GATE EE 2014 Set 2

A real symmetric matrix $A$ is a square matrix that is equal to its own transpose, $A = A^T$. Let $\lambda$ be an eigenvalue of $A$ and $v$ be its corresponding (possibly complex) eigenvector. The eigenvalue equation is $Av = \lambda v$.

Taking the conjugate transpose of this equation, we get $(Av)^* = (\lambda v)^*$, which simplifies to $v^*A^* = \bar{\lambda}v^*$. Since $A$ is a real symmetric matrix, its conjugate transpose $A^*$ is equal to $A$ itself. The equation thus becomes $v^*A = \bar{\lambda}v^*$.

Now, we can right-multiply by $v$ to obtain $v^*Av = \bar{\lambda}v^*v$. From our original equation $Av = \lambda v$, we can also left-multiply by $v^*$ to get $v^*Av = \lambda v^*v$. Therefore, we can equate the two expressions: $\lambda v^*v = \bar{\lambda}v^*v$.

Since $v$ is an eigenvector, it is a non-zero vector, which means the term $v^*v$ (the squared magnitude of $v$) is a positive real number. We can safely divide by $v^*v$ to find $\lambda = \bar{\lambda}$. This equality holds if and only if the imaginary part of $\lambda$ is zero, proving that all eigenvalues of a real symmetric matrix are real.

The phrase "proportional to n" means the probability of rolling a face with $n$ dots is $n$ times some base value. Let's call this base probability $k$. So, the probability of rolling a 1 is $1k$, a 2 is $2k$, a 3 is $3k$, and so on.

The sum of the probabilities of all possible outcomes must equal 1. We can express this with the equation:
$k + 2k + 3k + 4k + 5k + 6k = 1$

Solving for $k$, we find that $21k = 1$, so $k = \frac{1}{21}$.

The question asks for the probability of rolling a 3, which we defined as $3k$. Therefore, the probability is $3 \times \frac{1}{21} = \frac{3}{21} = \frac{1}{7}$. This is approximately $0.1428$.

Let's analyze the function $f(x)=(x-1)^{2/3}$. It is helpful to rewrite this expression as $f(x) = (\sqrt[3]{x-1})^2$.

The function's value is determined by squaring the term $\sqrt[3]{x-1}$. The square of any real number is always non-negative, meaning it is always greater than or equal to zero.

Therefore, the minimum possible value for $f(x)$ is $0$. This minimum occurs when the term being squared is zero.

We find the value of $x$ that makes this happen by solving $\sqrt[3]{x-1} = 0$. Cubing both sides, we get $x-1 = 0$, which means the minimum occurs at $x=1$.

To find the values of the multi-valued function $1^i$, we rely on the definition of complex exponentiation: $a^b = e^{b \log a}$.

For our problem, $a=1$ and $b=i$. The key is that the complex logarithm, $\log(a)$, is multi-valued. The number 1 can be expressed in polar form as $e^{i(2n\pi)}$ for any integer $n$.

Therefore, $\log(1) = \ln|1| + i \arg(1) = 0 + i(2n\pi) = 2n\pi i$.

Substituting this into the definition, we get:
$1^i = e^{i \cdot \log(1)} = e^{i \cdot (2n\pi i)}$

Simplifying the exponent gives $i^2(2n\pi) = -2n\pi$. Thus, all the possible values of $1^i$ are given by $e^{-2n\pi}$ where $n$ is any integer. Since $-2n\pi$ is always a real number, the values $e^{-2n\pi}$ are all positive real numbers, which are of course real and non-negative.

This problem presents a Cauchy-Euler equation, which can be simplified with the substitution $x = e^z$ (for $x>0$). This change of variable converts the equation with variable coefficients into one with constant coefficients. The derivative operators transform according to the rules $x\frac{d}{dx} = \frac{d}{dz}$ and $x^2\frac{d^2}{dx^2} = \frac{d^2}{dz^2} - \frac{d}{dz}$.

