Let $\{b_n\}$ and $\{c_n\}$ be sequences of real numbers. Then a necessary and sufficient condition for the sequence of polynomials $f_n(x) = b_n x + c_n x^2$ converge uniformly to $0$ on the real line is
$\sum_{n=1}^\infty |b_n| < \infty$ and $\sum_{n=1}^\infty |c_n| < \infty$
c.
There exists a positive integer $N$ such that $b_n = 0$ and $c_n = 0$ for all $n > N$
Correct Answer
d.
$\lim_{n \to \infty} c_n = 0$
Step-by-Step AI Explanation
Verified Method
Uniform convergence of $f_n(x) = b_n x + c_n x^2$ to $0$ on $\mathbb{R}$ means that $\sup_{x \in \mathbb{R}} |b_n x + c_n x^2| \to 0$ as $n \to \infty$.
If $c_n \neq 0$, then as $x \to \infty$, $|b_n x + c_n x^2| \to \infty$, so $\sup_{x \in \mathbb{R}} |f_n(x)| = \infty$. Hence, for the supremum to be finite (and sufficiently small, i.e., $< 1$ for large $n$), we must have $c_n = 0$.
If $c_n = 0$ but $b_n \neq 0$, then $\sup_{x \in \mathbb{R}} |b_n x| = \infty$ as well.
Thus, $\sup_{x \in \mathbb{R}} |f_n(x)| < \infty$ for a given $n$ if and only if $b_n = 0$ and $c_n = 0$.
Therefore, $\sup_{x \in \mathbb{R}} |f_n(x)| \to 0$ implies that there must exist an integer $N$ such that $b_n = 0$ and $c_n = 0$ for all $n > N$. Conversely, if $b_n = c_n = 0$ for all $n > N$, $f_n(x) = 0$ for all $n > N$, which trivially converges uniformly to $0$ on $\mathbb{R}$.
Q22mcq
Let $k$ be a positive integer. The radius of convergence of the series $\sum_{n=0}^{\infty} \frac{(n!)^k}{(kn)!} z^n$ is
a.
$k$
b.
$k^{-k}$
c.
$k^k$
Correct Answer
d.
$\infty$
Step-by-Step AI Explanation
Verified Method
Let $a_n = \frac{(n!)^k}{(kn)!}$. We can find the radius of convergence $R$ using the ratio test:
$$R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$$
We have:
$$\frac{a_n}{a_{n+1}} = \frac{(n!)^k}{(kn)!} \cdot \frac{(k(n+1))!}{((n+1)!)^k} = \frac{(kn+k)!}{(kn)! \cdot (n+1)^k}$$
Expanding the factorial in the numerator:
$$\frac{(kn+k)!}{(kn)!} = (kn+k)(kn+k-1)\cdots(kn+1)$$
This is a product of $k$ factors, each of which is asymptotically equal to $kn$ as $n \to \infty$. Therefore,
$$\lim_{n \to \infty} \frac{(kn+k)\cdots(kn+1)}{(n+1)^k} = \lim_{n \to \infty} \frac{(kn)^k}{n^k} = k^k$$
Hence, the radius of convergence is $R = k^k$.
Q23mcq
Suppose $p$ is a polynomial with real coefficients. Then which of the following Statements is necessarily true?
a.
There is no root of the derivative $p'$ between two real roots of the polynomial $p$
b.
There is exactly one root of the derivative $p'$ between any two real roots of $p$
c.
There is exactly one root of the derivative $p'$ between any two consecutive roots of $p$
d.
There is at least one root of the derivative $p'$ between any two consecutive roots of $p$
Correct Answer
Step-by-Step AI Explanation
Verified Method
By Rolle's Theorem, if $p$ is a real polynomial and $a < b$ are any two consecutive real roots of $p$, then $p$ is continuous on $[a, b]$ and differentiable on $(a, b)$ with $p(a) = p(b) = 0$. Thus, there exists at least one $c \in (a, b)$ such that $p'(c) = 0$.
Therefore, there is at least one root of the derivative $p'$ between any two consecutive roots of $p$.
It is not necessarily true that there is *exactly* one root (for instance, $p(x) = x^3 - x^5$ or $p(x) = x^4 - 2x^2$ can have multiple critical points between roots), so (B) and (C) are incorrect.
