Consider the sets of sequences
$$X = \{(x_n) : x_n \in \{0, 1\}, n \in \mathbb{N}\}$$
and
$$Y = \{(x_n) \in X : x_n = 1 \text{ for at most finitely many}\}$$
Then
a.
$X$ is countable, $Y$ is finite.
b.
$X$ is uncountable, $Y$ is countable.
Correct Answer
c.
$X$ is countable, $Y$ is countable.
d.
$X$ is uncountable, $Y$ is uncountable
Step-by-Step AI Explanation
Verified Method
The set $X = \{0, 1\}^{\mathbb{N}}$ is the set of all binary sequences, which is in bijection with the power set $\mathcal{P}(\mathbb{N})$. By Cantor's theorem, this set is uncountable.
The set $Y$ consists of all binary sequences with only finitely many 1s. This is in bijection with the collection of all finite subsets of $\mathbb{N}$:
$$\mathcal{P}_{\text{fin}}(\mathbb{N}) = \bigcup_{k=0}^{\infty} \{A \subset \mathbb{N} : |A| = k\}$$
Since each set of subsets of size $k$ is countable and a countable union of countable sets is countable, $Y$ is countably infinite (countable).
Therefore, $X$ is uncountable and $Y$ is countable, which corresponds to option B.
A symmetric real matrix is positive definite if and only if all its leading principal minors are strictly positive (Sylvester's criterion). For the matrix $A = \begin{pmatrix} 3 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 3 \end{pmatrix}$:
- First leading principal minor: $\Delta_1 = 3 > 0$.
- Second leading principal minor: $\Delta_2 = \det\begin{pmatrix} 3 & -1 \\ -1 & 2 \end{pmatrix} = 3(2) - (-1)(-1) = 6 - 1 = 5 > 0$.
- Third leading principal minor (determinant): $\Delta_3 = 3(6 - 1) - (-1)(-3 - 0) + 0 = 3(5) - 3 = 15 - 3 = 12 > 0$.
Since all leading principal minors are strictly positive, the matrix is positive definite.
Q3mcq
Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be given by
$$f(x, y) = (x^2, y^2 + \sin x)$$
Then the derivative of $f$ at $(x, y)$ is the linear transformation given by
$\begin{pmatrix} 2x & 0 \\ 2y & \cos x \end{pmatrix}$
c.
$\begin{pmatrix} 2y & \cos x \\ 2x & 0 \end{pmatrix}$
d.
$\begin{pmatrix} 2x & 2y \\ 0 & \cos x \end{pmatrix}$
Step-by-Step AI Explanation
Verified Method
Let $f(x, y) = (f_1(x, y), f_2(x, y)) = (x^2, y^2 + \sin x)$. The derivative of $f$ at $(x, y)$ is given by its Jacobian matrix:
$$Df(x, y) = \begin{pmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\[6pt] \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{pmatrix} = \begin{pmatrix} 2x & 0 \\ \cos x & 2y \end{pmatrix}.$$
Therefore, option A is correct.
Q4mcq
A function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ is defined by $f(x, y) = xy$. Let $v = (1, 2)$ and $a = (a_1, a_2)$ be two elements of $\mathbb{R}^2$. The directional derivative of $f$ in the direction of $v$ at $a$ is
a.
$a_1 + 2a_2$
b.
$a_2 + 2a_1$
Correct Answer
c.
$\frac{a_2}{2} + a_1$
d.
$\frac{a_1}{2} + a_2$
Step-by-Step AI Explanation
Verified Method
The directional derivative of a differentiable scalar field $f$ at a point $a$ in the direction of vector $v$ is given by $\nabla f(a) \cdot v$.
The gradient of $f(x, y) = xy$ is $\nabla f(x, y) = (y, x)$. At $a = (a_1, a_2)$, we have $\nabla f(a_1, a_2) = (a_2, a_1)$.
Taking the dot product with $v = (1, 2)$:
$$D_v f(a) = \nabla f(a) \cdot v = a_2(1) + a_1(2) = a_2 + 2a_1.$$
Thus, option B is correct.
