GATE EE 2015 Set 1

To solve this, we first need to find the values of the constants $a$ and $b$. We can do this by using two key properties of probability density functions (PDFs).

First, any valid PDF must integrate to 1 over its domain. This gives us our first equation:
\int_{0}^{1} (a+bx) dx = \[ax + \frac{bx^2}{2}]_0^1 = a + \frac{b}{2} = 1$.

Second, we are given the expected value, $E[X] = \frac{2}{3}$. Using the definition of expected value, we get our second equation:
E[X] = \int_{0}^{1} x(a+bx) dx = \[\frac{ax^2}{2} + \frac{bx^3}{3}]_0^1 = \frac{a}{2} + \frac{b}{3} = \frac{2}{3}$.

Solving the system of two equations ($2a+b=2$ and $3a+2b=4$) reveals that $a=0$ and $b=2$. Thus, the PDF is $f(x)=2x$ for $0 < x < 1$.

Finally, we calculate the probability $Pr[X < 0.5]$ by integrating our now-complete PDF from 0 to 0.5:
Pr[X < 0.5] = \int_{0}^{0.5} 2x \,dx = \[x^2]_0^{0.5} = (0.5)^2 - 0^2 = 0.25$.

Let's think about this visually. A continuous function is one you can draw without lifting your pen. If such a function has no root in the interval $[a,b]$, it means its graph never touches or crosses the x-axis within that domain.

This forces the entire portion of the graph from $x=a$ to $x=b$ to be either entirely above the x-axis or entirely below it.

In the first case, both $f(a)$ and $f(b)$ are positive. In the second case, both $f(a)$ and $f(b)$ are negative. In either scenario, the two values have the same sign. The product of two numbers with the same sign is always positive.

Therefore, we must have $f(a) \cdot f(b) > 0$. This is a direct consequence of the Intermediate Value Theorem.

Let the 2x2 matrix be

. We are given that its trace is $a+d=-6$, and we want to maximize its determinant, $\det(M) = ad-bc$.

For the determinant to have a well-defined maximum, we must constrain the $bc$ term. In the standard case of a symmetric matrix ($c=b$), the determinant becomes $ad-b^2$. To maximize this value, we must make the non-negative term $b^2$ as small as possible, which means setting $b=0$.

Our task simplifies to maximizing the product $ad$ subject to the sum $a+d=-6$. We can express the product as a function of one variable: let $d=-6-a$, so the product is $P(a) = a(-6-a) = -a^2 - 6a$. This function represents a downward-opening parabola, which attains its maximum value at its vertex.

The vertex occurs at $a = -(-6)/(2(-1)) = -3$. Substituting this back, we find $d=-6-(-3)=-3$. The maximum product, and thus the maximum determinant, is $ad = (-3)(-3) = 9$.

The problem asks for the integral of the divergence over the volume of the sphere, $\int_V (\nabla \cdot \vec{f}) dV$. Because the function $\vec{f}$ is singular at the origin ($r=0$), we cannot compute its divergence there. Instead, we use the Divergence Theorem, which equates the volume integral to the total flux through the enclosing surface:

$\int_V (\nabla \cdot \vec{f}) dV = \oint_S \vec{f} \cdot d\vec{S}$

On the surface of the sphere, the radius is a constant $r=R$. The differential area element in spherical coordinates is $d\vec{S} = R^2 \sin\theta \, d\theta \, d\phi \, \hat{r}$. We now compute the surface integral:

$\oint_S \left(\frac{1}{R^2}\hat{r}\right) \cdot (R^2 \sin\theta \, d\theta \, d\phi \, \hat{r})$

The $R^2$ terms cancel, and since $\hat{r} \cdot \hat{r} = 1$, the integral simplifies to the total solid angle of a sphere:

$\int_0^{2\pi} \int_0^{\pi} \sin\theta \, d\theta \, d\phi = 4\pi$

When the galvanometer shows zero current, the Wheatstone bridge is balanced. This occurs when the ratios of resistance in the two arms are equal: $\frac{R_1}{R_3} = \frac{R_2}{R_X}$. We can rearrange this equation to solve for the unknown resistor, $R_X = \frac{R_2 R_3}{R_1}$.

