113年政大轉學考-微積分(一)詳解

國立政治大學113學年度學士班轉學生招生考試

考試科目:微積分(一)
系所別:應用數學系二年級

解答: $$\textbf{a.}\;L=x^{\sqrt x} =e^{\sqrt x\ln x} \Rightarrow \ln L= \sqrt x\ln x \Rightarrow \lim_{x \to 0^+} \ln L=\lim_{x \to 0^+} {\ln x\over {1\over \sqrt x}} =\lim_{x \to 0^+} {(\ln x)'\over ({1\over \sqrt x})' } \\\quad =\lim_{x \to 0^+} {1/x\over  -x^{-3/2}/2}=\lim_{x \to 0^+} -2\sqrt x =0 \Rightarrow =\lim_{x \to 0^+} L=e^0=\bbox[red, 2pt] 1\\ \textbf{b.}\; \lim_{x \to 3}  \left({x \int_3^x {\sin t\over t}\,dt\over x-3} \right) =\lim_{x \to 3}   {(x \int_3^x {\sin t\over t}\,dt')\over (x-3)'}  =\lim_{x \to 3} \left(\int_3^x {\sin t\over t}\,dt +\sin x \right) =\bbox[red, 2pt]{\sin 3}$$

解答: $$F(x) =f(x f(x f(x))) \Rightarrow F'(x)= f'(x f(x f(x)))\cdot (f(xf(x))+ x f'(xf(x))\cdot (f(x)+ xf'(x))) \\\Rightarrow F'(1) =f'(f(2)) \cdot f(2)+ f'(2)(2+f'(1)) =f'(3)\cdot 3+ 5(2+4) =18+30= \bbox[red, 2pt]{48}$$

解答: $$ye^{\sin x} =x \cos y \Rightarrow y'e^{\sin x}+ y\cos xe^{\sin y} =\cos y-xy'\sin y \Rightarrow (e^{\sin x}+ x\sin y)y'= \cos y- y\cos x e^{\sin y} \\ \Rightarrow y'={\cos y- y\cos x e^{\sin y} \over e^{\sin x}+ x\sin y} \Rightarrow y'(0,0)={ 1\over 1}=1 \Rightarrow \text{ tangent line: }\bbox[red, 2pt]{y=x}$$

解答: $$\textbf{a. }\forall \alpha,\beta \in (a,b), \text{ where }\alpha\lt \beta, \text{ there exists }c \in[\alpha,\beta] \text{ such that} \\ f(\beta)-f(\alpha) =f'(c)(\beta-\alpha) =0 \Rightarrow f(\beta)= f(\alpha) \Rightarrow f(x) \text{ is constant, for }x \in(a,b) . \bbox[red, 2pt]{QED.}\\ \textbf{b. }\sin(\tan^{-1}x +\cot^{-1}x) = \sin(\tan^{-1}x) \cos(\cot^{-1}x) + \sin(\cot^{-1}x) \cos(\tan^{-1}x) \\={x\over \sqrt{x^2+1}} \cdot {x\over \sqrt{x^2+1}} +{1\over \sqrt{x^2+1}} \cdot {1\over \sqrt{x^2+1}}={x^2+1\over x^2+1}=1 =\sin({\pi\over 2}) \\\Rightarrow \tan^{-1}x +\cot^{-1}x ={\pi\over 2}. \bbox[red, 2pt]{QED.}$$


解答: $$\text{Let }h(x)= (g(b)-g(a))f(x)-(f(b)-f(a))g(x), \\\text{ then }h'(x)=(g(b)-g(a))f'(x)-(f(b)-f(a))g('x) \text{ and }\cases{ h(a)=g(b)f(a)-f(b)g(a) \\h(b)= f(a)g(b)-g(a)f(b)} \\ \Rightarrow h(a)-h(b)=0. \text{ By Mean Value Theorem, there exists }c \in (a,b) \text{ such that } \\h(b)-h(a)= h'(c)(b-a) =0 \Rightarrow h'(c)=(g(b)-g(a))f'(c)-(f(b)-f(a))g'(c) =0\\ \Rightarrow {f'(c) \over g'(c)} ={f(b)-f(a) \over g(b)-g(a)}, \bbox[red, 2pt]{QED.}$$

 解答: $$\textbf{a. }\cases{f_x(0,0) =\lim_{h\to 0}{f(h,0)-f(0,0) \over h}=\lim_{h\to 0}{0-0\over h} =0 \\f_y(0,0) =\lim_{h\to 0}{f(0,h)-f(0,0) \over h}=\lim_{h\to 0}{0-0\over h} =0} \Rightarrow \bbox[red, 2pt]{f_x(0,0)=f_y(0,0)=0} \\\textbf{b. }y=kx,k\ne 0 \Rightarrow f(x, y)= f(x,kx)={kx^2\over (k^2+1)x^2} ={k\over k^2+1} \Rightarrow \lim_{(x,y)\to (0,0)} f(x,kx) ={k\over k^2+1} \ne 0 \\ \Rightarrow f(x,y ) \text{ is NOT continuous at }(0,0) \Rightarrow \bbox[red, 2pt]{f(x,y) \text{ is continuous, for }(x,y)\ne (0,0)}$$

解答: $$f(x,y)= x^2+y^2+4x-4y =(x+2)^2+(y-2)^2-8 \ge -8 \Rightarrow \text{ absolute minimum =}f(-2,2)=-8 \\ \cases{f(x,y)= x^2+y^2+4x-4y \\g(x,y)=x^2+y^2-9} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{2x+4=2x\lambda\\ 2y-4= 2y\lambda} \Rightarrow {x+2\over y-2} ={x\over y} \Rightarrow x=-y\\ \Rightarrow g(x,-x)=2x^2=9 \Rightarrow x^2={9\over 2} \Rightarrow \cases{x=3/\sqrt 2 \Rightarrow y=-3\sqrt 2\\ x=-3\sqrt 2 \Rightarrow y=3\sqrt 2} \Rightarrow \cases{f(3/\sqrt 2,-3/\sqrt 2) =9+12\sqrt 2 \\ f(-3/\sqrt 2, 3/\sqrt 2) =9-12\sqrt 2\gt -8} \\ \Rightarrow  \bbox[red, 2pt]{\cases{\text{ absolute minimum =}-8\\{\text{ absolute maximum =}9+12\sqrt 2}}}$$

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解題僅供參考，轉學考歷年試題及詳解