Q24mcq
Let $G = \{(x, f(x)) : 0 \le x \le 1\}$ be the graph of a real valued differentiable function $f$. Assume that $(1, 0) \in G$. Suppose that the tangent vector to $G$ at any point is perpendicular to the radius vector at that point. Then which of the following is true?
a.
$G$ is the are of an ellipse
b.
$G$ is the arc of a circle
Correct Answer
c.
$G$ is a line segment
d.
$G$ is the are of a parabola
Step-by-Step AI Explanation
Verified Method
At any point $(x, y) = (x, f(x))$ on the graph $G$, the radius vector is $\mathbf{r} = (x, y)$ and the tangent vector is $\mathbf{t} = (1, f'(x)) = (1, y')$.
The condition that the tangent vector is perpendicular to the radius vector means:
$$\mathbf{r} \cdot \mathbf{t} = x \cdot 1 + y \cdot y' = 0$$
This differential equation can be rewritten as:
$$x + y \frac{dy}{dx} = 0 \implies x\,dx + y\,dy = 0$$
Integrating both sides yields:
$$x^2 + y^2 = C$$
where $C$ is a constant. Since $(1, 0) \in G$, we have $1^2 + 0^2 = C \implies C = 1$.
Thus, $x^2 + y^2 = 1$ for $0 \le x \le 1$, which represents an arc of the unit circle.
Q25mcq
Let $\Omega \subseteq \mathbb{R}^n$ be an open set and $f : \Omega \to \mathbb{R}$ be a differentiable function such that $(Df)(x) = 0$ for all $x \in \Omega$. Then which of the following is true?
a.
$f$ must be a constant function
b.
$f$ must be constant on connected components of $\Omega$
Correct Answer
c.
$f(x) = 0$ or $1$ for $x \in \Omega$
d.
The range of the function $f$ is a subset of $\mathbb{Z}$
Step-by-Step AI Explanation
Verified Method
A standard theorem in multivariable calculus states that if $\Omega$ is an open connected set in $\mathbb{R}^n$ and $(Df)(x) = 0$ for all $x \in \Omega$, then $f$ is constant on $\Omega$.
If $\Omega$ is not connected, it can be written as a disjoint union of its connected components (which are also open subsets of $\mathbb{R}^n$). On each connected component, $(Df)(x) = 0$ implies that $f$ is constant, although it may take different constant values on different components.
Thus, $f$ must be constant on each connected component of $\Omega$.
Q26mcq
Let $\{a_n : n \ge 1\}$ be a sequence of real numbers such that $\sum_{n=1}^\infty a_n$ is convergent and $\sum_{n=1}^\infty |a_n|$ is divergent. Let $R$ be the radius of convergence of the power series $\sum_{n=1}^\infty a_n x^n$. Then we can conclude that
a.
$0 < R < 1$
b.
$R = 1$
Correct Answer
c.
$1 < R < \infty$
d.
$R = \infty$
Step-by-Step AI Explanation
Verified Method
Since the power series $\sum_{n=1}^\infty a_n x^n$ converges at $x = 1$ (because $\sum_{n=1}^\infty a_n$ converges), by Abel's Theorem the radius of convergence must satisfy $R \ge 1$.
If $R > 1$, then the power series converges absolutely at every point inside the disc of convergence $|x| < R$, and in particular at $x = 1$, which would mean $\sum_{n=1}^\infty |a_n| < \infty$. However, it is given that $\sum_{n=1}^\infty |a_n|$ diverges.
Therefore, $R$ cannot be strictly greater than $1$. Combining $R \ge 1$ and $R \le 1$, we must have $R = 1$.
Q27mcq
Let $A, B$ be $n \times n$ matrices such that $BA + B^2 = I - BA^2$ where $I$ is then $n \times n$ identity matrix. Which of the following is always true ?
a.
$A$ is nonsingular
b.
$B$ is nonsingular
c.
$A+B$ is nonsingular
Correct Answer
d.
