GRE Mathematics Preparation: Advanced Quantitative Reasoning Quiz – Sept 2026 4th Edition

Because ##G## is a finite ##2##-group, its center ##Z(G)## is nontrivial. Thus ##|Z(G)|## can be ##2##, ##4##, or ##8##. If ##|Z(G)|=8##, then ##G## is abelian. If ##|Z(G)|=4##, then ##G/Z(G)## has order ##2## and is cyclic; whenever ##G/Z(G)## is cyclic, ##G## is abelian. Both cases contradict the hypothesis that ##G## is non-abelian. Therefore, ##|Z(G)|=2##.

The critical points are found by setting ##f'(x) = 3x^2 - 3 = 0##, giving ##x = \pm 1##. Evaluating ##f(-2) = -2##, ##f(-1) = 2##, ##f(1) = -2##, and ##f(3) = 18##, the maximum value on the closed interval is 18.

By the Heine-Borel theorem, a subset of ##\mathbb{R}^n## is compact if and only if it is closed and bounded. The interval ##[0, 1]## is both closed and bounded, making it compact. The other options are either not closed or not bounded.

The matrix representation of ##T## with respect to the given basis is a permutation matrix that cycles the basis vectors. Its diagonal entries are all zero, so the trace is 0.

The integrand has simple poles at ##z = \pm i##, both inside the contour ##|z| = 2##. By the residue theorem, the integral equals ##2\pi i## times the sum of residues. The residue at ##z = i## is ##e^i/(2i)## and at ##z = -i## is ##e^{-i}/(-2i)##. Summing gives ##(e^i - e^{-i})/(2i) = \sin(1)##. Thus, the integral is ##2\pi i \sin(1)##.

A homomorphism ##\varphi:\mathbb{Z}_6\to S_3## is completely determined by the image of the generator ##1##. The image ##g=\varphi(1)## must satisfy ##g^6=e##. Every element of ##S_3## has order ##1##, ##2##, or ##3##, so every element satisfies this condition. Since ##S_3## has ##6## elements, there are exactly ##6## distinct homomorphisms.

The series can be rewritten using partial fractions: ##\frac{1}{n^2 + n} = \frac{1}{n} - \frac{1}{n+1}##. This is a telescoping series. The partial sum is ##(1 - 1/2) + (1/2 - 1/3) + \cdots + (1/N - 1/(N+1)) = 1 - 1/(N+1)##. As ##N \to \infty##, the sum approaches 1.

The units in ##\mathbb{Z}[i]## are elements with norm 1. The norm of ##a + bi## is ##a^2 + b^2##. For ##i##, the norm is 1, so it is a unit. The other options have norms greater than 1 and are not units.

The gradient is ##\nabla f(x,y)=(2xy,x^2)##, so at ##(1,2)## we have ##\nabla f(1,2)=(4,1)##. The unit vector in the direction of ##(1,1)## is ##\mathbf{u}=\frac{1}{\sqrt{2}}(1,1)##. Therefore, the directional derivative is ###D_{\mathbf{u}}f(1,2)=\nabla f(1,2)\cdot\mathbf{u}=\frac{4+1}{\sqrt{2}}=\frac{5}{\sqrt{2}}=\frac{5\sqrt{2}}{2}.###

An automorphism of ##\mathbb{Z}_4## is determined by the image of the generator 1, which must be a generator of ##\mathbb{Z}_4##. The generators are 1 and 3. Thus, there are exactly 2 automorphisms: the identity and the map sending 1 to 3.

Assuming the sequence converges to ##L##, we have ##L = \sqrt{2 + L}##. Squaring gives ##L^2 = 2 + L##, so ##L^2 - L - 2 = 0##, which factors as ##(L - 2)(L + 1) = 0##. Since the sequence is positive, ##L = 2##.

Let ##z=1+iy## and write ##w=u+iv##. Then ##w=\frac{1}{1+iy}=\frac{1-iy}{1+y^2}##, so ##u=\frac{1}{1+y^2}## and ##v=-\frac{y}{1+y^2}##. Hence ###u^2+v^2=u,###or equivalently ###\left(u-\frac12\right)^2+v^2=\frac14.### Thus the image lies on the circle centered at ##(1/2,0)## with radius ##1/2##. The point ##w=0## is not attained by any finite ##z##, so the origin is omitted.