Using the sum of cubes formula $\sum_{j=0}^{m-1} j^3 = \frac{(m-1)^2 m^2}{4}$ with $m = 2n$:
$$\sum_{j=0}^{2n-1} j^3 = \frac{(2n-1)^2(2n)^2}{4} = n^2(2n-1)^2 = n^2(4n^2 - 4n + 1) = 4n^4 - 4n^3 + n^2.$$
Dividing by $n^4$ and taking the limit as $n \to \infty$:
$$\lim_{n \to \infty} \frac{4n^4 - 4n^3 + n^2}{n^4} = 4.$$
Alternatively, writing as a Riemann sum:
$$\lim_{n \to \infty} \frac{1}{n} \sum_{j=0}^{2n-1} \left(\frac{j}{n}\right)^3 = \int_0^2 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^2 = \frac{16}{4} = 4.$$
Thus, option A is correct.
Q6mcq
$f : \mathbb{R} \rightarrow \mathbb{R}$ is such that $f(0) = 0$ and $\left|\frac{df}{dx}(x)\right| \le 5$ for all $x$. We can conclude that $f(1)$ is in
a.
$(5, 6)$
b.
$[-5, 5]$
Correct Answer
c.
$(-\infty, -5) \cup (5, \infty)$
d.
$[-4, 4]$
Step-by-Step AI Explanation
Verified Method
By the Mean Value Theorem applied to $f$ on the interval $[0, 1]$, there exists $c \in (0, 1)$ such that:
$$f(1) - f(0) = f'(c)(1 - 0) = f'(c).$$
Since $f(0) = 0$, we have $f(1) = f'(c)$. Given that $|f'(x)| \le 5$ for all $x \in \mathbb{R}$, it follows that $|f(1)| = |f'(c)| \le 5$, meaning $f(1) \in [-5, 5]$.
Q7mcq
Which of the following subsets of $\mathbb{R}^4$ is a basis of $\mathbb{R}^4$?
$$B_1 = \{(1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)\}$$
$$B_2 = \{(1,0,0,0),(1,2,0,0),(1,2,3,0),(1,2,3,4)\}$$
$$B_3 = \{(1,2,0,0),(0,0,1,1),(2,1,0,0),(-5,5,0,0)\}$$
a.
$\text{B}_1$ and $\text{B}_2$ but not $\text{B}_3$
Correct Answer
b.
$\text{B}_1$, $\text{B}_2$ and $\text{B}_3$
c.
$\text{B}_1$ and $\text{B}_3$ but not $\text{B}_2$
d.
Only $\text{B}_1$
Step-by-Step AI Explanation
Verified Method
A set of 4 vectors forms a basis of $\mathbb{R}^4$ if and only if they are linearly independent, which is equivalent to the matrix formed by these vectors having a non-zero determinant.
- For $B_1$, the matrix formed by the vectors as rows is lower triangular with diagonal entries all equal to $1$. Its determinant is $1 \neq 0$, so $B_1$ is linearly independent and hence a basis.
- For $B_2$, the matrix formed with rows as vectors is lower triangular with diagonal entries $1, 2, 3, 4$. Its determinant is $1 \times 2 \times 3 \times 4 = 24 \neq 0$, so $B_2$ is also a basis.
- For $B_3$, notice that the four vectors only have non-zero entries in the first, second, and fourth/third components, but more directly: $(-5, 5, 0, 0) = -5(1, 2, 0, 0) + 0(0, 0, 1, 1) + 0\dots$ actually, $(-5, 5, 0, 0) = -5(1, 2, 0, 0) + 15(0, 1, 0, 0)$. Notice the first, third, and fourth vectors all have 0 in the 3rd and 4th components, meaning three vectors lie in a 2-dimensional subspace (span of $e_1, e_2$), so they are linearly dependent. Hence $B_3$ cannot be a basis.
Therefore, $B_1$ and $B_2$ are bases, but $B_3$ is not.