Applying this to the original equation gives:
$(\frac{d^2y}{dz^2} - \frac{dy}{dz}) + (\frac{dy}{dz}) - y = 0$
This simplifies nicely to $\frac{d^2y}{dz^2} - y = 0$. The corresponding characteristic equation is $m^2 - 1 = 0$, which has roots $m = 1$ and $m = -1$.

The general solution in terms of $z$ is $y(z) = C_1e^z + C_2e^{-z}$. To find the solution in terms of $x$, we substitute back $z = \ln x$.
This yields $y(x) = C_1e^{\ln x} + C_2e^{-\ln x} = C_1x + C_2x^{-1}$. The general solution is a linear combination of $x$ and $1/x$, so $1/x$ is a valid particular solution.

When coupled inductors are connected in series, the total inductance depends on whether their magnetic fields aid or oppose each other. The larger measured value corresponds to the series-aiding connection, while the smaller value is from the series-opposing connection.

We can set up two equations:
Series-aiding: $L_1 + L_2 + 2M = 380 \mu\text{H}$
Series-opposing: $L_1 + L_2 - 2M = 240 \mu\text{H}$

To isolate the mutual inductance, $M$, subtract the second equation from the first:
$(L_1 + L_2 + 2M) - (L_1 + L_2 - 2M) = 380 - 240$
This simplifies to $4M = 140 \mu\text{H}$.

Solving for $M$ gives $M = \frac{140}{4} = 35 \mu\text{H}$.

To solve for the capacitor voltage $v_c(t)$ for $t>0$, we analyze the circuit in three stages: the initial condition ($t=0^-$), the final condition ($t \to \infty$), and the transient response.

1. Initial Condition: For $t<0$, the switch has been at position '1' for a long time. The circuit is in DC steady state, so the capacitor acts as an open circuit. This means the initial voltage across the capacitor is equal to the source voltage it's connected to: $v_c(0^-) = V_{o1}$. Since the voltage across a capacitor cannot change instantaneously, the voltage at $t=0^+$ is $v_c(0^+) = v_c(0^-) = V_{o1}$.

2. Final Condition: At $t=0$, the switch moves to position '2'. As $t \to \infty$, the circuit reaches a new steady state. The capacitor again acts as an open circuit. In this new configuration, no current flows through the resistors, so there is no voltage drop across them. The capacitor voltage will equal the source voltage $V_{o2}$: $v_c(\infty ) = V_{o2}$.

3. Transient Response: The general solution for the voltage across a capacitor in a first-order RC circuit is given by the formula v_c(t) = v_c(\infty ) + \[v_c(0^+) - v_c(∞)]$e^{-t/\tau}$. For$t>0$, the capacitor "sees" an equivalent resistance$R_{th}$which is the series combination of the two resistors,$R_{th} = R + R = 2R$. The time constant is therefore$\tau = R_{th}C = 2RC$.

Substituting the initial condition, final condition, and time constant into the general solution, we get:
$v_c(t) = V_{o2} + (V_{o1} - V_{o2})e^{-t/2RC}$

To match this with the given options, we can rearrange the expression:
$v_c(t) = V_{o2} - (V_{o2} - V_{o1})e^{-t/2RC}$
$v_c(t) = V_{o2} - V_{o2}e^{-t/2RC} + V_{o1}e^{-t/2RC}$
$v_c(t) = (V_{o2} - V_{o1}) - V_{o2}e^{-t/2RC} + V_{o1}e^{-t/2RC} + V_{o1}$
$v_c(t) = (V_{o2}-V_{o1})(1-e^{-t/2RC}) + V_{o1}$

This capacitor can be treated as three capacitors connected in series. The two outer capacitors are identical, each with dielectric $\varepsilon_1$ and thickness $d/4$. The central capacitor has dielectric $\varepsilon_2$ and thickness $d/2$. The total voltage across the plates is 10 V. Since the voltage drop across one outer section is 2 V, by symmetry the drop across the other is also 2 V. Thus, the voltage across the central section is $V_2 = 10 - 2 - 2 = 6$ V.

