台灣聯合大學系統112學年度學士班轉學生考試
科目:微積分
類組別:A3/A4/A6
甲、填充題:共8題,每題8分
解答:L=(cos(2√x))2/x⇒lnL=2lncos(2√x)x⇒limx→0+lnL=limx→0+2lncos(2√x)x=limx→0+ddx(2lncos(2√x))ddxx=limx→0+−2sin(2√x)/√xcos(2√x)=limx→0+−2tan(2√x)√x=limx→0+ddx(−2tan(2√x))ddx√x=limx→0+−2sec2(2√x)/√x1/2√x=limx→0+−4sec2(2√x)=−4⇒limx→0+lnL=−4⇒limx→0+L=e−4=1e4
解答:g(x)=tan−1(x2)+∫x21sec(t−1)dt⇒g′(x)=2x1+x4+2xsec(x2−1)⇒{g(−1)=tan−1(1)+0=π/4g′(−1)=−1−2⋅1=−3⇒f′(x)=g′(x)eg(x)⇒f′(−1)=g′(−1)eg(−1)=−3eπ/4
解答:{x(t)=ln(sect+tant)−sinty(t)=cost⇒{x′(t)=sect−costy′(t)=−sint⇒(x′(t))2+(y′(t))2=sec2t−2+cos2+sin2t=sec2t−1 =tan2t⇒曲線長=∫π/30√(x′(t))2+(y′(t))2dt=∫π/30tantdt=[−ln(cost)]|π/30=−ln12=ln2
解答:∫e1(ln(x))2πdx=[π(x(lnx)2−2xln(x)+2x)]|e1=π(e−2)
解答:位置r(t)=(cos3t,sin3t,3t)⇒速度r′(t)=(−3sin3t,3cos3t,3)⇒{r(π/3)=(−1,0,π)r′(π/3)=(0,−3,3)又f(x,y,z)=xyz⇒∇f=(fx,fy,fz)=(yz,xz,xy)=(3tsin3t,3tcos3t,(sin6t)/2))⇒∇f(π/3)=(0,−π,0)⇒∇f(π/3)⋅r′(π/3)‖
解答:\int_0^8 \int_{\sqrt[3]x}^2 {1\over y^4+1}\,dydx= \int_0^2 \int_0^{y^3} {1\over y^4+1}\,dx dy = \int_0^2 {y^3\over y^4+1}dy =\left. \left[{1\over 4}\ln(y^4+1) \right]\right|_0^2 = \bbox[red, 2pt]{{1\over 4}\ln 17}
解答:
積分區域如上圖著色區域,令\cases{x= r\cos \theta\\ y=r\sin \theta} \Rightarrow x^2+(y-1)^2=1 \Rightarrow r^2=2r\sin \theta \Rightarrow r=2\sin \theta \\ 因此 \iint_R \sqrt{x^2+y^2}\,dA =2\int_0^{\pi/2} \int_{2\sin\theta}^2 r^2\,drd\theta =2\int_0^{\pi/2} \left. \left[ {1\over 3}r^3 \right] \right|_{2\sin \theta}^2 \,d\theta \\ ={16\over 3 }\int_0^{\pi/2} (1-\sin^3 \theta)\,d\theta= {16\over 3}\left. \left[ \theta-{1\over 12}(\cos 3\theta-9\cos \theta) \right] \right|_0^{\pi/2} ={16\over 3}({\pi\over 2}+{1\over 12}(-8)) \\=\bbox[red, 2pt]{{8\over 3}\pi -{32\over 9}}
解答:\int_{-1}^4 {dx\over \sqrt {|x|}} = \int_0^1{dx\over \sqrt x} +\int_0^4{dx\over \sqrt x}=2+4 = \bbox[red, 2pt]6
乙、計算、證明題: 共3題,每題12分,共36分
解答:\mathbf{(a)}\;令b_n={\ln n\over n-\ln n} \Rightarrow \lim_{n \to \infty}b_n = \lim_{n \to \infty}{\frac{\text{d}}{\text{d}n} \ln n\over \frac{\text{d}}{\text{d}n}(n-\ln n)} = \lim_{n \to \infty}{1\over n-1} =0\\ 又\frac{\text{d}}{\text{d}n}b_n ={1-\ln n\over (n-\ln n)^2} \lt 0, \forall n \ge 3; 因此由交錯級數審定法知該級數收斂\\ 現在要判斷是絕對收斂或條件收斂: 由於{\ln