GATE 2025 EC Paper: Worked Solutions to All 65 Questions

Key idea: The missing word has to be a stronger form of approval than the two after it.

Official answer: A

Group I goes from the harshest criticism to the mildest, so Group II must go from the warmest approval down to appreciation. To extol is to praise very highly, which fits before praise.

Prize means to value something, appropriate means to take or to be suitable, and espouse means to adopt a cause. None of them is a degree of praise. Option (A).

Key idea: A condition about the past that didn't happen takes a result with a modal plus have and a past participle.

Official answer: D

Had I learnt is the inverted form of if I had learnt: the speaker didn't learn acting as a child. The result of an unreal past condition is written with would have, could have or might have and a past participle, so could have been is right.

The other three are present or future forms, which don't match a condition that is already over. Option (D).

Key idea: A fixed ratio per step means the ratio over several steps is that factor raised to the number of steps.

Official answer: B

From C, the notes C#, D, D#, E, F and F# are six steps up. Six steps multiply the frequency by

The 130.8 Hz isn't needed, because a ratio doesn't depend on where the scale starts. Option (B).

Key idea: Find what one iteration does to the total length; a constant factor per iteration gives a power of n.

Official answer: B

Count the segments in the figure. Iteration 1 has 5 segments of \tfrac{1}{3}, so its length is \tfrac{5}{3}. Iteration 2 replaces each of those by 5 segments a third as long: 25 segments of \tfrac{1}{9}, length \tfrac{25}{9}.

Every iteration multiplies the number of segments by 5 and divides their length by 3, so the length grows by \tfrac{5}{3} each time:

Only (B) gives \tfrac{5}{3} at n = 1 and \tfrac{25}{9} at n = 2. Option (B).

Key idea: Treat positive and negative x separately, since |x| is x for one and -x for the other.

Official answer: A

For x > 0, |x| = x and f(x) = -1. For x < 0, |x| = -x and f(x) = +1. So the graph is the constant 1 left of the axis and the constant -1 right of it, with no value at x = 0.

That is plot (A). Plot (B) is the reverse, +|x|/x, and (C) and (D) aren't constant on either side. Option (A).

Key idea: Start with the sentence that refers back to nothing, then follow words such as this and such, which point to an earlier sentence.

Official answer: B

Q introduces the topic and leans on nothing before it, so it opens. S explains how adapting helps organisms survive as environments change.

R then names this process of adaptation as natural selection, and P, with such adaptations and over time, closes on the long-term result. The order Q, S, R, P is option (B).

The other orders put P or R, which both point back to something already said, ahead of Q.

Key idea: Three lengths with a fixed total make a triangle exactly when each one is less than half the total.

Official answer: D

The pieces are b_1, b_2 - b_1 and 1 - b_2, which add up to 1. Each must be shorter than the other two together, that is, shorter than 0.5:

Those are (A), (C) and (B). Condition (D) isn't needed: b_1 = 0.4 and b_2 = 0.7 give pieces of 0.4, 0.3 and 0.3, a valid triangle, yet b_1 + b_2 = 1.1. Option (D).

Key idea: A question given marks to all scores full marks for every candidate, whatever they answered.

Official answer: Marks to all

The final key awards marks to all for this question, and doesn't say why. We don't give an answer: as printed, the question doesn't say how far the students move in a round, or which way round the circle counts as behind, and the student who leaves each round depends on both.

In the exam this meant there was nothing to lose by answering it. As practice, treat it as an exercise in reading a puzzle's rules carefully before starting.

Ruling in the key: The final key gives marks to all (MTA) for this question: every candidate received its 2 marks.

Key idea: Every statement is about totals, so total each nation's medals first.

Official answer: C

Adding gold, silver and bronze:

• USA: 125
• Canada: 90
• France: 64
• Australia: 52
• Japan: 45
• All five: 376

(A) is true: France's 64 is third, ahead of Australia and Japan. (B) is true: the USA stays ahead of Canada. (D) is true: 90 is twice 45.