Q8mcq
Let $D_1 = \det \begin{pmatrix} a & b & c \\ x & y & z \\ p & q & r \end{pmatrix}$ and $D_2 = \det \begin{pmatrix} -x & a & -p \\ y & -b & q \\ z & -c & r \end{pmatrix}$. Then
a.
$D_1 = D_2$
b.
$D_1 = 2D_2$
c.
$D_1 = -D_2$
Correct Answer
d.
$2D_1 = D_2$
Step-by-Step AI Explanation
Verified Method
We are given:
$$D_1 = \det \begin{pmatrix} a & b & c \\ x & y & z \\ p & q & r \end{pmatrix}$$
Consider the matrix in $D_2$:
$$M = \begin{pmatrix} -x & a & -p \\ y & -b & q \\ z & -c & r \end{pmatrix}$$
Taking the transpose of $M$ (which leaves the determinant unchanged):
$$D_2 = \det(M^T) = \det \begin{pmatrix} -x & y & z \\ a & -b & -c \\ -p & q & r \end{pmatrix}$$
Factor $-1$ from the first row and $-1$ from the second row:
$$D_2 = (-1)(-1) \det \begin{pmatrix} x & -y & -z \\ -a & b & c \\ -p & q & r \end{pmatrix} = \det \begin{pmatrix} x & -y & -z \\ -a & b & c \\ -p & q & r \end{pmatrix}$$
Now factor $-1$ from the second column and $-1$ from the third column:
$$D_2 = (-1)(-1) \det \begin{pmatrix} x & y & z \\ -a & -b & -c \\ -p & -q & -r \end{pmatrix} = \det \begin{pmatrix} x & y & z \\ -a & -b & -c \\ -p & -q & -r \end{pmatrix}$$
Factor $-1$ from the second row and $-1$ from the third row:
$$D_2 = (-1)(-1) \det \begin{pmatrix} x & y & z \\ a & b & c \\ p & q & r \end{pmatrix} = \det \begin{pmatrix} x & y & z \\ a & b & c \\ p & q & r \end{pmatrix}$$
Interchanging the first row and the second row flips the sign:
$$D_2 = -\det \begin{pmatrix} a & b & c \\ x & y & z \\ p & q & r \end{pmatrix} = -D_1.$$
Thus, $D_1 = -D_2$.
Q9mcq
Consider the matrix $A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$, where $\theta = \frac{2\pi}{31}$. Then $A^{2015}$ equals
The matrix $A$ represents a rotation matrix by an angle $- heta$ (or its transpose/inverse depending on convention, specifically $A^k = \begin{pmatrix} \cos(k\theta) & \sin(k\theta) \\ -\sin(k\theta) & \cos(k\theta) \end{pmatrix}$).
Given $\theta = \frac{2\pi}{31}$, the order of rotation is $31$ because $31\theta = 2\pi$.
We divide $2015$ by $31$:
$$2015 = 31 \times 65$$
Thus, $2015\theta = 65 \times 2\pi$, which means:
$$A^{2015} = \begin{pmatrix} \cos(2015\theta) & \sin(2015\theta) \\ -\sin(2015\theta) & \cos(2015\theta) \end{pmatrix} = \begin{pmatrix} \cos(65 \times 2\pi) & \sin(65 \times 2\pi) \\ -\sin(65 \times 2\pi) & \cos(65 \times 2\pi) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$$
Therefore, option B is correct.
Q10mcq
Let $J$ denote the matrix of order $n \times n$ with all entries $1$ and let $B$ be a $(3n) \times (3n)$ matrix given by
$$B = \begin{pmatrix} 0 & 0 & J \\ 0 & J & 0 \\ J & 0 & 0 \end{pmatrix}$$
Then the rank of $B$ is
a.
$2n$
b.
$3n - 1$
c.
$2$
d.
$3$
Correct Answer
Step-by-Step AI Explanation
Verified Method
The matrix $J$ is an $n \times n$ matrix with all entries equal to $1$. Since all rows of $J$ are identical and non-zero, the row space of $J$ is spanned by the single row vector $\mathbf{1} = (1, 1, \dots, 1) \in \mathbb{R}^n$, so $\text{rank}(J) = 1$.