n\over n-\ln n} \gt {1\over n-\ln n} \gt {1\over n}\\ 而調和級數發散,因此原級數絕對值發散,即\bbox[red,2pt]{條件收斂} \\ \mathbf{(b)}\;\lim_{n\to \infty}\left|{a_{n+1} \over a_n} \right| =\lim_{n\to \infty}\left|{(x+4)^{n+1} \over (n+1)3^{n+1}} \cdot { n3^n\over (x+4)^n} \right| =\lim_{n\to \infty}\left|{x+4 \over 3} \cdot {n\over n+1} \right| =\left|{x+4 \over 3} \right| \lt 1 \\ \Rightarrow -7\lt x\lt -1 \\ x=-1 \Rightarrow \sum_{n=1}^\infty {(x+4)^n \over n3^n} =\sum_{n=1}^\infty {1 \over n} 為調和級數 \Rightarrow 發散\\ x=-7 \Rightarrow \sum_{n=1}^\infty {(x+4)^n \over n3^n} =\sum_{n=1}^\infty (-1)^n {1\over n} \Rightarrow \cases{\lim_{n\to \infty}{1\over n}=0 \\ {1\over n}\gt {1\over n+1}},交錯級數審定法 \Rightarrow 收斂\\ 因此級數收斂範圍為\bbox[red, 2pt]{-7\le x\lt -1}
解答:\mathbf{(a)}\;取y=mx \Rightarrow \lim_{(x,y)\to(0,0)} f(x,y)= \lim_{x\to 0} {x^2\cdot mx\over x^3+m^3x^3} =\lim_{x\to 0} {m\over 1+m^3} ={m\over 1+m^3} \ne f(0,0) \\ \quad \Rightarrow f(0,0) 不連續,\bbox[red, 2pt]{故得證} \\\mathbf{(b)}\; {\partial f\over \partial x} ={2xy \over x^3+y^3}-{x^2y \cdot 3x^2\over (x^3+y^3)^2} =\bbox[red, 2pt]{2xy^4-x^4y \over (x^3+y^3)^2}\\ \mathbf{(c)}\;\frac{\partial f}{\partial x} =\lim_{h\to 0}{f(0+h,0)-f(0,0)\over h} =\lim_{h\to 0}{0\over h}= \bbox[red, 2pt]0
解答:f(x,y)=x^2+3y^2+2y \Rightarrow \cases{f_x= 2x\\ f_y= 6y+2} \Rightarrow \cases{f_{xx}=2\\ f_{xy} =0\\ f_{yy}=6} \Rightarrow d(x,y)=f_{xx}f_{yy}-f_{xy}^2=12 \gt 0\\ 若\cases{f_x= 0\\ f_y= 0} \Rightarrow (x,y)=(0,-1/3) \Rightarrow f(0,-1/3)= -{1\over 3}為極小值\\ 再利用\text{ Lagrange }算子求極大值: \cases{f(x,y)=x^2+3y^2+2y\\ g(x,y)=x^2+y^2-1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y=\lambda g_y\\ g=0} \Rightarrow \cases{2x=\lambda(2x) \\ 6y+2=\lambda(2y) \\ x^2+y^2=1} \\ \Rightarrow \cases{x=0 \Rightarrow y=\pm 1\\ \lambda=1 \Rightarrow y=-1/2 \Rightarrow x=\pm \sqrt 3/2} \Rightarrow \cases{f(0,1)=5\\ f(0,-1)=1\\ f(\pm\sqrt 3/2,-1/2)=1/2} \Rightarrow 最大值5 \\ \Rightarrow \bbox[red, 2pt]{最大值5,最小值-{1\over 3}}
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第7題是(8/3)pi-(32/9)吧,不是+
回覆刪除已更正,謝謝!
刪除後面可以直接用wallis公式
回覆刪除第三題似乎是-ln2
回覆刪除應該還是ln 2,曲線長度不會負值
刪除5的r(π/3) 有誤 應為(-1,0,π)
回覆刪除已更正,謝謝
刪除