(C) is false: the USA and Canada have 215 of the 376 medals, about 57%, which is more than half. Option (C).

Key idea: With a slab rule, decide which slab the answer is in, then solve the single linear equation for that slab.

Official answer: B

At a fee of ₹5,00,000 the client pays 1.18 \times 1.2 \times 5{,}00{,}000 = 7{,}08{,}000 rupees, well under the limit, so the fee is in the upper slab. There, with the fee F in rupees, the overhead and the client's total are

The largest fee is ₹7,24,961. Option (B).

Key idea: The first two rows are always independent, so the rank is at least three as soon as the last two rows are not both zero.

Official answer: A

Rows 1 and 2 have pivots 2 and 6 in columns 1 and 2, so they give rank 2 whatever \alpha, \beta and \gamma are. Rows 3 and 4 have zeros in those columns, so any non-zero one of them adds to the rank.

• (i) Row 3 is (0, 0, 0, \beta), non-zero: rank 3.
• (ii) Rows 3 and 4 are both zero: rank 2.
• (iii) Row 3 is (0, 0, \alpha, 0), non-zero: rank 3.
• (iv) All four pivots are non-zero: rank 4.

So (i), (iii) and (iv) work. Option (A).

Key idea: Compare each series with one whose behaviour you know: a p-series, a telescoping sum, or the series for e.

Official answer: B

• (i) is a p-series with p = \tfrac{1}{2} \le 1, so it diverges: its partial sums grow like 2\sqrt{N}.
• (ii) telescopes, since \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}, and its sum is 1.
• (iii) is the series for e without its first term, so it converges to e - 1.

Only (ii) and (iii) converge. Option (B).

Key idea: Count unordered pairs: every pair of the four balls is equally likely.

Official answer: A

There are \binom{4}{2} = 6 pairs. A mixed pair takes one of 2 red and one of 2 blue: 2 \times 2 = 4 pairs.

Option (A). The same comes from the second draw: whatever the first ball, 2 of the 3 left are the other colour.

Key idea: Frequency deviation is the time derivative of the phase deviation divided by 2\pi, not the sum of the modulation indices.

Official answer: A

The phase deviation is \phi(t) = 3\sin(2\pi f_1 t) + 4\sin(6\pi f_1 t). Differentiating and dividing by 2\pi:

Both cosines equal 1 at t = 0, so the peak is 3f_1 + 12f_1 = 15f_1. Option (A).

The tone at 3f_1 contributes four times its index, 12, because deviation scales with the modulating frequency. Adding the indices, 3 + 4, is the usual slip.

Key idea: More bandwidth also lets in more noise, so the signal-to-noise ratio falls as W grows.

Official answer: A

By the Shannon-Hartley formula,

C is 0 at W = 0 and always increases with W, but ever more slowly. As W \to \infty, using \ln(1 + x) \approx x,

So the curve rises and flattens toward a finite limit: plot (A). It never falls, which rules out (D), and it doesn't grow without limit as (B) and (C) do.

Key idea: Gain crossover is where the plot meets the unit circle; phase crossover is where it meets the negative real axis.

Official answer: D

At the gain crossover frequency |G(j\omega)H(j\omega)| = 1, so the plot is on the unit circle. Of the four points, only S lies on the dashed unit circle.

At the phase crossover frequency the phase is -180°, so the plot is on the negative real axis. That is Q, where the branch crosses the axis inside the circle.

P and R are marked where the plot meets the 45° lines through the origin; they are neither crossover. So \omega_S is the gain crossover and \omega_Q the phase crossover. Option (D).

Key idea: An output term that doesn't depend on the input breaks linearity, and one fixed at n = 0 breaks time invariance.

