The given sequence is
$$a_n = \left(1 + \frac{(-1)^n}{n}\right)^n.$$
Consider the subsequences of even and odd indices:
1. For even indices $n = 2k$:
$$a_{2k} = \left(1 + \frac{1}{2k}\right)^{2k}.$$
As $k \to \infty$,
$$\lim_{k \to \infty} a_{2k} = \lim_{k \to \infty} \left(1 + \frac{1}{2k}\right)^{2k} = e.$$
2. For odd indices $n = 2k - 1$:
$$a_{2k-1} = \left(1 - \frac{1}{2k-1}\right)^{2k-1}.$$
As $k \to \infty$,
$$\lim_{k \to \infty} a_{2k-1} = \lim_{k \to \infty} \left(1 - \frac{1}{2k-1}\right)^{2k-1} = e^{-1} = \frac{1}{e}.$$
Since the sequence oscillates between these two subsequential limits, the set of limit points of $(a_n)$ is $\left\{\frac{1}{e}, e\right\}$.
Therefore,
$$\limsup_{n \to \infty} a_n = e \quad \text{and} \quad \liminf_{n \to \infty} a_n = \frac{1}{e}.$$
This matches Option D.
Q3mcq
For $a > 0$, the series
$$\sum_{n=1}^\infty a^{\ln n}$$
is convergent if and only if
a.
$0 < a < e$
b.
$0 < a \le e$
c.
$0 < a < \frac{1}{e}$
Correct Answer
d.
$0 < a \le \frac{1}{e}$
Step-by-Step AI Explanation
Verified Method
We can rewrite the general term of the series using the properties of logarithms and exponentials:
$$a^{\ln n} = \left(e^{\ln a}\right)^{\ln n} = e^{\ln a \cdot \ln n} = \left(e^{\ln n}\right)^{\ln a} = n^{\ln a} = \frac{1}{n^{-\ln a}}$$
Thus, the given series is:
$$\sum_{n=1}^\infty a^{\ln n} = \sum_{n=1}^\infty \frac{1}{n^{-\ln a}}$$
This is a standard $p$-series of the form $\sum_{n=1}^\infty \frac{1}{n^p}$ with $p = -\ln a$.
A $p$-series converges if and only if $p > 1$. Therefore, the series converges if and only if:
$$-\ln a > 1 \iff \ln a < -1 \iff a < e^{-1} = \frac{1}{e}$$
Since it is given that $a > 0$, the condition for convergence is:
$$0 < a < \frac{1}{e}$$
Hence, option C is the correct choice.
Q4mcq
Let $f : \mathbb{R} \to \mathbb{R}$ be defined by
$$f(x) = \begin{cases} \frac{\sin x}{x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$$
Then
a.
$f$ is not continuous
b.
$f$ is continuous but not differentiable
c.
$f$ is differentiable
Correct Answer
d.
$f$ is not bounded
Step-by-Step AI Explanation
Verified Method
First, consider the continuity of $f$ at $x = 0$:
$$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin x}{x} = 1 = f(0)$$
Since $\frac{\sin x}{x}$ is continuous for all $x \neq 0$, $f$ is continuous everywhere on $\mathbb{R}$.
Next, examine the differentiability of $f$ at $x = 0$ using the limit definition of the derivative:
$$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{\frac{\sin h}{h} - 1}{h} = \lim_{h \to 0} \frac{\sin h - h}{h^2}$$
Using the Taylor expansion $\sin h = h - \frac{h^3}{6} + O(h^5)$, we have:
$$f'(0) = \lim_{h \to 0} \frac{-\frac{h^3}{6} + O(h^5)}{h^2} = \lim_{h \to 0} \left(-\frac{h}{6} + O(h^3)\right) = 0$$
Since this limit exists and is finite, $f$ is differentiable at $x = 0$ with $f'(0) = 0$.
For $x \neq 0$, $f(x)$ is a quotient of differentiable functions, so it is differentiable with:
$$f'(x) = \frac{x\cos x - \sin x}{x^2}$$
Thus, $f$ is differentiable on all of $\mathbb{R}$.
Moreover, $|f(x)| \le 1$ for all $x \in \mathbb{R}$, so $f$ is also bounded. Therefore, option C is correct.
