Your Answer : (Not Answered)

Correct Answer : D

Neither \(d_{1}\) nor \(d_{2}\) is a metric on R.

A metric must vanish only on equal points. Both functions see only \(|x|\):

\[ \begin{aligned} d_{1}(1, -1) &= \big|\,|1| - |-1|\,\big| = 0 \\ d_{2}(1, -1) &= |1 - 1| = 0 \end{aligned} \]

yet \(1 \neq -1\). So both fail the identity of indiscernibles. (Each is symmetric and satisfies the triangle inequality, so each is a pseudometric; they become metrics only on \([0, \infty)\).)

That rules out every option claiming that one or both are metrics. The trap is that \(d_{1}\) looks like the familiar \(|x - y|\) and passes the triangle inequality check.

Your Answer : (Not Answered)

Correct Answer : D

Only S₂ is true.

Here \(x_{n}\) runs 1, 0, 1, 0, ... and \(y_{n} = 1 - x_{n}\) runs 0, 1, 0, 1, ..., so neither converges, but \(x_{n} + y_{n} = 1\) for every n. That makes S₂ true.

Put the values into (*):

\[ \begin{aligned} \underline{\lim}\, x_{n} + \underline{\lim}\, y_{n} &= 0 + 0 = 0 \\ \underline{\lim}\,(x_{n} + y_{n}) &= \overline{\lim}\,(x_{n} + y_{n}) = 1 \\ \overline{\lim}\, x_{n} + \overline{\lim}\, y_{n} &= 1 + 1 = 2 \end{aligned} \]

So the chain reads \(0 \lt 1 = 1 \lt 2\). The outer inequalities are strict, but the middle one is an equality, so S₁ ('strict inequalities hold in (*)') is false.

Your Answer : (Not Answered)

Correct Answer : B

The answer is for every \(\varepsilon \gt 0\) there is an open set U with \(E \subset U\) and \(m(U) \lt \varepsilon\).

Lebesgue measure is inner regular: for a measurable set, \(m(E) = \sup\{m(K) : K \subset E,\ K\) compact\(\}\). Since every compact subset has measure 0, \(m(E) = 0\). Outer regularity then gives, for each \(\varepsilon\), an open \(U \supset E\) with \(m(U) \lt m(E) + \varepsilon = \varepsilon\).

The other options do not follow. A null set need not be bounded (the integers, or the rationals). And not every null set is a \(G_{\delta}\) or an \(F_{\sigma}\): there are null subsets of the Cantor set that are not Borel at all.

Your Answer : (Not Answered)

Correct Answer : C

\(\lambda\) does not define a signed measure.

Take \(A = \mathbf{R}\). The positive part of \(\sin x\) has infinite integral over \(\mathbf{R}\), and so does the negative part, so \(\int_{\mathbf{R}} \sin x\,d\mu\) is of the form \(\infty - \infty\) and is undefined.

A signed measure must be defined on every measurable set and may take at most one of the values \(+\infty\) and \(-\infty\). Here \(\lambda\) would have to be \(+\infty\) on \(\{\sin x \gt 0\}\) and \(-\infty\) on \(\{\sin x \lt 0\}\), so it is not a signed measure, and certainly not a measure, since it is negative on sets like \((\pi, 2\pi)\).

Your Answer : (Not Answered)

Correct Answer : B

The image is an open set.

In complex notation \(f(z) = z^{2}\), since \((x + iy)^{2} = x^{2} - y^{2} + 2ixy\). The ball centred at 1 with radius 1/2 does not contain 0, so \(f'(z) = 2z \neq 0\) there. A non-constant analytic function is an open map (open mapping theorem), so the image of an open ball is open.

It is bounded, since \(|z| \lt 3/2\) gives \(|z^{2}| \lt 9/4\), so D is wrong. An open, non-empty, bounded proper subset of the plane cannot be closed, which rules out A and C.