The resistor $R_3$ has a nominal value of $100\,\Omega$ with a $\pm 5\%$ tolerance, which means its actual resistance lies in the range $[95\,\Omega, 105\,\Omega]$. To find the possible range of $R_X$, we calculate its value using these minimum and maximum values of $R_3$.

For the minimum value, $R_X = \frac{65\,\Omega \times 95\,\Omega}{50\,\Omega} = 123.50\,\Omega$.
For the maximum value, $R_X = \frac{65\,\Omega \times 105\,\Omega}{50\,\Omega} = 136.50\,\Omega$.

Therefore, the range of possible values for $R_X$ is $[123.50\,\Omega, 136.50\,\Omega]$.

To begin, we must find the internal resistance of the ammeter's moving coil, which we'll call $R_m$. Using the given full-scale deflection values and Ohm's Law, this resistance is the voltage drop divided by the current.
$R_m = \frac{V_m}{I_{fs}} = \frac{0.1 \, \text{V}}{50 \, \text{A}} = 0.002 \, \Omega$

Next, we determine the multiplying factor ($m$) for the new range, which is the ratio of the desired maximum current to the original maximum current.
$m = \frac{I_{new}}{I_{fs}} = \frac{500 \, \text{A}}{50 \, \text{A}} = 10$

A shunt resistor ($R_{sh}$) connected in parallel allows the ammeter's range to be extended. The formula for the required shunt resistance is $R_{sh} = \frac{R_m}{m-1}$.
$R_{sh} = \frac{0.002 \, \Omega}{10 - 1} = \frac{0.002}{9} \, \Omega \approx 0.000222 \, \Omega$

Finally, converting this resistance to milliohms as requested:
$0.000222 \, \Omega \times 1000 = 0.222 \, \text{m}\Omega$

An instrumentation amplifier is a precision differential amplifier. Its primary requirements stem from this function.

1. High Common Mode Rejection Ratio (P): Its main purpose is to amplify a small differential signal while rejecting large common-mode signals (noise). A high CMRR is essential for this noise immunity.
2. High Input Impedance (Q): To accurately measure a voltage from a sensor without "loading" it (drawing significant current), the amplifier's input impedance must be very high.
3. High Linearity (R): For precision measurements, the output voltage must be a directly proportional and undistorted replica of the input voltage over the entire operating range.
4. An ideal amplifier should have a low output impedance (not S) to efficiently drive subsequent stages or loads without signal voltage loss. Therefore, high output impedance is an undesirable characteristic.

Based on this, P, Q, and R are the major requirements.

This circuit is a step-down chopper, where the average DC output voltage, $V_0$, is determined by the input voltage, $V_s$, and the duty cycle, $\alpha$. The relationship is given by $V_0 = \alpha V_s$.
With the given values, the average output voltage is $V_0 = 0.4 \times 20\,\text{V} = 8\,\text{V}$.

Under steady-state conditions, this average voltage is applied across the load, which consists of the resistor $R$ and the battery $E$. The current flowing through this load is the charging current, $I_0$.

By applying KVL to the output loop, we have $V_0 = I_0 R + E$.
We can now solve for the charging current $I_0$:
$I_0 = \frac{V_0 - E}{R} = \frac{8\,\text{V} - 5\,\text{V}}{3\,\Omega} = \frac{3\,\text{V}}{3\,\Omega} = 1\,\text{A}$.