The matrix $B$ is a block matrix of size $(3n) \times (3n)$ given by:
$$B = \begin{pmatrix} 0 & 0 & J \\ 0 & J & 0 \\ J & 0 & 0 \end{pmatrix}$$
The rows of $B$ are formed by block rows of the form $(0, 0, \mathbf{1})$, $(0, \mathbf{1}, 0)$, and $(\mathbf{1}, 0, 0)$ repeated $n$ times each.
Thus, the row space of $B$ is spanned by the three linearly independent row vectors:
$$r_1 = (0, 0, \dots, 0, \; 0, 0, \dots, 0, \; 1, 1, \dots, 1)$$
$$r_2 = (0, 0, \dots, 0, \; 1, 1, \dots, 1, \; 0, 0, \dots, 0)$$
$$r_3 = (1, 1, \dots, 1, \; 0, 0, \dots, 0, \; 0, 0, \dots, 0)$$
These three vectors are clearly linearly independent. Hence, the dimension of the row space of $B$ is $3$, meaning $\text{rank}(B) = 3$.
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Which of the following sets of functions from $\mathbb{R}$ to $\mathbb{R}$ is a vector space over $\mathbb{R}$?
$$S_1 = \left\{f \mid \lim_{x \to 3} f(x) = 0\right\}$$
$$S_2 = \left\{g \mid \lim_{x \to 3} g(x) = 1\right\}$$
$$S_3 = \left\{h \mid \lim_{x \to 3} h(x) \text{ exists}\right\}$$
a.
Only $S_1$
b.
Only $S_2$
c.
$S_1$ and $S_3$ but not $S_2$
d.
All the three are vector spaces
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Let $A$ be an $n \times m$ matrix with each entry equal to $+1, -1$ or $0$ such that every column has exactly one $+1$ and exactly one $-1$. We can conclude that
a.
$\text{Rank } A \le n-1$
b.
$\text{Rank } A = m$
c.
$n \le m$
d.
$n - 1 \le m$
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Let $C$ be the circle $|z| = 3/2$ in the complex plane that is oriented in the counter clockwise direction. The value $a$ for which $$\int_C \left( \frac{z+1}{z^2-3z+2} + \frac{a}{z-1} \right) dz = 0$$ is
a.
$1$
b.
$-1$
c.
$2$
d.
$-2$
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Let $f$ be a holomorphic function on $0 < |z| < \epsilon, \epsilon > 0$, given by a convergent Laurent series $\sum_{n=-\infty}^\infty a_n z^n$. Given also that $\lim_{z \to 0} |f(z)| = \infty$, We can conclude that
a.
$a_{-1} \neq 0$ and $a_{-n} = 0$ for all $n \ge 2$
b.
$a_{-N} \neq 0$ for some $N \ge 1$ and $a_{-n} = 0$ for all $n > N$
c.
$a_{-n} = 0$ for all $n \ge 1$
d.
$a_{-n} \neq 0$ for all $n \ge 1$
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Let G be an open set in $\mathbb{R}^n$. Two points $x, y \in G$ are said to be equivalent if they can be joined by a continuous path completely lying inside $G$. Number of equivalence classes is
a.
only one
b.
at most finite.
c.
at most countable
d.
can be finite, countable or uncountable
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Let $(x(t), y(t))$ satisfy the system of ODEs
$$\frac{dx}{dt} = -x + ty$$
$$\frac{dy}{dt} = tx - y$$
If $(x_1(t), y_1(t))$ and $(x_2(t), y_2(t))$ are two solutions and $\Phi(t) = x_1(t)y_2(t) - x_2(t)y_1(t)$ then $\frac{d\Phi}{dt}$ is equal to
a.
$-2\Phi$
b.
$2\Phi$
c.
$-\Phi$
d.