Q5mcq
Let
$A = \{n \in \mathbb{N} : n = 1 \text{ or the only prime factors of } n \text{ are } 2 \text{ or } 3\}$,
for example, $6 \in A, 10 \notin A$. Let $S = \sum_{n \in A} \frac{1}{n}$. Then
a.
$A$ is finite
b.
$S$ is a divergent series
c.
$S = 3$
Correct Answer
d.
$S = 6$
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Q6mcq
For $n \ge 1$, let $f_n(x) = xe^{-nx^2}, x \in \mathbb{R}$.
Then the sequence $\{f_n\}$ is
a.
uniformly convergent on $\mathbb{R}$
Correct Answer
b.
uniformly convergent only on compact subsets of $\mathbb{R}$
c.
bounded and not uniformly convergent on $\mathbb{R}$
d.
a sequence of unbounded functions
No explanation saved for this question yet.
Q7mcq
Let $A$ be a $4\times 4$ matrix. Suppose that the null space $N(A)$ of $A$ is
$$\{(x, y, z, w) \in \mathbb{R}^4 : x + y + z = 0, x + y + w = 0\}.$$
Then
$S = \{(1, 1, 1, 0), (1, 1, 0, 1)\}$ is a basis of $N(A)$
Step-by-Step AI Explanation
Verified Method
The null space $N(A)$ of the matrix $A$ is given by the solution set to the system of linear equations in $\mathbb{R}^4$:
$$x + y + z = 0$$
$$x + y + w = 0$$
These two equations are linearly independent, so the number of free variables is $4 - 2 = 2$. Thus, the nullity of $A$ is:
$$\operatorname{nullity}(A) = \dim(N(A)) = 2$$
By the Rank-Nullity Theorem, for a $4 \times 4$ matrix $A$ acting on $\mathbb{R}^4$:
$$\operatorname{rank}(A) + \operatorname{nullity}(A) = 4$$
$$\operatorname{rank}(A) + 2 = 4 \implies \operatorname{rank}(A) = 2$$
Since the dimension of the column space of $A$ is equal to $\operatorname{rank}(A)$, we have:
$$\dim(\text{column space}(A)) = 2$$
Evaluating the options:
- A) Incorrect, as the dimension is $2$, not $1$.
- B) Correct, $\dim(\text{column space}(A)) = 2$.
- C) Incorrect, $\operatorname{rank}(A) = 2$, not $1$.
- D) Incorrect, because substituting $(1, 1, 1, 0)$ into $x + y + z = 0$ gives $1 + 1 + 1 = 3 \neq 0$, so the vectors in $S$ do not even lie in $N(A)$.
Q8mcq
Let $A$ and $B$ be real invertible matrices such that $AB = -BA$. Then
We are given that $A$ and $B$ are real invertible matrices satisfying the anticommutation relation:
$$AB = -BA$$
Since $B$ is invertible, we can multiply by $B^{-1}$ on the right:
$$A = -BAB^{-1}$$
Taking the trace of both sides and using the cyclic and linear properties of the trace function:
$$\operatorname{Trace}(A) = \operatorname{Trace}(-B A B^{-1}) = -\operatorname{Trace}(B A B^{-1}) = -\operatorname{Trace}(A B^{-1} B) = -\operatorname{Trace}(A)$$
This implies:
$$2\operatorname{Trace}(A) = 0 \implies \operatorname{Trace}(A) = 0$$
Similarly, since $A$ is invertible, we can multiply the original equation by $A^{-1}$ on the left:
$$B = -A^{-1}BA$$
Taking the trace of both sides:
$$\operatorname{Trace}(B) = \operatorname{Trace}(-A^{-1}BA) = -\operatorname{Trace}(A^{-1}BA) = -\operatorname{Trace}(B A A^{-1}) = -\operatorname{Trace}(B)$$
This gives:
$$2\operatorname{Trace}(B) = 0 \implies \operatorname{Trace}(B) = 0$$
Therefore, $\operatorname{Trace}(A) = \operatorname{Trace}(B) = 0$.
Q9mcq
Let $A$ be an $n \times n$ self-adjoint matrix with eigenvalues $\lambda_1, \dots, \lambda_n$.
Let $\|X\|_2 = \sqrt{|x_1|^2 + \dots + |x_n|^2}$ for $X = (x_1, \dots, x_n) \in \mathbb{C}^n$.