Let the input signal be $u(t) = \sin(\omega_0 t)$. The given frequency $f_0 = \frac{1}{2T}$ corresponds to an angular frequency of $\omega_0 = 2\pi f_0 = \frac{\pi}{T}$. We can now find the output $y(t)$ by solving the moving average integral:
$y(t) = \frac{1}{T}\int_{t-T}^{t} \sin(\omega_0 \tau) d\tau = \frac{-1}{T\omega_0}\left[\cos(\omega_0 \tau)\right]_{t-T}^{t}$
$y(t) = \frac{-1}{T\omega_0}\left[\cos(\omega_0 t) - \cos(\omega_0(t-T))\right]$
Using the key relationship $\omega_0 T = \pi$, we can simplify the second term: $\cos(\omega_0(t-T)) = \cos(\omega_0 t - \pi) = -\cos(\omega_0 t)$.
Substituting this back, the output becomes $y(t) = \frac{-1}{T\omega_0}\left[2\cos(\omega_0 t)\right]$. Since $\omega_0 = \pi/T$, this is $y(t) = -\frac{2}{\pi}\cos(\omega_0 t)$.
To compare this with the input $\sin(\omega_0 t)$, we use the identity $-\cos(\theta) = \sin(\theta - 90^\circ)$, giving $y(t) = \frac{2}{\pi}\sin(\omega_0 t - 90^\circ)$. Thus, the output lags the input by $90^\circ$.

The overall impulse response of two cascaded systems is the convolution of their individual impulse responses. Here, the new impulse response is $h(t) = g(t) * g(t)$, where $*$ denotes convolution.

The function $g(t)$ is a rectangular pulse with a height of 1 and a duration of 1. Convolving a rectangular pulse with itself results in a triangular pulse.

The peak value of this triangular pulse is equal to the area under the curve of the original rectangular pulse.

For the given $g(t)$, the area is calculated as height × width.
Area $= 1 \times 1 = 1$.

Therefore, the maximum value of the resulting impulse response $h(t)$ is 1.

A time-varying magnetic field induces an electromotive force (EMF) in the loop according to Faraday's Law. First, we find the magnetic flux, $\phi(t) = B(t)A$, where the loop's area is $A = 0.1\,\text{m} \times 0.05\,\text{m} = 0.005\,\text{m}^2$. The induced EMF is the negative time derivative of the flux: $e(t) = -d\phi/dt = -A (dB/dt)$.

Given $B(t) = 0.25 \sin(\omega t)$, the EMF is $e(t) = -A \cdot (0.25\omega \cos(\omega t))$. The peak value of this sinusoidal EMF is $E_{max} = A \cdot 0.25\omega$. Using $\omega = 2\pi \times 50 = 100\pi$ rad/s, we find the peak EMF to be $E_{max} = (0.005)(0.25)(100\pi) = 0.125\pi$ V.

The average power absorbed by the loop's resistance $R$ is $P_{avg} = \frac{e_{rms}^2}{R}$. For a sinusoidal voltage, $e_{rms} = E_{max}/\sqrt{2}$, so the average power is $P_{avg} = \frac{E_{max}^2}{2R}$. Plugging in our values gives:
$P_{avg} = \frac{(0.125\pi)^2}{2 \times 0.4} \approx 0.193$ W.

The total magnetic field at the point (0, L, 0) is the vector sum of the fields produced by each of the two wires. First, we determine the distance from the point to each wire along the y-axis. The distance to the wire at $y=L/2$ is $r_1 = L - L/2 = L/2$, and the distance to the wire at $y=-L/2$ is $r_2 = L - (-L/2) = 3L/2$.

Using the right-hand rule for a current flowing in the -x direction, we find that both wires generate a magnetic field pointing in the negative z-direction ($-\hat{z}$) at the location (0, L, 0). Therefore, we can sum the magnitudes of the individual fields:
$\vec{B}_{total} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0 I}{2\pi(L/2)}(-\hat{z}) + \frac{\mu_0 I}{2\pi(3L/2)}(-\hat{z})$
Factoring out common terms gives:
$\vec{B}_{total} = \frac{\mu_0 I}{\pi L} \left(1 + \frac{1}{3}\right)(-\hat{z}) = \frac{\mu_0 I}{\pi L} \left(\frac{4}{3}\right)(-\hat{z})$
Thus, the net magnetic field is $\vec{B} = -\frac{4\mu_0 I}{3\pi L} \hat{z}$.