$\Phi$
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The boundary value problem $x^2 y'' - 2xy' + 2y = 0$, subject to the boundary conditions $y(1) + \alpha y'(1) = 1$, $y(2) + \beta y'(2) = 2$, has a unique solution if
a.
$\alpha = -1, \beta = 2$
b.
$\alpha = -1, \beta = -2$
c.
$\alpha = -2, \beta = 2$
d.
$\alpha = -3, \beta = \frac{2}{3}$
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Let $u(x,t)$ satisfy the initial boundary value problem
$$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} ; x \in (0,1), t > 0$$
$$u(x,0) = \sin(\pi x) ; x \in [0,1]$$
$$u(0,t) = u(1,t) = 0, t > 0$$, Then for $x \in (0,1), u\left(x, \frac{1}{\pi^2}\right)$ is equal to
a.
$e \sin(\pi x)$
b.
$e^{-1} \sin(\pi x)$
c.
$\sin(\pi x)$
d.
$\sin(\pi^{-1} x)$
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The values of $\alpha$ and $\beta$, such that
$$x_{n+1} = \alpha x_n\left(3 - \frac{x_n^2}{a}\right) + \beta x_n\left(1 + \frac{a}{x_n^2}\right)$$
has $3^{\text{rd}}$ order convergence to $\sqrt{a}$, are
a.
$\alpha = \frac{3}{8}, \beta = \frac{1}{8}$
b.
$\alpha = \frac{1}{8}, \beta = \frac{3}{8}$
c.
$\alpha = \frac{2}{8}, \beta = \frac{2}{8}$
d.
$\alpha = \frac{1}{4}, \beta = \frac{3}{4}$
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A bead slides without friction on a frictionless wire in the shape of a cycloid with equation
$$x = a(\theta - \sin\theta), y = a(1+\cos\theta), 0 \le \theta \le 2\pi$$. Then the Lagrangian function is
There are two boxes. Box 1 contains 2 red balls and 4 green balls. Box 2 contains 4 red balls and 2 green balls. A box is selected at random and a ball is chosen randomly from the selected box. If the ball turns out to be red, what is the probability that Box 1 had been selected?
a.
$\frac{1}{2}$
b.
$\frac{1}{3}$
c.
$\frac{2}{3}$
d.
$\frac{1}{6}$
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Suppose customers arrive in a shop according to a Poisson process with rate 4 per hour. The shop opens at 10:00 am. If it is given that the second customer arrives at 10:40 am, what is the probability that no customer arrived before 10:30 am ?
a.
$\frac{1}{4}$
b.
$e^{-2}$
c.
$\frac{1}{2}$
d.
$e^{-1/2}$
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Suppose $X_1, X_2, \dots X_n$ is a random sample from a distribution with probability density function
$$f(x)=3x^2 I_{(0,1)}(x), \text{ where } I_{(0,1)}(z)=\begin{cases} 1 & \text{if } z\in(0,1) \\ 0 & \text{otherwise} \end{cases}.$$
What is the probability density function $g(y)$ of $Y = \min\{X_1, X_2, \dots X_n\}$ ?
a.
$g(y)=3ny^{3n-1}I_{(0,1)}(y)$
b.
$g(y)=1-(1-y^3)^n I_{(0,1)}(y)$
c.
$g(y)=(1-y^3)^n I_{(0,1)}(y)$
d.
$g(y)=3ny^2(1-y^3)^{n-1}I_{(0,1)}(y)$
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$X_1, X_2, \dots X_n$ are independent and identically distributed $N(\theta, 1)$ random variables, where $\theta$ takes only integer values i.e. $\theta \in \{\dots, -2, -1, 0, 1, 2, \dots\}$.
Which of the following is the maximum likelihood estimator of $\theta$ ?
a.
$\overline{X}$
b.
Integer closest to $\overline{X}$
c.
Integer part of $\overline{X}$ (Largest integer $\le \overline{X}$)
d.
median of $(X_1, X_2, \dots, X_n)$
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Let $A$ be a $n \times n$ non-singular matrix with real entries. Let $B = A^T$ denote the transpose of $A$. Which of the following matrices are positive definite?
a.
