Your Answer : (Not Answered)

Correct Answer : A

The direction is that of the gradient, \(\nabla\phi\).

The rate of change of \(\phi\) in the direction of a unit vector \(\hat{n}\) is the directional derivative \(\nabla\phi \cdot \hat{n} = |\nabla\phi|\cos\alpha\), where \(\alpha\) is the angle between \(\hat{n}\) and the gradient. It is largest when \(\alpha = 0\), that is, along \(\nabla\phi\), and its largest value is \(|\nabla\phi|\).

\(\nabla \times (\phi\vec{r})\) equals \(\nabla\phi \times \vec{r}\), which is perpendicular to the gradient, so along it the rate of change is zero. \(\phi\vec{r}\) points along the position vector, which has nothing to do with how \(\phi\) varies. \((\nabla\phi \cdot d\vec{r})\vec{r}\) is also a multiple of \(\vec{r}\), scaled by the change in \(\phi\), and again points radially rather than along the gradient.

Your Answer : (Not Answered)

Correct Answer : A

Both statements are true. Interpolation is a statistical method by which related known values are used to estimate an unknown value or set of values. In the Inverse Distance Weighted method the cell values are estimated by averaging the values of sample data points in the neighbourhood of each processing cell, with nearer points given more weight. So the first option is correct. In IDW a point close to the cell has more influence on its value than a point far away.

Your Answer : (Not Answered)

Correct Answer : B

Each term then carries the same power of \( x \), since \( x^{2} \) times the second derivative of \( x^{m} \) is \( m(m-1)x^{m} \), and the equation collapses to an algebraic one in \( m \). The exponential trial belongs to the constant-coefficient case, where it is the derivatives rather than the coefficients that carry the variable

Your Answer : (Not Answered)

Correct Answer : A

The first two hold. The third fails: two initial conditions, the initial displacement and the initial velocity, need two arbitrary functions to satisfy, so one travelling wave is not enough unless the two conditions happen to be compatible with it. That is exactly why d'Alembert's solution carries both terms

Your Answer : (Not Answered)

Correct Answer : D

With \( \mathbf{F} = \nabla\phi \) the integrand becomes an exact differential, \( \nabla\phi \cdot d\mathbf{r} = d\phi \), so

\[ \int_{A}^{B} \mathbf{F} \cdot d\mathbf{r} \;=\; \phi(B) - \phi(A). \]

It vanishes only when the two endpoints share a potential value, which for a closed path is automatic but for an open one is not

Your Answer : (Not Answered)

Correct Answer : A

Complex differentiability at an isolated point is too weak to be useful; analyticity demands it on a whole open disc about the point. That extra requirement is what buys the powerful results -- infinite differentiability, a convergent Taylor series, and Cauchy's theorem -- none of which follows from differentiability at one point alone

Your Answer : (Not Answered)

Correct Answer : B

The integrand is \( r \) and the element supplies another factor of \( r \):

\[ \int_{0}^{2\pi}\!\!\int_{0}^{a} r \cdot r\,dr\,d\theta \;=\; 2\pi\left[\frac{r^{3}}{3}\right]_{0}^{a} \;=\; \frac{2\pi a^{3}}{3}. \]

Two powers of \( r \) appear, one from the integrand and one from the area element, which is what raises the result to a cube

Your Answer : (Not Answered)

Correct Answer : C

The result is

\[ \frac{\Gamma(l)\Gamma(m)\Gamma(n)}{\Gamma(l+m+n+1)}, \]

and the extra unit in the denominator's argument is what distinguishes it from the beta-function pattern. Setting \( l = m = n = 1 \) gives \( 1/\Gamma(4) = 1/6 \), the volume of the simplex, which is the check that fixes the form

Your Answer : (Not Answered)

Correct Answer : C

A band of the surface has circumference \( 2\pi y \) and width equal to the arc element \( ds \), not \( dx \), so

\[ S \;=\; 2\pi\int y\,ds \;=\; 2\pi\int y\sqrt{1 + y'^{2}}\,dx. \]

Using \( dx \) in place of \( ds \) is the standard error: it measures the band along the axis rather than along the slanting surface, and understates the area

Your Answer : (Not Answered)

Correct Answer : B

The trapezoidal rule joins two ordinates by a chord and is exact for a straight line. Simpson's one-third fits a parabola through three ordinates but is exact up to a cubic, the cubic error terms cancelling. The three-eighths rule uses four ordinates and a cubic fit, so it too is exact to that degree

Your Answer : (Not Answered)

Correct Answer : A

Distinct eigenvalues do guarantee independence and so make the matrix diagonalisable, but they are sufficient rather than necessary. A repeated eigenvalue may still supply enough independent eigenvectors, as the identity matrix shows.

