115年台聯大碩士班-工程數學B詳解

 台灣聯合大學系統 115 學年度碩士班招生考試

類組: 電機類 科目: 工程數學 B
多選題, 每題 5 分, 每一選項 (ABCDE) 單獨計分, 每一選項個別分數為1 分, 答錯不倒扣。

解答:$$X, Y 為離散隨機變數,找出所有可能的整數座標點 (x, y) 並計算機率總和必須為 1\\ \cases{y=0 \Rightarrow P_{X,Y}(x,0)=0\\ y=1 \Rightarrow x=-3,-2,-1,0 \Rightarrow \sum P_{X,Y}=-6/a\\ y=2 \Rightarrow x=-4,-3,-2,-1,0 \Rightarrow \sum P_{X,Y}=-20/a} \Rightarrow -{6\over a}-{20\over a}=1 \Rightarrow a=-26 \\ (A)\times: a=-26\ne -9\\ (B)\times: P(Y=1)=-{6\over a}={3\over 13} \ne {1\over 3} \\(C)\times: E[Y]=1\cdot P(Y=1)+2\cdot P(Y=2)= {3\over 13}+2\cdot {10\over 13}={23\over 13} \ne{5\over 3} \\(D)\times: E[Y^2]=1\cdot {3\over 13}+2^2\cdot {10\over 13}={43\over 13} \Rightarrow Var(Y)= {43\over 13}- \left( {23\over 13} \right)^2={30\over 169} \ne {2\over 9}\\，故選\bbox[red, 2pt]{(ABCD)}$$

解答:$$X\sim U(1,1/2) \Rightarrow f_X(x)=2, 又Y\mid X \sim U[0,2X] \Rightarrow f_{Y\mid X}(y\mid x)={1\over 2x} \\(A)\bigcirc: f_{Y\mid X}(y\mid x)={1\over 2x} \\(B)\times: E[Y\mid X]= {2x\over 2} =x\ne 2x\\ (C)\bigcirc:  Y 的範圍受 X 影響 \Rightarrow X,Y是相依的\\ (D)\times: f_Y(y) = \int_{y/2}^{1/2} f_{Y\mid X}(y\mid x) \cdot f_X(x)\,dx  = \int_{y/2}^{1/2} {1\over x}\,dx =\ln{1\over 2}-\ln {y\over 2} =-\ln y \ne {1\over 6}\ln{2\over y}\\，故選\bbox[red, 2pt]{(BD)}$$

解答:$$\int f_X(x)\,dx =1 \Rightarrow \int_{-\infty}^{-1} ce^x \,dx +\int_1^\infty ce^{-x}\,dx =2ce^{-1}=1 \Rightarrow c={e\over 2} \\ (A)\bigcirc: c={e\over 2} \\(B) \bigcirc: E[Y] =E[X^2]= 2\int_1^\infty x^2 \cdot {e\over 2}e^{-x}\,dx =5 \\(C)\bigcirc: f_X為偶函數\Rightarrow E[X]=0 \Rightarrow Var(X)= E[X^2]=5\\ (D)\times: |x|\ge 1 \Rightarrow y\ge 1 \not \ge 0\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$三角形三頂點\cases{A(-2,0) \\B(2,0) \\C(0,1)} \Rightarrow \cases{L_1= \overleftrightarrow{AC}: 2y=x+2\\ L_2=\overleftrightarrow{BC} : 2y=-x+2} \Rightarrow \cases{L_1\cap (y=2/3) =(-1/2,2/3) \\ L_2\cap (y=2/3)=(1/2,2/3)} \\ \Rightarrow 所求區間 [-1, 1] 完全涵蓋了 [-1/2, 1/2]，故機率為 1，故選\bbox[red, 2pt]{(A)}$$