If ##\lambda## is an eigenvalue of ##A##, then ##\lambda^2 + \lambda + 1## is an eigenvalue of ##A^2 + A + I##. Thus, the eigenvalues of ##A^2 + A + I## are ##1^2 + 1 + 1 = 3##, ##2^2 + 2 + 1 = 7##, and ##3^2 + 3 + 1 = 13##. The determinant is the product of the eigenvalues: ##3 \cdot 7 \cdot 13 = 273##.

Elements of order 5 in ##S_5## are 5-cycles. The number of 5-cycles in ##S_5## is ##5!/5 = 24##.

The antiderivative of ##1/(1 + x^2)## is ##\arctan(x)##. Evaluating from ##-\infty## to ##\infty## gives ##\arctan(\infty) - \arctan(-\infty) = \pi/2 - (-\pi/2) = \pi##.

The multiplicative group of a finite field ##\mathbb{F}_q## has order ##q - 1##. By Lagrange's theorem, the order of any element divides the group order, so it divides ##q - 1##.

Using polar coordinates, ##x = r\cos\theta##, ##y = r\sin\theta##, and the region is ##0 \leq r \leq 1##, ##0 \leq \theta \leq 2\pi##. The integral becomes ##\int_0^{2\pi} \int_0^1 r^2 \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^1 r^3 \, dr \, d\theta = \int_0^{2\pi} 1/4 \, d\theta = \pi/2##.

The matrix has the single eigenvalue ##2## with algebraic multiplicity ##2##. Because it is a nontrivial Jordan block, ##A-2I\neq 0## but ##(A-2I)^2=0##. Therefore, the monic polynomial of least degree that annihilates the matrix is ##(x-2)^2##.

This is the Taylor series for ##e^x##. Using the ratio test, ##\lim_{n \to \infty} |a_{n+1}/a_n| = \lim_{n \to \infty} 1/(n+1) = 0##, so the radius of convergence is ##\infty##.

If ##a^2 = e## for all ##a##, then for any ##a, b \in G##, ##(ab)^2 = e##, so ##abab = e##. Multiplying on the left by ##a## and on the right by ##b## gives ##ba = ab##. Thus, ##G## is abelian.

The product can be written as ##\prod_{n=2}^{\infty} \frac{(n-1)(n+1)}{n^2}##. This telescopes: ##\frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2 \cdot 4}{3 \cdot 3} \cdot \frac{3 \cdot 5}{4 \cdot 4} \cdots##. The partial product up to ##N## is ##\frac{1 \cdot (N+1)}{2 \cdot N}##, which tends to ##1/2## as ##N \to \infty##.

The matrix is upper triangular with all diagonal entries equal to 1, so it is invertible. Therefore, the kernel is trivial, and its dimension is 0.

The equation ##z^4 = 16## has exactly 4 distinct complex roots, given by ##z = 2e^{i(2\pi k)/4}## for ##k = 0, 1, 2, 3##.

The word has 11 letters: M appears once, I appears 4 times, S appears 4 times, and P appears 2 times. The number of distinct arrangements is ##11! / (4! \cdot 4! \cdot 2!)##.

A continuous solution of Cauchy's functional equation ##f(x+y)=f(x)+f(y)## must be linear. Therefore there exists a constant ##c## such that ##f(x)=cx## for every ##x\in\mathbb{R}##.

The Euler characteristic of a torus is ##\chi = V - E + F = 0##. For a torus, it is 0, as it can be represented as a quotient of a square with opposite sides identified, giving ##V = 1##, ##E = 2##, ##F = 1##, so ##\chi = 1 - 2 + 1 = 0##.

The units in ##\mathbb{Z}_{12}## are the elements relatively prime to 12: 1, 5, 7, 11. Thus, there are 4 units.

Using the standard limit ##\lim_{x \to 0} \frac{\sin(x)}{x} = 1##, we have ##\lim_{x \to 0} \frac{\sin(3x)}{x} = 3 \cdot \lim_{x \to 0} \frac{\sin(3x)}{3x} = 3 \cdot 1 = 3##.

The determinant of a product is the product of the determinants, which is a fundamental property. The other statements are not generally true.

In any tree with ##n## vertices, there are ##n - 1## edges. By the handshaking lemma, the sum of the degrees of all vertices is ##2(n - 1) = 2n - 2##.

The function has a simple pole at ##z = i##. The residue is ##\lim_{z \to i} (z - i) f(z) = \lim_{z \to i} \frac{1}{z + i} = \frac{1}{2i}##.

