114年政大科管與智財碩士班-微積分詳解

國立政治大學114學年度碩士班招生考試 

考試科目：微積分
系所別：科技管理與智慧財產研究所

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

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

解答: $$e^x =1+x+{x^2\over 2}+ {x^3\over 6} +{x^4\over 24} +\cdots \Rightarrow e^{x^2} = 1+x^2+{x^4\over 2}+ {x^6\over 6}+ {x^8\over 24} +\cdots\\ f(x) =\int_0^x {e^{t^2}-1\over t^2}\,dt \Rightarrow f'(x)= {e^{x^2}-1\over x^2}  =1+{x^2\over 2}+{x^4\over 6} + {x^6\over 24}+ \cdots \\ \Rightarrow f''(x)= x+{2\over 3}x^3 +{1\over 4}x^5+\cdots \Rightarrow f'''(x)=1+2x^2 +{5\over 4}x^4 +\cdots \\ \Rightarrow f'''(0) = \bbox[red, 2pt]1$$

解答: $$y^4e^{2x}+{dy\over dx}=0 \Rightarrow \int{1\over y^4}\,dy =-\int e^{2x}\,dx \Rightarrow -{1\over 3y^3} =-{1\over 2}e^{2x}+c_1 \\ \Rightarrow {1\over y^3} ={3\over 2}e^{2x}+c_2 \Rightarrow \bbox[red, 2pt]{y= \sqrt[3]{1\over {3\over 2}e^{2x}+c_2}}$$

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

解答: $$\textbf{(1) } \int f(x)\,dx =1 \Rightarrow \int_0^1 c(1-x)\,dx ={1\over 2}c=1 \Rightarrow \bbox[red, 2pt]{c=2} \\\textbf{(2) }F_Y(y) =f(Y\le y) =f(X^2 \le y) =f(X\le \sqrt y) =\int_0^\sqrt{y} 2(1-x)\,dx = 2\sqrt y-y \\\quad \Rightarrow f_Y(y) = {d\over dy}(2\sqrt y-y) \Rightarrow \bbox[red, 2pt]{f_Y(y)={1\over \sqrt y}-1} \\\textbf{(3) }四千元=0.04十萬元 \Rightarrow P(y\le 0.04) =2\sqrt{0.04}-0.04 =0.4-0.04= \bbox[red, 2pt]{0.36}$$

解答: $$\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow (r^2)^2 =a^2(r^2\cos^2\theta -r^2\sin^2 \theta) \Rightarrow r^2 =a^2\cos(2\theta) \\ \Rightarrow {1\over 4}面積={1\over 2}\int_0^{\pi/4} a^2\cos(2\theta)\,d\theta ={1\over 4}a^2 \Rightarrow 面積= \bbox[red, 2pt]{a^2}$$

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

解答: $$總花費100x+200y\le 80000 \Rightarrow x+2y\le 800 \Rightarrow \cases{f(x,y)=200x^{3/4} y^{1/4} \\ g(x,y) =x+2y-800}\\ \text{Lagrange算子: }\cases{f_x= \lambda g_x\\ f_y=\lambda g_y \\ g=0} \Rightarrow \cases{150x^{-1/4}y^{1/4 } =\lambda \\ 50x^{3/4}y^{-3/4} =2\lambda \\ x+2y=800} \Rightarrow {150x^{-1/4}y^{1/4 }\over  50x^{3/4}y^{-3/4}  } ={\lambda\over 2\lambda} \Rightarrow {3y\over x}={1\over 2} \\ \Rightarrow x=6y \Rightarrow 6y+2y=8y=800 \Rightarrow y=100 \Rightarrow x=600 \Rightarrow f(600,100) =200\cdot600^{3/4}\cdot 100^{1/4} \\=200\cdot 6^{3/4}\cdot 100^{3/4}\cdot 100^{1/4} = \bbox[red, 2pt]{20000\cdot 6^{3/4}}$$

解答: $$\int_0^{\sqrt{\pi/2}} \int_x^{\sqrt{\pi/2}} \int_1^5 \sin(y^2)\,dzdydx = \int_0^{\sqrt{\pi/2}} \int_x^{\sqrt{\pi/2}} 4 \sin(y^2)\, dydx = \int_0^{\sqrt{\pi/2}} \int_0^{y} 4 \sin(y^2)\, dxdy \\=\int_0^{\sqrt{\pi/2}}4y\sin(y^2)\,dy =\int_0^{\pi/2} 2 \sin u\,du =\bbox[red, 2pt]2$$

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