解答:$$(A)\bigcirc: f_{X,Y}(x,y)= 12x(1-x)\cdot y 可分離\Rightarrow 兩者獨立\\(B) \bigcirc: f_Y(y)= \int_0^1 12xy(1-x)\,dx = \int_0^1 \left( 12xy-12x^2y \right)\,dx = \left. \left[ 6x^2y-4x^3y \right] \right|_0^1=2y \\(C) \bigcirc: X,Y獨立\Rightarrow  P \left( Y\le{1\over 4}\mid X\gt {1\over 4} \right) = P \left( Y\le{1\over 4}  \right) = \int_0^{1/4} 2y\,dy= {1\over 16} \\(D)\times:X,Y獨立\Rightarrow    E[Y\mid X]=E[Y]= \int_0^1 2y^2\,dy={2\over 3} \ne {1\over 3}\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$(A)\times: \int_0^1 \int_0^1 cxy\,dx\,dy= {c\over 4}=1 \Rightarrow c=4\ne {1\over 4} \\(B)\bigcirc: X,Y獨立\Rightarrow f_{X|Y}(x\mid 1/2)= f_X(x)= \int_0^1 4xy\,dy= \left. \left[ 2xy^2 \right] \right|_0^1=2x \\(C)\bigcirc: X,Y獨立\Rightarrow  E[Y\mid X\le 0.5]= E[Y]= \int_0^1 y\cdot 2y\,dy ={2\over 3} \\(D) \times: P(Y\ge X)= \int_0^1 \int_x^1 4xy\,dy\,dx \int_0^1 (2x-2x^3)\,dx= \left. \left[ x^2-{1\over 2}x^4 \right] \right|_0^1 ={1\over 2} \ne {1\over 3}\\，故選\bbox[red, 2pt]{(BC)}$$

解答:$$E[X\mid Y]= E[X] +{Cov(X,Y)\over Var(Y)} (Y-E[Y]) \Rightarrow 2=2 +{Cov(X,Y)\over 1} (Y-(-2)) \\ \Rightarrow Cov(X,Y)\cdot(Y+2)=0 \Rightarrow Cov(X,Y) =0\Rightarrow X,Y獨立\\(A) \times: X,Y獨立 \Rightarrow E[XY\mid X=1]=E[Y]=-2 \ne -4 \\(B) \bigcirc: W=X+Y \Rightarrow E[W]=E[X]+E[Y]=2-2=0 \Rightarrow Var(W)= Var(X+Y) \\\quad =Var(X)+Var(Y)+2 Cov(X,Y)=2+1+0=3 \Rightarrow E[W^2]=Var(W)+(E[W])^2=3\\ \qquad \Rightarrow E[(W-4)^2] =Var(W-4)+(E[W-4])^2 =3+(0-4)^2=19 \Rightarrow 19\ge 3 \\(C) \times: Cov(X+2Y, X-Y)=Cov(X,X)-Cov(X,Y)+2 Cov(Y,X)-2Cov(Y,Y) \\ \qquad =Var(X)-0+0-2Var(Y)=2-2\cdot 1=0 \Rightarrow 相關係數=0\ne {\sqrt 2\over 6} \\(D)\times: \cases{E[U]=E[X]-E[Y]=4\\ Var(U)=Var(X)+Var(-Y)= Var(X)+(-1)^2Var(Y) =2+1=3} \Rightarrow U\sim N(4,3) \\ \qquad \Rightarrow M(s)=e^{ \mu_s+ \sigma^2 s^2/2} \Rightarrow M_U(s)= e^{4s+3s^2/2} \Rightarrow M_U(2)= e^{8+6} =e^{14}\ne e^{12}\\，故選\bbox[red, 2pt]{(B)}$$

解答:$$Y= \begin{cases}0,& \text{if }{1\over X}\ge 3,\\1,& \text{if }{1\over X}\lt 3.  \end{cases} = \begin{cases} 0,& \text{if }0\lt X\le{1\over 3},\\1,& \text{if } {1\over 3}\lt X\le 1.  \end{cases} \Rightarrow \cases{P(Y=0) =1/3\\P(Y=1)=2/3} \\(A)\bigcirc: E[Y]=0\cdot {1\over 3}+1\cdot {2\over 3}={2\over 3} \\(B)\times: E[Y^2]=0^2\cdot {1\over 3}+1^2\cdot {2\over 3}={2\over 3} \Rightarrow Var(Y)=E[Y^2]-(E[Y])^2={2\over 3}-{4\over 9} ={2\over 9} \ne{2\over 3} \\(C) \bigcirc: E[X\mid Y=1] =X在(1/3,1]的期望值={1/3+1\over 2}={2\over 3} \\(D) \times: E[XY] =\int_{1/3}^1 x\cdot 1\,dx ={4\over 9} \ne {2\over 3}\\，故選\bbox[red, 2pt]{(AC)}$$