Your Answer : (Not Answered)

Correct Answer : A

The curvatures satisfy \(k_{1}, k_{2} \ge 0\).

For a graph \(z = f(x, y)\), the Gaussian curvature at a critical point (where \(f_{x} = f_{y} = 0\)) is

\[ K = f_{xx}f_{yy} - f_{xy}^{2} \]

the determinant of the Hessian. At a local maximum the Hessian is negative semi-definite and at a local minimum positive semi-definite; in both cases its two eigenvalues have the same sign (or one is 0), so the determinant is \(\ge 0\). Both surfaces bend the same way in every direction, like a cap or a bowl.

A negative curvature would mean a saddle, which is neither a maximum nor a minimum, so the options with \(\le 0\) are wrong.

Your Answer : (Not Answered)

Correct Answer : C

Both (a) and (b) are true.

(a) A non-empty open subset of R contains an open interval \((x - \varepsilon, x + \varepsilon)\), and every non-degenerate interval is uncountable, so the set is uncountable.

(b) The connected subsets of R are exactly the intervals. A connected set containing two points \(a \lt b\) therefore contains the whole interval \([a, b]\), which is uncountable.

Since both statements hold, the options saying only one is true, or both false, are wrong.

Your Answer : (Not Answered)

Correct Answer : B

The non-empty set is \(\bigcap J_{n}\), which equals {0}.

Every closed interval \(J_{n} = [0, 1/n]\) contains 0, so 0 lies in the intersection. No positive number \(x\) does, because \(x \gt 1/n\) once \(n \gt 1/x\). Hence \(\bigcap J_{n} = \{0\}\).

The other three are empty. The open intervals \((0, 1/n)\) exclude 0 and shrink past every positive number. For \(K_{n} = (n, \infty)\) and \(L_{n} = [n, \infty)\), any real \(x\) fails to belong once \(n \gt x\), so no point survives in either intersection. The contrast between (A) and (B) is the point of the question: nested closed bounded intervals keep a common point, open ones need not.

Your Answer : (Not Answered)

Correct Answer : A

\(f \in L(F, \mu)\) and \(\int_{E} f\,d\mu = \int_{F} f\,d\mu\).

\(\mu(E \,\Delta\, F) = 0\) means E and F differ by a null set: \(E \setminus F\) and \(F \setminus E\) both have measure zero. Integrals ignore null sets, so

\[ \begin{aligned} \int_{F} f &= \int_{E \cap F} f + \int_{F \setminus E} f \\ &= \int_{E \cap F} f + 0 \end{aligned} \]

and similarly \(\int_{E} f = \int_{E \cap F} f\). Since \(|f|\) is integrable over E, it is integrable over \(E \cap F\) and hence over F. A strict inequality between the two integrals, or non-integrability over F, is therefore impossible.

Your Answer : (Not Answered)

Correct Answer : C

The integral is 3/2.

\(f(x) = x^{-1/3}\) is non-negative and the single point \(x = 0\) has measure zero, so the Lebesgue integral equals the improper Riemann integral:

\[ \begin{aligned} \int_{0}^{1} x^{-1/3}\,dx &= \left[\frac{x^{2/3}}{2/3}\right]_{0}^{1} \\ &= \frac{3}{2} \end{aligned} \]

The integral is finite because the exponent \(1/3\) is less than 1. The value 3/4 is \(\int_{0}^{1} x^{1/3}dx\), the wrong sign of exponent; 2/3 inverts the answer; and 4/3 does not come from either.

Your Answer : (Not Answered)

Correct Answer : D

\(f\) is Lebesgue integrable and \(g\) is not Lebesgue integrable.

\(f\) is the Dirichlet function. It equals 0 except on the rationals, a countable set of measure zero, so it is Lebesgue integrable with integral 0. (It is not Riemann integrable, since every upper sum is 1 and every lower sum 0.)

