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(2020-2021期中高数上

I.(1)B

( $f(x)=|x|\sin x=\{\begin{array}{ll}x\sin x,& x⩾0\\ -x\sin x,& x<0\end{array}{f}^{\prime }(x)=\{\begin{array}{l}\sin x+x\cos x,x⩾0\\ -\sin x\times \cos x,x<0\end{array}$

$$f^{\prime }(0)=\underset{x\to 0}{lim}\frac{f(x)-f(0)}{x}=\underset{x\to 0}{lim}\frac{|x|}{x}\sin x=0$$

$\Rightarrow f_{+}^{\prime \prime }(0)=\underset{x\to 0^{+}}{lim}\frac{f^{\prime }(x)-f^{\prime }(0)}{x}=\underset{x\to 0}{lim}\frac{\sin x+x\cos x}{x}=\underset{x\to 0}{lim}(\frac{\sin x}{x}+\cos x)=2$

ล $f(0)=\underset{x\to 0^{-}}{lim}(-\frac{\sin x}{x}-\cos x)=-2$

$\to f^{(2)}(0)$ does not exist

$\Rightarrow f^{(n)}(0)$ exists, $n$ 最大是 $n=1$

$$\begin{array}{rl}& \text{ (2) A. }\\ & \underset{x\to \mathrm{\infty }}{lim}(\frac{x^2+1}{x+1}-ax-b)=\frac{1}{2}\Rightarrow \text{ 户े }{\text{ i土 }}^{y=ax+b+\frac{1}{2}}y=\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1}\\ & \text{ 的: 斩近线 }\Rightarrow ax+b+\frac{1}{2}=x-1\\ & \therefore \{\begin{array}{c}a=1\\ b+\frac{1}{2}=-1\end{array}\Rightarrow \{\begin{array}{c}a=1\\ b=-\frac{3}{2}\end{array}\\ & \text{ 忶 }=:a=\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{x^2+1}{x+1}}{x}=1\\ & b=\underset{x\to \mathrm{\infty }}{lim}(\frac{x^2+1}{x+1}-x-\frac{1}{2})=\underset{x\to \mathrm{\infty }}{lim}(\frac{1-x}{x+1}-\frac{1}{2})=-1-\frac{1}{2}=-\frac{3}{2}\text{. }\end{array}$$

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13. B

$$g(c)=\frac{{\int }_0^6(x^2+6)dx}{6}=\frac{{\frac{1}{3}{x}^3+6x]}_0^6}{6}=\frac{1}{3}\times 36+6=18$$

(4) . $D$

c). $f(x)=|x|\sin (x)\Rightarrow f^{\prime }(0)=\underset{x\to 0}{lim}\frac{f(x)-f(0)}{x}=\underset{x\to 0}{lim}\frac{|x|}{x}\sin |x|=0$

$$f(x)=|x|\sin \sqrt{|x|}\to f^{\prime }(0)=\underset{x\to 0}{lim}\frac{f(x)-f(0)}{x}=\underset{x\to 0}{lim}\frac{|x|}{x}\sin \sqrt{|x|}=0$$$$f(x)=\cos |x|\to f^{\prime }(0)=\underset{x\to 0}{lim}\frac{f(x)-f(0)}{x}=\underset{x\to 0}{lim}\frac{\cos |x|-1}{x}=\underset{x\to 0}{lim}\frac{{\sin }^2|x|}{x}\cdot \frac{1}{\cos |x|+1}$$$$=\frac{-1}{2}\underset{x\to 0}{lim}{\left(\frac{\sin |x|}{|x|}\right)}^2\frac{|x{|}^2}{x\to 0}=0$$$$\text{begin{aligned} f(x)=cos sqrt{|x|} Rightarrow f^{prime}(0) \& =limlimits\_{x rightarrow 0} frac{f(x)-f(0)}{x}=limlimits\_{x rightarrow 0} frac{cos sqrt{x mid}-1}{x}=limlimits\_{x rightarrow 0} frac{-sin ^2 sqrt{|x|}}{x} \& =frac{-1}{2} limlimits\_{x rightarrow 0}left(frac{sin sqrt{|x|}}{sqrt{|x|}}right)^2 cdot frac{|x|}{x}=left{begin{array}{l} frac{1}{2}=f^{prime}(0) frac{1}{2}=f^{prime}(0) end{array}right end{aligned}}$$