If $p(A) = a_0 I + a_1 A + \dots + a_n A^n$ then $\sup_{\|X\|_2 = 1} \|p(A)X\|_2$ is equal to
Since $A$ is self-adjoint, by the spectral theorem there exists an orthonormal basis of eigenvectors of $A$ corresponding to the real eigenvalues $\lambda_1, \dots, \lambda_n$. Thus, $p(A)$ is a normal matrix with eigenvalues $p(\lambda_j) = a_0 + a_1\lambda_j + \dots + a_n\lambda_j^n$ for $j = 1, \dots, n$. The operator norm induced by the $l_2$-norm, $\sup_{\|X\|_2 = 1} \|p(A)X\|_2$, is the spectral norm of $p(A)$, which equals its largest singular value. For a normal matrix, the singular values are the absolute values of its eigenvalues. Hence, the supremum is $\max_{1 \le j \le n} |p(\lambda_j)| = \max\{|a_0 + a_1\lambda_j + \dots + a_n\lambda_j^n| : 1 \le j \le n\}$.
Q10mcq
Let $p(x) = \alpha x^2 + \beta x + \gamma$ be a polynomial, where $\alpha, \beta, \gamma \in \mathbb{R}$. Fix $x_0 \in \mathbb{R}$.
Let $S = \left\{(a, b, c) \in \mathbb{R}^3 : p(x) = a(x - x_0)^2 + b(x - x_0) + c \text{ for all } x \in \mathbb{R}\right\}$
Then the number of elements in $S$ is
a.
0
b.
1
Correct Answer
c.
strictly greater than 1 but finite
d.
infinite
Step-by-Step AI Explanation
Verified Method
By Taylor's expansion of a polynomial about the point $x_0$, any polynomial of degree at most 2 can be written uniquely as:
$$p(x) = p(x_0) + p'(x_0)(x - x_0) + \frac{p''(x_0)}{2}(x - x_0)^2$$
Equating this to $a(x - x_0)^2 + b(x - x_0) + c$, we get:
$$c = p(x_0) = \alpha x_0^2 + \beta x_0 + \gamma$$
$$b = p'(x_0) = 2\alpha x_0 + \beta$$
$$a = \frac{p''(x_0)}{2} = \alpha$$
Since the polynomials are identical for all $x \in \mathbb{R}$, the coefficients $a, b, c$ are uniquely determined in terms of $\alpha, \beta, \gamma,$ and $x_0$.
Thus, the set $S$ contains exactly $1$ element.
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Let $C$ denote the unit circle centered at the origin in $\mathbb{C}$. Then $\frac{1}{2\pi i} \int_C \left|1 + z + z^2\right|^2 dz$, where the integral is taken anti-clockwise along $C$, equals
a.
0
b.
1
c.
2
d.
3
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For an odd integer $k \ge 1$, let $F$ be the set of all entire function $f$ such that $f(x) = |x|^k$ for all $x \in (-1, 1)$. Then the cardinality of $F$ is
a.
0
b.
1
c.
strictly greater than 1 but finite
d.
infinite
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Suppose $f$ is holomorphic in an open neighbourhood of $z_0 \in \mathbb{C}$. Given that the series $$\sum_{n=0}^\infty f^{(n)}(z_0)$$ converges absolutely, we can conclude that
a.
$f$ is constant
b.
$f$ is a polynomial
c.
$f$ can be extended to an entire function
d.
$f(x) \in \mathbb{R}$ for all $x \in \mathbb{R}$
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Suppose $x\colon [0, \infty) \to [0, \infty)$ is continuous and $x(0) = 0$. If $(x(t))^2 \le 2 + \int_0^t x(s) \, ds$, $\forall t \ge 0$. Then which of the following is TRUE?
Suppose $u \in C^2(B)$, $B$ is the unit ball in $\mathbb{R}^2$, satisfies
$$\Delta u = f \text{ in } B$$
$$au + \frac{\partial u}{\partial n} = g \text{ on } \alpha > 0$$ where $n$ is the unit outward normal to $B$. If a solution exists then
a.
it is unique
b.
there are exactly two solutions
c.
there are exactly three solutions.
d.
there are infinitely many solutions
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The infimum of $\int_0^1 (u'(t))^2 \, dt$ on the class of functions $\{u \in C^1[0,1] \text{ such that } u(0) = 0 \text{ and } \max_{[0,1]} |u| = 1\}$ is equal to
a.
