Your Answer : (Not Answered)

Correct Answer : A

The complex Fourier transform is \(\frac{2}{1 + n^{2}}\).

Using \(F(n) = \int_{-\infty}^{\infty} f(x)e^{inx}dx\) and splitting the range at zero, where \(|x|\) changes sign:

\[ \begin{aligned} F(n) &= \int_{-\infty}^{0} e^{(1 + in)x}dx \\ &\quad + \int_{0}^{\infty} e^{-(1 - in)x}dx \\ &= \frac{1}{1 + in} + \frac{1}{1 - in} \\ &= \frac{2}{1 + n^{2}} \end{aligned} \]

The paper uses the convention without the factor \(1/\sqrt{2\pi}\). With it, the result would be \(\sqrt{2/\pi}/(1 + n^{2})\), which is not offered. The other numerators come from dropping one of the two halves (1) or mishandling the constants (4 or \(\pi\)).

Your Answer : (Not Answered)

Correct Answer : D

The series contains odd sine terms only.

The square wave is an odd function about \(\theta = 0\) (shifting the second half to \(-\pi \lt \theta \lt 0\) gives \(f(-\theta) = -f(\theta)\)), so every cosine coefficient and the constant term vanish. For the sine coefficients:

\[ \begin{aligned} b_{n} &= \frac{2}{\pi}\int_{0}^{\pi}\sin n\theta\,d\theta \\ &= \frac{2}{n\pi}(1 - \cos n\pi) \end{aligned} \]

This is \(4/n\pi\) for odd n and 0 for even n, giving \(f = \frac{4}{\pi}(\sin\theta + \frac{1}{3}\sin 3\theta + \dots)\). 'Sine terms only' is incomplete because the even harmonics are absent, and the cosine options fail the odd symmetry.

Your Answer : (Not Answered)

Correct Answer : D

The integral is \(I = \pi/3\).

The integrand has simple poles at \(z = \pm i\) and \(z = \pm 2i\); the contour encloses \(z = i\) and \(z = 2i\). Their residues are:

\[ \begin{aligned} \text{Res}_{z=i} &= \frac{i^{2}}{(2i)(i^{2}+4)} = \frac{-1}{6i} \\ \text{Res}_{z=2i} &= \frac{(2i)^{2}}{(4i)((2i)^{2}+1)} = \frac{1}{3i} \end{aligned} \]

\[ \begin{aligned} I &= 2\pi i\left(\frac{-1}{6i} + \frac{2}{6i}\right) \\ &= \frac{2\pi i}{6i} = \frac{\pi}{3} \end{aligned} \]

The option \(-\pi/3\) is the contribution of the pole at i alone, missing the pole at 2i; \(\pi/6\) is half the correct value; and 0 would hold only if the contour enclosed no pole.

Your Answer : (Not Answered)

Correct Answer : A

The unit vector is \(\frac{-3\hat{i} + 5\hat{j} + 11\hat{k}}{\sqrt{155}}\).

A vector perpendicular to both is their cross product:

\[ \begin{aligned} \bar{A} \times \bar{B} &= \hat{i}[(-1)(-1) - (1)(4)] \\ &\quad - \hat{j}[(2)(-1) - (1)(3)] \\ &\quad + \hat{k}[(2)(4) - (-1)(3)] \\ &= -3\hat{i} + 5\hat{j} + 11\hat{k} \end{aligned} \]

Its magnitude is \(\sqrt{9 + 25 + 121} = \sqrt{155}\), and dividing by it gives the unit vector. A quick check: its dot product with \(\bar{A}\) is \(-6 - 5 + 11 = 0\).

The other options fail that check; for example \((\hat{i} - \hat{j} + 2\hat{k})\cdot\bar{A} = 2 + 1 + 2 = 5\), so it is not perpendicular to \(\bar{A}\).

Your Answer : (Not Answered)

Correct Answer : C

The Laplacian of the field is 0.

The field is the position vector, \(\overline{A} = x\hat{e}_{x} + y\hat{e}_{y} + z\hat{e}_{z}\) (the paper prints the last unit vector as \(\hat{e}_{j}\)). In Cartesian coordinates the vector Laplacian acts on each component separately, and each component is linear in the coordinates:

\[ \nabla^{2}x = \nabla^{2}y = \nabla^{2}z = 0 \]

so \(\nabla^{2}\overline{A} = 0\). The value 3 is the divergence \(\nabla \cdot \overline{A}\), a common confusion; 1 and -3 do not arise from either operation.