This circuit is a differential integrator. For an ideal op-amp in a negative feedback configuration, we can apply the virtual short concept, meaning the voltages at the inverting ($V_-$) and non-inverting ($V_+$) terminals are equal. By writing Kirchhoff's Current Law at both input nodes and treating the input $V_i$ as a balanced differential signal, we can derive the circuit's transfer function.

Let's analyze the circuit in the frequency domain using phasors, where the impedance of each capacitor is $Z_C = 1/(j\omega C)$. The KCL equations for the inverting and non-inverting nodes can be combined to show that the overall transfer function is given by the ratio of the feedback to input impedances:
$\frac{V_o}{V_i} = -\frac{Z_{feedback}}{Z_{input}} = -\frac{R}{Z_C}$
Substituting the expression for the capacitor's impedance gives:
$\frac{V_o}{V_i} = -\frac{R}{1/(j\omega C)} = -j\omega RC$
The term $-j$ in the transfer function represents a phase shift of $-90^\circ$, or $-\pi/2$ radians. Therefore, the output voltage $V_o$ lags the input voltage $V_i$ by this angle. The specific values of R, C, and frequency do not affect the phase for this ideal integrator circuit, only the gain magnitude.

This circuit uses a collector feedback configuration. We can find the base current, $I_B$, by analyzing the voltage drop across the base resistor $R_B$. Assuming a standard base-emitter voltage drop of $V_{BE} = 0.7V$ for a silicon transistor, we get $I_B = \frac{V_C - V_{BE}}{R_B} = \frac{9 - 0.7}{R_B} = \frac{8.3}{R_B}$. The current flowing through the collector resistor $R_C$ is the emitter current $I_E$, which is $I_E = \frac{15 - V_C}{R_C} = \frac{15 - 9}{R_C} = \frac{6}{R_C}$. We can relate these currents using the formula $I_E = (\beta + 1)I_B$. Substituting our expressions gives $\frac{6}{R_C} = (75 + 1) \frac{8.3}{R_B}$. Rearranging to solve for the required ratio yields $\frac{R_B}{R_C} = \frac{76 \times 8.3}{6} \approx 105.13$.

The output of a 4x1 multiplexer is determined by its select lines and data inputs. With $A$ and $B$ as select lines ($S_1=A, S_0=B$), the MUX's general output equation is $F = I_0\bar{A}\bar{B} + I_1\bar{A}B + I_2A\bar{B} + I_3AB$.

Our goal is to produce the output $F = A \oplus B$. We can expand this XOR function into its sum-of-products form, which is $F = \bar{A}B + A\bar{B}$.

To find the required inputs ($I_0, I_1, I_2, I_3$), we simply compare the coefficients of the minterms in the MUX equation with those in our desired function. For a clearer comparison, we can write the target function as $F = (0)\bar{A}\bar{B} + (1)\bar{A}B + (1)A\bar{B} + (0)AB$.

Matching the coefficients term-by-term, we find:

• The coefficient of $\bar{A}\bar{B}$ ($AB=00$) is 0, so $I_0=0$.
• The coefficient of $\bar{A}B$ ($AB=01$) is 1, so $I_1=1$.
• The coefficient of $A\bar{B}$ ($AB=10$) is 1, so $I_2=1$.
• The coefficient of $AB$ ($AB=11$) is 0, so $I_3=0$.

Therefore, the required input sequence is $I_3I_2I_1I_0 = 0110$.

In a thyristor-based HVDC system, the valves are unidirectional, meaning they can only conduct current in one direction. Therefore, to reverse power flow, the direction of current $I$ cannot be changed and it remains $1.5$ kA.