0
b.
1/2
c.
1
d.
2
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Consider a spherical pendulum consisting of a particle of mass $m$ which moves under gravity on a smooth sphere of radius $a$. In terms of spherical polar angles $\theta, \phi$, with $\theta$ measured up from the downward vertical, the Lagrangian is given by
A box contains 40 numbered red balls and 60 numbered black balls. From the box, balls are drawn one by one at random without replacement till all the balls are drawn. The probability that the last ball drawn is black equals
a.
$1/100$
b.
$1/60$
c.
$3/5$
d.
$2/3$
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$X_1, X_2, \dots$ are independent identically distributed random variables having common density $f$. Assume $f(x) = f(-x)$ for all $x \in \mathbb{R}$. Which of the following statements is correct
a.
$\frac{1}{n}(X_1 + \dots + X_n) \to 0$ in probability
b.
$\frac{1}{n}(X_1 + \dots + X_n) \to 0$ almost surely
Let $N_t$ denote the number of accidents up to time $t$. Assume that $\{N_t\}$ is a Poisson process with intensity $2$. Given that there are exactly $5$ accidents during the time period $[20, 30]$, what is the conditional probability that there is exactly one accident during the time period $[15, 25]$?
a.
$\frac{15}{32} e^{-10}$
b.
$20e^{-20}$
c.
$\frac{10^5}{5!} e^{-30}$
d.
$\frac{1}{5}$
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$X$ and $Y$ are independent random variables each having the density
$$f(t) = \frac{1}{\pi} \frac{1}{1+t^2}, -\infty < t < \infty.$$
Then the density function of $\frac{X+Y}{3}$ for $-\infty < t < \infty$ is given by
a.
$\frac{6}{\pi} \frac{1}{4+9t^2}$
b.
$\frac{6}{\pi} \frac{1}{9+4t^2}$
c.
$\frac{3}{\pi} \frac{1}{1+9t^2}$
d.
$\frac{3}{\pi} \frac{1}{9+t^2}$
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Suppose $\{X_1, \dots, X_n\}, n \geq 2$, is a random sample from the distribution with probability density function
$$f(x; \theta) = \begin{cases} \frac{\theta^\theta}{\Gamma(\theta)} x^{\theta-1} e^{-x\theta} & ; x > 0 \\ 0 & ; x \leq 0 \end{cases}$$
with $\theta > 0$. Then the method of moments estimator of $\theta$
a.
does not exist
b.
is $\frac{n}{\sum_{i=1}^n (x_i - 1)^2}$
c.
is $\frac{n}{\sum_{i=1}^n (X_i - \bar{X})^2}$
d.
is $\frac{n-1}{\sum_{i=1}^n (X_i - 1)^2}$
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Let $X_1, X_2, \dots, X_n$ for $n \geq 5$ be a random sample from the distribution with probability density function
$$f(x; \theta) = \begin{cases} e^{-(x-\theta)} & \text{if } x > \theta \\ 0 & \text{otherwise} \end{cases}$$
for $\theta > 0$. The confidence coefficient of the confidence interval
$$\left[ \min\{X_1, \dots, X_n\} - \frac{\ln 4}{n}, \min\{X_1, \dots, X_n\} + \frac{\ln 2}{n} \right]$$
for $\theta$, is
a.
$0.5$
b.
$0.75$
c.
$0.95$
d.
$1 - \frac{1}{2^n}$
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Let $X$ be a random sample from an exponential distribution with mean $1/\lambda$. If $\lambda$ has a prior distribution with probability density function $g(\lambda) = \begin{cases} \lambda e^{-\lambda} & ; \lambda > 0 \\ 0 & \lambda \le 0 \end{cases}$
then the Bayes estimator of $1/\lambda$ with respect to the squared error loss function is
a.
$\frac{2}{X + 1}$
b.
$\frac{1}{X}$
c.
$X$
d.
$\frac{X + 1}{2}$
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Consider the linear statistical model
$$y_{ij} = \mu + \tau_i + \varepsilon_{ij}; i = 1, 2, \dots, a; j = 1, 2, \dots, n$$
where $\mu$ is unknown, $\tau_i$ are independently and identically distributed as $N(0, \sigma_\tau^2)$, independently and identically $\varepsilon_{ij}$ are distributed as $N(0, \sigma^2)$, $\tau_i$ and $\tau_{ij}$ are independent for all $i$ and $j$. Note that $\tau_i$ is the $i^{\text{th}}$ treatment effect. Suppose $\text{SS}_{\text{total}}, \text{SS}_{\text{treatment}}, \text{SS}_{\text{error}}$ are total sum of squares, total treatment sum of squares and error sum of squares, respectively.