For capacitors in series, the charge $Q$ on each is the same. Using the relationship $Q = CV$, we can write $C_1V_1 = C_2V_2$. The capacitance of a parallel plate capacitor is $C = \frac{\varepsilon\varepsilon_0 A}{t}$, where $t$ is the thickness.

Substituting this into our equation:
$\left(\frac{\varepsilon_1 \varepsilon_0 A}{d/4}\right) (2 \text{ V}) = \left(\frac{\varepsilon_2 \varepsilon_0 A}{d/2}\right) (6 \text{ V})$
After canceling common terms ($\varepsilon_0, A, d$), we get:
$4\varepsilon_1(2) = 2\varepsilon_2(6) \implies 8\varepsilon_1 = 12\varepsilon_2$
This simplifies to $\frac{\varepsilon_1}{\varepsilon_2} = \frac{12}{8} = \frac{3}{2}$, so the ratio $\varepsilon_1:\varepsilon_2$ is 3:2.

For a stable LTI system, when the input is a sinusoid, the steady-state output is also a sinusoid of the same frequency. Its amplitude is the input amplitude (which is 1 for $\cos(3t)$) multiplied by the magnitude of the system's frequency response at the input frequency.

The input signal is $\cos(3t)$, so the angular frequency is $\omega=3$ rad/s. We find the frequency response by evaluating the transfer function $H(s)$ at $s=j\omega$.

The magnitude of the frequency response at $\omega=3$ is:
$|H(j3)| = \left|\frac{1}{j3(j3+4)}\right| = \frac{|\,1\,|}{|j3| \cdot |4+j3|}$

Using the fact that $|a+jb|=\sqrt{a^{2+}b^2}$, we can calculate the magnitudes:
$|j3| = 3$ and $|4+j3| = \sqrt{4^{2+}3^2} = \sqrt{16+9} = \sqrt{25} = 5$.

The output amplitude $A$ is therefore equal to this magnitude:
$A = |H(j3)| = \frac{1}{3 \times 5} = \frac{1}{15}$.

To find the unknown input $x(t)$, we can work in the Laplace domain. The relationship between the input $X(s)$, output $Y(s)$, and the system's transfer function $H(s)$ is given by $Y(s) = H(s)X(s)$.

First, let's find the Laplace transforms of the given signals.
The output signal is $y(t) = e^{-3t}u(t) - e^{-5t}u(t)$, so its transform is $Y(s) = \frac{1}{s+3} - \frac{1}{s+5} = \frac{2}{(s+3)(s+5)}$.
The transfer function $H(s)$ is the Laplace transform of the impulse response $h(t) = e^{-5t}u(t)$, which is $H(s) = \frac{1}{s+5}$.

To find the input's transform, we solve for $X(s) = \frac{Y(s)}{H(s)}$.
Substituting our expressions gives $X(s) = \frac{2/[(s+3)(s+5)]}{1/(s+5)} = \frac{2}{s+3}$.
Taking the inverse Laplace transform of $X(s)$ gives us the input signal, $x(t) = 2e^{-3t}u(t)$.

First, let's find the Thevenin voltage ($V_{Th}$), which is the open-circuit voltage across terminals x and y. With the secondary circuit open, no current flows in the ideal transformer. Therefore, there's no voltage drop across the $1\Omega$ resistor, and the voltage across the primary winding is the source voltage, $\sin(\omega t)$. This voltage is stepped up by the turns ratio, so $V_{Th} = \sin(\omega t) \times \frac{2}{1} = 2\sin(\omega t)$.

Next, we find the Thevenin impedance ($Z_{Th}$) by looking into terminals x-y with the voltage source shorted. The $1\Omega$ resistance on the primary side must be referred to the secondary side. Impedance is referred by the square of the turns ratio, giving us $Z_{Th} = 1\Omega \times (\frac{2}{1})^2 = 4\Omega$.

First, we find the voltage ratio of the autotransformer, $a_{auto}$, by dividing the high-side voltage by the low-side voltage. Based on the figure, this is $a_{auto} = \frac{1100}{1000} = \frac{11}{10}$.

