GATE EE 2025 Set 1

The sentence refers to a past event indicated by the word "yesterday". Therefore, option (D) "couldn't; wasn't" is the most appropriate choice.

The relationship in the first pair is that of antonyms: "Good" is the opposite of "Evil".
Following the same logic for the second pair, the opposite of "Genuine" (authentic/real) is "Counterfeit" (fake/forged).

• Contraband refers to illegal goods.
• Counterfoil is a part of a check or ticket kept as a record.
• Counterpart refers to a person or thing holding a corresponding position.

Thus, (A) is the correct answer.

The given equation of the line is $x + py + q = 0$.
From the provided graph, we can identify two points on the line: $-2 + p(0) + q = 0 \Rightarrow q = 2$ $0 + p(4) + 2 = 0 \Rightarrow 4p = -2 \Rightarrow p = -\frac{2}{4} = -\frac{1}{2}$
So, $p = -\frac{1}{2}$ and $q = 2$, which matches option (A).

Let's analyze the relationships in Column-I:Matching these gives: (1)-(Q), (2)-(R), (3)-(S), (4)-(P), which corresponds to option (B).

Given expression: $(\frac{3^{81}}{27^4})^{\frac{1}{3}}$Step 1: Express 27 as a power of 3.
$27 = 3^3$
So, $27^4 = (3^3)^4 = 3^{3 \times 4} = 3^{12}$Step 2: Simplify the fraction inside the parentheses.
$\frac{3^{81}}{3^{12}} = 3^{81 - 12} = 3^{69}$Step 3: Apply the outer exponent.
$(3^{69})^{\frac{1}{3}} = 3^{69 \times \frac{1}{3}} = 3^{23}$Thus, the correct option is (C).

To form a coherent paragraph, we look for logical flow:The logical sequence is Q → S → R → P, which is option (C).

The coding logic is based on the product of the alphabetical positions of the letters.

• E is the 5th letter and F is the 6th letter, so EF = $5 \times 6 = 30$.
• H is the 8th letter and J is the 10th letter, so HJ = $8 \times 10 = 80$.

In the circular diagram, opposite sectors are related:

• EF (30) is opposite the sector containing 30.
• HJ (80) is opposite the sector containing 80.
• FG is opposite the sector marked (ii). Since F is the 6th letter and G is the 7th, FG = $6 \times 7 = 42$. Thus, (ii) = 42.
• The sector containing 35 is opposite the sector marked (i). Since $35 = 5 \times 7$, and the 5th letter is E and the 7th letter is G, (i) must be EG.

Therefore, (i) = EG and (ii) = 42, which corresponds to option (C).

First, let's calculate the required statistics for both students:Student P:

• Sorted scores: 22, 39, 45, 50, 60, 78, 89
• Sum = $22 + 89 + 50 + 45 + 78 + 60 + 39 = 383$
• Average = $383 / 7 \approx 54.71$
• Median = 50 (the 4th value in the sorted list)
• Range (Max - Min) = $89 - 22 = 67$

Student Q:

• Sorted scores: 35, 45, 50, 56, 60, 65, 81
• Sum = $35 + 65 + 60 + 56 + 81 + 45 + 50 = 392$
• Average = $392 / 7 = 56$
• Median = 56 (the 4th value in the sorted list)
• Range (Max - Min) = $81 - 35 = 46$

Evaluating the statements:
(A) Average P (54.71) < Average Q (56). This is TRUE.
(B) Median P (50) is same as Median Q (56). This is FALSE.
(C) Range P (67) > Range Q (46). This is TRUE.
(D) Median Q (56) and Average Q (56) are same. This is TRUE.The incorrect statement is (B).

The centers of the spheres form a simple cubic lattice with a lattice constant $a = 1$ (since the centers are at integer coordinates).Consider a sphere centered at the origin $(0, 0, 0)$:The ratio $\frac{y}{x} = \frac{\sqrt{2}}{1} = \sqrt{2}$.

In $\triangle PQR$, since $ST \parallel QR$, by the Basic Proportionality Theorem (or similarity of triangles $\triangle PST$ and $\triangle PQR$):$$\frac{PS}{PQ} = \frac{PT}{PR}$$Given $PT:TR = 3:2$, we have:$$\frac{PT}{PR} = \frac{PT}{PT + TR} = \frac{3}{3 + 2} = \frac{3}{5}$$Thus, $\frac{PS}{PQ} = \frac{3}{5} \implies PQ = \frac{5}{3} PS$.Let $A_{PS}$ be the area of the semicircle with diameter $PS$:$$A_{PS} = \frac{1}{2} \pi \left(\frac{PS}{2}\right)^2 = \frac{\pi PS^2}{8}$$Let $A_{PQ}$ be the area of the semicircle with diameter $PQ$:$$A_{PQ} = \frac{1}{2} \pi \left(\frac{PQ}{2}\right)^2 = \frac{\pi PQ^2}{8} = \frac{\pi \left(\frac{5}{3} PS\right)^2}{8} = \frac{25}{9} \left(\frac{\pi PS^2}{8}\right) = \frac{25}{9} A_{PS}$$The shaded area $PQS$ is the region between the two semicircles:$$\text{Shaded Area} = A_{PQ} - A_{PS} = \frac{25}{9} A_{PS} - A_{PS} = \left(\frac{25}{9} - 1\right) A_{PS} = \frac{16}{9} A_{PS}$$Therefore, the shaded area is $\frac{16}{9}$ times the area of the semicircle with diameter $PS$.

