这是份西交利物浦大学的MTH019/20-21 calculus with Applications exam辅导案例，学生在我们的帮助下获得了满意的成绩。
Mathematical Sciences Institute
EXAMINATION: Semester 1
MATH1014_Semester 1Mathematics and Applications 2
This practise questions for calculus
I. Multiple Choice Questions. (3 marks for each question, 60 marks in total)
- $\lim _{x \rightarrow-1} \frac{x^{2}+x}{x^{2}-2 x-3}=\left[\begin{array}{l}\text { D }\end{array}\right]$
(A) 2 .
(B) $-\frac{3}{2}$.
(C) $-\frac{1}{3}$.
(D) $\frac{1}{4}$ - If the function $f(x)=\left{\begin{array}{cl}\frac{1-e^{2 x}}{\sin x}, & x>0 \ a e^{2 x}, & x \leq 0\end{array}\right.$ is continuous at $x=0$, then $a=[$ B ]
(A) 2 .
(B) $-2$.
(C) 1 .
(D) $-1$. - If $f$ is differentiable at $x=c$, then $\lim _{h \rightarrow 0} \frac{f(c+h)-f(c-h)}{h}=[\mathbf{A}]$
(A) $2 f^{\prime}(c)$.
(B) $f^{\prime}(c)$.
(C) 0 .
(D) $f^{\prime}(2 c)$ - If $f(x)=\frac{\cos x}{1-\sin x}$, then $f^{\prime}\left(\frac{\pi}{6}\right)=[$ A $]$
(A) 2 .
(B) $\frac{1}{2}$.
(C) $-4$.
(D) 4 . - Find the slope of the tangent line to the curve $x y+e^{2 y}=x+1$ at the point $(0,0)$. [ $\left.\mathbf{B}\right]$
(A) $-\frac{1}{2}$.
(B) $\frac{1}{2}$.
(C) $-1$.
(D) 1 . - Let $g$ be a differentiable function and $g(100)=10, g^{\prime}(100)=\frac{1}{20}$. Use differential to approximate $g(100.5)$. [ A ]
(A) $g(100.5) \approx 10.025$.
(B) $g(100.5) \approx 10.25$.
(C) $g(100.5) \approx 10.05$.
(D) $g(100.5) \approx 10.075$. - If $f^{\prime}(x)=x(x-1)(x-2)$, then which of the following statements about $f(x)$ must be true? [ B ]
I. $f$ has a local maximum at $x=0$.
II. $f$ has a local maximum at $x=1$.
III. $f$ has a local maximum at $x=2$.
(A) Only III.
(B) Only II.
(C) Only I and III.
(D) I, II, III. - The $x$-coordinate of the inflection point on the graph of $y=\frac{1}{3} x^{3}+5 x^{2}+24$ is $[\mathbf{D}]$
(A) 5 .
(B) 0 .
(C) $-\frac{10}{3}$.
(D) $-5$. - Let $f$ be a differentiable function such that $f^{\prime}(x) \leq 3$ for all real numbers $x$ and $f(-1)=-1$. Then [ D]
(A) $f(1) \leq 4$.
(B) $f(2) \leq 7$.
(C) $f(3) \leq 10$.
(D) $f(4) \leq 14$. - $\lim _{x \rightarrow 0^{+}} x^{2} \ln x=[$ A $]$
(A) 0 .
(B) 1 .
(C) e.
(D) $-\infty$.
II. Comprehensive problems. (10 marks for each question, 40 marks in total)
(10 marks) Suppose that the function $f$ is continuous at $x=2$ and $\lim {x \rightarrow 0} \frac{f(2-x)-3}{2 \sin x}=$ (1) Find $f(2)$. (2) Show that $f$ differentiable at $x=2$ and find $f^{\prime}(2)$. (3) Find an equation of the tangent line to the curve $y=f(x)$ at $x=2$.
Solution (1) Since $\lim {x \rightarrow 0} \frac{f(2-x)-3}{2 \sin x}=5$, it follows that
$$
\lim {x \rightarrow 0}(f(2-x)-3)=0 . $$ Then $\lim {x \rightarrow 0} f(2-x)=3$. On the other hand, $f$ is continuous at $x=2$, then
$$
\lim {x \rightarrow 0} f(2+x)=\lim {x \rightarrow 0} f(2-x)=f(2) .
$$
Therefore, $f(2)=3$. $\quad[1 \mathrm{mark}]$
(b) Since $\lim {x \rightarrow 0} \frac{f(2-x)-3}{2 \sin x}=5$, then we have $$ \begin{aligned} 10=\lim {x \rightarrow 0} \frac{f(2-x)-3}{\sin x} &=\lim {x \rightarrow 0} \frac{f(2-x)-3}{x} \frac{x}{\sin x} \ &=\lim {x \rightarrow 0} \frac{f(2-x)-f(2)}{x} \
(h=-x) &=-\lim _{h \rightarrow 0} \frac{f(2+h)-3}{h}=-f^{\prime}(2) .
\end{aligned}
$$
So $f^{\prime}(2)=-10$ and $f$ is differentiable at $x=2$.
(3) The equation of the tangent line is $y-3=-10(x-2)$, i.e. $y=-10 x+23$.