Since power is the product of voltage and current ($P=VI$), the only way to reverse the power flow direction is to reverse the polarity of the DC link voltage. This is accomplished by controlling the firing angles of the thyristors, causing both $V_1$ and $V_2$ to become negative.

For power to flow from System 2 to System 1, System 2 must now act as the rectifier and System 1 as the inverter. The rectifier's voltage magnitude must always be greater than the inverter's to drive the current. Using the original magnitudes for these new roles gives $|V_2| = 500$ kV and $|V_1| = 485$ kV. Applying the reversed polarity, we get $V_2 = -500$ kV and $V_1 = -485$ kV.

Base load power plants are designed to provide a continuous, reliable supply of electricity to meet the minimum demand.

Nuclear power plants (R) are classic examples of base load generators. They have high initial costs but low fuel costs, making them economical to run continuously at full capacity. It is also technically difficult and inefficient to frequently start and stop them.

Run-of-river plants (Q) also provide a consistent power output, as they utilize the natural, continuous flow of a river. This makes them suitable for meeting base load demand.

In contrast, wind farms (P) are intermittent sources, as they only generate power when the wind blows. Diesel plants (S) have high fuel costs and are typically used as peaking power plants to meet short-term high demand, not for continuous base load operation. Therefore, only Q and R are base load plants.

To determine the power delivered by the 5V source, we first need to find the current, $I$, that it supplies. We can use the formula $P = VI$.

Let's apply Kirchhoff's Current Law (KCL) to the currents leaving the "Network N1" block. The current in the top branch is $I_{3\Omega} = V/R = 6\text{ V} / 3\text{ } \Omega = 2\text{ A}$. Based on the polarity, this current flows from right to left, into Network N1.

The current in the middle branch is $I_{2\Omega} = V/R = 2\text{ V} / 2\text{ } \Omega = 1\text{ A}$. The polarity indicates this current flows from left to right, out of Network N1.

Let $I$ be the current from the 5V source, which flows out of N1. KCL states the sum of currents leaving the block must be zero: $(-2 \text{ A}) + (1 \text{ A}) + I = 0$. Solving this gives $I = 1$ A.

Finally, the power delivered by the source is $P = (5 \text{ V}) \times (1 \text{ A}) = 5 \text{ W}$.

The Thevenin voltage, $V_{Th}$, is the voltage $V_{AB}$ across the terminals in the given configuration. Let's use nodal analysis, with the bottom wire as our 0V reference. Let the voltage at the node above the central 1Ω resistor be $V_C$. The controlling current is then $i = V_C / 1\Omega = V_C$.

The voltage at terminal A, $V_{A}$, is the sum of the voltage at node $C$ and the voltage from the dependent source. So, $V_A = V_C + 20i = V_C + 20V_C = 21V_C$.

Now, we apply Kirchhoff's Current Law (KCL) at node $C$. The sum of currents leaving the node is zero:
$\frac{V_C - 2\text{V}}{1\Omega} + \frac{V_C}{1\Omega} + \frac{V_A}{2\Omega} = 0$
Substituting our expressions for $i$ and $V_A$: $(V_C - 2) + V_C + \frac{21V_C}{2} = 0$.
Solving for $V_C$ gives $2.5V_C + 10.5V_C = 2$, which means $12.5V_C = 2$, so $V_C = 0.16$ V.
The required voltage is $V_{Th} = V_{AB} = V_A = 21V_C = 21(0.16) = 3.36$ V.

The impedance of the inductor is $Z_L = j\omega L$ and the capacitor is $Z_C = 1/(j\omega C)$. Since they are connected in parallel, the total impedance $Z$ is given by the formula $Z = \frac{Z_L Z_C}{Z_L + Z_C}$. Substituting the expressions for $Z_L$ and $Z_C$ and simplifying, we get the overall impedance of the network as $Z = \frac{j\omega L}{1 - \omega^2 LC}$.