The kVA rating of an autotransformer ($S_{auto}$) is related to its original two-winding rating ($S_{two-winding}$) by the formula:
$\frac{S_{auto}}{S_{two-winding}} = \frac{a_{auto}}{a_{auto}-1}$
Substituting the voltage ratio we found:
$\frac{S_{auto}}{S_{two-winding}} = \frac{11/10}{(11/10) - 1} = \frac{11/10}{1/10} = 11$
This means the autotransformer can handle 11 times the power. Therefore, its kVA rating is:
$S_{auto} = 11 \times S_{two-winding} = 11 \times 50 \text{ kVA} = 550 \text{ kVA}$

An induction machine must be driven faster than its synchronous speed ($N_r > N_s$) to operate as a generator. For a 4-pole machine, the synchronous speed at 50 Hz is $N_s = \frac{120 \times 50}{4} = 1500$ rpm. Since the generator is maintained at $N_r = 1500$ rpm, the synchronous speed in both load cases must be less than 1500 rpm. This implies that both operating frequencies, $f_1$ and $f_2$, must be less than 50 Hz.

The amount of power generated is determined by the slip, $S = \frac{N_s - N_r}{N_s}$. A larger load requires a larger magnitude of negative slip. When the load is partially removed (decreased), the required slip magnitude becomes smaller. For the slip to decrease with a constant rotor speed $N_r$, the synchronous speed $N_s$ must increase. Since frequency is directly proportional to synchronous speed ($f \propto N_s$), a decrease in load leads to an increase in frequency. Therefore, $f_2 > f_1$.

The motor's real power draw, $P$, remains constant at 12 MW. First, let's find the initial reactive power ($Q_1$) when the power factor is 0.6 lagging. The power triangle gives us $Q_1 = P \tan(\phi_1) = P \tan(\cos^{-1}0.6) = 12 \tan(53.13^\circ) = 16$ MVAR.

Next, we calculate the desired final reactive power ($Q_2$) after improving the power factor to 0.8 lagging. This new reactive power is $Q_2 = P \tan(\phi_2) = P \tan(\cos^{-1}0.8) = 12 \tan(36.87^\circ) \approx 9.0$ MVAR.

The capacitor's role is to supply reactive power, thereby reducing the total reactive power drawn from the source. The reactive power delivered by the capacitor ($Q_C$) is simply the difference between the initial and final reactive power values.
$Q_C = Q_1 - Q_2 = 16 - 9.0 = 7.0$ MVAR. Using more precise intermediate values, $16 - 8.999...$ MVAR gives approximately 7.0 MVAR. (Using the original explanation's rounded intermediate value: $16 - 8.99 = 7.01$ MVAR).

To find the actual load voltage in kV, we must first determine the system's base voltage. The base line-to-line voltage ($V_{base, L-L}$) can be calculated using the given three-phase base power ($S_{base}$) and base line current ($I_{base}$).

The formula for three-phase power is $S_{base} = \sqrt{3} \times V_{base, L-L} \times I_{base}$.
Rearranging to solve for the base voltage, we get:
$V_{base, L-L} = \frac{S_{base}}{\sqrt{3} \times I_{base}} = \frac{100 \times 10^6 \text{ VA}}{\sqrt{3} \times 437.38 \text{ A}} \approx 132 \text{ kV}$.

Finally, the actual line-to-line voltage is the product of its per-unit value and the base voltage:
Actual Voltage $= V_{pu} \times V_{base, L-L} = 0.9 \times 132 \text{ kV} = 118.8 \text{ kV}$.

High-voltage transmission lines act like large capacitors due to their physical structure. Under lightly loaded or no-load conditions, this capacitance generates reactive power, causing a leading current to flow. This phenomenon, known as the Ferranti effect, causes the voltage at the receiving (load) end to rise, sometimes to dangerously high levels.

A shunt reactor is a large inductor connected in parallel with the line. It absorbs reactive power by drawing a lagging current, which effectively cancels out the leading current from the line's capacitance. This compensation stabilizes the line's voltage and prevents over-voltages during periods of low load.