$\Rightarrow f^{\prime }(0)$ does not expst

(5) C

$$\begin{array}{rl}f(x)=\frac{1-\sin x}{1+\sin x}\Rightarrow f^{\prime }(x)& =\frac{-\cos x(1+\sin x)-\cos x(1-\sin x)}{(1+\sin x{)}^2}=\frac{-2\cos x}{(1+\sin x{)}^2}\\ f^{\prime }\left(\frac{\pi }{6}\right)& =\frac{-2x\frac{\sqrt{3}}{2}}{{\left(\frac{3}{2}\right)}^2}=-\frac{4\sqrt{3}}{9}=-\frac{4}{3\sqrt{3}}\end{array}$$

2(1) $0.{\int }_{-2}^{\frac{\pi }{2}}\frac{{\sin }^{5}x{\cos }^{3}x}{\text{ 大数 }}dx=0$

(2) $\frac{1}{1^2}\cdot {\int }_0^{x^3-1}f(t)dt=x\Rightarrow 3x^{2}f(x^3-1)=1$ (et $x=2\Rightarrow f(7)=\frac{1}{12}$

(3) $\frac{1}{3}{x}^3+2x-\frac{1}{x}-\frac{4}{3},f^{\prime }(x)={(x+\frac{1}{x})}^2=x^2+2+\frac{1}{x^2}$

$$\Rightarrow f(x)=\frac{1}{3}{x}^3+2x-\frac{1}{x}+c,f(1)=1\Rightarrow c=-\frac{4}{3}$$

(4) $-2,x^2+y^2+z^2={13}^2\Rightarrow 2x\cdot x^{\prime }+2y\cdot y^{\prime }+2z\cdot z^{\prime }=0$

let $t=x_0.x\left(t_0\right)=3,x^{\prime }\left(t_0\right)=4,y\left(t_0\right)=4,y^{\prime }\left(t_0\right)=3$

$$z\left(t_0\right)=12$$$$\Rightarrow 24+24+24z^{\prime }\left(t_0\right)=0\Rightarrow z^{\prime }\left(t_0\right)=-2$$

5. $\frac{a}{\sqrt{a^2+1}}\cdot \underset{s\to a}{lim}\frac{\sqrt{s^2+1}-\sqrt{a^2+1}}{s-a}={\frac{d}{ds}\left(\sqrt{s^2+1}\right)|}_{s=a}=\frac{a}{\sqrt{a^2+1}}$

6. (1) $\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{3+x}-\sqrt{x+1}}{x^2+x-2}=\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{\sqrt{3+x}}{x^2}-\frac{\sqrt{x+1}}{x^2}}{1+\frac{1}{x}-\frac{2}{x^2}}=0$

12. $\underset{x\to 0}{lim}\frac{\cos x-{\sec }^{2}x}{x\sin x}=\underset{x\to 0}{lim}\frac{\cos x-\frac{1}{{\cos }^{2}x}}{x\sin x}=\underset{x\to 0}{lim}\frac{{\cos }^{3}x-1}{x\sin x}=\underset{x\to 0}{lim}\frac{(\cos x-1)}{x\sin x}\cdot {\cos }^{2}x+\cos x+1)$

$=3\underset{x\to 0}{lim}\frac{\cos x-1}{x\sin x}=3\underset{x\to 0}{lim}\frac{-{\sin }^{2}x}{x\sin x}\cdot \frac{1}{\cos x+1}=\frac{3}{-2}\underset{x\to 0}{lim}\frac{\sin x}{x}=-\frac{3}{2}$