$AB$ is nonsingular
Step-by-Step AI Explanation
Verified Method
We are given the matrix equation:
$$BA + B^2 = I - BA^2$$
Rearranging the terms to one side:
$$BA^2 + BA + B^2 = I$$
Factor $B$ from the left:
$$B(A^2 + A + B) = I$$
This shows that $B$ is invertible and its inverse is $A^2 + A + B$. Since matrices commute with their inverses, we also have:
$$(A^2 + A + B)B = I$$
Now, rearrange the original equation differently:
$$BA^2 + BA + B^2 = I \implies B(A+B) + BA^2 = I$$
Alternatively, consider adding $A+B$ terms or factoring $(I - A)$:
Notice that:
$$B(A^2 + A + B) = I \implies B(A(A+I) + B) = I$$
Wait, let's rearrange $BA^2 + BA + B^2 = I$ as:
$$B(A^2) + B(A+B) = I$$
Wait, let's factor $(A+B)$ instead. Can we express $I$ as a product involving $(A+B)$?
Look at $(B + I - BA)(A + B)$ or similar. Let's rewrite the equation:
$$I = B(A^2 + A + B) = BA(A+I) + B^2$$
Wait, $I - BA^2 = B(A+B)$.
Also, $I = B A^2 + B(A+B)$. Multiply both sides by something?
Let's test an algebraic identity:
$$(I - BA)(A+B) = A + B - BA^2 - B^2 = A + B - (BA^2 + B^2)$$
From the given equation, $BA^2 + B^2 = I - BA$.
Substitute this in:
$$(I - BA)(A+B) = A + B - (I - BA) = A + B - I + BA$$
This doesn't immediately give $I$.
Wait, notice that:
$$BA + B^2 = I - BA^2 \implies B(A+B) + BA^2 = I$$
What about $(B + I)(A+B)$? No.
Let's check if $B(A+B) = I - BA^2$ implies anything. What if $A = -B$? Then $0 = I - B^3$, so $B^3 = I$, which is possible, but then $A+B = 0$ which is singular!
Wait! If $A = -B$, then $BA + B^2 = -B^2 + B^2 = 0$, while $I - BA^2 = I - B^3$.
If $B^3 = I$, then $I - B^3 = 0$. So $BA + B^2 = I - BA^2$ holds!
Let $B = I$ and $A = -I$. Then:
$BA + B^2 = -I + I = 0$.
$I - BA^2 = I - I(-I)^2 = I - I = 0$.
So the relation holds! But here $A+B = -I + I = 0$, which is singular! So (C) CANNOT be always true!
Wait, then is $B$ always nonsingular?
Wait, in our first line: $B(A^2 + A + B) = I$. If the product of two square matrices is the identity $I$, then BOTH matrices are invertible! That is, $B$ is invertible!
Let's double check:
$BA + B^2 = I - BA^2 \implies BA^2 + BA + B^2 = I \implies B(A^2 + A + B) = I$.
Since $B$ and $(A^2 + A + B)$ are $n \times n$ matrices and their product is the identity matrix $I$, $B$ MUST be nonsingular (invertible)!
Therefore, $B$ is nonsingular is ALWAYS true!
Q28mcq
Which of the following matrices has the same row space as the matrix $\begin{pmatrix} 4 & 8 & 4 \\ 3 & 6 & 1 \\ 2 & 4 & 0 \end{pmatrix}$?
Let $M = \begin{pmatrix} 4 & 8 & 4 \\ 3 & 6 & 1 \\ 2 & 4 & 0 \end{pmatrix}$. We find the row-reduced echelon form (RREF) of $M$:
1. Divide Row 3 by 2: $R_3 \to \frac{1}{2}R_3 = \begin{pmatrix} 1 & 2 & 0 \end{pmatrix}$.
2. Divide Row 1 by 4: $R_1 \to \frac{1}{4}R_1 = \begin{pmatrix} 1 & 2 & 1 \end{pmatrix}$.
3. Subtract $R_3$ from $R_1$: $R_1 - R_3 = \begin{pmatrix} 1 & 2 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \end{pmatrix}$.
4. For Row 2, $R_2 = \begin{pmatrix} 3 & 6 & 1 \end{pmatrix} = 3\begin{pmatrix} 1 & 2 & 0 \end{pmatrix} + 1\begin{pmatrix} 0 & 0 & 1 \end{pmatrix} = 3R_3 + (R_1 - R_3)$, which is a linear combination of the other two rows.
Thus, the row space of $M$ has dimension 2 (rank is 2), and a basis for the row space is given by the rows of:
$$\begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
Therefore, this matrix has the exact same row space as $M$.
Q29mcq
The determinant of the $n \times n$ permutation matrix
$$\begin{bmatrix} & & & & 1 \\ & & & 1 & \\ & & \cdot & & \\ & \cdot & & & \\ \cdot & & & & \\ 1 & & & & \end{bmatrix}$$
Here $[x]$ denotes the greatest integer not exceeding $x$.
a.