解答:$$誤差e =X-\hat X =X-(bY_1+cY_2) =X-b(X+N_1)-c(X+N_2) =(1-b-c)X- bN_1-cN_2\\ \Rightarrow 取f(b,c)=E[e^2] =E \left[ ((1-b-c)X- bN_1-cN_2)^2 \right] \\\qquad=(1-b-c)^2 E[X^2]+(-b)^2E[N_1^2] +(-c)^2 E[N_2^2] =2(1-b-c)^2+b^2+c^2 \\ \Rightarrow \cases{{\partial \over \partial b}f=0 \Rightarrow -4(1-b-c) +2b= 6b+4c-4=0 \Rightarrow 3b+2c=2 \\ {\partial \over \partial c}f=0 \Rightarrow 2b+3c=2} \Rightarrow b =c = {2\over 5}，故選\bbox[red, 2pt]{(D)}$$

解答:$$X_i \sim \text{Unifrom}[-1,1] \Rightarrow \cases{E[X_i]=0\\ Var(X_i)=(1-(-1))^2/12=1/3} \\(A)\bigcirc: \text{Assume }0\lt \epsilon \lt 1 \Rightarrow P(|Z_n| \gt \epsilon) = P(|X_n|^n \gt \epsilon) =P(|X_n| \gt \epsilon^{1/n}) =1-\epsilon^{1/n} \\ \quad \Rightarrow \lim_{n\to \infty } P(|Z_n|\gt \epsilon) = \lim_{n\to \infty} (1-\epsilon^{1/n})=1-1=0 \Rightarrow Z_n \text{ converges to 0} \\(B)\bigcirc: X_1+X_2=1 \Rightarrow X_2=1-X_1 \Rightarrow -1\le 1-X_1\le 1 \Rightarrow 0\le X_1\le 2 \Rightarrow 0\le X_1\le 1\\ \qquad \Rightarrow E[X_1\mid X_1+X_2]={1\over 2} \\(C)\times:   MGF \text{ for a single variable }X_1 \sim \text{Uniform}[-1,1] \text{ is }\\\qquad M_X(s)=E[e^{sX}] = \int_{-1}^1 {1\over 2}e^{sx}\,dx ={e^{s}-e^{-s} \over 2s} \\ \Rightarrow M_{S_n}(s)= (M_X(s))^n = \left( {e^s-e^{-s}\over 2s} \right)^n \Rightarrow M_{S_n}(1)= \left( {e-e^{-1}\over 2} \right)^n \ne \left( {e^2-e^{-2}\over 2} \right)^n \\(D)\bigcirc: \text{Central Limit Theorem: }S_n \approx N(n\mu, n\sigma^2) \Rightarrow S_{100} \approx N(0,100/3) \\ \Rightarrow P(S_{100}\le 100) = P \left( {S_{100}-0\over 10/\sqrt 3}\le {10-0\over 10/\sqrt 3} \right) = P(Z\le \sqrt 3) =F(\sqrt 3)\\，故選\bbox[red, 2pt]{(ABD)}$$

解答:$$A= \begin{bmatrix}1 & 2 & 4 & 1\\1 & 1 & 3 & 2\\2 & 3 & 7 & 3\\4 & 5 & 13 & 7 \end{bmatrix} \Rightarrow RREF(A) = \begin{bmatrix}1 & 0 & 2 & 3\\0 & 1 & 1 & -1\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0 \end{bmatrix} \Rightarrow \rank(A)=2\\(A)\bigcirc: rank(A)+null(A)=4 \Rightarrow 2+null(A)=4 \Rightarrow null(A)=2\\ (B)\times : \text{The dimension of the row space equals the rank, which is 2.} \\(C)\times: \text{The dimension of the column space equals the rank, which is 2.} \\(D)\bigcirc: x\in  \text{ null space of }BA \Rightarrow (BA)x=0 \Rightarrow B(Ax)=0 \Rightarrow B^{-1}B(Ax)=B^{-1}0\\ \qquad \Rightarrow Ax=0 \Rightarrow x \in \text{ null space of }A \Rightarrow null(BA)=null(A)=2\\，故選\bbox[red, 2pt]{(AD)}$$

解答:$$(A)\bigcirc: \dim(\mathbb R^3) = \dim(R_2[x]) =3\\ \qquad T(a_1,a_2,a_3)=0 \Rightarrow \cases{a_1=0\\a_1+a_2=0\\ a_1+a_2+a_3=0} \Rightarrow \cases{a_1=0\\a_2= 0 \\ a_3=0} \Rightarrow T \text{ is injective.} \\(B)\times: T(0,0,1) =0 \Rightarrow T \text{ is not injective} \\(C)\times: \dim(\mathbb R^3) =3 \ne \dim(\mathbb R^4)=4 \\(D)\times: T(1,-1,0)=0 \Rightarrow T \text{ is not injective}\\，故選\bbox[red, 2pt]{(A)}$$