To test : $H_0 : \sigma_\tau^2 = 0 \text{ vs } H_A : \sigma_\tau^2 > 0$ which of the following statements is not true ?
a.
The sum of squares identity is $\text{SS}_{\text{total}} + \text{SS}_{\text{treatment}} + \text{SS}_{\text{error}}$
Suppose $(X_1, X_2)$ follows a bivariate normal distribution with $E(X_1) = E(X_2) = 0$, $V(X_1) = V(X_2) = 2$ and $\text{Cov}(X_1, X_2) = -1$. If $\phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{y^2}{2}} dy$ then $P[X_1 - X_2 > 6]$ is equal to
a.
$\phi(-1)$
b.
$\phi(-3)$
c.
$\phi(\sqrt{6})$
d.
$\phi(-\sqrt{6})$
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Consider the problem of drawing a sample of size 2 from a finite population of size 20. The sampling is done with replacement using probability proportional to size sampling scheme. The normed size measures $P_1, \dots, P_{20}$ are given by $p_i = \frac{1}{40}, i = 1, \dots, 10$, $p_i = \frac{3}{40}, i = 11, \dots, 20$. The expected number of distinct units drawn is
a.
$\frac{83}{80}$
b.
$\frac{157}{80}$
c.
$\frac{17}{16}$
d.
$\frac{31}{16}$
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Let $\alpha = 0.10110111011110...$ be a given real number written in base $10$, that is, the $n$-th digit of $\alpha$ is $1$, unless $n$ is of the form $\frac{k(k + 1)}{2} - 1$ in which case it is $0$. Choose all the correct statements from below
a.
$\alpha$ is a rational number
b.
$\alpha$ is an irrational number
c.
For every integer $q \ge 2$, there exists an integer $r \ge 1$ such that $\frac{r}{q} < \alpha < \frac{r + 1}{q}$
d.
$\alpha$ has no periodic decimal expansion
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For $a, b \in \mathbb{N}$, consider the sequence $$d_n = \frac{\begin{pmatrix} n \\ a \end{pmatrix}}{\begin{pmatrix} n \\ b \end{pmatrix}}$$ for $n > a, b$. Which of the following statements are true? As $n \to \infty$,
a.
$\{d_n\}$ converges for all values of $a$ and $b$
b.
$\{d_n\}$ converges if $a < b$
c.
$\{d_n\}$ converges if $a = b$
d.
$\{d_n\}$ converges if $a > b$
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Let $\{a_n\}$ be a sequence of real numbers satisfying $\sum_{n=1}^\infty |a_n - a_{n-1}| < \infty$. Then the series $\sum_{n=0}^\infty a_n x^n, x \in \mathbb{R}$ is convergent
a.
nowhere on $\mathbb{R}$
b.
everywhere on $\mathbb{R}$
c.
on some set containing $(-1, 1)$
d.
only on $(-1, 1)$
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Consider the vector space $V$ of real polynomials of degree less than or equal to $n$. Fix distinct real numbers $a_0, a_1, \dots, a_k$. For $p \in V$, $\max \left\{\left|p(a_j)\right| : 0 \le j \le k\right\}$ defines a norm on $V$
a.
only if $k < n$
b.
only if $k \ge n$
c.
if $k + 1 \le n$
d.
if $k \ge n + 1$
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Let $V$ be the vector space of polynomials of degree at most $3$ in a variable $x$ with coefficients in $\mathbb{R}$. Let $T = d/dx$ be the linear transformation of $V$ to itself given by differentiation. Which of the following are correct?
a.
$T$ is invertible
b.
$0$ is an eigenvalue of $T$
c.
There is a basis with respect to which the matrix of $T$ is nilpotent.
d.
The matrix of $T$ with respect to the basis $\{1, 1 + x, 1 + x + x^2, 1 + x + x^2 + x^3\}$ is diagonal
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Let $m, n, r$ be natural numbers. Let $A$ be an $m \times n$ matrix with real entries such that $(AA')^r = I$, where $I$ is the $m \times m$ identity matrix and $A'$ is the transpose of the matrix $A$. We can conclude that
a.