$(-1)^n$
b.
$(-1)^{\left[\frac{n}{2}\right]}$
Correct Answer
c.
$-1$
d.
$1$
Step-by-Step AI Explanation
Verified Method
The given matrix is the anti-diagonal (exchange) matrix $J_n$, which corresponds to the permutation $\sigma(i) = n - i + 1$ for $i = 1, \dots, n$. This permutation consists of $\lfloor n/2 \rfloor$ disjoint transpositions: $(1, n), (2, n-1), \dots, (k, n-k+1)$ where $k = [n/2]$. Each transposition has a sign of $-1$. Therefore, the determinant is the sign of the permutation, which is $(-1)^{\left[\frac{n}{2}\right]}$. Hence, option B is correct.
Q30mcq
The determinant
$$\begin{vmatrix} 1 & 1+x & 1+x+x^2 \\ 1 & 1+y & 1+y+y^2 \\ 1 & 1+z & 1+z+z^2 \end{vmatrix}$$
is equal to
a.
$(z-y)(z-x)(y-x)$
Correct Answer
b.
$(x-y)(x-z)(y-z)$
c.
$(x-y)^2(y-z)^2(z-x)^2$
d.
$(x^2-y^2)(y^2-z^2)(z^2-x^2)$
Step-by-Step AI Explanation
Verified Method
Applying elementary column operations: replace Column 2 by $C_2 - C_1$ and Column 3 by $C_3 - C_2$:
$$\begin{vmatrix} 1 & 1+x & 1+x+x^2 \\ 1 & 1+y & 1+y+y^2 \\ 1 & 1+z & 1+z+z^2 \end{vmatrix} = \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}$$
This is the classic Vandermonde determinant, which equals $(y-x)(z-x)(z-y) = (z-y)(z-x)(y-x)$. Thus, option A is correct.
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Let $P$ be a $2\times 2$ complex matrix such that $P^* P$ is the identity matrix, where $P^*$ is the conjugate transpose of $P$. Then the eigenvalues of $P$ are
a.
real
b.
complex conjugates of each other
c.
reciprocals of each other
d.
of modulus 1
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Let $p(z)=a_0 + a_1 z + \dots + a_n z^n$ and $q(z)=b_1 z + b_2 z^2 + \dots + b_n z^n$ be complex polynomials. If $a_0, b_1$ are non-zero complex numbers then the residue of $\frac{p(z)}{q(z)}$ at $0$ is equal to
a.
$\frac{a_0}{b_1}$
b.
$\frac{b_1}{a_0}$
c.
$\frac{a_1}{b_1}$
d.
$\frac{a_0}{a_1}$
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Let $\sum_0^\infty a_n z^n$ be a convergent power series such that $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = R > 0$. Let $p$ be a polynomial of degree $d$. Then the radius of convergence of the power series $\sum_{n=0}^\infty p(n)a_n z^n$ equals
a.
R
b.
d
c.
Rd
d.
R+d
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Let $\mathbb{Z}_+ = \{1, 2, \dots\}$ be the set of positive integers. Let $\tau_1 :=$ subspace topology on $\mathbb{Z}_+$ induced from the usual topology of $\mathbb{R}$, $\tau_2 :=$ order topology on $\mathbb{Z}_+$ i.e., the topology with base
$$\{\{x: 1 \le x < b\} : b \in \mathbb{Z}_+\} \cup \{\{x : a < x < b\} : a, b \in \mathbb{Z}_+\}$$
$\tau_3 =$ discrete topology. Then
a.
$\tau_1 \ne \tau_3$ and $\tau_1 = \tau_2$
b.
$\tau_1 \ne \tau_2$ and $\tau_1 = \tau_3$
c.
$\tau_1 \ne \tau_3$ and $\tau_2 = \tau_3$
d.
$\tau_1 = \tau_2 = \tau_3$
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Let $G$ be the Galois group of a field with nine elements over its subfield with three elements. Then the number of orbits for the action of $G$ on the field with nine elements is
a.
3
b.
5
c.
6
d.