Official answer: D

Follow a signal through the box:

The second term is the same for every input, and it is not zero because h_2[n] \neq \delta[n]. A zero input therefore gives a non-zero output, so the system is not linear.

That term also stays put at n = 0 when the input is delayed, so a delayed input doesn't give a delayed output: the system is time variant. Option (D). (If h_2[n] = \delta[n], the two b\,\delta[n] terms would cancel and the system would be LTI.)

Key idea: The magnitude of an integral is at most the integral of the magnitude.

Official answer: A

Since |e^{-j\omega t}| = 1,

So (A) holds for every f, and (B) never does. (C) and (D) compare with the signed integral, which can fail either way:

• (C) fails for f(t) = -1 on |t| < \tfrac{1}{2}: |F(0)| = 1 but \int f\,dt = -1.
• (D) fails for f(t) = 1 on |t| < \tfrac{1}{2}: at \omega = \pi, |F| = 2/\pi \approx 0.64 < 1.

Option (A).

Key idea: The known current through the 3 Ω fixes the junction voltage, and then the rest of the network doesn't matter for node X.

Official answer: A

The 3 Ω resistor drops 1 \times 3 = 3 V, so the junction is at 9 - 3 = 6 V.

Node X has only two connections, so the current in from the 8 V node equals the current out to the junction:

Option (A). The resistors to the right of the junction only carry whatever current is left over there.

Key idea: With admittances, each branch current is I = YV: multiply, don't divide.

Official answer: A

Every branch has the full source voltage V = 1\angle 90° = j across it, so

Option (A). Option (B) divides by the admittances, as if they were impedances.

Key idea: Find i_b from the input loop alone, then let the dependent source drive the load.

Official answer: A

The input side is R_S and r_\pi in series, so

The source \beta i_b pulls current down out of the output node, so that current flows up through R_L and makes v_o negative:

Option (A). Dropping r_\pi gives (C), and missing the inversion gives (D).

Key idea: With R_B taken from the collector, the collector resistor carries the base current as well as the collector current.

Official answer: C

With \beta = 100 and I_B = 10 µA, I_C = 1 mA and I_E = 1.01 mA.

The 5 kΩ carries I_C + I_B = 1.01 mA, because I_B leaves the collector node through R_B. So

Option (C). Options (A) and (B) are V_C and V_E on their own, and using only I_C in the 5 kΩ gives 1.97 V.

Key idea: F is 1 exactly when some pair of inputs are both 1.

Official answer: A

Two or more inputs are 1 exactly when at least one of the pairs XY, YZ or ZX is 1, so F = XY + YZ + ZX. Option (A).

The others fail on one row each: X \oplus Y \oplus Z is 0 for X = Y = 1, Z = 0; X + Y + Z is 1 with a single 1; and XYZ is 0 with two 1s.

Key idea: Adding the complement of Y plus 1 is subtracting Y in two's complement.

Official answer: B

With Z = 1, the XOR gives \overline{Y} and the carry-in is 1, so the full adder computes

That is X - Y in two's complement: the sum bit is the difference bit X \oplus Y, and the carry-out is 1 when no borrow is needed (X \ge Y). The circuit is a subtractor. Option (B). With Z = 0 it would add.

Key idea: A cubic is unbounded both ways, and the sign of f'' says which stationary point is which.

Official answer: A and B

As x \to \infty, f \to \infty, and as x \to -\infty, f \to -\infty. So f has neither a global maximum nor a global minimum: (A) and (B) are correct.

f''(-1) = -18 < 0, so x = -1 is a local maximizer, and f''(2) = 18 > 0, so x = 2 is a local minimizer. Both (C) and (D) have them the wrong way round. Options (A) and (B).

Key idea: A point is a closed-loop pole for some K > 0 only if it lies on the root locus, which the angle condition decides.

