Answer

2021-2022 Calculus I Mid

1.Multiple Choice Questions:

(1) .$B$

(2) .A

(3) .C

(4) .A

(5) .C

2.Fill in the blanks:

(1) . 3

(2) .$n$ !

(3) .$\frac{2\sin x+x\cos x}{4\sqrt{x\sin x\sqrt{\sin x}}}$

(4) .$\frac{1}{6}$

(5) .$y=x+1$

3.$S_{\mathrm{\Delta }}\le 100$, the equal sign holds when $a=b=10\sqrt{2}$

4.${\frac{dy}{dx}|}_{x=0}=0,{\frac{d^{2}y}{d{x}^2}|}_{x=0}=-\frac{1}{2}$

5. $2\pi -\frac{(\pi {)}^2}{2}$

6.(1)$\frac{4}{3}$

(2) $\frac{11}{3}$

7.(1)$\frac{1}{24}$

(2) $\frac{1}{2}$

8.$y=2x+1$

9.when $x=\frac{\pi }{4},f(x)$ takes its minimum and $f\left(\frac{\pi }{4}\right)=2\sqrt{2}-1$

2020-2021 Calculus I Mid

1.Multiple Choice Questions:

(1) .B

(2) .A

(3) .B

(4) .D

(5) .C

2.Fill in the blanks:

(1) . 0

(2) .$\frac{1}{12}$

(3) .$\frac{1}{3}{x}^3+2x-\frac{1}{x}-\frac{1}{3}$

(4) .-2

(5) .$\frac{a}{\sqrt{a^2+1}}$

3.(1) 0 .(2)$-\frac{3}{2}$

4.(1) 4 .$(2)\frac{1}{3}+\frac{\pi }{2}$

5.(1)local maxima at $x=\sqrt{2}$, local minima at $x=-\sqrt{2}$

$◻$

$y=\frac{(x^3+x-2)}{x-x^2}$

(2) horizontal asymptote: not exist；vertical asymptote: $\mathrm{x}=0$ ；oblique asymptotes: $y=-x-1$

6.(1)$\frac{dy}{dx}=4\sqrt{x^2+x}$

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

7.The area is $\frac{1}{3}$

8.The volume is $\frac{9\pi }{5}$

9.You can easily show the conclusion by the mean value theorem of integration for continuous functions

2019-2020 Calculus I Mid

1.True or False:

(1) .TRUE

(2) .FALSE

(3) .TRUE

(4) .TRUE

(5) .FALSE

(counter-example: $f(x)=1$ can be considered as peridic function with period 1, then $g(x)=x$ is not a periodic function with $f(x)\mathrm{\&}g(x)$ satisying such condition)

2.Multiole Choice Questions:

(1) .D

(2) .B

(3) .C

(4) .D(keep ur eyes open: B is a Common Mistake )

(5) .B

3.(1)Deravatives is nonnegative, and does not exist zero point which changes sign, so local maxima and local minima do not exist.inflection point: $\mathrm{x}=0$

(2) horizontal asymptote: not exist；vertical asymptote: NE；oblique asymptotes: $y=x$

(3)

4.$(1)\frac{5}{4}+\sin 5$

(2) $\frac{\pi }{4}$

5.(1)$\frac{1}{6}{a}^4+\frac{1}{2}$

(2) $\frac{4\sqrt{2}}{3}$

6.Write the explicit function respectively as $r_1(x)$(y from $\frac{2}{3}to1$ )and $r_2$(y from $0to\frac{2}{3}$ )on $[0,\frac{16}{9}]$ .Obviously, volume is ${\int }_0^{\frac{16}{9}}\left(\pi {(2-r_1(x))}^2\right)dx-{\int }_0^{\frac{16}{9}}\left(\pi {(2-r_2(x))}^2\right)dx=$ $\underset{―}{}$

此类题不考, 无需担心.

7.$f\left(\frac{11\pi }{60}\right)=\frac{\pi }{45}+\frac{\sqrt{3}}{3}$

8.The area is ${\int }_0^2(3-x-\frac{1}{4}{x}^2)dx-{\int }_0^1(3-x-(\sqrt{x}+1))dx=\frac{5}{2}$

9.All critical points on $\mathbb{R}$ can be found, $x=0/\pm {(\frac{\pi }{4}+\frac{k\pi }{2})}^{\frac{1}{4}}$, where $\mathrm{k}=0,1,2\dots$ . In the interval $[-1,1],x=0/\pm {\left(\frac{\pi }{4}\right)}^{\frac{1}{4}}$

10.Structure $\varphi (x)=f(x)-\alpha x-{\beta }_{\text{then }}$ let $\alpha =\frac{f(b)-f(a)}{b-a}$, hence, by Rolle＇s theorem, qed

2022-2023 Calculus I Mid

1. Multiple Choice Questions:

(1) .C

(2) . $B$

(3) .A

(4) . C

(5) .A

2. Fill in the blanks:

(1) . $\frac{2}{\pi }$

(2) . 16 or -16

(3) .more than 2 methods, $\frac{11}{2^{15/4}}$

(4) . 2

(5) . 0

3. Proof. recall how we show the existence of root. And we assume there only 1 root and deduce a contradiction algebraically

I draw 2 figures for illustration

Plot $[x^{\wedge }5+2x-100,\{x,-1,3\}]$

绘图

4. $\frac{37}{168}$ substitution: $x\to \tan \theta$

5. we know linear approximate part $\mathcal{L}(x)=f\left(x_0\right)+f^{\prime }\left(x_0\right)\cdot (x-x_0)$ answer is $\frac{3}{2}x+3$

6. hint: rewrite $2(x-1)+3+\frac{b+1}{x-1},\mathrm{b}=-1.\mathrm{a}=3$

7. nothing to write, you just take the derivative of that, and then solve the equation

8. $y=\frac{1}{7}(x-1)+1$,hint: $2y^3-y^2+3y-4=0\to (y-1)(\dots )$

9. $F(x)={\int }_0^{x}xtf(x^2-t^2)dt=\dots =\dots =\dots =-\frac{1}{2}x{\int }_0^{x}f(-t^2+x^2)d(-t^2+x^2)$ then you can compute it..