$A+B$
b.
$A^{-1} + B^{-1}$
c.
$AB$
d.
$ABA$
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Let $\chi_A(x)$ denote the function which is 1 if $x \in A$ and 0 otherwise. Consider
$$f(x)=\sum_{n=1}^{200} \frac{1}{n^6} \chi_{\left[0, \frac{n}{200}\right]}(x), \, x \in [0,1]$$
Then $f(x)$ is
a.
Riemann integrable on $[0,1]$
b.
Lebesgue integrable on $[0,1]$
c.
is a continuous function on $[0,1]$
d.
is a monotone function on $[0,1]$
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A function $f(x,y)$ on $\mathbb{R}^2$ has the following partial derivatives
$$\frac{\partial f}{\partial x}(x,y)=x^2, \frac{\partial f}{\partial y}(x,y)=y^2$$
Then
a.
$f$ has directional derivatives in all directions everywhere.
b.
$f$ has a derivative at all points.
c.
$f$ has directional derivative only along the direction $(1,1)$ everywhere.
d.
$f$ does not have directional derivatives in any direction everywhere
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Let $d_1, d_2$ be the following metrics on $\mathbb{R}^n$.
$$d_1(x,y)=\sum_{i=1}^n |x_i-y_i|, \quad d_2(x,y)=\left(\sum_{i=1}^n |x_i-y_i|^2\right)^{1/2}$$
Then decide which of the following is a metric on $\mathbb{R}^n$.
Let $A$ be the following subset of $\mathbb{R}^2$.
$$A = \left\{(x,y):(x+1)^2+y^2 \leq 1\right\} \cup \left\{(x,y):y=x\sin\frac{1}{x}, x>0\right\}$$
Then
a.
$A$ is connected
b.
$A$ is compact
c.
$A$ is path connected
d.
$A$ is bounded
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Let $A = [a_{ij}]$ be an $n \times n$ matrix such that $a_{ij}$ is an integer for all $i, j$. Let $AB = I$ with $B = [b_{ij}]$ (where $I$ is the identity matrix). For a square matrix $C$, $\det C$ denotes its determinant. Which of the following statements is true?
a.
If $\det A = 1$ then $\det B = 1$
b.
A sufficient condition for each $b_{ij}$ to be an integer is that $\det A$ is an integer
c.
$B$ is always an integer matrix
d.
A necessary condition for each $b_{ij}$ to be an integer is $\det A \in \{-1, +1\}$
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Let $A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ and let $\alpha_n$ and $\beta_n$ denote the two eigenvalues of $A^n$ such that $|\alpha_n| \ge |\beta_n|$. Then
a.
$\alpha_n \to \infty$ as $n \to \infty$
b.
$\beta_n \to 0$ as $n \to \infty$
c.
$\beta_n$ is positive if $n$ is even.
d.
$\beta_n$ is negative if $n$ is odd
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Let $n$ be an odd number $\geq 7$. Let $A = [a_{ij}]$ be an $n \times n$ matrix with $a_{i, i+1} = 1$ for all $i = 1, 2, \ldots, n-1$ and $a_{n, 1} = 1$. Let $a_{ij} = 0$ for all the other pairs $(i, j)$. Then we can conclude that
a.
A has $1$ as an eigenvalue.
b.
A has $-1$ as an eigenvalue.
c.
A has at least one eigenvalue with multiplicity $\geq 2$.
d.
A has no real eigenvalues
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Let $W_1, W_2, W_3$ be three distinct subspaces of $\mathbb{R}^{10}$ such that each $W_i$ has dimension $9$. Let $W = W_1 \cap W_2 \cap W_3$. Then we can conclude that
a.
$W$ may not be a subspace of $\mathbb{R}^{10}$
b.
$\dim W \leq 8$
c.
$\dim W \geq 7$
d.