(10 marks) A man launches his boat from point $A$ on a bank of a straight river, $3 \mathrm{~km}$ wide, and wants to reach point $B, 8 \mathrm{~km}$ downstream on the opposite bank, as quickly as possible. He could proceed in any of three ways (as shown in Figure 1):
Row his boat directly across the river to point $C$ and then run to $B$;
Row directly to $B$;
Row to some point $D$ between $C$ and $B$ and then run to $B$.
If he can row $6 \mathrm{~km} / \mathrm{h}$ and run $8 \mathrm{~km} / \mathrm{h}$, where should he land to reach $B$ as soon as possible?
Solution Let $x$ be the distance from $C$ to $D$, then the running distance is $|D B|=8-x$ and rowing distance is $|A D|=\sqrt{x^{2}+9}$. So the total time $T$ as a function of $x$ is
$$
T(x)=\frac{\sqrt{x^{2}+9}}{6}+\frac{8-x}{8}
$$
The domain of this function $T$ is $[0,8]$.
The derivative of $T$ is
$$
T^{\prime}(x)=\frac{x}{6 \sqrt{x^{2}+9}}-\frac{1}{8} .
$$
Thus, using the fact that $x \geq 0$, we have
$$
T^{\prime}(x)=0 \Leftrightarrow \frac{x}{6 \sqrt{x^{2}+9}}=\frac{1}{8} \Leftrightarrow 4 x=3 \sqrt{x^{2}+9} \Leftrightarrow 7 x^{2}=81,
$$
so the only stationary paint on $(0,8)$ is $\bar{x}=9 / 7$.
Now we evaluate $T$ at the stationary point and two end points:
$$
T(0)=1.5, \quad T\left(\frac{9}{\sqrt{7}}\right)=1+\frac{\sqrt{7}}{8} \approx 1.33, \quad T(8)=\frac{\sqrt{73}}{6} \approx 1.42 .
$$
So the minimum value of $T$ must occur when $x=9 / \sqrt{7}$. Thus the man should land the boat at a point $9 / \sqrt{7} \mathrm{~km}(\approx 3.4 \mathrm{~km})$ downstream from his starting point.
(10 marks) Let $R$ be the region in the first quadrant enclosed by the graphs of $y=6 x-x^{2}$ and $y=2 x$
(1) Find the area of $R$.
(2) Find the volume of the solid of revolution generated by revolving $R$ about the $y$-axis.
(3) Find the volume of the solid of revolution generated by revolving $R$ about the $x$-axis.
Solution (1) The points of intersection of these two curves are $(0,0)$ and $(4,8)$. Then the area of $R$ is
$$
\begin{aligned}
A &=\int_{0}^{4}\left(6 x-x^{2}-2 x\right) d x \
&=\int_{0}^{4}\left(4 x-x^{2}\right) d x \
&=\left[2 x^{2}-\frac{1}{3} x^{3}\right]{x=0}^{x=4}=\frac{32}{3} \end{aligned} $$ (2) The volume of the solid revolving about the $y$-axis is $$ \begin{aligned} V &=2 \pi \int{0}^{4} x\left(6 x-x^{2}-2 x\right) d x \
&=2 \pi \int_{0}^{4}\left(4 x^{2}-x^{3}\right) d x \
&=2 \pi\left[\frac{4}{3} x^{3}-\frac{1}{4} x^{4}\right]{x=0}^{x=4}=\frac{128 \pi}{3} \end{aligned} $$ (3) The volume of the solid revolving about the $x$-axis is $$ \begin{aligned} V &=\pi \int{0}^{4}\left(\left(6 x-x^{2}\right)^{2}-(2 x)^{2}\right) d x \
&=\pi \int_{0}^{4}\left(x^{4}-12 x^{3}+32 x^{2}\right) d x \
&=\pi\left[\frac{1}{5} x^{5}-3 x^{4}+\frac{32}{3} x^{3}\right]_{x=0}^{x=4}=\frac{1792 \pi}{15} .
\end{aligned}
$$
CALCULUS FOR BUSINESS
MODULE TITLECalculus for Business
MODULE LEVELLevel 0
MODULE CREDITS5.00
ACADEMIC YEAR2021/22
SEMESTERSEM1
AIMS AND FIT OF MODULE
To give students a broad education in calculus, which include the topics usually covered in a course on single-variable calculus;
To give students an appreciation of the application of mathematics to business, finance and social sciences;
To train the students’ ability to work independently and to acquire the skill of problem solving.
LEARNING OUTCOMES
A To understand all the key concepts of Limit, Continuity, Derivative and Integration for the function of polynomial, exponential, logarithmic, and trigonometric functions;
B To have a good appreciation of the link between mathematics and business, finance and social sciences;
C To be able to calculate the limits, derivatives and integrations of the polynomial, exponential, logarithmic, and trigonometric functions.
D To be able to establish simple mathematical models using the idea of differential, integration and first order differential equation.
METHOD OF TEACHING AND LEARNING
Students will be expected to attend 3 hours of formal lectures and 1 hour of tutorial in a typical week. In lectures, teachers will introduce the academic content and practical skills which are the subject of the module. In the tutorials, the students can practice those skills.
In addition to the formal lectures and tutorials, students are expected to devote the unsupervised time to study the lecture materials and background readings. Online resources will be provided to the students to promote their active learning and self-leaning. Continuous assessment including online home assignments will be used to assess the learning outcomes. Written examinations in the middle and at the end of the semester constitute the major part of the assessment of the academic achievement of students.
