Introductory test for the course Mathematical Analysis II

To proficiently attend the class Mathematical Analysis II, each student is expected to know the basics of Mathematical Analysis I and Geometry.
This page includes a set of questions and exercises that helps you to test your skills on these topics. You are supposed to answer these questions, and solve these exercises without using notes or textbooks.
If you are not able to solve these questions, you are invited to review your skills following the suggestions reported in the prerequisites document.

Question #1

$ab\geq1$ implies that

$a\geq1$ and $b\geq1$;
$a\geq1$ or $b\geq1$;
$a>0$ and $b>0$;
$a\neq0$ and $b\neq0$;
no answer is correct.

Solution

$a\neq0$ and $b\neq0$

Question #2

$|x-3|<|x+1|$

if and only if $x>1$;
for any $x\in\mathbb{R}$;
never;
if and only if $x<1$;
no answer is correct.

Solution

If and only if $x>1$

Question #3

Let $\alpha, \beta \in \mathbb{R}$, with $\alpha < \beta < 0$. Then, surely

$\log |\alpha| > \log |\beta|$;
$\tan \alpha > \tan \beta$;
$\tan \alpha < \tan \beta$;
$\alpha ^2 < \beta ^2$;
no answer is correct.

Solution

$\log |\alpha| > \log |\beta|$

Question #4

The system $$\displaystyle{\left\{\begin{array}{l} x^2-y^2=1\\ x-y=1 \end{array}\right.}$$ has

$1$ solution;
$2$ solutions;
$3$ solutions;
more then $3$ solutions;
no solutions.

Solution

$1$ solution

Question #5

The plane orthogonal to $\mathbf n=(1,-1,1)$ and passing through $P=(1,0,0)$ has equation

$x-y+z=1$;
$x-y+z=0$;
$x=1$;
$(x-1)+(y+1)+(z-1)=1$;
no answer is correct.

Solution

$x-y+z=1$

Question #6

The equation $$\ x^2+y^2+y=0$$ describes in the plane

the circle with radius $1$ and center $(0,-1)$;
the circle with radius $\frac12$ and center $\big(0,-\frac12\big)$;
the point $(0,-1)$;
an hyperbole;
no answer is correct.

Solution

The circle with radius $\frac12$ and center $\big(0,-\frac12\big)$

Question #7

$\sin x+\cos x=0$ in $[0,2\pi]$ has

$4$ solutions;
no solution;
$2$ solutions;
$1$ solution;
no answer is correct.

Solution

$2$ solutions

Question #8

$\displaystyle\lim_{x \to 0^-} e^{\frac {1}{x}}$

$= 1$;
does not exixt;
$ = 0$;
$ = +\infty$;
no answer is correct.

Solution

$ = 0$

Question #9

$$\displaystyle\lim_{x\to0^+}\frac{1-\cos(x^{\alpha})}{x^2}=\frac12$$

for any $\alpha>1$;
if and only if $\alpha=1$;
for a suitable $\alpha\in(0,1)$;
never;
no answer is correct.

Solution

If and only if $\alpha=1$

Question #10

The domain of $$f(x)=\sqrt{\log(|x-1|)}$$ is

$(-\infty,0]\cup[2,+\infty)$;
$\mathbb{R}\setminus\{1\}$;
$[2,+\infty)$;
$\mathbb{R}$;
no answer is correct.

Solution

$(-\infty,0]\cup[2,+\infty)$

Question #11

The derivative of $$\displaystyle f(x)=\log\big(e^{\cos x}+1\big)$$ is

$\displaystyle \frac{-(\sin x)e^{\cos x}}{e^{\cos x}+1}$;
$\displaystyle \frac1{e^{\cos x}+1}$;
$\displaystyle \frac{-\sin x}{e^{\cos x}+1}$;
$\displaystyle -\sin x$;
no answer is correct.

Solution

$\displaystyle \frac{-(\sin x)e^{\cos x}}{e^{\cos x}+1}$

Question #12

The function $$\displaystyle f(x) = \sqrt{\frac{x^2}{x+1}}$$ is strictly increasing

in $(-1,+\infty)$;
in $(-\infty,-2]$ and in $[0,+\infty)$;
only in $[0,+\infty)$;
in $(-\infty,-1)$;
no answer is correct.

Solution

only in $[0,+\infty)$

Question #13

The function $$f(x)=x^4-2x^3+1$$ has a minimum in

$x=0$ and $x=\frac32$;
$x=\frac32$;
$x=0$;
$x=1$;
no answer is correct.

Solution

$x=\frac32$

Question #14

The Taylor expansion of order $3$ at $x=0$ of $f(x)=xe^x$ is

$f(x)=x+x^2+\frac{x^3}2+o(x^3)$;
$f(x)=1+2x+\frac{x^2}2+\frac{x^3}6+o(x^3)$;
$f(x)=\frac{x^2}2+\frac{x^3}6+o(x^3)$;
$f(x)=x+o(x)$;
no answer is correct.