$m = n$
b.
$AA'$ is invertible
c.
$A'A$ is invertible
d.
if $m = n$, then $A$ is invertible
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For any $n \times n$ matrix $B$, let $N(B) = \{X \in \mathbb{R}^n : BX = 0\}$ be the null space of $B$. Let $A$ be a $4 \times 4$ matrix with $\dim(N(A - 2I)) = 2$, $\dim(N(A - 4I)) = 1$ and $\operatorname{rank}(A) = 3$. Then
a.
$0, 2$ and $4$ are eigenvalues of $A$
b.
$\operatorname{determinant}(A) = 0$
c.
$A$ is not diagonalizable
d.
$\operatorname{trace}(A) = 8$
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Let $H$ be a real Hilbert space and $M \subseteq H$ be a closed linear subspace. Let $x_0 \in H \setminus M$. Let $y_0 \in M$ be such that
$$\|x_0 - y_0\| = \inf \{\|x_0 - y\| : y \in M\}$$
Then
a.
such a $y_0$ is unique
b.
$x_0 \perp M$
c.
$y_0 \perp M$
d.
$x_0 - y_0 \perp M$
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Suppose that $f : \mathbb{R}^n \to \mathbb{R}$ is given by $$f(\underline{x}) = a_1 x_1^2 + a_2 x_2^2 + \dots + a_n x_n^2,$$ where $\underline{x} = (x_1, x_2, \dots, x_n)$ and at least one $a_j$ is not zero. Then we can conclude that
a.
$f$ is not everywhere differentiable
b.
the gradient $(\nabla f)(\underline{x}) \neq 0$ for every $\underline{x} \in \mathbb{R}^n$
c.
if $\underline{x} \in \mathbb{R}^n$ is such that $(\nabla f)(\underline{x}) = 0$ then $f(\underline{x}) = 0$
d.
if $\underline{x} \in \mathbb{R}^n$ is such that $f(\underline{x}) = 0$ then $(\nabla f)(\underline{x}) = 0$
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Let $S$ be the set of $(\alpha, \beta) \in \mathbb{R}^2$ such that
$$\frac{x^\alpha y^\beta}{\sqrt{x^2 + y^2}} \to 0 \text{ as } (x, y) \to (0,0)$$
Then $S$ is contained in
a.
$\{(\alpha, \beta): \alpha > 0, \beta > 0\}$
b.
$\{(\alpha, \beta): \alpha > 2, \beta > 2\}$
c.
$\{(\alpha, \beta): \alpha + \beta > 1\}$
d.
$\{(\alpha, \beta): \alpha + 4\beta > 1\}$
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Let $f = u + iv$ be an entire function where $u, v$ are the real and imaginary parts of $f$ respectively. If the Jacobian matrix
$$J_a = \begin{bmatrix} u_x(a) & u_y(a) \\ v_x(a) & v_y(a) \end{bmatrix}$$
is symmetric for all $a \in \mathbb{C}$, then
a.
$f$ is a polynomial
b.
$f$ is a polynomial of degree $\le 1$
c.
$f$ is necessarily a constant function
d.
$f$ is a polynomial of degree strictly greater than $1$.
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Consider the Mobius transformation $f(z) = \frac{1}{z}$, $z \in \mathbb{C}$, $z \neq 0$. If $C$ denotes a circle with positive radius passing through the origin, then $f$ maps $C \setminus \{0\}$ to
a.
a circle.
b.
a line.
c.
a line passing through the origin.
d.
a line not passing through the origin.
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For which among the following functions $f(z)$ defined on $G = \mathbb{C} \setminus \{0\}$, is there no sequence of polynomials approximating $f(z)$ uniformly on compact subsets of $G$?
a.
$\exp(z)$
b.
$1/z$
c.
$z^2$
d.
$1/z^2$
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For an integer $n \ge 2$, let $S_n$ be the permutation group on $n$ letters and $A_n$ the alternating group. Let $\mathbb{C}^*$ be the group of non-zero complex numbers under multiplication. Which of the following are correct statements ?
a.
For every integer $n \ge 2$, there is a nontrivial homomorphism $\chi: S_n \to \mathbb{C}^*$.
b.