To find the steady-state error, we first need to derive the system's open-loop transfer function, $G(s)$, from the given closed-loop function $T(s)$. For a unity feedback system, the relationship is $T(s) = \frac{G(s)}{1+G(s)}$. We can solve for $G(s)$ as $G(s)=\frac{T(s)}{1-T(s)}$.

Substituting the given $T(s)$, we get $G(s) = \frac{4/(s^{2+}0.4s+4)}{1-4/(s^{2+}0.4s+4)} = \frac{4}{s^{2+}0.4s} = \frac{4}{s(s+0.4)}$.

The steady-state error, $e_{ss}$, for a unit step input ($R(s)=1/s$) is found using the Final Value Theorem: $e_{ss} = \lim_{s \to 0}\frac{sR(s)}{1+G(s)}$.

Plugging in our functions for $R(s)$ and $G(s)$, we get:
$e_{ss} = \lim_{s \to 0}\frac{s(1/s)}{1+\frac{4}{s(s+0.4)}} = \lim_{s \to 0}\frac{1}{\frac{s(s+0.4)+4}{s(s+0.4)}} = \lim_{s \to 0}\frac{s(s+0.4)}{s^{2+}0.4s+4}$.

Evaluating the limit as $s$ approaches zero results in $\frac{0}{4}$, so the steady-state error is 0.

The state transition matrix, $\Phi(t) = e^{At}$, can be found using the Laplace transform method, where $\Phi(t) = \mathcal{L}^{-1}\{(sI - A)^{-1}\}$. The system matrix from the given equation is

.

First, we construct the matrix

. Next, we find its inverse. For a 2x2 matrix, this is straightforward:

.

Finally, we take the inverse Laplace transform of each element. Using the standard transform pairs $\mathcal{L}^{-1}\{\frac{1}{s-1}\} = e^t$ and $\mathcal{L}^{-1}\{\frac{1}{(s-1)^2}\} = te^t$, we arrive at the state transition matrix:

A moving iron voltmeter measures the Root Mean Square (RMS) value of the input voltage. The signal shown is a sawtooth waveform with a period of $T = 20 \text{ ms}$ and a peak voltage of $V_p = 100 \text{ V}$. To find the meter's reading, we calculate the RMS value over one period.

First, define the voltage function for one cycle ($0 \le t < 0.02$ s):
$V(t) = \frac{V_p}{T}t = \frac{100}{0.02}t = 5000t$

Next, apply the standard formula for RMS voltage:
$V_{rms} = \sqrt{\frac{1}{T}\int_{0}^{T} V(t)^2 dt} = \sqrt{\frac{1}{0.02}\int_{0}^{0.02} (5000t)^2 dt}$

Solving the integral gives $V_{rms}^2 = \frac{5000^2}{0.02} \left[ \frac{t^3}{3} \right]_0^{0.02} = \frac{(5000 \times 0.02)^2}{3} = \frac{100^2}{3}$.
Therefore, the voltmeter reading is $V_{rms} = \frac{100}{\sqrt{3}} \approx 57.73 \text{ V}$.

To determine the power factor from the two-wattmeter method, we first find the power factor angle, $\phi$. The relationship between the angle and the wattmeter readings ($W_1$ and $W_2$) is given by:
$\tan(\phi) = \frac{\sqrt{3}(W_1 - W_2)}{W_1 + W_2}$
Substituting the given readings, $W_1 = 250$ W and $W_2 = 100$ W:
$\tan(\phi) = \frac{\sqrt{3}(250 - 100)}{250 + 100} = \frac{150\sqrt{3}}{350} \approx 0.742$
Solving for the angle gives $\phi = \tan^{-1}(0.742) \approx 36.58^\circ$.
The power factor is the cosine of this angle:
$\text{Power Factor} = \cos(\phi) = \cos(36.58^\circ) \approx 0.802$

An 8-4-2-1 Binary Coded Decimal (BCD) counter is designed to represent the decimal digits 0 through 9. Each digit is encoded using its 4-bit binary equivalent. The valid states therefore range from $0000_2$ (for decimal 0) up to $1001_2$ (for decimal 9).