Solution

$f(x)=x+x^2+\frac{x^3}2+o(x^3)$

Question #15

$\displaystyle\int_2^3\frac1{t^2-1}dt$

$= \arctan3-\arctan2$;
$= \frac12\ln\frac32$;
$= \ln\frac83$;
$= 0$;
no answer is correct.

Solution

$= \frac12\ln\frac32$

Question #16

$\displaystyle\int{\,\frac1{\sqrt t}\, dt}$

$=2\sqrt t+c$, with $c\in\mathbb{R}$;
$=\displaystyle{-\frac2{t^{3/2}}+c}$, with $c\in\mathbb{R}$;
$=\displaystyle{-\frac1{2t^{3/2}}+c}$, with $c\in\mathbb{R}$;
$=\log(\sqrt t)+c$, with $c\in\mathbb{R}$;
no answer is correct.

Solution

$=2\sqrt t+c$, with $c\in\mathbb{R}$

Question #17

The series $\ \displaystyle{\sum_{n=0}^{+\infty}\frac1{n^{\alpha}}}$

converges if and only if $\alpha>0$;
converges if and only if $\alpha>1$;
converges for a suitable $\alpha<0$;
does not converge for any $\alpha\in\mathbb{R}$;
no answer is correct.

Solution

Converges if and only if $\alpha>1$

Question #18

The series $\ \displaystyle{\sum_{n=0}^{+\infty} x^n}$

converges if and only if $|x|<1$;
does not converge for any $x\in\mathbb{R}$;
converges if and only if $x=0$;
converges for a suitable $x>1$;
no answer is correct.

Solution

Converges if and only if $|x|<1$.

Question #19

Let $\ \displaystyle A=\begin{pmatrix} 2 & 0\\ 1 & 3 \end{pmatrix}$. Then

$\not\exists A^{-1}$;
$2, 3$ are eigenvalues of $A$ and $\mathbf v_2=(1,-1), \mathbf v_3=(0,1)$ are two corresponding eigenvectors;
$A$ is not diagonalizable;
$1, 3$ are eigenvalues of $A$ and $\mathbf v_1=(1,0), \mathbf v_3=(0,1)$ are two corresponding eigenvectors;
no answer is correct.

Solution

$2, 3$ are eigenvalues of $A$ and $\mathbf v_2=(1,-1), \mathbf v_3=(0,1)$ are two corresponding eigenvectors

Question #20

Let $\ \displaystyle A=\begin{pmatrix}1 & 0 & 1\\ 2 & 3 & -2\\ 2 & 2 & 1 \end{pmatrix}$. Then,

$1, 2+i, 2-i$ are eigenvalues of $A$;
$1, 1, 3$ are eigenvalues of $A$;
$A$ is diagonalizable in $\mathbb{R}$;
$\text{det}(A)=0$ ($\text{det}$ stands for the determinant of $A$);
no answer is correct.

Solution

$1, 2+i, 2-i$ are eigenvalues of $A$

Exercise #1

Calculate the integrals $$\int_{1}^{2}\frac{x}{\sqrt[3]{x-2}}\, dx\, ,\qquad\int_0^{\pi/4}\frac{\sqrt{\tan x}-1}{\cos^2 x}\, dx\, .$$

Solution

In the first integral we use the change of variable $t=\sqrt[3]{x-2}$, so that $x=2+t^3$. Hence,
$$\int\frac{x}{\sqrt[3]{x-2}}\, dx=\int\big(6t+3t^4\big)\, dt=3t^2+\frac35t^5+C=3(x-2)^{2/3}+\frac35(x-2)^{5/3}+C=F(x)+C,\quad\text{with $C\in\mathbb{R}$},$$
implying
$$\int_{1}^{2}\frac{x}{\sqrt[3]{x-2}}\, dx=F(2)-F(1)=-3+\frac35=-\frac{12}5.$$
In the second integral we use the change of variable $t=\tan x$, so that
$$\int\frac{\sqrt{\tan x}-1}{\cos^2 x}\, dx=\int\big(\sqrt{t}-1\big)\, dt=\frac23t^{3/2}-t+C=\frac23(\tan x)^{3/2}-\tan x+C=F(x)+C,$$
hence
$$\int_0^{\pi/4}\frac{\sqrt{\tan x}-1}{\cos^2 x}\, dx=F(\pi/4)-F(0)=\frac23-1=-\frac13.$$

Exercise #2

Study the function
$$f(x)=|x^2-1|e^x$$
and plot a qualitative graph of $f$.