Your Answer : (Not Answered)

Correct Answer : A

One solution is sech x.

Try \(f = \operatorname{sech} x\). Using \((\operatorname{sech} x)' = -\operatorname{sech} x \tanh x\) and \((\tanh x)' = \operatorname{sech}^{2} x\):

\[ \begin{aligned} f' &= -\operatorname{sech} x \tanh x \\ f'' &= \operatorname{sech} x \tanh^{2} x - \operatorname{sech}^{3} x \end{aligned} \]

Substituting:

\[ \begin{aligned} &f'' + 2\operatorname{sech}^{2}x\, f - f \\ &= \operatorname{sech} x\,(\tanh^{2} x + \operatorname{sech}^{2} x - 1) = 0 \end{aligned} \]

since \(\tanh^{2} x + \operatorname{sech}^{2} x = 1\). The same substitution with \(\operatorname{sech}^{2} x\), \(\tanh x\) or \(\operatorname{sech}^{3} x\) leaves terms that do not cancel; for example \(f = \tanh x\) has \(f'' = -2\operatorname{sech}^{2}x\tanh x\), which cancels the middle term and leaves \(-\tanh x \neq 0\).

Your Answer : (Not Answered)

Correct Answer : A

The inverse transform is \(\frac{1}{\pi}\frac{1}{1+x^{2}}\).

With the convention \(f(x) = \frac{1}{2\pi}\int F(k) e^{ikx} dk\), split the integral at \(k = 0\):

\[ \begin{aligned} f(x) &= \frac{1}{2\pi}\Big[\int_{0}^{\infty} e^{-k(1-ix)}dk + \int_{0}^{\infty} e^{-k(1+ix)}dk\Big] \\ &= \frac{1}{2\pi}\Big[\frac{1}{1-ix} + \frac{1}{1+ix}\Big] \\ &= \frac{1}{2\pi}\cdot\frac{2}{1+x^{2}} = \frac{1}{\pi}\frac{1}{1+x^{2}} \end{aligned} \]

This is the Lorentzian (Cauchy) profile. A Gaussian \(e^{-x^{2}}\) is the transform of a Gaussian, not of \(e^{-|k|}\); \(e^{-|x|}\) simply copies the input; and \(1/(1+x)\) is not even an even function, while the transform of an even function must be even.

Your Answer : (Not Answered)

Correct Answer : B

The integral is zero.

The integrand \(1/(z+2)\) has a single simple pole, at \(z = -2\). The contour is the unit circle \(|z| = 1\), and since \(|-2| = 2 \gt 1\) the pole lies outside it. Inside and on the contour the function is analytic, so by Cauchy's theorem

\[ \oint_{|z|=1} \frac{dz}{z+2} = 0 \]

The value \(2\pi i\) would be the answer only if the pole were enclosed (residue 1 times \(2\pi i\)); \(-2\pi i\) is that with the contour traversed clockwise, and \(\pi i\) arises for a pole lying on a contour, in the principal-value sense. None of these applies here.

Your Answer : (Not Answered)

Correct Answer : B

The correct option is that all \(a_n\) are zero.

On \(0 < x < \pi\) the graph is \(f(x) = \frac{\pi}{2} - x\), repeated with period \(\pi\). Shifting one period back, on \(-\pi < x < 0\) it is \(f(x) = -\frac{\pi}{2} - x\). So for \(0 < x < \pi\):

\[ f(-x) = -\tfrac{\pi}{2} + x = -f(x) \]

The function is odd. The cosine coefficients \(a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos nx\,dx\) integrate an odd function over a symmetric interval, so every one of them, \(a_0\) included, is zero; only sine terms survive.

Since \(a_0\) is twice the average value and the sawtooth averages to zero, \(a_0 = 3\pi\) and \(a_0 = 3\pi/2\) are wrong. The \(b_n\) cannot all vanish, or \(f\) would be identically zero.

Your Answer : (Not Answered)

Correct Answer : A

The answer is \(a_{0} \neq 0, b_{n} = 0\).

The function equals \(-|x|\): it is even, since \(f(-x) = f(x)\). An even function has no sine terms, so every \(b_{n} = 0\).

The constant term is the mean value over the period:

\[ \begin{aligned} a_{0} &= \frac{1}{2\pi}\int_{-\pi}^{\pi} (-|x|)\,dx \\ &= -\frac{\pi}{2} \neq 0 \end{aligned} \]

Treating the function as odd (because each piece is linear in x) would wrongly give \(a_{0} = 0\) and nonzero \(b_{n}\).