Official answer: C

From the figure, G(s) = \dfrac{(s+3)(s+4)}{(s+1)(s+2)} up to a positive constant. At s = -1 + j, the angle of G is

A point on the root locus needs \pm 180°, so -1 + j is not on it. Solving for K directly confirms it:

That isn't a real number, so no K works. The figure agrees: the circle only reaches about \pm j0.87. Option (C). Option (B) is the size of that complex value, 0.2, taken as if it were a gain.

Key idea: The DTFT is the z-transform on the unit circle, and the RoC can never contain a pole but may contain zeros.

Official answer: C and D

• (A) is false: x[n] = 2^n u[n] has X(z) = \frac{1}{1 - 2z^{-1}} for |z| > 2, but no DTFT.
• (B) is false: the RoC has no poles, but it can contain zeros. x[n] = \delta[n] - \delta[n-1] has X(z) = 1 - z^{-1}, with a zero at z = 1 inside its RoC.
• (C) is true: on the unit circle, X(e^{j\omega}) is the DTFT, and it converges there when the circle is in the RoC.
• (D) is true: X(z) = \alpha for every z, so the RoC is the whole plane.

Options (C) and (D).

Key idea: An envelope detector recovers whatever multiplies the carrier, provided that factor never goes negative.

Official answer: A and D

The envelope is |A_c\, m(t)| for DSB-SC and |A_c\,(1 + \mu\, m(t))| for AM.

• (i): if m(t) > 0 always, |A_c\, m(t)| = A_c\, m(t), so the envelope is the message. (i) is TRUE.
• (ii): AM needs 1 + \mu\, m(t) > 0, not m(t) > 0. A tone m(t) = \cos(2\pi W t) with \mu = 0.5 goes negative, yet 1 + 0.5\cos(2\pi W t) stays between 0.5 and 1.5, and the envelope detector recovers it. So the "only if" is FALSE.

Options (A) and (D).

Key idea: An ideal op-amp amplifies only the difference between its inputs, draws no input current and drives any load.

Official answer: A and C

An ideal op-amp has infinite input resistance (A) and infinite differential gain (C). Its output resistance is zero, not infinite, so it can drive any load, and its common-mode gain is zero, since it ignores what the two inputs share (infinite CMRR). Options (A) and (C).

Key idea: Whether the signal goes into the gate or the source decides the input resistance, and which terminal it comes out of decides the phase.

Official answer: B and D

• (A) is false: a common-gate stage takes its input at the source, where the input resistance is about 1/g_m.
• (B) is true: the common-source input is the gate, which draws no current.
• (C) is false: the common-source gain is -g_m R_D, so the output is inverted.
• (D) is true: the common-drain stage (source follower) has a positive gain just under 1.

Options (B) and (D).

Key idea: An n-type dopant has one more valence electron than silicon: it comes from group 15.

Official answer: A and D

Silicon has four valence electrons. Arsenic and phosphorus (group 15) have five, so each donates a free electron: they are donors, and make silicon n-type.

Boron and gallium (group 13) have three and accept an electron, making silicon p-type. Options (A) and (D).

Key idea: An equation whose coefficients are powers of t matching the order of each derivative has solutions of the form t^m.

Official answer: 0.25

Try y = t^m. Substituting gives

So y = c_1 t + c_2 t^2 and y' = c_1 + 2c_2 t. Then y'(0) = c_1 = 1 and y'(1) = 1 + 2c_2 = -1, so c_2 = -1:

Its maximum on [0, 1] is at t = \tfrac{1}{2}, where y = \tfrac{1}{4} = 0.25.

Key idea: For a linear code, the minimum distance equals the smallest weight of a non-zero codeword.

Official answer: 3

List the 7 non-zero codewords, the sums of the rows of G in every combination, with their weights:

• Single rows: 100101, 010011, 001110 (weight 3 each)
• Two rows: 110110, 101011, 011101 (weight 4 each)
• All three: 111000 (weight 3)

The smallest weight is 3, so d_{min} = 3.