9
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Let $P : \mathbb{R} \to \mathbb{R}$ be a polynomial of the form $P(x) = a_0 + a_1 x + a_2 x^2$, with $a_0, a_1, a_2 \in \mathbb{R}$ and $a_2 \neq 0$
Let $E_1 = \int_0^1 P(x)dx - \frac{1}{2}(P(0) + P(1))$, $E_2 = \int_0^1 P(x)dx - P\left(\frac{1}{2}\right)$. If $|x|$ is the absolute value of $x \in \mathbb{R}$, then
a.
$|E_1| > |E_2|$
b.
$|E_2| > |E_1|$
c.
$|E_2| = |E_1|$
d.
$|E_2| = 2|E_1|$
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For $\lambda \in \mathbb{R}$, consider the boundary value problem
$$\left. \begin{array}{c} x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} + \lambda y = 0, \quad x \in [1, 2] \\ y(1) = y(2) = 0 \end{array} \right\} - (P_\lambda)$$
Which of the following statement is true ?
a.
There exists a $\lambda_0 \in \mathbb{R}$ such that $(P_\lambda)$ has a nontrivial solution for any $\lambda > \lambda_0$.
b.
$\{\lambda \in \mathbb{R} : (P_\lambda) \text{ has a nontrivial solution}\}$ is a dense subset of $\mathbb{R}$.
c.
For any continuous function $f : [1, 2] \to \mathbb{R}$ with $f(x) \neq 0$ for some $x \in [1, 2]$, there exists a solution $u$ of $(P_\lambda)$ for some $\lambda \in \mathbb{R}$ such that $\int_1^2 fu \neq 0$
d.
There exists a $\lambda \in \mathbb{R}$ such that $(P_\lambda)$ has two linearly independent solutions
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Let $y : \mathbb{R} \to \mathbb{R}$ be differentiable and satisfy the ODE:
$$\left.\begin{aligned} \frac{dy}{dx} &= f(y), x \in \mathbb{R} \\ y(0) &= y(1) = 0 \end{aligned}\right\}$$
where $f : \mathbb{R} \to \mathbb{R}$ is a Lipschitz continuous function. Then
a.
$y(x) = 0$ if and only if $x \in \{0,1\}$
b.
$y$ is bounded
c.
$y$ is strictly increasing
d.
$\frac{dy}{dx}$ is unbounded
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Consider a body of unit mass falling freely from rest under gravity with velocity $v$. If the air resistance retards the acceleration by $cv$ where $c$ is a constant, then
a.
$v = \frac{g}{c}[1+e^{ct}]$
b.
$v = \frac{g}{c}[1+e^{-ct}]$
c.
$v = \frac{g}{c}[1-e^{-ct}]$
d.
$v = \frac{g}{c}[1-e^{ct}]$
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Suppose that $P$ is a monic polynomial of degree $n$ in one variable with real coefficients and $K$ is a real number. Then which of the following statements is/are necessarily true ?
a.
If $n$ is even and $K > 0$, then there exists $x_0 \in \mathbb{R}$ such that $P(x_0) = K e^{x_0}$
b.
If $n$ is odd and $K < 0$, then there exists $x_0 \in \mathbb{R}$ such that $P(x_0) = K e^{x_0}$
c.
For any natural number $n$ and $0 < K < 1$, there exists $x_0 \in \mathbb{R}$ such that $P(x_0) = K e^{x_0}$
d.
If $n$ is odd and $K \in \mathbb{R}$, then there exists $x_0 \in \mathbb{R}$ such that $P(x_0) = K e^{x_0}$
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Let $\{a_k\}$ be an unbounded, strictly increasing sequence of positive real numbers and $x_k = (a_{k+1} - a_k) / a_{k+1}$
Which of following statements is/are correc
a.
For all $n \ge m$, $\sum_{k=m}^n x_k > 1 - \frac{a_m}{a_n}$
b.
There exists $n \ge m$ such that $\sum_{k=m}^n x_k > \frac{1}{2}$
c.
$\sum_{k=1}^\infty x_k$ converges to a finite limit.
d.
$\sum_{k=1}^\infty x_k$ diverges to $\infty$
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For a non-empty subset $S$ and a point $x$ in a connected metric space $(X, d)$, let $d(x, S) = \inf\{d(x, y) : y \in S\}$. Which of the following statements is/are correct ?
a.
If $S$ is closed and $d(x, S) > 0$ then $x$ is not an accumulation point of $S$
b.
If $S$ is open and $d(x, S) > 0$ then $x$ is not an accumulation point of $S$
c.
