國立臺北大學111學年度日間學士班轉學生招生考試
學制系級：統計學系日間學士班二年級
科目：微積分
一、單選題
解答：$$f(x,y)=x^3-y^3+33xy \Rightarrow \cases{f_x = 3x^2+33y\\ f_y=-3y^2+33x} \Rightarrow \cases{f_{xx}= 6x\\ f_{xy}= 33\\ f_{yy}= -6y} \\ \Rightarrow D(x,y)=f_{xx}f_{yy}-f_{xy}^2 =-36xy-33^2\\ 若\cases{f_x=0 \\ f_y=0} \Rightarrow \cases{x^2+11y=0\\ -y^2+11x =0} \Rightarrow (x,y)=(0,0),(11,-11) \Rightarrow \cases{D(0,0)=-33^2\lt 0\\ D(11,-11)= 3267 \gt 0}\\ \Rightarrow f_{xx}(11,-11)= 66 \gt 0 \Rightarrow \cases{(0,0)為\text{saddle point}\\ f(11,-11) 為相對極小值}，故選\bbox[red, 2pt]{(C)}$$
解答：$$f(x)={1\over \sqrt{3\pi}} e^{-x^2/2} \Rightarrow f'(x)=-{x\over \sqrt{3\pi}} e^{-x^2/2} \Rightarrow f''(x)=-{1\over \sqrt{3\pi}} e^{-x^2/2} +{x^2\over \sqrt{3\pi}} e^{-x^2/2}\\ 因此若f'(x)=0 \Rightarrow x=0 \Rightarrow f''(0)=-{1\over \sqrt{3\pi}} \lt 0 \Rightarrow f(0)={1\over \sqrt{3\pi}} \approx 0.3為極大值，故選\bbox[red, 2pt]{(D)}$$
解答：$$假設有x人，需支付f(x)= x(65-(x-34)= -x^2+99x \Rightarrow f'(x)=-2x+99=0 \\\Rightarrow x=49或50(x\in \mathbb N)有極大值 \Rightarrow \cases{f(34)= 2210元\\ f(49)=f(50)=2450元}，故選\bbox[red, 2pt]{(B)}$$
解答：$$\lim_{x\to 3}\sqrt{2x-6} =\lim_{x\to 3^+} \sqrt{2x-6}=0，故選\bbox[red, 2pt]{(B)}$$
解答：$$H(t)=-16t^2+bt+c \Rightarrow H(0)=240 \Rightarrow c=240 \Rightarrow H(t)=-16t^2+bt+240\\ 速度V(t)=H'(t)=-32t+b \Rightarrow V(0)=32 \Rightarrow b=32 \Rightarrow H(t)=-16t^2+32t+240\\ 最高點發生在V(t)=0 \Rightarrow -32t+32=0 \Rightarrow t=1，故選\bbox[red, 2pt]{(C)}$$
解答：$$9ye^x=2通過(0,2/9)只有(B),(C)符合此條件；又y={2\over 9}e^{-x} \Rightarrow y'=-{2\over 9}e^{-x}\lt 0 \Rightarrow 圖形遞減\\，故選\bbox[red, 2pt]{(C)}$$
二、計算題
解答：$$改變積分順序：\int_0^1 \int_{x^2}^1 xe^{y^2}\,dydx = \int_0^1 \int_0^{\sqrt y}xe^{y^2}\,dxdy = \int_0^1 {1\over 2} ye^{y^2}\,dy =\left.\left[ {1\over 4}e^{y^2} \right]\right|_0^1 =\bbox[red, 2pt]{{1\over 4}(e-1)}$$
解答：$$f(x,y)=2x^2+4y^2 \Rightarrow \cases{f_x=4x\\ f_y=8y} \Rightarrow L(x,y)=f(a,b)+f_x(a,b)(x-a)+ f_y(a,b)(y-b)\\ \Rightarrow L(1.2,2.1)=f(1,2)+4\cdot 1(1.2-1)+8\cdot 2(2.1-2) = 18+0.8+1.6 = \bbox[red, 2pt]{20.4}$$
解答：$$取\cases{u=\ln x\\ dv=2x^3\,dx} \Rightarrow \cases{du=dx/x\\ v={1\over 2}x^4} \Rightarrow \int 2x^3\ln x\,dx = {1\over 2}x^4\ln x-{1\over 2}\int x^3\,dx \\ = \bbox[red,2pt]{{1\over 2}x^4\ln x-{1\over 8}x^4+C}$$
解答：$$\int_0^1 (\sqrt x)^2 \pi\,dx-\int_0^1 (x^2)^2\pi\,dx= \pi \int_0^1 x-x^4\,dx= \pi \left.\left[ {1\over 2}x^2-{1\over 5}x^5 \right]\right|_0^1 =\bbox[red,2pt]{{3\over 10}\pi}$$
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解答：$$\int_0^1 (\sqrt x)^2 \pi\,dx-\int_0^1 (x^2)^2\pi\,dx= \pi \int_0^1 x-x^4\,dx= \pi \left.\left[ {1\over 2}x^2-{1\over 5}x^5 \right]\right|_0^1 =\bbox[red,2pt]{{3\over 10}\pi}$$
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