115年中山大學應數碩士班-基礎數學詳解

國立中山大學 115學年度碩士班考試入學招生考試試題

科目名稱:基礎數學【應數系碩士班甲組】

解答:$$\lim_{x\to 0^+} {\int_0^x \sin(t^2)\,dt\over x\sin (x^2)} =\lim_{x\to 0^+} {{d\over dx}\int_0^x \sin(t^2)\,dt\over {d\over dx}x\sin (x^2)} =  \lim_{x\to 0^+} {  \sin(x^2) \over \sin (x^2)+2x^2 \cos(x^2)} \\ =  \lim_{x\to 0^+} {1 \over 1+2\cos(x^2)\cdot {x^2\over \sin(x^2)}} ={1\over 1+2} =\bbox[red, 2pt]{1\over 3} \quad \left( \lim_{u\to 0}{u\over \sin u} =1 \right)$$

解答:$$\textbf{(a) }u=x+3 \Rightarrow \int {1\over (x+3) \sqrt{x^2+6x-7}} dx = \int {1\over u\sqrt{u^2-16}}\,du ={1\over 4}\sec^{-1}{|u|\over 4}+C \\\qquad = \bbox[red, 2pt]{{1\over 4}\sec^{-1}{|x+3|\over 4}+C}\\ \textbf{(b) }\int_0^4 \int_{\sqrt x}^2 \sin {x\over y} \,dydx = \int_0^2 \int_0^{y^2} \sin {x\over y}\,dx\,dy =\int_0^2 \left. \left[ -y \cos{x\over y}  \right] \right|_0^{y^2} \,dy  =\int_0^2  (y-y\cos y)\,dy \\\qquad = \left. \left[ {1\over 2}y^2 -y\sin y-\cos y\right] \right|_0^2= \bbox[red, 2pt]{3-2\sin 2-\cos 2}$$

解答:$$f(x)=\cos x \Rightarrow f'(x)=-\sin x \Rightarrow f''(x)=-\cos x \Rightarrow f'''(x)=\sin x=f(x) \\ \Rightarrow \cases{f(0)=1\\f'(0)=0\\ f''(0)= -1\\ f'''(0)=1} \Rightarrow f(x)=1-{x^2\over 2!}+{x^4\over 4!}-{x^6\over 6!}+\cdots \Rightarrow \bbox[red, 2pt]{f(x)= \sum_{n=0}^\infty {(-1)^n\over (2n)!}x^{2n}} \\ g(x)=\cos(3x^2)=f(3x^2)= \sum_{n=0}^\infty {(-1)^n\over (2n)!}(3x^2)^{2n} \Rightarrow \bbox[red, 2pt]{g(x) = \sum_{n=0}^\infty {(-9)^n\over (2n)!}x^{4n}} \\ \Rightarrow \text{Coefficient of }x^{12} ={g^{(12)}(0)\over 12!} ={(-9)^3\over 6!}=-{729\over 720} \Rightarrow g^{(12)}(0)=\bbox[red, 2pt]{(-729)\cdot {12!\over 6!}}$$

解答:

$$\text{Let }f(x,y)={x+y\over x^2+y^2}, \text{ then } \\\qquad I= \int_{1}^{\sqrt{2}} \int_{0}^{\sqrt{2-y^{2}}}\frac{x+y}{x^{2}+y^{2}}dxdy + \int_{0}^{1}\int_{1-y }^1 {x+y \over x^{2}+y^{2}}dxdy + \int_{1}^{\sqrt{2}} \int_{0}^{\sqrt{2-x^{2}}}\frac{x+y}{x^{2}+y^{2}}dydx \\ \qquad = \iint_{D_1} f\,d A+ \iint_{D_2} f\,dA+ \iint_{D_3} f\,dA= \iint_{D_1\cup D_2\cup D_3\cup D_4} f\,dA- \iint_{D_4}f\,dA\\ \text{ where }\cases{ D_1=\{1\le y\le \sqrt 2 \text{ and }0\le x\le \sqrt{2-y^2}\} \\D_2=\{ 0\le y\le 1 \text{ and }1-y\le x\le 1\} \\D_3 =\{1\le x\le \sqrt 2 \text{ and }0\le y\le \sqrt{2-x^2}\} \\D_4=\{x,y\ge 0\text{ and }x+y\le 1\}} \text{ and }\\D_1\cup D_2\cup D_3\cup D_4 =D\text{ represents the entire quarter circle in the first quadrant} \\ \cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \iint_D f\,dA= \int_0^{\pi/2} \int_0^\sqrt 2 (\cos \theta+ \sin \theta)\,dr\,d\theta = \sqrt 2\int_0^{\pi/2}  (\cos \theta+ \sin \theta)\,dr \,d\theta \\ = \sqrt 2 \left. \left[ \sin \theta-\cos \theta \right] \right|_0^{\pi/2} =2\sqrt 2 \\ \iint_{D_4} f\,dA= \int_0^{\pi/2} \int_0^{1/(\cos \theta+\sin \theta)} (\cos\theta+ \sin \theta)\,dr\,d\theta = \int_0^{\pi/2} 1\,d\theta={\pi\over 2} \\ \Rightarrow I= \bbox[red, 2pt]{2\sqrt 2-{\pi\over 2}}$$