Your Answer : (Not Answered)

Correct Answer : C

The residues are \(-\frac{3}{16}\) and \(\frac{3}{16}\).

At \(z = 0\) (a pole of order 2), the residue is the first derivative of \(\frac{1}{(z-2)^{3}}\) at 0. At \(z = 2\) (order 3), it is half the second derivative of \(\frac{1}{z^{2}}\) at 2:

\[ \begin{aligned} \text{Res}_{0} &= \left.\frac{-3}{(z-2)^{4}}\right|_{0} = -\frac{3}{16} \\ \text{Res}_{2} &= \frac{1}{2}\left.\frac{6}{z^{4}}\right|_{2} = \frac{3}{16} \end{aligned} \]

Check: the function falls off as \(z^{-5}\), so the residues must sum to zero, as they do. The options with \(\frac{3}{8}\) forget the factor \(\frac{1}{2!}\) or the power of 2 in the denominator.

Your Answer : (Not Answered)

Correct Answer : B

The correct statement is the order of the pole is 3; it is the only option the key accepts in this multiple-select question.

Put \(w = z - \pi\). Near \(z = \pi\), \(\sin z = -\sin w = -w + \frac{w^{3}}{6} - \ldots\) has a simple zero, which cancels one power of the \((z - \pi)^{4}\) in the denominator, while \(z^{2} \rightarrow \pi^{2} \neq 0\). So

\[ f(z) \approx \frac{-\pi^{2}}{(z - \pi)^{3}} \]

and the pole is of order 3, not 4.

The residue is the coefficient of \(w^{3}\) in \(z^{2}\sin z = (\pi^{2} + 2\pi w + w^{2})(-w + w^{3}/6 - \ldots)\), which is

\[ \text{Res} = \frac{\pi^{2}}{6} - 1 \approx 0.64 \]

so neither \(\pi/6\) nor \(2\pi/3\) is the residue.

Your Answer : (Not Answered)

Correct Answer : B

The coefficients satisfy \(c = -2a \ne 0\) and \(b = 0\).

Substitute \(y = a + bx + cx^{2}\), \(y' = b + 2cx\), \(y'' = 2c\):

\[ \begin{aligned} 2c - 2x(b + 2cx) + 4(a + bx + cx^{2}) \\ = (2c + 4a) + 2bx = 0 \end{aligned} \]

The \(x^{2}\) terms cancel. For this to vanish for all \(x\), \(b = 0\) and \(c = -2a\), so the solution is \(a(1 - 2x^{2})\), proportional to the Hermite polynomial \(H_{2}\). The option with \(a = 0\) and \(c = -2b\) fails the constant term, \(b = -2a\) with \(c = 0\) fails because \(b\) must vanish, and \(c = 2a\) has the wrong sign.

Your Answer : (Not Answered)

Correct Answer : C

The number is \(\frac{n(n - 1)}{2}\).

An \(n \times n\) real matrix has \(n^{2}\) entries. Orthogonality, \(O^{T}O = I\), imposes one condition for each entry of the symmetric matrix \(O^{T}O\), that is \(\frac{n(n + 1)}{2}\) independent conditions. The free parameters are

\[ n^{2} - \frac{n(n + 1)}{2} = \frac{n(n - 1)}{2} \]

For \(n = 2\) this gives 1 (a rotation angle) and for \(n = 3\) it gives 3 (the Euler angles), as expected. \(\frac{n(n + 1)}{2}\) is the number of constraints, not of free parameters, and \((n - 1)^{2}\) and \(n(n - 2)\) fail the \(n = 2\) and \(n = 3\) checks.

Your Answer : (Not Answered)

Correct Answer : D

The unit tangent is \(\frac{\hat{\imath} + 2\hat{\jmath} + 3\hat{k}}{\sqrt{14}}\).

The tangent is the derivative of the position vector, divided by its length:

\[ \begin{aligned} \frac{d\vec{r}}{dt} &= \hat{\imath} + 2t\hat{\jmath} + 3t^{2}\hat{k} \\ \text{at } t = 1: &\ \hat{\imath} + 2\hat{\jmath} + 3\hat{k} \\ \left|\,\cdot\,\right| &= \sqrt{1 + 4 + 9} = \sqrt{14} \end{aligned} \]

The first option is the position vector itself normalised, not its derivative. The others use wrong coefficients for the \(t^{2}\) and \(t^{3}\) terms; for example \(\hat{\imath} + 2\hat{\jmath} + 2\hat{k}\) forgets that \(d(t^{3})/dt = 3t^{2}\).