Solution

$f$ is defined on $\mathbb{R}$. $f$ has no symmetry. $f(x)\geq0$ for any $x\in\mathbb{R}$, $f(x)=0$ if and only if $x=\pm1$. $$\lim_{x\to+\infty}f(x)=+\infty\qquad\text{e}\qquad\lim_{x\to-\infty}f(x)=\lim_{t\to+\infty}t^2e^{-t}=0.$$ The line of equation $y=0$ is an asymptote for $f$ as $x\to-\infty$. There are not asymptote for $x\to\infty$, since $$\lim_{x\to+\infty}\frac{f(x)}x=\lim_{x\to+\infty}xe^x=+\infty.$$ $$f(x)=\left\{\begin{array}{ll} (x^2-1)e^x & \text{ per } x\in(-\infty,-1]\cup[1,+\infty)\\ (1-x^2)e^x & \text{ per } x\in(-1,1) \end{array}\right.$$ so $$f'(x)=\left\{\begin{array}{ll} (x^2+2x-1)e^x & \text{ per } x\in(-\infty,-1)\cup(1,+\infty)\\[10pt] -(x^2+2x-1)e^x & \text{ per } x\in(-1,1) \end{array}\right.$$ Since $$\lim_{x\to1^+}f'(x)=2e,\quad \lim_{x\to1^-}f'(x)=-2e\quad\lim_{x\to-1^-}f'(x)=\frac2e,\quad=\lim_{x\to-1^+}f'(x)=-\frac2e,\quad \not\exists f'(\pm1).$$ Since $x^2+2x-1>0$ if and only if $x<-1-\sqrt2$ or $x>-1+\sqrt2$, then $f'(x)>0$ in $(-\infty,-1-\sqrt2)\cup(-1,-1+\sqrt2)\cup(1,+\infty)$, $f'(x)<0$ in $(-1-\sqrt2,-1)\cup(-1+\sqrt2,1)$, $f'(-1\pm\sqrt2)=0$. So $f$ is strictly increasing in $(-\infty,-1-\sqrt2)$, in $(-1,-1+\sqrt2)$, and in $(1,+\infty)$, and $f$ is strictly decreasing in $(-1-\sqrt2,-1)$ and in $(-1+\sqrt2,1)$. $x=-1-\sqrt2$ is a point of local maximum, and $f(-1-\sqrt2)=(2+2\sqrt2)e^{-1-\sqrt2}$, while $x=-1+\sqrt2$ is a point of local minimum, and $f(-1+\sqrt2)=(2\sqrt2-2)e^{-1+\sqrt2}$. From the limits, we deduce that $f$ has not global maximum nor global minimum. $$f”(x)=\left\{\begin{array}{ll} (x^2+4x+1)e^x & \text{ per } x\in(-\infty,-1)\cup(1,+\infty)\\[10pt] -(x^2+4x+1)e^x & \text{ per } x\in(-1,1) \end{array}\right.$$ Since $x^2+4x+1>0$ if and only if $x<-2-\sqrt3$ or $x>-2+\sqrt3$, $f”(x)>0$ in $(-\infty,-2-\sqrt3)\cup(-1,-2+\sqrt3)\cup(1,+\infty)$, $f”(x)<0$ in $(-2-\sqrt3,-1)\cup(-2+\sqrt3,1)$, $f”(-2\pm\sqrt3)=0$. Then, $f$ is convex in $(-\infty,-2-\sqrt3)$, in $(-1,-2+\sqrt3)$, and in $(1,+\infty)$, $f$ is concave in $(-2-\sqrt3,-1)$ and in $(-2+\sqrt3,1)$. The points $x=-2\pm\sqrt3$ are flex, and $f'(-2\pm\sqrt3)=(2\sqrt3\pm2)e^{-2\mp\sqrt3}$. A plot of the graph of $f$ is

Exercise #3

Find the Taylor expansion of order $3$ at $x=0$ of the function
$$f(x)=e^{x-x^3}-\cos(\sin x).$$
and calculate the limit
$$\lim_{x\to0}\frac{e^{x-x^3}-\cos(\sin x)-x-x^2}{\ln(1+x^3)}.$$

Solution

We know that $$e^t=1+t+\frac12t^2+\frac16t^3+o(t^3),\quad\cos t=1-\frac12{t^2}+o(t^3),\qquad \sin x=x+o(x^2),$$ so that \begin{align*} f(x)&=1+x-x^3+\frac12\big(x-x^3\big)^2+\frac16\big(x-x^3\big)^3+o\big(\big(x-x^3\big)^3\big) -1+\frac12{(\sin x)^2}+o((\sin x)^3)\\ &=x-x^3+\frac12x^2+\frac16x^3+o(x^3) +\frac12x^2+o(x^3)=x+x^2-\frac5{6}x^3+o(x^3). \end{align*} Further, since $\ln(1+x^3)\sim x^3$ if $x\to0$, we get $$\lim_{x\to0}\frac{e^{x-x^3}-\cos(\sin x)-x-x^2}{\ln(1+x^3)}=\lim_{x\to0}\frac{-\frac56x^3}{x^3}=-\frac56.$$

Exercise #4

For which $x \in \mathbb{R}$ does the series
$$\displaystyle {\sum _{n=1}^{+\infty} \frac {1}{n}}\left(\frac{x}{3}\right)^n\,$$
converge?