If $S$ is closed and $d(x, S) > 0$ then $S$ does not contain $x$
d.
If $S$ is open and $d(x, S) = 0$ then $x \in S$
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Let $f$ be a continuously differentiable function on $\mathbb{R}$. Suppose that $$L = \lim_{x \to \infty} (f(x) + f'(x))$$ exists. If $0 < L < \infty$, then which of the following statements is/are correct?
a.
If $\lim_{x \to \infty} f'(x)$ exists, then it is $0$
b.
If $\lim_{x \to \infty} f(x)$ exists then it is $L$
c.
If $\lim_{x \to \infty} f'(x)$ exists then $\lim_{x \to \infty} f(x) = 0$
d.
If $\lim_{x \to \infty} f(x)$ exists then $\lim_{x \to \infty} f'(x) = L$
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Consider the normed linear spaces $X_1 = (C[0, 1], \|\cdot\|_1)$ and $X_\infty = (C[0, 1], \|\cdot\|_\infty)$, where $C[0, 1]$ denotes the vector space of all continuous real valued functions on $[0, 1]$ and $\|f\|_1 = \int_0^1 |f(t)| dt$, $\|f\|_\infty = \sup \{|f(t)| \mid t \in [0, 1]\}$. Let $U_1$ and $U_\infty$ be the open unit balls in $X_1$ and $X_\infty$ respectively. Then
a.
$U_\infty$ is a subset of $U_1$
b.
$U_1$ is a subset of $U_\infty$
c.
$U_\infty$ is equal to $U_1$
d.
Neither $U_\infty$ is a subset of $U_1$ nor $U_1$ is a subset of $U_\infty$
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Let $A$ be a real $n \times n$ orthogonal matrix that is, $A^t A = A A^t = I_n$, the $n \times n$ identity matrix. Which of the following statements are necessarily true?
a.
$\langle Ax, Ay \rangle = \langle x, y \rangle \,\forall x, y \in \mathbb{R}^n$
b.
All eigenvalues of $A$ are either $+1$ or $-1$
c.
The rows of $A$ form an orthonormal basis of $\mathbb{R}^n$
d.
$A$ is diagonalizable over $\mathbb{R}$
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Let $f$ be a non-zero symmetric bilinear form on $\mathbb{R}^3$. Suppose that there exist linear transformations $t_i : \mathbb{R}^3 \to \mathbb{R}$, $i = 1, 2$ such that for all $\alpha, \beta \in \mathbb{R}^3$, $f(\alpha, \beta) = T_1(\alpha) T_2(\beta)$. Then
a.
$\operatorname{rank} f = 1$
b.
$\dim \{\beta \in \mathbb{R}^3 : f(\alpha, \beta) = 0 \text{ for all } \alpha \in \mathbb{R}^3\} = 2$
c.
$f$ is a positive semi-definite or negative semi-definite
d.
$\{\alpha : f(\alpha, \alpha) = 0\}$ is a linear subspace of dimension 2
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Let $A$ be $5\times 5$ matrix and let $B$ be obtained by changing one element of $A$. Let $r$ and $s$ be the ranks of $A$ and $B$ respectively. Which of the following statements is/are correct ?
a.
$s \le r+1$
b.
$r-1 \le s$
c.
$s = r-1$
d.
$s \neq r$
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Let $M_n(K)$ denote the space of all $n\times n$ matrices with entries in a field $K$. Fix a non-singular matrix $A = (A_{ij}) \in M_n(K)$, and consider the linear map $T : M_n(K) \to M_n(K)$ given by: $T(X) = AX$ Then
Let A be a $4 \times 7$ real matrix and B be a $7 \times 4$ real matrix such that $AB = I_4$ where $I_4$ is the $4 \times 4$ identity matrix. Which of the following is/are always true ?
a.
$\operatorname{rank}(A) = 4$
b.
$\operatorname{rank}(B) = 7$
c.
$\operatorname{nullity}(B) = 0$
d.
$BA = I_7$, where $I_7$ is the $7 \times 7$ identity matrix
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Let $f$ be an entire function on $\mathbb{C}$ and let $\Omega$ be a bounded open subset of $\mathbb{C}$.
Let $S = \{\operatorname{Re} f(z) + \operatorname{Im} f(z) \mid z \in \Omega\}$. Which of the following statements is/are necessarily correct ?
a.