For \(g\), \(\int_{\epsilon}^{1} dx/x = -\log\epsilon \to \infty\), so \(\int_{[0,1]} g = \infty\) and \(g\) is not Lebesgue integrable. That rules out the options calling \(g\) integrable, and option A, which also wrongly denies that \(f\) is integrable.

Your Answer : (Not Answered)

Correct Answer : B

The coefficients are E = 1, F = 0, G = 1.

For \(\bar{x}(u, v) = (\cos u, \sin u, v)\): \(\bar{x}_{u} = (-\sin u, \cos u, 0)\) and \(\bar{x}_{v} = (0, 0, 1)\). So

\[ \begin{aligned} E &= \bar{x}_{u} \cdot \bar{x}_{u} = \sin^{2}u + \cos^{2}u = 1 \\ F &= \bar{x}_{u} \cdot \bar{x}_{v} = 0 \\ G &= \bar{x}_{v} \cdot \bar{x}_{v} = 1 \end{aligned} \]

This is why the cylinder is locally isometric to the plane: its first fundamental form is \(du^{2} + dv^{2}\). \(\sin^{2}u\) and \(\cos u\) are single components of \(\bar{x}_{u}\), not its squared length, and F = 1 would mean the coordinate curves are not perpendicular.

Your Answer : (Not Answered)

Correct Answer : B

The true statement is there exists \(x \in [0, 1]\) with \(f(x) = x\).

Let \(g(x) = f(x) - x\). Since \(f\) maps into [0, 1], \(g(0) = f(0) \ge 0\) and \(g(1) = f(1) - 1 \le 0\). By the intermediate value theorem \(g\) has a zero, which is a fixed point of \(f\). This is the one-dimensional Brouwer fixed point theorem.

The other statements fail for \(f(x) = 1 - x\). It is not increasing, which refutes A. C fails because its hypothesis holds trivially for \(x = y = 0\) while \(f\) is decreasing. And its only fixed point is 1/2, so having one fixed point does not make \(f\) the identity (D).

Your Answer : (Not Answered)

Correct Answer : B

The series converges for \( s \gt 1 \).

This is the p-series (the Riemann zeta series). By the integral test it behaves like \( \int_{1}^{\infty} x^{-s} dx \), which is finite only when \( s \gt 1 \).

At \( s = 1 \) it is the harmonic series, which diverges, so it does not converge for all \( s \). For \( s \lt 0 \) the terms \( n^{-s} \) actually grow, so it certainly diverges there. And it does not diverge for all \( s \): for \( s = 2 \) its sum is \( \pi^{2}/6 \).

Your Answer : (Not Answered)

Correct Answer : A

The statement that is NOT true is \(g\) has a local minimum at \(x = 1\); \(g\) has a local maximum there.

By the fundamental theorem of calculus and the chain rule:

\[ \begin{aligned} g'(x) &= (x^{6} - 9x^{3} + 8) \cdot 3x^{2} \\ &= 3x^{2}(x^{3} - 1)(x^{3} - 8) \end{aligned} \]

So the critical points are \(x = 0, 1, 2\), which makes the third statement true. Near \(x = 1\) the factor \(x^{3} - 8\) is negative while \(x^{3} - 1\) changes from negative to positive, so \(g'\) changes from positive to negative: a local maximum, not a minimum.

Near \(x = 0\), \(g'(x) \approx 24x^{2} \geq 0\), so \(g'\) has a local minimum of 0 there and \(g''\) changes sign: \(x = 0\) is a point of inflection. Since \(g'(0) = g'(1) = g'(2) = 0\), Rolle's theorem puts a zero of \(g''\) in each of \((0,1)\) and \((1,2)\), so \(g'' = 0\) has at least two solutions.

Your Answer : (Not Answered)

Correct Answer : B

The necessarily true statement is \(f'(x) = 0\) for infinitely many \(x \in \mathbb{R} \setminus \mathbb{Z}\).