Key idea: The input network sets the lower corner and the feedback network the upper one, each through its own time constant.

Official answer: 50

The gain is -Z_f / Z_i, with

The input network is a high-pass with its corner at 1/(2\pi \cdot 10RC), and the feedback network a low-pass with its corner at 1/(2\pi \cdot 0.2RC). Taking each corner as a -3 dB frequency, their ratio is

Note: The exact -3 dB points come out at about 54 MHz, because the two poles interact (the peak gain is 1.96, not 2). The key uses the standard corner-frequency approximation, which is the method expected in the exam.

Key idea: Each closed switch adds V_{REF} times the feedback-to-input resistance ratio, so the output is a weighted sum of the bits.

Official answer: 250

• 1110: V_O = -2\,(1 + 0.5 + 0.25) = -3.5 V
• 1101: V_O = -2\,(1 + 0.5 + 0.125) = -3.25 V

The output changes by 0.25 V, that is 250 mV.

Key idea: Write the closed-loop characteristic polynomial: a missing power of s rules out stability.

Official answer: D

The closed-loop poles are the roots of 1 + M(s)G(s) = 0.

• (i) With M = K_I/s: 10s^3 + K_I = 0. The s^2 and s terms are missing, so some root is in the right half-plane (or on the axis) for every K_I > 0. No such controller works: (i) is TRUE.
• (ii) With M = K_P + sK_D: 10s^2 + K_D s + K_P = 0. A second-order polynomial with all coefficients positive has both roots in the left half-plane, so any positive K_P and K_D work: (ii) is TRUE.

Option (D).

Key idea: Every option has imaginary roots, so look for a factor s^2 + a and divide it out.

Official answer: A

Try s = j, so s^2 = -1, s^3 = -j, s^4 = 1, s^5 = j:

So s^2 + 1 is a factor, and dividing it out leaves a cubic that factors further:

The roots are \pm j, 2, -4 and -5: L = 2, I = 2, R = 1. Option (A). A Routh array gets the same counts, with a row of zeros at the s^2 + 1 factor.

Key idea: Shifting the sampling instants changes where the samples are taken, not which signal they are samples of.

Official answer: A

The impulses sit at t = \tau + nT_s and carry f's own values there, f(\tau + nT_s). In frequency, the shifted impulse train is

The k = 0 copy is F(\omega)/T_s with no phase factor: the offset only multiplies the other copies by a phase. If the copies don't overlap, \omega_s > 2\omega_c, that is T_s < \pi/\omega_c, the filter keeps just the k = 0 copy and its gain T_s restores F(\omega). The output is f(t). Option (A).

Option (B) is the usual slip: the samples are taken later, but they are still samples of f(t), not of f(t - \tau).

Key idea: The two meter readings give the branch's impedance magnitude without needing anything about the rest of the circuit.

Official answer: A

The branch has 25 V across it and 5 A through it, so

So the reactance is zero: the branch is at resonance at \omega = 200 rad/s.

Option (A). The unknown circuit doesn't matter, because the meters measure the branch directly.

Key idea: Each Z-parameter is a ratio measured with the other port open, so the open port's series resistor carries no current.

Official answer: A

Call the two shunt nodes A (near port 1) and B. With port 2 open, the last 2 Ω carries nothing, so V_2 = V_B. From node A the network looks like 2 Ω in parallel with 2 + 2 = 4 Ω:

The current into node A splits so that V_A = \tfrac{4}{3} I_1, and the divider of 2 Ω and 2 Ω gives V_B = \tfrac{1}{2} V_A:

The network is symmetric and reciprocal, so Z_{22} = \tfrac{10}{3} and Z_{12} = \tfrac{2}{3}. Option (A).

Key idea: With equal priors and Gaussian noise, the ML thresholds sit halfway between neighbouring symbols, and an inner symbol can be pushed out either way.