For every integer $n \ge 2$, there is a unique nontrivial homomorphism $\chi: S_n \to \mathbb{C}^*$.
c.
For every integer $n \ge 3$, there is a nontrivial homomorphism $\chi: A_n \to \mathbb{C}^*$.
d.
For every integer $n \ge 5$, there is no nontrivial homomorphism $\chi: A_n \to \mathbb{C}^*$.
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Let $R = \{f : \{1, 2, \dots, 10\} \to \mathbb{Z}_2\}$ be the set of all $\mathbb{Z}_2$-valued functions on the set $\{1, 2, \dots, 10\}$ of the first ten positive integers. Then $R$ is commutative ring with pointwise addition and pointwise multiplication of functions. Which of the following statements are correct?
a.
$R$ has a unique maximal ideal
b.
Every prime ideal of $R$ is also maximal
c.
Number of proper ideals of $R$ is $511$
d.
Every element of $R$ is idempotent
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Let $X$ be any topological space. Let $A \subseteq X$ be nonempty. For $x, y \in A$, define $x \sim y$ if there is a connected subset $C \subseteq A$ such that $x, y \in C$. For $x \in A$, define $C(x) = \{y \in A : y \sim x\}$. Then
Consider the solution of the ordinary differential equation $y'(t) = -y^3 + y^2 + y$ subject to $y(0) = y_0 \in (0, 2)$.
Then $\lim_{t \to \infty} y(t)$ belongs to
a.
$\{-1, 0\}$
b.
$\{-1, 2\}$
c.
$\{0, 2\}$
d.
$\{0, +\infty\}$
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If the solution to
$$\begin{cases} \frac{du}{dx} = y^2 + x^2 &, x > 0 \\ y(0) = 2 \end{cases}$$
exists in the interval $[0, L_0)$ and the maximal interval of existence of
$$\begin{cases} \frac{dz}{dx} = z^2 &, x > 0 \\ z(0) = 1 \end{cases}$$
is $[0, L_1)$, then which of the following statements are correct
a.
$L_1 = 1, L_0 > 1$
b.
$L_1 = 1, L_0 \le 1$
c.
$L_1 < 2, L_0 \le 1$
d.
$L_1 > 2, L_0 < 1$
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If a complete integral of the partial differential equation
$$x(p^2 + q^2) = zp; p = \frac{\partial z}{\partial x}, q = \frac{\partial z}{\partial y}$$
passes through the curve $x = 0, z^2 = 4y$, then the envelope of this family passing through $x = 1$ and $y = 1$ has
a.
$z = -2$
b.
$z = 2$
c.
$z = \sqrt{2 + 2\sqrt{2}}$
d.
$z = -\sqrt{2 + 2\sqrt{2}}$
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For a differentiable function $f: \mathbb{R} \to \mathbb{R}$ define the difference quotient
$$(D_x f)(h) = \frac{f(x+h)-f(x)}{h}; h > 0$$
Consider numbers of the form $\hat{h} = h(1+\epsilon)$ for a fixed $\epsilon > 0$ and let
$$e_1(h) = f'(x) - (D_x f)(h),$$
$$e_2(h) = (D_x f)(h) - (D_x f)(\hat{h})$$
$$e(h) = e_1(h) + e_2(h)$$
if $f(x+h) = f(x+h)$, then
a.
$e_1(h) \to 0$ as $h \to 0$
b.
$e_2(h) \to 0$ as $h \to 0$
c.
$e_2(h) \to \epsilon f'(x)/(1+\epsilon)$ as $h \to 0$
d.
$e(h) \to 0$ as $h \to 0$
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Let $y(x)$ be the solution of the integral equation $y(x) = x - \int_0^x xt^2 y(t)\,dt,\; x > 0$. Then the value of the function $y(x)$ at $x = \sqrt{2}$ is equal to
a.
$\frac{1}{\sqrt{2e}}$
b.
$\frac{e}{2}$
c.
$\frac{\sqrt{2}}{e^2}$
d.
$\frac{\sqrt{2}}{e}$
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The solutions for $\lambda = -1$ and $\lambda = 3$ of the integral equation $y(x) = 1 + \lambda \int_0^1 K(x,t) y(t)\,dt$, where $K(x,t) = \begin{cases} \cosh x \sinh t &, 0 \le x \le t \\ \cosh t \sinh x &, t \le x \le 1 \end{cases}$ are, respectively,
