115年清華生醫碩士班-微積分詳解

 國立清華大學115學年度碩士班考試入學

系所班組別:生命科學暨醫學院丙組
考試科目:微積分

解答:$$\textbf{(A) }L= \lim_{n \to \infty} \sum_{j=1}^n \frac{n}{(n+j)^2} \sin\left(2 - \frac{n}{n+j}\right) = \lim_{n \to \infty} \sum_{j=1}^n \frac{1}{n} \left[ \frac{1}{(1 + \frac{j}{n})^2} \sin\left(2 - \frac{1}{1 + \frac{j}{n}}\right) \right] \\=  \int_{0}^{1} \frac{1}{(1+x)^2} \sin\left(2 - \frac{1}{1+x}\right) dx \\ 取u=2-{1\over 1+x} \Rightarrow du ={1\over (1+x)^2}\,dx \Rightarrow L=  \int_{1}^{\frac{3}{2}} \sin(u) du = \left. \left[ -\cos (u) \right] \right|_1^{3/2}= \bbox[red, 2pt]{\cos 1-\cos {3\over 2}} \\\textbf{(B) }L= x^{1/x} \Rightarrow \ln L={\ln x\over x} \Rightarrow \lim_{x\to \infty} \ln L= \lim_{x\to \infty} {\ln x\over x} =\lim_{x\to \infty} {{d\over dx}\ln x\over {d\over dx}x} =\lim_{x\to \infty} {1\over x} =0 \\ \quad \Rightarrow \lim_{x\to \infty }L=e^0= \bbox[red, 2pt]1$$

解答:$$\textbf{(A) }舉反例,假設\cases{f(x)=x\\ g(x)=x+1} 滿足條件，但g(x)\gt f(x)違反f(x)\ge g(x),因此敘述\bbox[red, 2pt]{不正確} \\ \textbf{(B)}舉反例,假設g(x)= \sqrt[3]x \Rightarrow \lim_{x\to 0}{g(x)-g(0) \over x-0} =\lim_{x\to 0}{1\over x^{2/3}} =\infty \Rightarrow g(x)在x=0不可微分\\ 反函數g^{-1}(y)=y^3 \Rightarrow {d\over dy}g^{-1}(y) =3y^2 \Rightarrow {d\over dy}g^{-1}(0) =0 \Rightarrow g^{-1}可微,因此敘述\bbox[red, 2pt]{不正確}$$

解答:$$\textbf{(A) }u=a+b-x \Rightarrow du=-dx \Rightarrow  \int_a^b f(a+b-x)\,dx = \int_b^a f(u)(-du) =\int_a^b f(u)\,du \\ \quad = \int_a^b f(x)\,dx \Rightarrow \int_a^b f(a+b-x)\,dx=\int_a^b f(x)\,dx\quad \bbox[red, 2pt]{故得證} \\ \textbf{(B) } I=\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx =  \int_0^{\frac{\pi}{2}} \frac{\sin(\frac{\pi}{2} - x)}{\sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x)} dx   = \int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} dx \\ \quad \Rightarrow I+I=\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx+ \int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} dx \Rightarrow 2I= \int_0^{\frac{\pi}{2}} \frac{\sin x+\cos x}{\cos x + \sin x} dx \\=  \int_0^{\frac{\pi}{2}} 1\,dx ={\pi\over 2} \Rightarrow I= \bbox[red, 2pt]{\pi\over 4}$$

解答:$$\textbf{(A) } 細菌數量 P(t)=P_0\times 3^{t/2},其中P_0為原本數量, t的單位為天數\\ 依題意P(14)= 2843.1萬=P_0 \times 3^{14/2} =P_0\times 3^7=2187\cdot P_0 \Rightarrow P_0={2843.1 \over 2187}= \bbox[red, 2pt]{1.3萬} \\ \textbf{(B)} P(16)= P(14)\times 3=2843.1\times 3 = \bbox[red, 2pt]{8529.3萬}$$