解答:$$(A)\times: \cases{ A= \begin{bmatrix}1&0\\ 0& 1 \end{bmatrix} \Rightarrow \det(A)=1 \\[1ex] B= \begin{bmatrix}-1&0\\0& -1 \end{bmatrix} \Rightarrow \det(B)=1} \Rightarrow A+B = 0 \Rightarrow \det(A+B)=0 \ne 2=\det(A)+ \det(B) \\ (B)\bigcirc: \det(AA^T) =\det(A)\cdot \det(A^T)= \left[  \det(A) \right]^2 \ge 0 \\(C)\bigcirc: A^T=-A \Rightarrow \det(A^T) =\det(-A) \Rightarrow \det(A) =(-1)^5\det(A)=-\det(A) \Rightarrow 2\det(A)=0 \\\qquad \Rightarrow \det(A)=0 \\(D) \bigcirc: A= \begin{bmatrix}1&2& 4& 8\\1& 3& 9 & 27\\1& 5& 25& 125\\ 1& 7& 49& 343 \end{bmatrix} \xrightarrow{R_2- R_1 \to R_2,R_3-R_1 \to R_3, R_4-R_1\to R_4} \begin{bmatrix} 1 & 2 & 4 & 8\\0 & 1 & 5 & 19\\0 & 3 & 21 & 117\\0 & 5 & 45 & 335\end{bmatrix} \\ \qquad \Rightarrow \det(A)= \det \left( \begin{bmatrix}1 & 5 & 19\\3 & 21 & 117\\5 & 45 & 335 \end{bmatrix} \right) \xrightarrow{R_2-3R_1\to R_2,R_3-5R_1\to R-3} \det \left( \begin{bmatrix}1 & 5 & 19\\0 & 6 & 60\\0 & 20 & 240 \end{bmatrix} \right) \\ \qquad \Rightarrow \det(A) = \det \left( \begin{bmatrix}6&60\\20& 240 \end{bmatrix} \right)=1440-1200=240\\，故選\bbox[red, 2pt]{(BCD)}$$

解答:$$(A) \bigcirc: 2A^{-1}+I的特徵值:\cases{2\cdot 1^{-1}+1=3\\ 2\cdot 2^{-1}+1= 2 \\2\cdot 3^{-1}+1=5/3} \\ (B) \times: A+I的特徵值:\cases{ 1+1=2\\ 2+1=3\\3+1=4} \Rightarrow \det(A+I)= 2\cdot 3\cdot 4=24 \ne 6 \\(C)\bigcirc: \det(2AA^T) =2^3\cdot \det(A)\cdot \det(A^T) =8\cdot \det(A)\cdot \det(A) =8\cdot 6^2=288 \\ (D)\bigcirc: A^3的特徵值為1^2,2^3,3^3 三相異且不為零 \Rightarrow rank(A^3)=3\\，故選\bbox[red, 2pt]{(ACD)}$$

解答:$$(A)\times:  A(v_1+v_2)=Av_1+ Av_2=\lambda_1 v_1+ \lambda_2v_2 \ne \lambda(v_1+v_2), \text{ for }\lambda_1\ne \lambda_2 \\(B)\times: \text{A symmetric matrix }A= \begin{bmatrix} 1&0 \\0 &2\end{bmatrix} \text{ and }P = \begin{bmatrix}1& 1\\0& 1 \end{bmatrix}\\ \qquad  \Rightarrow  B = P^{-1}AP = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} \text{ is not NOT symmetric} \\(C) \times: \cases{Av= \lambda v\\ B= P^{-1}AP} \Rightarrow A=PBP^{-1} \Rightarrow Av= (PBP^{-1})v= \lambda v \Rightarrow B(P^{-1}v)= \lambda(P^{-1}v) \\\qquad \Rightarrow P^{-1}v \ne v, \text{ for all } P\ne I \Rightarrow A,B \text{ do not have the same eigenvectors.} \\(D)\times: D= \begin{bmatrix}1 & 0\\0& 1 \end{bmatrix} \text{ is real symmetric matrix, but its eigenvalues are }\lambda_1=\lambda_2=1, \\\qquad \text{ which are not distinct}\\，故選\bbox[red, 2pt]{(E)}$$