Any 4-bit binary number that represents a value greater than 9 is considered an invalid state. Let's analyze the binary number $1100_2$. Converting this to decimal gives us $8 + 4 + 0 + 0 = 12$. Since 12 is outside the valid 0-9 range for BCD, the state $1100_2$ is invalid.

First, we determine the base current ($I_b$) by applying KVL to the input loop: $I_b = \frac{5V - V_{BE}}{R_s} = \frac{5V - 0.7V}{2k\Omega} = 2.15 \text{ mA}$.

For the transistor to be in the active region, the collector current ($I_c$) is determined by the gain: $I_c = \beta \times I_b = 100 \times 2.15 \text{ mA} = 0.215 \text{ A}$.

Now, let's analyze the output loop using KVL: $V_{CE} = 5V - I_c R_c$. The condition for staying in the active region is $V_{CE} > V_{CE(sat)}$, or $V_{CE} > 0.2V$.

The maximum value of $R_c$ corresponds to the edge of the active region, where $V_{CE}$ is at its minimum allowed value. By setting $V_{CE} = 0.2V$, we find the maximum resistance:
$R_{c,max} = \frac{5V - 0.2V}{I_c} = \frac{4.8V}{0.215A} \approx 22.32 \Omega$.

The peak voltage from the sinusoidal source is $V_p = V_{rms} \times \sqrt{2} = \frac{20}{\sqrt{2}} \times \sqrt{2} = 20 V. The full-wave rectifier and filter capacitor provide this as a nearly constant DC input, $V_s \approx 20 V, to the shunt regulator.

The Zener diode has a voltage of $V_Z = 5 V and a power rating of $P_Z = 1/4 W. Its maximum safe current is therefore $I_{Z,max} = \frac{P_Z}{V_Z} = \frac{1/4 \text{ W}}{5 \text{ V}} = \frac{1}{20} A.

To find the minimum resistance $R_s$ that avoids burnout, we must consider the worst-case scenario. This occurs when the Zener diode carries the maximum possible current, which happens when the load is disconnected ($R_L \to \infty $) and the load current $I_L = 0$. In this case, the entire source current $I_s$ flows through the Zener.

This source current is given by $I_s = \frac{V_s - V_Z}{R_s} = \frac{20 - 5}{R_s} = \frac{15}{R_s}$. To prevent damage, we must ensure `$I_s \le I_{Z,max}$`.

Substituting our values gives the inequality $\frac{15}{R_s} \le \frac{1}{20}$. Solving for $R_s$, we find $R_s \ge 15 \times 20 = 300 \, \Omega$. Thus, the minimum value for $R_s$ is $300 \, \Omega$.

For a step-up chopper operating with continuous inductor current, the relationship between the output voltage ($V_0$), input voltage ($V_s$), total switching period ($T$), and switch 'off' time ($T_{OFF}$) is given by:
$V_0 = V_s \left(\frac{T}{T_{OFF}}\right)$

We are given $V_0 = 400$ V, $V_s = 250$ V, and $T_{OFF} = 20 \, \mu \text{s}$. We can rearrange the formula to solve for the total period $T$:
$T = T_{OFF} \left(\frac{V_0}{V_s}\right) = (20 \times 10^{-6} \, \text{s}) \left(\frac{400}{250}\right) = 32 \times 10^{-6} \, \text{s}$

The switching frequency ($f$) is the inverse of the total period:
$f = \frac{1}{T} = \frac{1}{32 \times 10^{-6} \, \text{s}} = 31250 \, \text{Hz}$

Converting this to kilohertz, the switching frequency is $31.25$ kHz.

The primary goal of V/f control is to maintain a constant magnetic flux ($\phi$) in the motor. The flux is directly proportional to the ratio of the induced back EMF ($E$) and the frequency ($f$), so $\phi \propto E/f$.

Ideally, we approximate the applied stator voltage ($V$) as being equal to the back EMF ($E$). Following this, keeping the $V/f$ ratio constant would keep the flux constant.