This is a purely reactive impedance, $Z=jX$, where the reactance is $X(\omega) = \frac{\omega L}{1 - \omega^2 LC}$. At the resonant frequency, $\omega_0 = 1/\sqrt{LC}$, the denominator becomes zero, causing the impedance to diverge. For frequencies below resonance ($\omega < \omega_0$), the denominator is positive, so the reactance $X$ is positive (inductive) and approaches $+\infty$. For frequencies above resonance ($\omega > \omega_0$), the denominator is negative, making the reactance $X$ negative (capacitive) and approaching from $-\infty$. This behavior is accurately represented by graph B.

The generated EMF of a DC generator ($E_g$) is directly proportional to both the field flux (controlled by field current) and the rotor speed. The initial open-circuit voltage is given as $200 \text{ V}$. When both speed and field current are halved, the new EMF is reduced by a factor of four: $E'_g = 200 \text{ V} / 4 = 50 \text{ V}$.

This new EMF acts as a constant voltage source charging the capacitor through the armature resistance, forming an RC circuit. The voltage across the capacitor, $v_c(t)$, follows the standard charging equation: $v_c(t) = V_{final}(1 - e^{-t/\tau})$. Here, the final voltage is $V_{final} = E'_g = 50 \text{ V}$ and the time constant is $\tau = R_a C = 0.1 \Omega \times 1000 \mu\text{F} = 10^{-4} \text{ s}$.

We need to find the time $t$ when the capacitor voltage reaches $25 \text{ V}$.
$25 = 50(1 - e^{-t/10^{-4}})$
Solving this equation gives $0.5 = 1 - e^{-t/10^{-4}}$, which means $e^{-t/10^{-4}} = 0.5$.
Taking the natural logarithm of both sides yields $t = 10^{-4} \ln(2) \approx 0.693 \times 10^{-4} \text{ s}$, or $69.3 \mu\text{s}$.

The effective inductance seen by the source is determined by the total impedance of the primary circuit, which includes the effect of the short-circuited secondary. Using phasor analysis, we can write the voltage equations for the primary ($V_1$) and secondary ($V_2$) windings:
$V_1 = j\omega L_1 I_1 + j\omega M I_2$
$V_2 = j\omega L_2 I_2 + j\omega M I_1$

Since the secondary is short-circuited, $V_2 = 0$. We can use the second equation to express the secondary current, $I_2$, in terms of the primary current, $I_1$:
$0 = j\omega L_2 I_2 + j\omega M I_1 \implies I_2 = -\frac{M}{L_2}I_1$

Now, substitute this relationship back into the primary voltage equation:
$V_1 = j\omega L_1 I_1 + j\omega M \left(-\frac{M}{L_2}I_1\right) = j\omega \left(L_1 - \frac{M^2}{L_2}\right)I_1$

The term in the parenthesis is the effective inductance ($L_{eff}$) seen by the source. Plugging in the given values:
$L_{eff} = 800 \text{ mH} - \frac{(480 \text{ mH})^2}{600 \text{ mH}} = 800 - 384 = 416 \text{ mH}$

The primary magnetomotive force (MMF) is given by the product of the primary turns and the primary current, $MMF_p = N_p I_p$. In power, potential, and distribution transformers, the primary is connected across a voltage source, and the primary current $I_p$ is drawn in response to the load on the secondary. Therefore, their primary MMF is heavily dependent on the secondary conditions.

In a current transformer (CT), the primary winding is connected in series with the main power line. The primary current $I_p$ is the line current itself, which is determined by the load on the main circuit, not the burden (load) connected to the CT's secondary. Consequently, the primary MMF, $N_p I_p$, remains nearly constant and is least affected by the CT's secondary terminal conditions.

The Bode plot's low-frequency asymptote is a flat line at +20 dB. From this, we can find the system's DC gain, $K$, using the formula $20 \log_{10}(K) = 20$, which yields $K=10$.