$\dim W \leq 3$
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Consider the polynomial $P(z) = \left(\sum_{n=0}^5 a_n z^n\right)\left(\sum_{n=0}^9 b_n z^n\right)$ where $a_n, b_n \in \mathbb{R}$ $\forall n$, $a_5 \neq 0, b_9 \neq 0$. Then counting roots with multiplicity we can conclude that $P(z)$ has
a.
at least two real roots
b.
$14$ complex roots
c.
no real roots
d.
$12$ complex roots
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Let $\mathbb{D}$ be the open unit disc in $\mathbb{C}$. Let $g : \mathbb{D} \to \mathbb{D}$ be holomorphic, $g(0) = 0$, and let
$$h(z) = \begin{cases} g(z)/z, & z \in \mathbb{D}, z \neq 0 \\ g'(0), & z = 0 \end{cases}$$
Which of the following statements are true?
a.
$h$ is holomorphic in $\mathbb{D}$.
b.
$h(\mathbb{D}) \subseteq \overline{\mathbb{D}}$
c.
$|g'(0)| > 1$
d.
$|g(1/2)| \leq 1/2$
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Consider the following subsets of the group of $2\times 2$ non-singular matrices over $\mathbb{R}$:
$$G = \left\{\begin{pmatrix} a & b \\ 0 & d \end{pmatrix} : a, b, d \in \mathbb{R},\ ad = 1\right\}$$
$$H = \left\{\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} : b \in \mathbb{R}\right\}$$
Which of the following statements are correct?
a.
$G$ forms a group under matrix multiplication.
b.
$H$ is a normal subgroup of $G$.
c.
The quotient group $G/H$ is well-defined and is Abelian.
d.
The quotient group $G/H$ is well defined and is isomorphic to the group of $2\times 2$ diagonal matrices (over $\mathbb{R}$) with determinant $1$
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Let $\mathbb{C}$ be the field of complex numbers and $\mathbb{C}^*$ be the group of non zero complex numbers under multiplication. Then which of the following are true ?
a.
$\mathbb{C}^*$ is cyclic
b.
Every finite subgroup of $\mathbb{C}^*$ is cyclic
c.
$\mathbb{C}^*$ has finitely many finite subgroups
d.
Every proper subgroup of $\mathbb{C}^*$ is cyclic
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Let $R$ be a commutative ring with unity and $R[x]$ be the polynomial ring in one variable. For a non zero $f = \sum_{n=0}^N a_n x^n$, define $\omega(f)$ to be the smallest $n$ such that $a_n \neq 0$. Also $\omega(0) = +\infty$.
Then which of the following statements is/are true ?
a.
$\omega(f+g) \ge \min(\omega(f), \omega(g))$
b.
$\omega(fg) \ge \omega(f) + \omega(g)$
c.
$\omega(f+g) = \min(\omega(f), \omega(g))$ if $\omega(f) \neq \omega(g)$
d.
$\omega(fg) = \omega(f) + \omega(g)$ if $R$ is an integral domain
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Let $S = \left\{x, y \in \mathbb{R}^2 \mid -1 \leq x \leq 1 \text{ and } -1 \leq y \leq 1\right\}$
Let $T = S \setminus \{(0, 0)\}$, the set obtained by removing the origin from $S$.
Let $f$ be a continuous function from T to $\mathbb{R}$. Choose all correct options.
a.
Image of $f$ must be connected.
b.
Image of $f$ must be compact.
c.
Any such continuous function $f$ can be extended to a continuous function from S to $\mathbb{R}$.
d.
If $f$ can be extended to a continuous function from S to $\mathbb{R}$ then the image of $f$ is bounded
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Let $x : [0, 3\pi] \to \mathbb{R}$ be a nonzero solution of the ODE
$$x''(t) + e^{t^2} x(t) = 0, \text{ for } t \in [0, 3\pi].$$
Then the cardinality of the set $\{t \in [0, 3\pi] : x(t) = 0\}$ is
a.
equal to 1
b.
greater than or equal to 2
c.
equal to 2
d.
greater than or equal to 3
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Consider the initial value problem
$$y'(t) = f(y(t)), \quad y(0) = a \in \mathbb{R} \text{ where } f : \mathbb{R} \to \mathbb{R}.$$
Which of the following statements are necessarily true ?
a.