By Sylow's theorems, the number of Sylow 3-subgroups, ##n_3##, must divide 5 and be congruent to 1 mod 3. The divisors of 5 are 1 and 5. 5 is congruent to 2 mod 3, so ##n_3 = 1##.

A symmetric ##2 \times 2## matrix has the form ##\begin{pmatrix} a & b \\ b & c \end{pmatrix}##, which has 3 independent entries: ##a##, ##b##, and ##c##. Thus, the dimension is 3.

By Green's theorem, ##\oint_C y \, dx + x \, dy = \iint_D \left(\frac{\partial x}{\partial x} - \frac{\partial y}{\partial y}\right) \, dA = \iint_D (1 - 1) \, dA = 0##.

The number of abelian groups of order ##p^n## is the number of partitions of ##n##. For ##16 = 2^4##, the partitions of 4 are 5: 4, 3+1, 2+2, 2+1+1, 1+1+1+1. Thus, there are 5 non-isomorphic abelian groups of order 16.

The derivatives of ##\ln(x)## are ##f'(x) = 1/x##, ##f''(x) = -1/x^2##, ##f'''(x) = 2/x^3##, and in general, ##f^{(n)}(x) = (-1)^{n-1}(n-1)!/x^n##. At ##x = 1##, this is ##(-1)^{n-1}(n-1)!##.

Using the principal logarithm, ##\operatorname{Log}(i)=i\pi/2##. Hence ###i^i=e^{\,i\operatorname{Log}(i)}=e^{\,i(i\pi/2)}=e^{-\pi/2}.### Without the word "principal," ##i^i## is multivalued because the complex logarithm has infinitely many branches.

The rows are linearly dependent; the second row is twice the first, and the third row is three times the first. Thus, the rank is 1.

The sequence converges pointwise to ##f(x) = 0## for ##x \in [0, 1)## and ##f(1) = 1##. The limit function is discontinuous at ##x = 1##, so the convergence is not uniform on ##[0, 1]##.

Lagrange's theorem states that the order of a subgroup divides the order of the group. Since the cyclic subgroup generated by an element has order equal to the element's order, the element's order divides the group order.

The solutions are found by solving ##x^2 \equiv 1 \pmod{3}## and ##x^2 \equiv 1 \pmod{5}##. For mod 3, solutions are ##x \equiv \pm 1##. For mod 5, solutions are ##x \equiv \pm 1##. By the Chinese remainder theorem, there are ##2 \times 2 = 4## solutions modulo 15.

The kernel of ##L## consists of all polynomials ##p## such that ##L(p) = p(1) = 0##. This is the set of polynomials that vanish at ##x = 1##.

By the divergence theorem, ##\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V \nabla \cdot \mathbf{F} \, dV##. The divergence of ##\mathbf{F}## is ##1 + 1 + 1 = 3##. The volume of the unit sphere is ##4\pi/3##, so the integral is ##3 \cdot 4\pi/3 = 4\pi##.

The norm of ##2 + i## is ##2^2 + 1^2 = 5##. The quotient ring ##\mathbb{Z}[i]/(a + bi)## has ##a^2 + b^2## elements. Thus, the quotient ring has 5 elements.

The derivative at 0 is ##\lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} h \sin(1/h) = 0##, since ##|h \sin(1/h)| \leq |h|##. Thus, ##f## is differentiable at 0 with derivative 0.

The commutator subgroup of ##S_3## is the alternating group ##A_3##, which consists of the identity and the two 3-cycles. This is because ##S_3/A_3## is abelian, and ##A_3## is the smallest normal subgroup with abelian quotient.

This is the Gaussian integral. The standard result is ##\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}##. By symmetry, ##\int_0^{\infty} e^{-x^2} \, dx = \sqrt{\pi}/2##.

We have ##\det(A^{-1}) = 1/\det(A) = 1/3##. For a ##2 \times 2## matrix, ##\det(2A^{-1}) = 2^2 \det(A^{-1}) = 4 \cdot 1/3 = 4/3##.

The space ##\mathbb{R} \setminus \{0\}## consists of two disjoint open intervals: ##(-\infty, 0)## and ##(0, \infty)##. Each is connected, and they are disconnected from each other, so there are 2 connected components.

We need to find ##x## such that ##3x \equiv 1 \pmod{7}##. Testing values, ##3 \cdot 5 = 15 \equiv 1 \pmod{7}##, so the inverse is 5.