分 系所別 科技管理與智慧財產研究所 

解答：$$\textbf{(1) }\int \sin(4x) \cos(3x)\,dx ={1\over 2}\int \left( \sin(7x)+\sin x \right)\,dx ={1\over 2} \left( -{1\over 7}\cos(7x)-\cos x \right)+C\\ \qquad = \bbox[red, 2pt]{-{1\over 14}\cos(7x)-{1\over 2}\cos x+C} \\\textbf{(2) }\lim_{x\to 0} {\sin(a+4x)-2\sin(a+2x)+ \sin a\over x^2} =\lim_{x\to 0} {(\sin(a+4x)-2\sin(a+2x)+ \sin a)' \over (x^2)'} \\\qquad =\lim_{x\to 0} {4\cos(a+4x)-4\cos(a+2x) \over 2x} = \lim_{x\to 0} {(4\cos(a+4x)-4\cos(a+2x))' \over (2x)'} \\\qquad = \lim_{x\to 0} {-16\sin(a+4x)+8\sin(a+2x) \over 2} =-8\sin a+4\sin a=\bbox[red, 2pt]{-4\sin a}$$

解答：$$\cases{u= \cos^{-1} x\\ dv=dx} \Rightarrow \cases{du =-dx/\sqrt{1-x^2} \\v=x} \Rightarrow \int_0^1 \cos^{-1}x\,dx = \left. \left[ x\cos^{-1}x \right] \right|_0^1+\int_0^1 {x\over \sqrt{1-x^2}}\,dx \\= 0+\int_0^1 {x\over \sqrt{1-x^2}}\,dx =\int_1^0 {-1/2\over \sqrt u}du = \left. \left[ -\sqrt u \right] \right|_1^0 =\bbox[red, 2pt]1$$

解答：$$f(x)= \int_0^x {e^{t^2}-1\over t^2}dt \Rightarrow f'(x)={e^{x^2}-1\over x^2} ={x^2+{1\over 2}x^4+{1\over 6}x^6+\cdots \over x^2} =1+{1\over 2}x^2+{1\over 6}x^4+\cdots\\ \Rightarrow f''(x)=x+{2\over 3}x^3+\cdots \Rightarrow f^{(3)}(x)=1+2x^2+\cdots \Rightarrow f^{(3)}(0)=\bbox[red, 2pt]1$$

解答：$$y^4 e^{2x}+{dy\over dx}=0 \Rightarrow {dy\over dx}=-y^4e^{2x} \Rightarrow \int {1\over y^4}\,dy = \int -e^{2x}\,dx \Rightarrow -{1\over 3y^3} =-{1\over 2}e^{2x}+C_1 \\ \Rightarrow {1\over y^3}={3\over 2}e^{2x}+C_2 ={3e^{2x}+C_3\over 2} \Rightarrow \bbox[red, 2pt]{y=\sqrt[3]{2\over 3e^{2x}+C_3}}$$

解答：$$f(x)=4+ \int_1^{x^2} \sec(t-1)\,dt \Rightarrow f(-1)=4 \Rightarrow f'(x)=2x \sec(x^2-1) \Rightarrow f'(-1)=-2 \\ \Rightarrow 切線方程式: y=-2(x+1)+4 \Rightarrow \bbox[red, 2pt]{2x+y=2}$$

解答：$$\textbf{(1) } \int f(x)\,dx=1 \Rightarrow \int_0^1 c(1-x)\,dx =1 \Rightarrow \left. \left[ cx-{1\over 2}cx^2 \right] \right|_0^1= {1\over 2}c=1 \Rightarrow c= \bbox[red, 2pt]2 \\\textbf{(2) }0\lt x\lt 1\Rightarrow 0\lt x^2\lt 1 \Rightarrow 0\lt y\lt 1\\ y=x^2 \Rightarrow x=\sqrt y \Rightarrow {dx\over dy} ={1\over 2\sqrt y} \Rightarrow f_Y(y)=f_X(x) \left| {dx\over dy}\right| =2(1-\sqrt y)\cdot {1\over 2\sqrt y} ={1\over \sqrt y}-1 \\ \qquad \Rightarrow \bbox[red, 2pt]{f_Y(y) =\begin{cases} \displaystyle {1\over \sqrt y}-1,& 0\lt y\lt 1\\ 0, &其他\end{cases}} \\\textbf{(3) }四千元={4000\over 100000} =0.04十萬元 \Rightarrow P(Y\lt 0.04) =\int_0^{0.04} \left( {1\over \sqrt y}-1 \right)\,dy \\\qquad = \left. \left[ 2\sqrt y-y \right] \right|_0^{0.04} =0.4-0.04= \bbox[red, 2pt]{0.36}$$

解答：

$$\cases{x=r\cos \theta \\ y= r\sin \theta} \Rightarrow r^4=a^2(r^2\cos^2\theta-r^2\sin ^2\theta) =a^2r^2 \cos 2\theta \Rightarrow r^2=a^2\cos 2\theta\gt 0 \\ \Rightarrow \cos 2\theta \gt 0 \Rightarrow -{\pi\over 2}\lt  2\theta \lt {\pi\over 2} \Rightarrow -{\pi\over 4} \lt \theta\lt {\pi\over 4} \Rightarrow A=\int_0^{\pi/4} {1\over 2}r^2\,d\theta \\ \Rightarrow 欲求之面積=4A=\int_0^{\pi/4} 2r^2\,d\theta =2a^2 \int_0^{\pi/4} \cos 2\theta \,d\theta =2a^2 \left. \left[ {1\over 2}\sin 2\theta \right] \right|_0^{\pi/4}= \bbox[red, 2pt]{a^2}$$

解答：$$u=e^{2x} \Rightarrow du=2e^{2x}\,dx \Rightarrow \int{2\over e^{2x}+e^{-2x}} \,dx =\int{2e^{2x}\over e^{4x}+1} \,dx \\= \int{du\over 1+u^2} =\tan^{-1}u+C = \bbox[red, 2pt]{\tan^{-1}e^{2x}+C}$$

解答：$$100x+200y=80000 \Rightarrow x+2y=800 \Rightarrow \cases{f(x,y) =200x^{3/4} y^{1/4} \\ g(x,y)=x+2y-800} \Rightarrow \cases{f_x=\lambda g_x\\ f_y=\lambda g_y\\ g=0} \\ \Rightarrow \cases{150x^{-1/4}y^{1/4}=\lambda \cdots(1)\\ 50x^{3/4}y^{-3/4}=2\lambda \cdots(2)\\ x+2y=800 \cdots(3)} \Rightarrow {(1) \over (2)} ={3y\over x}={1\over 2} \Rightarrow x=6y代入(3) \Rightarrow 8y=800 \Rightarrow y=100 \\ \Rightarrow x=600 \Rightarrow f(600,100) = \bbox[red, 2pt]{20000\cdot 6^{3/4}}$$

解答：$$\int_0^{\sqrt{\pi/2}} \int_x^{\sqrt{\pi/2}} \int_1^5 \sin y^2\,dz\,dy\,dx =\int_0^{\sqrt{\pi/2}} \int_0^{y} \int_1^5 \sin y^2\,dz\,dx\,dy =\int_0^{\sqrt{\pi /2}} \int_0^{y} 4 \sin y^2\, dx\,dy \\= \int_0^{\sqrt{\pi/2}}   4y \sin y^2 \,dy = \left. \left[  -2\cos y^2 \right] \right|_0^{\sqrt{\pi/2}} = \bbox[red, 2pt]2$$

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解題僅供參考，碩士班歷年試題及詳解