4. (1) ${{\int }_0^{2\pi }|{\sin }^{2}x-{\cos }^{2}x|dx={\int }_0^{2\pi }|\cos 2x|dx=8{\int }_0^{\frac{2}{4}}\cos 2xdx=4\sin 2x]}_0^{\frac{2}{4}}=4$

12. ${\int }_0^1(x+2)\sqrt{1-x^2}dx={\int }_0^{1}x\sqrt{1-x^2}dx+2{\int }_0^1\sqrt{1-x^2}dx=-\frac{1}{3}{(1-x^2)}^{\frac{3}{2}}{\int }_0^1+2x\frac{x}{4}=\frac{1}{3}+\frac{\pi }{2}=$

5. $f(x)=\frac{x^3+x-2}{x-x^2}$

(i) $f(x)=\frac{x^3+x-2}{x-x^2}=\frac{(x-1)(x^2+x+2)}{x(1-x)}=\frac{x^2+x+2}{x}=-x-1-\frac{2}{x}$ domain. $x\ne 0$. $x=1$

$f^{\prime }(x)=-1+\frac{2}{x^2},f^{\prime \prime }(x)=-\frac{4}{x^3}$

let $f^{\prime }(x)=0\Rightarrow x=\pm \sqrt{2}{x}^{-1}x-\sqrt{2}\Rightarrow f^{\prime }(x)<0\downarrow f(\sqrt{2})=-\sqrt{2}+-\frac{2}{\sqrt{2}}=-2\sqrt{2}-1$

citifical points at $x=\pm \sqrt{2}$

$f^{\prime \prime }(x)=\frac{4}{-x^3}\Rightarrow x<0,f^{\prime \prime }(x)>0$ concave up but $x\ne 0\Rightarrow$ hes inflection print $x>0,f^{\prime }(x)<0$ concave down

(2) horizontal agunptote: no. vertical asymptote: $x=0$. oblique asymptote: $y=-x-1$

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$\underset{―}{}$

$\underset{―}{}$

6. (1) $y={\int }_1^{1+2x}\sqrt{t^2-1}dt\to \frac{dy}{dx}=2\sqrt{(1+2x{)}^2-1}=2\sqrt{4x^2+4x}$

(2) $x^2-y^2=9\Rightarrow 2x-2y\cdot \frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{x}{y}=-\frac{5}{4}$

$$y-(-4)=-\frac{5}{4}(x-5)$$

T. $\{\begin{array}{l}y=x^2\\ y=2x-x^2\end{array}\Rightarrow 2x^2=2x\Rightarrow x=0,x=1$

${s={\int }_0^1[(2x-x^2)-x^2]dx={\int }_0^1(2x-2x^2)dx=x^2-\frac{2}{3}{x}^3]}_0^{11}=1-\frac{2}{3}=\frac{1}{3}$

$$\text{ 8. }\begin{array}{rl}& \begin{array}{c}y=\sqrt{x}\\ y=2x-1\end{array}\Rightarrow x^2=4x^2-4x+1(3x-1)(x-1)=0\\ & 3x^2-4x+1=0x=\frac{1}{3},x=1\end{array},\begin{array}{rl}V& =2\pi {\int }_0^1(x+1)(\sqrt{x}-(2x-1))dx\\ & =2\pi {\int }_0^1(x^{\frac{3}{2}}-2x^2+x+\sqrt{x}-2x+1)dx\\ & {=2\pi (\frac{2}{5}{x}^{\frac{5}{2}}-\frac{2}{3}{x}^3-\frac{1}{2}{x}^2+\frac{2}{3}{x}^{\frac{3}{2}}+x)]}_0^1\\ & =2\pi (\frac{2}{5}-\frac{2}{3}-\frac{1}{2}+\frac{2}{3}+1)=\frac{9\pi }{5}\end{array}$$

9. $f(x)$ is continuous on 6,1$]\Rightarrow f(x)$ has max, min., $f(\alpha )=M=f(x)$

$$f(\beta )⩽{\int }_e^{1}f(x)dx⩽f(\alpha )$$