113年嘉義大學轉學考-微積分詳解

 國立嘉義大學113學年度轉學生招生考試

科目: 微積分 (每題10分，共100分)

解答: $$\lim_{x\to 1} {x-1\over x^2-1} = \lim_{x\to 1} {x-1\over (x-1)(x+1)} = \lim_{x\to 1} {1\over x+1} = \bbox[red, 2pt]{1\over 2}$$

解答: $$\cases{\lim_{h\to 0^+} {f(h)-f(0)\over h} =\lim_{h\to 0^+}{|h|\over h} =1 \\\lim_{h\to 0^-} {f(h)-f(0)\over h} =\lim_{h\to 0^-}{|h|\over h} =-1} \Rightarrow f'(0)不存在\Rightarrow f(x)在x=0不可微分\\ \Rightarrow f(x)\bbox[red, 2pt]{在x\in \mathbb R-\{0\}可微分}$$

解答: $$\lim_{x\to 0} x\cot x =\lim_{x\to 0} {x\cos x \over \sin x}= \lim_{x\to 0} {(x\cos x)' \over (\sin x)'} =\lim_{x\to 0} {\cos x -x\sin x\over \cos x} =\bbox[red, 2pt]1$$

解答: $$\lim_{x\to \infty} {3x^7-x^2-2\over 5x^7+4x^4+1} =\lim_{x\to \infty} {3 -1/x^5-2/x^7\over 5 +4/x^3+1 /x^7} = \bbox[red, 2pt]{3\over 5}$$

解答: $$\frac{d }{dx} \int_1^{x^3} \sec t\,dt =\sec x^3\cdot \frac{d }{dx}x^3 = \bbox[red, 2pt]{3x^2 \sec x^3}$$

解答: $$\cases{u=\ln(x^2+9) \\ dv=dx} \Rightarrow \cases{du={2x\over x^2+9}dx \\ v=x} \Rightarrow \int \ln(x^2+9)\,dx = x\ln(x^2+9)-\int {2x^2\over x^2+9}\,dx \\=x\ln(x^2+9)-\int \left(2-{18\over x^2+9} \right) \,dx  = x\ln(x^2+9)-2x+ 2\int{1\over (x/3)^2+1}\,dx \\= \bbox[red, 2pt]{x\ln(x^2+9)-2x+6\tan^{-1} {x\over 3} +c_1}$$

解答: $$\cases{u=\sin x\\ dv=e^x\,dx} \Rightarrow \cases{du= \cos x\,dx\\ v=e^x} \Rightarrow I=\int e^x\sin x\,dx = e^x \sin x-\int e^x \cos x\,dx \\ 再一次,\cases{u=\cos x\\ dv=e^x\,dx} \Rightarrow \cases{du=-\sin x\,dx\\ v=e^x} \Rightarrow I=e^x\sin x-e^x\cos x-I\\ \Rightarrow I= \bbox[red, 2pt]{{1\over 2}e^x(\sin x-\cos x)+c_1}$$

解答: $$\cases{u=\arcsin x\\ dv=dx} \Rightarrow \cases{du={1\over \sqrt{1-x^2}}dx \\v=x} \Rightarrow I=\int \arcsin x\,dx = x\arcsin x-\int {x\over \sqrt{1-x^2}}\,dx\\ w=1-x^2 \Rightarrow dw=-2x\,dx \Rightarrow I=x\arcsin x+{1\over 2}\int {1\over \sqrt w}\,dw=x\arcsin x+  w^{1/2} +c_1\\= \bbox[red, 2pt]{x\arcsin x+ \sqrt{1-x^2}+c_1}$$

解答:

$$\cos(3\theta)=0 \Rightarrow \theta=\pm {\pi\over 6} \Rightarrow 一片花瓣面積=\int_{-\pi/6}^{\pi/6} {1\over 2}\cos^2 (3\theta)\,d\theta= {1\over 4}\left. \left[ {1\over 6} \sin(6\theta)+\theta \right] \right|_{-\pi/6}^{\pi/6} ={1\over 12}\pi \\ \Rightarrow 三瓣面積= \bbox[red, 2pt]{{1\over 4}\pi}$$

解答: $$\lim_{(x,y)\to (0,0)} {x^2y^2\over x^2+y^2} =0 \ne 1 \Rightarrow f在(0,0)不連續\Rightarrow f在(0,0)\bbox[red, 2pt]{不可微}$$

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解題僅供參考，轉學考歷年試題及詳解