解答:$$\textbf{(a) }T(x^3) =3x^2 = \begin{bmatrix}0\\3\\0\\0 \end{bmatrix}, T(x^2)=2x = \begin{bmatrix}0\\0\\2\\0 \end{bmatrix} , T(x)=1 = \begin{bmatrix}0\\0\\0\\1 \end{bmatrix}, T(1)=0 = \begin{bmatrix}0\\0\\0\\0 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{A= \begin{bmatrix}0&0& 0& 0\\ 3&0&0&0\\0&2& 0 & 0\\ 0&0&1&0 \end{bmatrix}} \\\textbf{(b) }\begin{bmatrix}0&0& 0& 0\\ 3&0&0&0\\0&2& 0 & 0\\ 0&0&1&0 \end{bmatrix} \begin{bmatrix}3\\-2\\5\\10 \end{bmatrix} = \begin{bmatrix}0\\9\\ -4\\ 5\end{bmatrix} \Rightarrow T(3x^3-2x^2+5x-10) =\bbox[red, 2pt]{9x^2-4x+5}$$

解答:$$\textbf{(a) } \cases{\lambda_1=0\\ x_1=(1,0,0)^T} \Rightarrow Ax_1=0\cdot x_1=0 \Rightarrow \bbox[red, 2pt]{\text{Basis for }N(A)=\{(1,0,0)^T\}} \\\Rightarrow \text{Basis for }C(A) =\{x_2,x_3\} \Rightarrow  \bbox[red, 2pt]{\text{Basis for }C(A) =\{(0,1,2)^T, (0,1,1)^T\}} \\\textbf{(b) } x=c_1x_1+ c_2x_2+c_3x_4 \Rightarrow A(x)=A(c_1x_1+ c_2x_2 +c_3x_4) =x_2-3x_3 \\ \Rightarrow c_1A(x_1)+ c_2A(x_2)+c_3 A(x_3) =0+c_2x_2+ 2c_3x_3=x_2-3x_3 \Rightarrow \cases{c_2=1\\ c_3=-3/2} \\ \Rightarrow x=c_1x_1+x_2-{3\over 2}x_3 =c_1 \begin{bmatrix}1\\0\\0 \end{bmatrix}+ \begin{bmatrix}0\\1\\2 \end{bmatrix}-{3\over 2} \begin{bmatrix}0\\1\\1 \end{bmatrix} =  \bbox[red, 2pt]{\begin{bmatrix}c_1\\-1/2\\1/2 \end{bmatrix}, c_1\in \mathbb R}$$

解答:$$\textbf{(a) }\cases{f(x)=\displaystyle {1\over x^b(1+x^3)} \\ g(x)=\displaystyle {1\over x^{b+3}}} \Rightarrow L= \lim_{x \to \infty}{f(x) \over g(x)} = \lim_{x \to \infty}{x^{b+3} \over x^b(1+x^3)} = \lim_{x \to \infty}{x^{3} \over 1+x^3} =1\\\quad \text{By the Limit Comparison Test, }\int_1^\infty f(x)\,dx \text{ converges iff }\int_1^\infty g(x)\,dx \text{ converges}\\ \quad \int_1^\infty {1\over x^{b+3}}\,dx \text{ converges iff }b+3\gt 1 \Rightarrow \bbox[red, 2pt]{b\gt -2} \\\textbf{(b) }\cases{f(x)=\displaystyle {e^{-ax}\over x^b(1+x^3)} \\g(x)=\displaystyle {1\over x^2}} \Rightarrow  L= \lim_{x \to \infty}{f(x) \over g(x)} = \lim_{x \to \infty} {x^2 e^{-ax} \over x^b(1+x^3)} = \lim_{x \to \infty} {x^{-b-1}  \over (1/x^3+1)e^{ax}} =0\\\text{By the Limit Comparison Test, since} \int_1^\infty g(x)\,dx \text{ converges, then } \int_1^\infty f(x) \text{ converges} \\ \quad \Rightarrow \text{The integral converges for all real values of }b, i.e., \bbox[red, 2pt]{b\in \mathbb R}$$

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