解答:$$(A)\times: \text{ $\text{FCF}_{16} + \text{FA3}_{16} = \text{1F72}_{16}$ } \Rightarrow \text{the sum of two 3-digit hexadecimal numbers results in a}\\\qquad \text{4-digit hexadecimal number. It is not closed under addition. The set is not a subspace.} \\(B)\times: \text{the set fails to be a subspace is the lack of closure under addition} \\(C)\times: \text{Multiplication is also defined in the usual way.} \\(D)\times: 0.6\times CA3 \text{ cannot yield a valid integer hexadecimal representation}\\，故選\bbox[red, 2pt]{(E)}$$

解答:$$(B)\bigcirc: A\ne cB, \text{ for all }c\ne 0\\ (C)\bigcirc: \dim(S) \text{ cannot be determined from the given information alone}\\，故選\bbox[red, 2pt]{(BC)}$$

解答:$$(A)\times : X=[X_1\; X_2\; X_3], \text{ where }\cases{X_1=(1/\sqrt 3, 1/\sqrt 3,1/\sqrt 3)^T\\ X_2=(1/\sqrt 6,-2/\sqrt 6, 1/\sqrt 6)^T\\ X_3=(-1/\sqrt 2,0, 1/\sqrt 2)^T} \Rightarrow \cases{ \Vert X_1\Vert =\Vert X_2\Vert =\Vert X_3\Vert=1 \\ X_1\cdot X_2= 0\\X_2\cdot X_3 =0\\X_3\cdot X_1=0} \\\qquad \Rightarrow X \text{ is  an orthonormal matrix. } \\(B)\times: v_1= \begin{bmatrix}-1/\sqrt 2\\ 0\\ 1/\sqrt 2 \end{bmatrix} \Rightarrow A^Tv_1= \begin{bmatrix}0\\0 \end{bmatrix} \Rightarrow v_1\in \text{null}(A^T) \\ \qquad v_2= \begin{bmatrix}1\\0\\-1 \end{bmatrix} \Rightarrow A^Tv_2= \begin{bmatrix}0\\0 \end{bmatrix} \Rightarrow v_2 \in \text{null}(A^T)\\ \qquad \text{That, null$(A^T)$ contains more than one element.} \\(C)\times:  \text{col$(Y)$ is spanned by $\begin{bmatrix}\sqrt 3\\0\\ 0 \end{bmatrix}$ and $\begin{bmatrix}2/\sqrt 3\\ 2/\sqrt 3\\0 \end{bmatrix}$}\\ \qquad \Rightarrow \text{Every vector in $\text{col}(Y)$ has a zero in its third component. This is different from col(A).}\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$\text{Let }E=X-(\theta_1A_1+ \theta_2A_2 \Rightarrow  f(\theta_1, \theta_2)= \Vert X-(\theta_1A_1+ \theta_2A_2) \Vert_F^2 = \Vert E\Vert_F^2 =tr (E^TE)\\ \text{By the principle of orthogonal projection,  matrix $E$ must be orthogonal to the basis matrices}\\ \text{ $A_1$ and $A_2$ under the Frobenius inner product $\langle A, B \rangle = \text{tr}(A^T B)$.} \\ \Rightarrow \cases{\langle E,A_1 \rangle =0 \\\langle E,A_2 \rangle =0} \Rightarrow \cases{tr(E^TA_1)=0 \\ tr(E^TA_2)=0} \Rightarrow \cases{\text{tr}((X - (\theta_1 A_1 + \theta_2 A_2))^T A_1) = 0 \\\text{tr}((X - (\theta_1 A_1 + \theta_2 A_2))^T A_2) = 0} \\ \cases{tr(E^TA_1)= tr((E^TA_1)^T) =tr(A_1^TE) = tr(A_1^T(X- (\theta_1A_1+\theta_2A_2))) =0\\ tr(E^TA_2) = tr((E^TA_2)^T) = tr(A_2^TE) = tr(A_2^T (X-(\theta_1A_1+ \theta_2A_2))) =0}\\，故選\bbox[red, 2pt]{(AD)}$$

解答:$$f(b_1)= \int_0^{2\pi} (b_1 \sin x-y(x))^2\,dx \\ f'(b_1)=0\Rightarrow f'(b_1)= \int_0^{2\pi} 2(b_1\sin x-y(x)) \sin x\,dx  =0 \Rightarrow \int_0^{2\pi} (b_1\sin x-y(x)) \sin x\,dx  =0  \\ \Rightarrow b_1\int_0^{2\pi} \sin^2 x\,dx - \int_0^{2\pi} y(x)\sin x\,dx =0 \Rightarrow b_1={\int_0^{2\pi} y(x)\sin x\,dx \over \int_0^{2\pi}  \sin^2 x\,dx}，故選\bbox[red, 2pt]{(B)}$$

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