Official answer: D

The decision thresholds are at -1 and 1. Using the symmetry of N, with P = P(N > 1):

• S = -2 is wrong only if N > 1: probability P.
• S = 0 is wrong if |N| > 1: probability 2P.
• S = 2 is wrong only if N < -1: probability P.

So \alpha = \tfrac{4}{3}. Option (D).

Key idea: Zero-mean coefficients make the mean zero everywhere, but the pulses fill only half of each slot, so the power switches on and off with time.

Official answer: A

(i) E[f(t)] = \sum_n E[a_n]\, p(t - nT) = 0 for every t: TRUE.

(ii) Since E[a_n a_m] is 1 for n = m and 0 otherwise,

At \tau = 0 this is 1 when t is in the first half of a slot and 0 in the second half, so it depends on t: FALSE. Option (A).

Key idea: The tail current splits equally at rest, and each transistor's current changes by g_m times half the differential input.

Official answer: D

At rest each transistor carries half of 2 mA, so V_3 = 5 - 1\,\text{k}\Omega \times 1\,\text{mA} = 4 V.

The differential input is V_1 - V_2 = 0.02\sin\omega t. Each transistor sees half of it, so M_1's current rises by

Option (D). The drain falls when the current rises, hence the minus sign; using the whole differential input gives the 0.1 in (B) and (C).

Key idea: A full-scale sine quantized to N bits has an SNR of 6.02N + 1.76 dB, whatever the full-scale voltage.

Official answer: A

The full-scale voltage cancels out of the ratio. Each of the 10^6 samples per second produces 10 bits, so the data rate is 10 Mbps. Option (A).

5 Mbps would come from using the 500 kHz signal frequency instead of the 1 MHz sampling rate.

Key idea: The flip-flops change only on rising clock edges, so read the inputs at each rising edge and the output at the instant after it.

Official answer: A

Y = \overline{Q_1} \oplus \overline{Q_2} = Q_1 \oplus Q_2. The marks T_0 to T_3 are falling edges, so each comes just after a rising edge. At those rising edges:

• Before T_0: SEL = 0, so Q_1 = P_0 = 0 and Q_2 = P_1 = 1. Y = 1.
• Before T_1: SEL = 1, S = 0, so Q_1 = S = 0 and Q_2 takes the old Q_1 = 0. Y = 0.
• Before T_2: SEL = 1, S = 1, so Q_1 = 1 and Q_2 takes the old Q_1 = 0. Y = 1.
• Before T_3: SEL = 0, so Q_1 = 0 and Q_2 = 1. Y = 1.

The sequence is 1011. Option (A). The first edge loads both flip-flops, so their starting state doesn't matter.

Key idea: In an intrinsic semiconductor both carriers have concentration n_i, so their mobilities add.

Official answer: B

Option (B).

Key idea: The diode drop and the Zener voltage fix Q_1's base voltage, and the base current adds to the Zener current in R_1.

Official answer: A

Q_2 drops 0.7 V, so the Zener's top is at 19.3 V and its bottom node at 19.3 - 5 = 14.3 V. That node is Q_1's base.

I_L = I_{C1} = 12 mA, so I_B = 12/120 = 0.1 mA and I_E = 12.1 mA. The base current flows out of Q_1's base into the node, joining the Zener current in R_1:

Option (A).

Key idea: The Einstein relation ties diffusivity to mobility through the thermal voltage kT/e.

Official answer: B

To the nearest integer, 98. Option (B). Forgetting that 1 m² is 10^4 cm² gives 0.0098, and 26 is the thermal voltage in mV.

Key idea: Integrate the density over the area; the absolute value doubles the half where y > 0.

Official answer: C

Option (C). Dropping the absolute value would make the two halves cancel to 0, and 16 is the total without the factor of 4.

Key idea: Ohm's law in point form, J = \sigma E, times the cross-sectional area gives the current.