For each integer \(n\), \(f(n) = f(n+1) = 1\), so Rolle's theorem gives a point \(c_n\) strictly between \(n\) and \(n+1\) with \(f'(c_n) = 0\). Each \(c_n\) is a non-integer, and they lie in different intervals, so there are infinitely many.

The other three can fail. With \(f(x) = 1 + \sin(\pi x)\), \(f(n) = 1\) but \(f'(n) = \pi\cos(n\pi) = \pm\pi \neq 0\), so the first option fails; and \(f''(x) = -\pi^{2}\sin(\pi x)\) vanishes only at integers, so the last option fails too. With \(f(x) = 1 + \sin^{2}(\pi x)\), \(f''(x) = 2\pi^{2}\cos(2\pi x)\), which equals \(2\pi^{2} \neq 0\) at every integer, so \(f''(n) = 0\) is not forced.

Your Answer : (Not Answered)

Correct Answer : A

The equal integral is \(\int_{0}^{2}\int_{0}^{y}\int_{y}^{2} f(x,y,z)\,dx\,dz\,dy\).

Read the region from the given limits: \(z\) runs from 0 to \(y\), \(y\) from 0 to \(x\), and \(x\) from 0 to 2, so the region is

\[ 0 \leq z \leq y \leq x \leq 2. \]

Describe it again with \(y\) outermost: \(y\) runs over \([0,2]\); for fixed \(y\), \(z\) runs over \([0,y]\) and, independently, \(x\) over \([y,2]\). That is exactly the first option, with \(dx\) innermost, then \(dz\), then \(dy\).

The others each lose one inequality. The \(dy\,dz\,dx\) option lets \(y\) run up to 2 instead of stopping at \(x\). The \(dz\,dx\,dy\) option lets \(z\) run up to \(x\) instead of \(y\). The \(dy\,dx\,dz\) option lets \(y\) start at 0, so \(y\) can fall below \(z\).

Your Answer : (Not Answered)

Correct Answer : B

The relation holds on the line 2x + 3y + 4 = 0.

By Euler's theorem for a homogeneous function of degree 4, \(xf_{x} + yf_{y} = 4f\). Then

\[ \begin{aligned} xg_{x} + yg_{y} &= e^{2x+3y}\,[\,xf_{x} + yf_{y} + f(2x + 3y)\,] \\ &= f\,e^{2x+3y}\,(4 + 2x + 3y) \end{aligned} \]

The exponential never vanishes, so the expression is zero wherever \(2x + 3y + 4 = 0\), whatever the value of f. It does not vanish on all of \(\mathbb{R}^{2}\), and the option excluding that line describes the opposite set. The line \(2x + 3y = 0\) would come from forgetting the degree 4.

Your Answer : (Not Answered)

Correct Answer : C

0 is a saddle point of g.

For \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), \(\text{Trace}(A^{2}) = a^{2} + 2bc + d^{2}\). This quadratic form has gradient zero at the origin, so 0 is a critical point.

Along \(b = c = t\) (others zero) \(g = 2t^{2} \gt 0\), and along \(b = t, c = -t\) it is \(-2t^{2} \lt 0\). The form takes both signs in every neighbourhood of 0, so 0 is neither a local minimum nor a local maximum: it is a saddle point.

Your Answer : (Not Answered)

Correct Answer : C

\(f\) is uniformly continuous on both \([0, 1)\) and \((0, \infty)\).

By the fundamental theorem of calculus, \(f'(x) = \sin^{2}(x^{2})\), so \(0 \le f'(x) \le 1\) for all \(x \ge 0\). By the mean value theorem,

\[ |f(x) - f(y)| \le |x - y| \]

so \(f\) is Lipschitz with constant 1 on the whole of \([0, \infty)\), and therefore uniformly continuous on every subset of it, including both \([0,1)\) and \((0, \infty)\).

The options denying uniform continuity on either set are ruled out by this bound; the growth of \(x^{2}\) inside the sine does not matter because the derivative stays bounded.