$S$ is an open set in $\mathbb{R}$
b.
$S$ is a closed set in $\mathbb{R}$
c.
$S$ is an open set of $\mathbb{C}$
d.
$S$ is a discrete set in $\mathbb{R}$
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Let $f$ be an entire function on $\mathbb{C}$. Let $g(z) = \overline{f(\bar{z})}$. Which of the following statements is/are correct ?
a.
if $f(z) \in \mathbb{R}$ for all $z \in \mathbb{R}$ then $f = g$
b.
if $f(z) \in \mathbb{R}$ for all $z \in \{z \mid \operatorname{Im} z = 0\} \cup \{z \mid \operatorname{Im} z = a\}$ for some $a > 0$, then $f(z+ia) = f(z-ia)$ for all $z \in \mathbb{C}$
c.
if $f(z) \in \mathbb{R}$ for all $z \in \{z \mid \operatorname{Im} z = 0\} \cup \{z \mid \operatorname{Im} z = a\}$, for some $a > 0$, then $f(z+2ia) = f(z)$ for all $z \in \mathbb{C}$.
d.
if $f(z) \in \mathbb{R}$ for all $z \in \{z \mid \operatorname{Im} z = 0\} \cup \{z \mid \operatorname{Im} z = a\}$, for some $a > 0$, then $f(z+ia) = f(z)$ all $z \in \mathbb{C}$.
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Let $X = \{(a,b) \in \mathbb{R}^2 : a^2 + b^2 = 1\}$ be the unit circle inside $\mathbb{R}^2$. Let $f : X \to \mathbb{R}$ be a continuous function. Then:
a.
$\text{Image }(f)$ is connected,
b.
$\text{Image }(f)$ is compact.
c.
The given information is not sufficient to determine whether $\text{Image }(f)$ is bounded.
d.
$f$ is not injective
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Let $\mathbb{R}[x]$ denote the vector space of all real polynomials. Let $D: \mathbb{R}[x] \to \mathbb{R}[x]$ denote the map $Df = \frac{df}{dx}, \forall f$. Then,
a.
$D$ is one-one.
b.
$D$ is onto.
c.
There exists $E: \mathbb{R}[x] \to \mathbb{R}[x]$ so that $D(E(f)) = f, \forall f$.
d.
There exists $E: \mathbb{R}[x] \to \mathbb{R}[x]$ so that $E(D(f)) = f, \forall f$.
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Given any positive integer $n$, there exists a field extension of $\mathbb{Q}$ of degree $n$.
b.
Given a positive integer $n$, there exist fields $F$ and $K$ such that $F \subseteq K$ and $K$ is Galois over $F$ with $[K : F] = n$,
c.
Let $K$ be a Galois extension of $\mathbb{Q}$ with $[K : \mathbb{Q}] = 4$. Then there is a field $L$ such that $K \supseteq L \supseteq \mathbb{Q}, [L : \mathbb{Q}] = 2$ and $L$ is a Galois extension of $\mathbb{Q}$.
d.
There is an algebraic extension $K$ of $\mathbb{Q}$ such that $[K : \mathbb{Q}]$ is not finite.
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The system of ODE
$$\begin{cases} \frac{dx}{dt} = (1+x^2)y, & t \in \mathbb{R} \\ \frac{dy}{dt} = -(1+x^2)x, & t \in \mathbb{R} \end{cases}$$
$(x(0), y(0)) = (a, b)$
has a solution:
a.
only if $(a, b) = (0, 0)$
b.
for any $(a, b) \in \mathbb{R} \times \mathbb{R}$
c.
such that $x^2(t) + y^2(t) = a^2 + b^2$ for all $t \in \mathbb{R}$
d.
such that $x^2(t) + y^2(t) \to \infty$ as $t \to \infty$ if $a > 0$ and $b > 0$
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Let $y: \mathbb{R} \to \mathbb{R}$ be a solution of the ODE
$$\begin{cases} \frac{d^2 y}{dx^2} - y = e^{-x}, & x \in \mathbb{R} \\ y(0) = \frac{dy}{dx}(0) = 0 \end{cases}$$
then
a.
$y$ attains its minimum on $\mathbb{R}$
b.
$y$ is bounded on $\mathbb{R}$
c.
$\lim_{x \to \infty} e^{-x} y(x) = \frac{1}{4}$
d.