The plot exhibits two corner frequencies. The first, at $f=10$ Hz, introduces a slope of -20 dB/decade, indicating a pole at $\omega_p=10$ rad/s. The second, at $f=1000$ Hz, cancels this slope, indicating a zero at $\omega_z=1000$ rad/s.

We can assemble the transfer function using the time-constant form: $G(s) = K \frac{1+s/\omega_z}{1+s/\omega_p}$.
Substituting the values we found: $G(s) = 10 \frac{1+s/1000}{1+s/10}$.

To match the format of the options, we rearrange the expression:
$G(s) = 10 \frac{(1000+s)/1000}{(10+s)/10} = 10 \cdot \frac{10}{1000} \cdot \frac{s+1000}{s+10} = \frac{1}{10} \frac{s+1000}{s+10} = \frac{s+1000}{10(s+10)}$.

To find the transfer function, we'll use Mason's Gain Formula, remembering to set the input $X_1(s)$ to zero.

First, identify the direct forward path from the input $X_2(s)$ to the output $Y(s)$. There is only one, with a gain of $P_1 = G_2$.

Next, find all the feedback loops. The system has two: one with gain $L_1 = -G_1$ and a larger one with gain $L_2 = -G_1G_2$.

The determinant, $\Delta$, is calculated as 1 - (\text{\sum of all individual loop gains}). This gives $\Delta = 1 - (L_1 + L_2) = 1 - (-G_1 - G_1G_2) = 1 + G_1 + G_1G_2$.

Since both loops touch the forward path $P_1$, the path's cofactor $\Delta_1$ is 1.

Finally, the transfer function is $\frac{P_1\Delta_1}{\Delta} = \frac{G_2 \cdot 1}{1+G_1+G_1G_2}$. Factoring the denominator gives the result: $\frac{G_2}{1+G_1(1+G_2)}$.

For a matrix to have three linearly independent real eigenvectors, it must be diagonalizable over the real numbers. A sufficient condition for this is that it has three distinct real eigenvalues. The characteristic equation is given by $\det(A - \lambda I) = 0$, which for this matrix is $(\lambda+1)(\lambda+2)(\lambda+3) = -2a$.

Let $g(\lambda) = (\lambda+1)(\lambda+2)(\lambda+3)$. For the characteristic equation to have three distinct real roots, the value of the constant $-2a$ must lie strictly between the local minimum and maximum of the cubic function $g(\lambda)$. The local extrema of $g(\lambda)$ are found by setting its derivative to zero: $g'(\lambda) = 3\lambda^{2+}12\lambda+11=0$. The critical points are $\lambda_c = -2 \pm \frac{\sqrt{3}}{3}$.

The extremal values of $g(\lambda)$ at these points are $g(\lambda_c) = \mp \frac{2\sqrt{3}}{9}$. Therefore, for distinct eigenvalues, we must have $-\frac{2\sqrt{3}}{9} < -2a < \frac{2\sqrt{3}}{9}$. Dividing by $-2$ and reversing the inequalities gives $-\frac{\sqrt{3}}{9} < a < \frac{\sqrt{3}}{9}$.

If $a$ reaches these boundary values, the matrix has repeated eigenvalues and can be shown to not be diagonalizable. The maximum value (supremum) for $a$ is the upper bound of this interval, which is $\frac{\sqrt{3}}{9} = \frac{1}{3\sqrt{3}}$.

To solve this second-order linear homogeneous differential equation, we first find the roots of its characteristic equation, $m^{2+}5m+6=0$. The roots are $m=-2$ and $m=-3$, leading to a general solution of the form $y(t) = c_1e^{-2t} + c_2e^{-3t}$.