Official answer: A

To two decimal places, 0.46 A. Option (A). Using the 1 mm diameter as the radius gives 1.82, option (B).

Key idea: The integral lies between the interval length times the minimum and the interval length times the maximum.

Official answer: A

The interval has length 6, so

A pair is guaranteed to work if \beta \le 6m and \alpha \ge 6M for every such f. Since m, M \ge 0:

• (A) 5m \le 6m and 7M \ge 6M: always holds.
• (B) \alpha = 5M fails for a constant f = 1, whose integral is 6 > 5.
• (C) and (D): \beta = 7m fails for the same constant, since 7 > 6.

Option (A) only.

Key idea: By Cauchy's theorem the integral is zero unless the function has a singularity inside the circle with a non-zero residue.

Official answer: A and B

• (A) e^z is analytic everywhere, so the integral is 0: TRUE.
• (B) For n \ge 0, z^n is analytic. For negative even n there is a pole at 0, but \oint_C z^n\,dz is 2\pi j only for n = -1 and 0 for every other integer. TRUE.
• (C) \cos z is analytic, so the integral is 0, and the statement is FALSE.
• (D) \sec z = 1/\cos z has poles at \pm\pi/2, \pm 3\pi/2, \dots, all outside the unit circle, so this integral is 0 too: FALSE.

Options (A) and (B).

Key idea: The matrix is diagonal, so each state is a separate first-order system, and each one's pole decides what a bounded input can do to it.

Official answer: B

Each state obeys its own equation (from rest):

• \dot{x}_1 = 2x_1 + u: pole at +2. A constant u = 1 gives x_1 = (e^{2t} - 1)/2, unbounded.
• \dot{x}_2 = -2x_2 + u: pole at -2. For any input with |u| \le U, |x_2| \le U/2, so x_2 is always bounded.
• \dot{x}_3 = u: an integrator. u = 1 gives x_3 = t, unbounded.

So u = 1 makes x_1 unbounded: (B) is TRUE and (A) is FALSE. x_2 can never be unbounded, so (C) is FALSE, and u = 0 keeps everything at zero, so (D) is FALSE. Option (B).

Key idea: Entropy is largest when the three outcomes are equally likely, at \alpha = \tfrac{1}{3}, and falls away on either side.

Official answer: B and C

• g(0.25) = 1.5 bits (probabilities 0.25, 0.5, 0.25)
• g(0.3) \approx 1.571 bits (0.3, 0.4, 0.3)
• g(0.4) \approx 1.522 bits (0.4, 0.2, 0.4)

So g(0.3) > g(0.4) and g(0.3) > g(0.25): options (B) and (C). Both make sense: 0.3 is the closest of the three to \tfrac{1}{3}.

Key idea: Compressing a signal in time shrinks its period and raises every harmonic's frequency, but leaves the weight of each harmonic unchanged.

Official answer: A and D

Substitute \alpha t for t in the series for f:

That is exactly the given form for y, so d_k = c_k: (A) is TRUE and (C) is FALSE. y repeats when \alpha t advances by T_0, so its fundamental period is T_0 / \alpha: (D) is TRUE and (B) is FALSE. Options (A) and (D).

Key idea: Find where the feedback is taken: here it is after G_1 but before G_2's path joins, so the loop gain is -G_1 alone.

Official answer: B

From the block diagram, E = R - G_1 E, so E = R/(1 + G_1), and

Graph (B) has a single loop, G_1 and the -1 branch, with forward paths G_1 and G_2 that both touch it, so Mason's rule gives (G_1 + G_2)/(1 + G_1). It matches.

In (A) and (D) the -1 branch returns from a node where G_2 has already joined, so the loop gain is -(G_1 + G_2) and the result is (G_1 + G_2)/(1 + G_1 + G_2). Graph (C) has positive feedback. Option (B) only.

Key idea: A bridge routes current through the load the same way for either input polarity, but the source current still reverses with the source.

Official answer: A and D