解答:$$I=\int \tan^3 x\,dx = \int \tan x\cdot \tan^2 x\,dx = \int \tan x\cdot( \sec^2 x-1)\,dx = \int \tan x \sec^2 x dx - \int \tan x dx \\取u=\tan x \Rightarrow du=\sec^2 x\,dx \Rightarrow \int \tan x \sec^2 x dx =\int u\,du={1\over 2}u^2+C_1={1\over 2}\tan^2 x+C_1\\ \int \tan x\,dx =\int{\sin x\over \cos x}\,dx = -\ln |\cos x|+C_2 \\因此I= \bbox[red, 2pt]{{1\over 2}\tan^2 x+\ln|\cos x|+C}$$

解答:$$\textbf{(A) }取\cases{a_k={k\over (1+k^2)^3}\\ b_k={1\over k^5}} \Rightarrow  \lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{\frac{k}{(1+k^2)^3}}{\frac{1}{k^5}} = \lim_{k \to \infty} \frac{k^6}{(1+k^2)^3} =    \lim_{k \to \infty} \frac{1}{(1/k^2 + 1)^3} =  1 \\ 由於 \sum_{k= 1}^{\infty} \frac{1}{k^5} 收斂，依極限比較審斂法 \text{(Limit Comparison Test)}, 級數\sum_{k= 1}^{\infty}a_k\bbox[red, 2pt]{收斂} \\ \textbf{(B) }  a_n = \left(1 - \frac{1}{n+1}\right)^{n^2} \Rightarrow  \lim_{n \to \infty} \sqrt[n]{\vert{}a_n\vert{}} = \lim_{n \to \infty} \left( \left(1 - \frac{1}{n+1}\right)^{n^2} \right)^{\frac{1}{n}} = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)^n \\= \lim_{u \to \infty} \left(1 - \frac{1}{u}\right)^{u-1} = \lim_{u \to \infty} \left[ \left(1 + \frac{-1}{u}\right)^u \cdot \left(1 - \frac{1}{u}\right)^{-1} \right] =e^{-1}\cdot 1={1\over e} \lt 1\\ \quad 依據根值審斂法\text{(Root Test)},該級數絕對\bbox[red, 2pt]{收斂}$$

解答:$$ f(x,y) = x^3 - 3x + 3xy^2 \Rightarrow \cases{f_x= 3x^2-3+3y^2\\ f_y=6xy} \Rightarrow \cases{f_{xx} =6x\\ f_{xy}=6y\\ f_{yy}=6x}\\ \Rightarrow D(x,y)=f_{xx}f_{yy}-(f_{xy})^2 =36(x^2-y^2) \\ \cases{f_x=0\\ f_y=0} \Rightarrow (x,y)=(\pm 1,0), (0,\pm 1) \Rightarrow \cases{D(1,0)=36\gt 0 \Rightarrow f_{xx}(1,0)=6\gt 0\\ D(-1,0)=36\gt 0\Rightarrow f_{xx}(-1,0)=-6\lt 0 \\ D(0,1)= -36\lt 0\\ D(0,-1) =-36\lt 0} \\ \Rightarrow \bbox[red, 2pt]{\cases{相對極大值:f(-1,0)=2\\ 相對極小值:f(1,0)=-2\\ 鞍點:(0,1),(0,-1)}}$$

解答:$$ 由於 f(x,y,z) = x^2z + y^2z + z^3 \Rightarrow f(x,y,-z)= -(x^2z+y^2z+z^3) =-f(x,y,z)\\ 對變數z而言, f是一個奇函數, 因此最內層積分\int_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}} (x^2z + y^2z + z^3) dz= \int_{-c}^{c} f(x,y,z) dz  =0\\ \Rightarrow  \int_{-a}^{a} \int_{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}} \int_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}} (x^2z + y^2z + z^3) dz dx dy = \int_{-a}^{a} \int_{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}} 0 \, dx dy = \bbox[red, 2pt]0 $$

 

解答:$$\cases{P(x,y)=x^2\\ Q(x,y)=y^2} \Rightarrow P_y=0=Q_x \Rightarrow 存在位勢函數 f(x,y)= \int x^2\,dx=\int y^2\,dy \\ \Rightarrow f(x,y)={1\over 3}x^3+{1\over 3}y^3\Rightarrow  \int_C x^2 dx + y^2 dy = f(\text{終點}) - f(\text{起點}) = f(0,3) - f(3,0) =9-9= \bbox[red, 2pt]0$$

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