However, this approximation fails at low frequencies. The voltage drop across the stator resistance ($I_s R_s$) becomes significant compared to the small applied voltage ($V$). The back EMF is more accurately given by the relationship $E \approx V - I_s R_s$. Because of this uncompensated voltage drop, the actual back EMF is lower than intended. As a result, the flux ($\phi \propto E/f$) decreases from its rated value.

We can transform the integral by applying the substitutions one at a time.

First, let's address the inner integral with respect to $x$. We use the substitution $u = \frac{2x-y}{2}$. Treating $y$ as a constant, the differential is $du = dx$. The limits for $x$ change as follows: the lower limit $x=y/2$ becomes $u = \frac{2(y/2)-y}{2} = 0$, and the upper limit $x=(y/2)+1$ becomes $u = \frac{2((y/2)+1)-y}{2} = 1$. The integral is now $\int_{0}^{8} (\int_{0}^{1} u \, du) dy$.

Next, we transform the outer integral using $v=y/2$. This gives us the differential $dv = \frac{1}{2}dy$, or $dy = 2dv$. The integration bounds for $y$ from $0$ to $8$ become new bounds for $v$ from $v=0/2=0$ to $v=8/2=4$.

Substituting for the outer integral gives $\int_{0}^{4} (\int_{0}^{1} u \, du) 2dv$. Since the constant $2$ can be moved inside the inner integral, we arrive at the final expression: $\int_{0}^{4} (\int_{0}^{1} 2u \, du) dv$.

To find the probability $P(0.5 < X < 5)$, we need to find the area under the probability density function, $f(x)$, over the interval from $0.5$ to $5$. Since $f(x)$ is a piecewise function, we must break our integral into parts that correspond to each definition of the function.

The interval $[0.5, 5]$ spans three different regions for $f(x)$:

1. From $0.5$ to $1$, where $f(x)=0.2$.
2. From $1$ to $4$, where $f(x)=0.1$.
3. From $4$ to $5$, where $f(x)=0$.

The total probability is the sum of the integrals over these sub-intervals:
$P(0.5 < X < 5) = \int_{0.5}^{1} 0.2 \,dx + \int_{1}^{4} 0.1 \,dx + \int_{4}^{5} 0 \,dx$
Evaluating these simple integrals gives us:
$(0.2 \times (1 - 0.5)) + (0.1 \times (4 - 1)) + 0 = 0.1 + 0.3 = 0.4$

To find the absolute minimum of a function on a closed interval, we must evaluate the function at the interval's endpoints and at any critical points that fall within the interval.

First, we find the critical points by setting the derivative, $f'(x) = 3x^2 - 6x - 24$, equal to zero. Factoring the derivative gives $3(x-4)(x+2) = 0$, so the critical points are $x=4$ and $x=-2$. Since $x=4$ is outside the interval $[-3, 3]$, we only need to consider $x=-2$.

Now, we test the function at the endpoints ($x=-3, x=3$) and the one valid critical point ($x=-2$):
$f(-3) = (-3)^3 - 3(-3)^2 - 24(-3) + 100 = 118$
$f(-2) = (-2)^3 - 3(-2)^2 - 24(-2) + 100 = 128$
$f(3) = (3)^3 - 3(3)^2 - 24(3) + 100 = 28$

Comparing these three results, the lowest value is 28.

This circuit is a double-ended clipper, which limits the output voltage $v_o$ between two levels. Based on the circuit's configuration and the problem's logic, one diode branch clips the output at a ceiling of $-1V$, while the other clips it at a floor of $-2V$.

The output is only unclipped when $v_o$ is between these two levels: $-2V < v_o < -1V$. In this range, both diodes are off, and the circuit acts as a voltage divider with two equal resistors, so $v_o = v_i/2$.

To find the input range where the output is unclipped, we substitute $v_o = v_i/2$ into the inequality: $-2V < v_i/2 < -1V$. Solving for $v_i$ gives $-4V < v_i < -2V$. The output will be clipped whenever the input voltage $v_i$ is outside this specific range.