Mathematics-II: Multiple Integrals, Differential Equations & Complex Analysis: 21 MCQs with answers

D · Green's theorem

Green's theorem states $\oint_C (M\,dx + N\,dy) = \iint_R (N_x - M_y)\,dA$, converting a line integral to a double integral.

D · Compulsory 4-mark part (a) and choice of two 8-mark parts

Each question has a compulsory 4-mark part (a) and a choice of two 8-mark parts from (b), (c), (d).

D · Redraw the region and rewrite the limits

Changing the order requires redrawing the region of integration and rewriting the limits for the new order.

B · $v = xy$

The substitution $v = xy$ collapses the equation into a separable form.

D · Complementary function and particular integral

The general solution is the complementary function (from the auxiliary equation) plus a particular integral.

B · $x = e^t$

Substituting $x = e^t$ converts $x^2 y'' + axy' + by = f(x)$ into a constant-coefficient equation in $t$.

D · $u_x = v_y$ and $u_y = -v_x$

A function is analytic iff the Cauchy–Riemann equations $u_x = v_y$ and $u_y = -v_x$ hold.

D · $\frac{2\pi i}{n!}f^{(n)}(a)$

Cauchy's integral formula gives $\oint_C \frac{f(z)}{(z-a)^{n+1}}dz = \frac{2\pi i}{n!}f^{(n)}(a)$.

A · It equals $2\pi i$ times the sum of enclosed residues

The residue theorem states $\oint_C f(z)\,dz = 2\pi i \times$ (sum of residues at enclosed poles).

B · $z = e^{i\theta}$

The substitution $z = e^{i\theta}$ converts the real integral into a contour integral on the unit circle.

A · True

Cauchy's integral formula $\oint_C \frac{f(z)}{(z-a)^{n+1}}dz = \frac{2\pi i}{n!}f^{(n)}(a)$ lets you evaluate certain contour integrals without any integration.

A · Variation of parameters

The right-hand side $\tan 2x$ is non-standard, so variation of parameters is used.

C · Each annulus requires a different geometric series for convergence

Each annulus demands a different geometric series choice for $\frac{1}{z-1}$ and $\frac{1}{z-2}$ to ensure convergence.

B · Milne–Thomson's method

Milne–Thomson's method uses the Cauchy–Riemann equations to find $v$ from $u$.

B · Compute both sides independently and show equality

Verification means computing both sides of the theorem separately and demonstrating they are equal.

B · Differentiate with respect to $x$

Clairaut's equation is solved by differentiating with respect to $x$, solving for $p$, then substituting back.

C · 1/3

Changing the order of integration requires redrawing the region. The correct limits yield the result of 1/3 after evaluating the integral.

C · y = e^{-2x}(C + e^{-x})

The integrating factor is $e^{2x}$. Multiplying through gives the left side as a derivative, allowing us to solve for $y$ and find the particular solution.

C · 1

The residue at $z = -1$ is found by evaluating the limit of $(z + 1)f(z)$ as $z$ approaches -1, which equals 1. By the residue theorem, the integral evaluates to $2\pi i$ times this residue.

B · It is not analytic anywhere.

The Cauchy-Riemann equations do not hold for this function, indicating it is not analytic anywhere in the complex plane.

C · y = C_1 x^{-1} + C_2 x^{-4}

The substitution $x = e^t$ transforms the equation into a constant-coefficient form, leading to the characteristic equation with roots that yield the general solution.