Your Answer : (Not Answered)

Correct Answer : B

Correct answer: a unique solution

The two ranks are equal, so the system is consistent, and their common value 3 equals the number of unknowns, so there are no free variables. The fourth equation is therefore not an extra constraint but a consequence of the other three - more equations than unknowns does not imply inconsistency, only that some are redundant.

Your Answer : (Not Answered)

Correct Answer : D

Interchanging the identical rows must change the sign while leaving the determinant unaltered, and only zero satisfies both. The same conclusion follows from the rows being linearly dependent, which makes the matrix singular.

Your Answer : (Not Answered)

Correct Answer : A

Clairaut's theorem gives equality only when the mixed derivatives are continuous at the point. Counterexamples exist in which \( f_{xy} \) and \( f_{yx} \) differ at the origin, both being discontinuous there. Mere existence of the first derivatives, or a domain covering the plane, is not enough

Your Answer : (Not Answered)

Correct Answer : C

Differentiate twice and set the result to zero:

\[ \frac{dy}{dx} = 3x^{2} - 12x + 9, \qquad \frac{d^{2}y}{dx^{2}} = 6x - 12 \;=\; 0. \]

So \( x = 2 \), and the second derivative changes from negative to positive there, confirming a genuine inflexion. The values \( x = 1 \) and \( x = 3 \) are the stationary points

Your Answer : (Not Answered)

Correct Answer : C

The four first derivatives are \( u_{x} = 1 \), \( u_{y} = 1 \), \( v_{x} = 1 \), \( v_{y} = -1 \), so

\[ \frac{\partial(u,v)}{\partial(x,y)} \;=\; \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} \;=\; (1)(-1) - (1)(1) \;=\; -2. \]

The magnitude two is the area stretch of this rotation and scaling, and the sign records that the map reverses orientation

Your Answer : (Not Answered)

Correct Answer : C

Repeated use of the product rule generates binomial coefficients, exactly as in the binomial theorem:

\[ (uv)^{(n)} \;=\; \sum_{r=0}^{n} \binom{n}{r} u^{(n-r)} v^{(r)}. \]

For \( n = 1 \) this is \( u'v + uv' \). The two-term forms offered omit every mixed derivative between the extremes, which for \( n = 2 \) loses the \( 2u'v' \) term

Your Answer : (Not Answered)

Correct Answer : A

The form must be recognised before anything is done to it, since a determinate power needs no such treatment. Logarithms then turn the power into \( g \ln f \), a product that becomes a quotient. That quotient is evaluated, and only at the end is the exponential taken to recover the limit itself rather than its logarithm

Your Answer : (Not Answered)

Correct Answer : B

Cauchy's form states that \( [f(b)-f(a)]/[g(b)-g(a)] = f'(c)/g'(c) \) for some interior \( c \), so two functions are compared at once. Taking \( g(x) = x \) recovers Lagrange's theorem. No second derivative is needed, monotonicity is not required, and the vanishing derivative is Rolle's conclusion

Your Answer : (Not Answered)

Correct Answer : B

Replace each factor by its leading term, \( e^{2x} - 1 \approx 2x \) and \( \tan 3x \approx 3x \):

\[ \lim_{x \to 0}\frac{e^{2x}-1}{\tan 3x} \;=\; \lim_{x \to 0}\frac{2x}{3x} \;=\; \frac{2}{3}. \]

L'Hopital's rule gives the same result, the derivatives being \( 2e^{2x} \) and \( 3\sec^{2}3x \), both equal to their coefficients at the origin