There exists a continuous function $f : \mathbb{R} \to \mathbb{R}$ and $a \in \mathbb{R}$ such that the above problem does not have a solution in any neighbourhood of 0.
b.
The problem has a unique solution for every $a \in \mathbb{R}$ when $f$ is Lipschitz continuous.
c.
When $f$ is twice continuously differentiable, the maximal interval of existence for the above initial value problem is $\mathbb{R}$.
d.
The maximal interval of existence for the above problem is $\mathbb{R}$ when $f$ is bounded and continuously differentiable
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Consider the wave equation for $u(x, t)$
$$\left. \begin{aligned} &\frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial x^2} = 0, \quad (x, t) \in \mathbb{R} \times (0, \infty) \\ &u(x, 0) = f(x), \quad x \in \mathbb{R} \\ &\frac{\partial u}{\partial t}(x, 0) = g(x), \quad x \in \mathbb{R} \end{aligned} \right\}$$
Let $u_i$ be the solution of the above problem with $f = f_i$ and $g = g_i$ for $i = 1, 2$, where $f_i : \mathbb{R} \rightarrow \mathbb{R}$ and $g_i : \mathbb{R} \rightarrow \mathbb{R}$ are given $C^2$ functions satisfying $f_1(x) = f_2(x)$ and $g_1(x) = g_2(x)$, for every $x \in [-1, 1]$. Which of the following statements are necessarily true
Let $u : \mathbb{R}^2 \setminus \{(0, 0)\} \rightarrow \mathbb{R}$ be a $C^2$ function satisfying $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$, for all $(x, y) \neq (0, 0)$. Suppose $u$ is of the form $u(x, y) = f\left(\sqrt{x^2 + y^2}\right)$, where $f : (0, \infty) \rightarrow \mathbb{R}$, is a nonconstant function, then
The Cauchy problem
$$\left. \begin{aligned} y\frac{\partial u}{\partial x} - x\frac{\partial u}{\partial y} = 0 \\ u = g \text{ on } \Gamma \end{aligned} \right\}$$
has a unique solution in a neighbourhood of $\Gamma$ for every differentiable function $g : \Gamma \rightarrow \mathbb{R}$ if
a.
$\Gamma = \{(x, 0) : x > 0\}$
b.
$\Gamma = \{(x, y) : x^2 + y^2 = 1\}$
c.
$\Gamma = \{(x, y) : x + y = 1, x > 1\}$
d.
$\Gamma = \{(x, y) : y = x^2, \ x > 0\}$
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Let $y(x)$ be a piecewise continuously differentiable function on $[0,4]$. Then the functional
$$J[y] = \int_0^4 (y'-1)^2 (y'+1)^2 dx$$
attains minimum if $y = y(x)$ is
a.
$y = \frac{x}{2}, \quad 0 \leq x \leq 4$
b.
$y = \begin{cases} -x & 0 \leq x \leq 1 \\ x-2 & 1 \leq x \leq 4 \end{cases}$
c.
$y = \begin{cases} 2x & 0 \leq x \leq 2 \\ -x+6 & 2 \leq x \leq 4 \end{cases}$
d.
$y = \begin{cases} x & 0 \leq x \leq 3 \\ -x+6 & 3 \leq x \leq 4 \end{cases}$
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Which of the following are the characteristic numbers and the corresponding eigenfunctions for the Fredholm homogeneous equation whose kernel is
$$K(x,t) = \begin{cases} (x+1)t, & 0 \leq x \leq t \\ (t+1)x, & t \leq x \leq 1 \end{cases} \;?$$
a.
$1, e^x$
b.
$-\pi^2, \pi \sin \pi x + \cos \pi x$
c.
$-4\pi^2, \pi \sin \pi x + \pi \cos 2\pi x$
d.
$-\pi^2, \pi \cos \pi x + \sin \pi x$
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