$\lim_{x \to \infty} e^x y(x) = \frac{1}{4}$
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Let $u \in C^2([0,1])$ satisfy for some $\lambda \neq 0$ and $a \neq 0$
$$u(x) + \frac{\lambda}{2}\int_0^1 |x-s|u(s)\,ds = ax+b$$
Then $u$ also satisfies
Let $f : \mathbb{R} \to \mathbb{R}$ be a smooth function with non vanishing derivative. The Newton's method for finding a root of $f(x) = 0$ is the same as
a.
fixed point iteration for the map $g(x) = x - f(x)/f'(x)$
b.
Forward Euler method with unit step length for the differential equation $\frac{dy}{dt} + \frac{f(y)}{f'(y)} = 0$
c.
fixed point iteration for $g(x) = x + f(x)$
d.
fixed point iteration for $g(x) = x - f(x)$
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Consider a particle moving with coordinates $(x(t), y(t))$ on a smooth curve $\phi(x,y) = 0$. If the particle moves from $(x(0), y(0))$ to $(x(\tau), y(\tau))$ for $\tau > 0$ such that its kinetic energy is minimized, then
Consider the Cauchy problem of finding $u = u(x, t)$ such that
$$\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} = 0 \text{ for } x \in \mathbb{R}, t > 0$$
$$u(x, 0) = u_0(x), \quad x \in \mathbb{R}$$
Which choice(s) of the following functions for $u_0$ yield a $C^1$ solution $u(x, t)$ for all $x \in \mathbb{R}$ and $t > 0$.
a.
$u_0(x) = \frac{1}{1+x^2}$
b.
$u_0(x) = x$
c.
$u_0(x) = 1+x^2$
d.
$u_0(x) = 1+2x$
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Let $P, Q$ be continuous real valued functions defined on $[-1, 1]$ and $u_i: [-1, 1] \rightarrow \mathbb{R}$, $i = 1, 2$ be solutions of the ODE $\frac{d^2u}{dx^2} + P(x)\frac{du}{dx} + Q(x)u = 0$, $x \in [-1, 1]$ satisfying $u_1 \ge 0$, $u_2 \le 0$ and $u_1(0) = u_2(0) = 0$.
Let $w$ denote the Wronskian of $u_1$ and $u_2$ then
a.
$u_1$ and $u_2$ are linearly independent
b.
$u_1$ and $u_2$ are linearly dependent
c.
$w(x) = 0$ for all $x \in [-1, 1]$
d.
$w(x) \neq 0$ for some $x \in [-1, 1]$
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Which of the following approximations for estimating the derivative of a smooth function $f$ at a point $x$ is of order 2 (i.e. the error term is $O(h^2)$)
a.
$f'(x) \approx \frac{f(x+h)-f(x)}{h}$
b.
$f'(x) \approx \frac{f(x+h)-f(x-h)}{2h}$
c.
$f'(x) \approx \frac{3f(x)-4f(x-h)+f(x-2h)}{2h}$
d.
$f'(x) \approx \frac{-3f(x)+4f(x+h)-f(x+2h)}{2h}$
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Let $u(x, t)$ satisfy for $x \in \mathbb{R}, t > 0$
$$\frac{\partial^2 u}{\partial t^2} + \frac{\partial u}{\partial t} + 2\frac{\partial^2 u}{\partial x^2} = 0$$
A solution of the form $u = e^{ix}v(t)$ with $v(0) = 0$ and $v'(0) = 1$
a.
is necessarily bounded
b.
satisfies $|u(x, t)| < e^t$
c.
is necessarily unbounded
d.
is oscillatory in $x$.
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Let $u = u(x, t)$ be the solution of the Cauchy problem
$$\frac{\partial u}{\partial t} + \left(\frac{\partial u}{\partial x}\right)^2 = 1 \quad x \in \mathbb{R}, t > 0$$
$$u(x, 0) = -x^2 \quad x \in \mathbb{R}$$
then
a.
$u(x, t)$ exists for all $x \in \mathbb{R}$ and $t > 0$.
b.
$|u(x, t)| \rightarrow \infty$ as $t \rightarrow t^*$ for some $t^* > 0$ and $x \neq 0$
c.
$u(x, t) \le 0$ for all $x \in \mathbb{R}$ and for all $t < 1/4$.
d.
$u(x, t) > 0$ for some $x \in \mathbb{R}$ and $0 < t < 1/4$.
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