We now use the given conditions to determine the constants $c_1$ and $c_2$. The first condition, $y(0)=2$, gives us the equation $c_1+c_2=2$. The second condition, $y(1) = -\frac{1-3e}{e^3}$, becomes $c_1e^{-2} + c_2e^{-3} = \frac{3e-1}{e^3}$, which simplifies to $ec_1+c_2 = 3e-1$. Solving this system of two equations yields $c_1=3$ and $c_2=-1$.

With the constants found, our specific solution is $y(t) = 3e^{-2t} - e^{-3t}$. To find the value of the derivative at $t=0$, we first differentiate this expression: $\frac{dy}{dt} = -6e^{-2t} + 3e^{-3t}$. Finally, substituting $t=0$ gives $\frac{dy}{dt}(0) = -6e^0 + 3e^0 = -6+3=-3$.

To find the components of the Fourier series for $f(t) = \text{sgn}(\cos(t))$, we should analyze the function's symmetry properties.

First, let's check if the function is even or odd. Since $\cos(-t) = \cos(t)$, we have $f(-t) = \text{sgn}(\cos(-t)) = \text{sgn}(\cos(t)) = f(t)$. The function is even, which means its Fourier series will only contain cosine terms.

Next, let's test for half-wave symmetry. A function has half-wave symmetry if $f(t + T/2) = -f(t)$. Here the period is $T=2\pi$, so we check $f(t+\pi)$. We find $f(t+\pi) = \text{sgn}(\cos(t+\pi)) = \text{sgn}(-\cos(t)) = -f(t)$. This symmetry means the series contains only odd-numbered harmonics.

Combining these two properties, the Fourier series must consist exclusively of cosine terms with odd harmonics.

Player A can win on their first turn, second turn, third turn, and so on. We can find the total probability by summing the probabilities of these individual, mutually exclusive events.

The probability A wins on their first roll is $\frac{1}{6}$. For A to win on their second turn, A must first fail, then B must fail, and then A must succeed. The probability for this sequence is $(\frac{5}{6})(\frac{5}{6})(\frac{1}{6})$. This pattern continues for all of A's subsequent turns.

The total probability is an infinite geometric series:
$P(\text{A wins}) = \frac{1}{6} + (\frac{5}{6})^2 \frac{1}{6} + (\frac{5}{6})^4 \frac{1}{6} + \dots$

This series has a first term $a = \frac{1}{6}$ and a common ratio $r = (\frac{5}{6})^2 = \frac{25}{36}$. Using the formula for the sum of an infinite geometric series, $S = \frac{a}{1-r}$, we find the probability:
$P(\text{A wins}) = \frac{1/6}{1 - 25/36} = \frac{1/6}{11/36} = \frac{1}{6} \times \frac{36}{11} = \frac{6}{11}$.

The output voltage $V_o$ is the difference between the potentials at the two central nodes of the bridge. Applying the voltage divider rule to each arm, we find the expression for $V_o$ as a function of the variable resistance parameter $p$:
$V_o = E \left[ \frac{1+x}{p+1+x} - \frac{1}{p+1} \right]$
Combining the terms into a single fraction simplifies the expression to:
$V_o = E \left[ \frac{px}{(p+1)(p+1+x)} \right]$
To find the maximum value of $V_o$, we need to find the value of $p$ for which the derivative $\frac{dV_o}{dp}$ equals zero. Since $E$ and $x$ are constants, we only need to maximize the part of the expression that depends on $p$. Using the quotient rule, the derivative is zero when the numerator of the derivative is zero. For a function $\frac{u(p)}{v(p)}$, this condition is $u'v - uv' = 0$.

Here, $u=p$ and $v=(p+1)(p+1+x) = p^{2+}(2+x)p+(1+x)$.
This gives the condition $1 \cdot (p^{2+}(2+x)p+(1+x)) - p \cdot (2p+2+x) = 0$.
Expanding and simplifying this equation gives $-p^2 + 1 + x = 0$, which rearranges to $p^2 = 1+x$. Since resistance must be positive, we take the positive root, yielding $p=\sqrt{1+x}$.