Your Answer : (Not Answered)

Correct Answer : C

The curl is non-zero and points into the plane.

The sketch shows a shear field: \(\vec{V} = V_{x}(y)\hat{x}\), with \(V_{x}\) positive above the x axis, negative below it, and growing with \(|y|\). Take it as \(V_{x} = ky\) with k > 0. Then

\[ \begin{aligned} \vec{\nabla} \times \vec{V} &= \left(\frac{\partial V_{y}}{\partial x} - \frac{\partial V_{x}}{\partial y}\right)\hat{z} \\ &= (0 - k)\hat{z} = -k\hat{z} \end{aligned} \]

With x to the right and y up, \(-\hat{z}\) points into the page. A paddle wheel placed anywhere would turn clockwise, because the flow above it is faster to the right than the flow below. The curl is the same everywhere, so it is not zero and does not reverse across the x axis, and its direction is into, not out of, the plane.

Your Answer : (Not Answered)

Correct Answer : B

The tensor becomes matrix (B):

\[ \begin{pmatrix} \sin 2\theta & \cos 2\theta \\ \cos 2\theta & -\sin 2\theta \end{pmatrix} \]

A rank 2 contravariant tensor transforms as \(T' = R\,T\,R^{T}\). For axes rotated anticlockwise by \(\theta\),

\[ R = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \]

First

\[ R\,T = \begin{pmatrix} \sin\theta & \cos\theta \\ \cos\theta & -\sin\theta \end{pmatrix} \]

then multiplying by \(R^{T}\) gives the entries \(2\sin\theta\cos\theta = \sin 2\theta\), \(\cos^{2}\theta - \sin^{2}\theta = \cos 2\theta\) (twice) and \(-\sin 2\theta\).

A quick check: the original matrix is symmetric with zero trace, and a rotation preserves both properties. Only matrix (B) is symmetric and traceless; (A) and (C) are not symmetric, and the last has trace \(\sin 2\theta - \cos 2\theta\).

Your Answer : (Not Answered)

Correct Answer : B

The quantities that change sharply are \(\rho, C, \chi\).

• Resistivity drops abruptly to zero at \(T_{c}\), the defining property of a superconductor.
• Specific heat shows a discontinuous jump at \(T_{c}\), the signature of a second-order phase transition.
• Magnetic susceptibility changes abruptly to \(\chi = -1\), perfect diamagnetism, because of the Meissner effect.

Thermal conductivity does not jump at \(T_{c}\); it changes gradually below it as electrons condense into pairs that carry no entropy. Young’s modulus, a lattice property, is essentially unaffected. So options including \(\kappa\) or \(Y\) are wrong, and only \(\rho, C, \chi\) matches.

Your Answer : (Not Answered)

Correct Answer : A

The area element is \(\frac{1}{13}\,d\xi\,d\eta\).

Areas transform with the Jacobian of the change of variables:

\[ \begin{aligned} \frac{\partial(\xi, \eta)}{\partial(x, y)} &= \begin{vmatrix} 2 & 3 \\ 3 & -2 \end{vmatrix} = -13 \\ d\xi\,d\eta &= 13\,dx\,dy \\ dx\,dy &= \tfrac{1}{13}\,d\xi\,d\eta \end{aligned} \]

Only the absolute value of the Jacobian matters for an area. The trap is to use 13 directly instead of its inverse; the options 5 and 3/5 come from adding coefficients instead of taking the determinant.

Your Answer : (Not Answered)

Correct Answer : D

The scale factors are \(1, r, r\sin\theta\).

The line element in spherical polar coordinates is

\[ \begin{aligned} ds^{2} &= dr^{2} + r^{2}d\theta^{2} \\ &\quad + r^{2}\sin^{2}\theta\,d\phi^{2} \end{aligned} \]

so the covariant metric is diagonal with \(g_{rr} = 1\), \(g_{\theta\theta} = r^{2}\), \(g_{\phi\phi} = r^{2}\sin^{2}\theta\). The scale (Lamé) factors are the square roots, \(h_{i} = \sqrt{g_{ii}}\): 1, \(r\) and \(r\sin\theta\).

\(1, r^{2}, r^{2}\sin^{2}\theta\) are the metric components themselves, not the scale factors. \(1, r^{2}, \sin^{2}\theta\) is a garbled version of them, and 1, 1, 1 belongs to Cartesian coordinates.