Solution

Notice that
$$\lim_{n\to\infty}\frac {1}{n}\left(\frac{x}{3}\right)^n=0
\qquad\text{if and only if}\qquad \left|\frac{x}{3}\right|\leq1\qquad\text{if and only if}
\qquad -3\leq x\leq3.$$
So, if $x<-3$ or $x>3$, the series does not converge.
Let us analyze the absolute converge of the series if $-3\leq x\leq 3$. Since
$$\lim_{n\to+\infty}\frac{|a_{n+1}|}{|a_n|}=\lim_{n\to+\infty}\frac {1}{n+1}\left|\frac{x}{3}\right|^{n+1}\frac{n}{\left|\frac{x}{3}\right|^{n}}
=\lim_{n\to+\infty}\frac {n}{n+1}\left|\frac{x}{3}\right|=\left|\frac{x}3\right|.$$
Then, if $x\in(-3,3)$, the series converges.
If $x=3$, then $a_n=\frac1n$, so the series diverges.
If $x=-3$, then $a_n=(-1)^n\frac1n$, so by Leinbiz criterion the series converges.
Finally: the series converges if and only if $-3\leq x<3$.

Exercise #5

Find the plane $\pi$ passing through $O=(0,0,0)$ and $A=(0,-1,1)$, and parallel to the line $r$ with equation
$$\left\{\begin{array}{l}
x=t\\
y=-t\\
z=1
\end{array}\right.,\qquad \,t\in\mathbb R.$$

Solution

The vector $\mathbf n=(a,b,c)$ orthogonal to $\pi$ has to be orthogonal to $\mathbf v=(1,-1,0)$. Hence, $a-b=0$, so $\mathbf n=(a,a,c)$, with $(a,c)\neq(0,0)$. We get that $\pi$ has equation: $$ax+ay+cz=d.$$ Since $A, O\in\pi$, we obtain that $$\left\{\begin{array}{l} -a+c=d\\ d=0 \end{array}\right.\qquad\Longleftrightarrow \qquad a=c,\ d=0.$$ We finally obtain the equation of $\pi$: $$x+y+z=0.$$

Exercise #6

Let $$A_k= \begin{pmatrix} 1&0&k\\ 0&2&0\\ k&0&1 \end{pmatrix}. $$

Find $k$ such that $\textbf{v}=(1,1,1)^T$ is an eigenvactor of $A_k$.
For such a $k$, find a basis for $\mathbb{R}^3$ made by eigenvectors of $A_k$.

Solution

a. We have that $A_k(1,1,1)^T=\lambda(1,1,1)^T$, for some $\lambda$ if and only if $1+k=\lambda$, $2=\lambda$, $k+1=\lambda$. So, $\lambda=2$ and $k=1$.

b. Let $k=1$. We look for the eigenvalues and corresponding eigenvectors of $A_1$. \begin{align*} {det}\,(A_1-\lambda I)&=det\,\begin{pmatrix} 1-\lambda&0&1\\ 0&2-\lambda&0\\ 1&0&1-\lambda \end{pmatrix}=(2-\lambda){det}\,\begin{pmatrix}1-\lambda&1\\1&1-\lambda\end{pmatrix}=(2-\lambda)[(1-\lambda)^2-1] =-\lambda(\lambda-2)^2. \end{align*} So, $\lambda_1=0$, $\lambda_{2,3}=2$ are eigenvalues of $A_1$. For $\lambda=2$, we have that $$ A_1-2I=\begin{pmatrix} -1&0&1\\ 0&0&0\\ 1&0&-1 \end{pmatrix}, $$ So, rk$\,(A_1-2I)=1$, hence dim$\,\textrm{ker}\,(A_1-2I)=3-1=2$ and $2$ is a regular eigenvalue. Analyzing $A_1-2I$, we notice that $\textbf{v}=(x,y,z)^\top$ is an eigenvector of $A_1$ relative to the eigenvalue $2$ if and only if $x=z$, hence we can take the two linear independent eigenvalues $(1,0,1)^\top$ and $(0,1,0)^\top$. Concerning the eigenvalue $0$, $A_1\textbf{v}=\textbf{0}$ if and only if $y=0$ and $z=-x$, so we can take the eigenvector $(1,0,-1)^\top$. Finally, we have obtained the basis $$\big\{\textbf{v}_1=(1,0,1)^\top, \textbf{v}_2=(0,1,0)^\top, \textbf{v}_3=(1,0,-1)^\top\big\}.$$