Let $u = x + y$. The function can be rewritten as a function of a single variable $u$:$$g(u) = (u - 1)^2 + u^2 = u^2 - 2u + 1 + u^2 = 2u^2 - 2u + 1$$To find the minimum, we take the derivative with respect to $u$ and set it to zero:$$g'(u) = 4u - 2 = 0 \implies u = \frac{1}{2}$$Since $g''(u) = 4 > 0$, this value of $u$ corresponds to a minimum. Thus, the function $f(x, y)$ is minimized whenever $x + y = \frac{1}{2}$.
The set $S = \{(x, y) \in \mathbb{R}^2 \mid x + y = \frac{1}{2}\}$ represents a line in the 2D plane. A line contains an infinite number of points. Therefore, the number of elements in the set $S$ is infinite.

A fundamental property of real symmetric matrices is that eigenvectors corresponding to distinct eigenvalues are orthogonal. For two vectors $\mathbf{v_1}$ and $\mathbf{v_2}$, orthogonality implies that their inner product (dot product) is zero:$$\mathbf{v_1} \cdot \mathbf{v_2} = \mathbf{v_1}^T \mathbf{v_2} = 0$$Thus, option (B) is correct.

The system of linear equations $Ax = b$ can be analyzed using the rank of the coefficient matrix $A$ and the augmented matrix $[A|b]$.Matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 0 & 1 & -1 \end{bmatrix}$.Performing row operations on $A$:
$R_2 \to R_2 + R_1 \implies \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 1 & -1 \end{bmatrix}$.Rearranging rows:
$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}$.The rank of $A$ is $\rho(A) = 2$.Now, consider the augmented matrix $[A|b] = \begin{bmatrix} 1 & 1 & 1 & 1/3 \\ -1 & -1 & -1 & -1/3 \\ 0 & 1 & -1 & 0 \end{bmatrix}$.Performing the same row operation $R_2 \to R_2 + R_1$:
$[A|b] \sim \begin{bmatrix} 1 & 1 & 1 & 1/3 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 1 & 1/3 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.The rank of the augmented matrix is $\rho(A|b) = 2$.Since $\rho(A) = \rho(A|b) = 2$, which is less than the number of variables ($n = 3$), the system has infinitely many solutions.

To find $P^2$, we multiply matrix $P$ by itself:$P = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$P^2 = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} (2)(2) + (1)(-1) + (0)(0) & (2)(1) + (1)(0) + (0)(0) & (2)(0) + (1)(0) + (0)(1) \\ (-1)(2) + (0)(-1) + (0)(0) & (-1)(1) + (0)(0) + (0)(0) & (-1)(0) + (0)(0) + (0)(1) \\ (0)(2) + (0)(-1) + (1)(0) & (0)(1) + (0)(0) + (1)(0) & (0)(0) + (0)(0) + (1)(1) \end{bmatrix}$$P^2 = \begin{bmatrix} 3 & 2 & 0 \\ -2 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$Now, let's evaluate option (A) $2P - I$:$2P - I = 2 \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 0 \\ -2 & 0 & 0 \\ 0 & 0 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 & 0 \\ -2 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$Since $P^2 = 2P - I$, option (A) is correct.

The conditional expectation of $Y$ given $X = 1$ is$$E[Y \mid X = 1] = \sum_{y} y \, P(Y = y \mid X = 1)$$Step 1 - marginal probability of $X = 1$:$$P(X = 1) = P(X = 1 \text{ and } Y = 0) + P(X = 1 \text{ and } Y = 1) = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$$Step 2 - conditional probabilities:$$P(Y = 0 \mid X = 1) = \frac{1/8}{1/4} = \frac{1}{2}, \qquad P(Y = 1 \mid X = 1) = \frac{1/8}{1/4} = \frac{1}{2}$$Step 3 - expected value:$$E[Y \mid X = 1] = 0 \times \frac{1}{2} + 1 \times \frac{1}{2} = \frac{1}{2}$$Since $Y$ takes only the values $0$ and $1$, no other terms contribute, so the required expectation is $\frac{1}{2}$, which is option (B).Answer: (B) $\frac{1}{2}$

Consider small-signal voltage gain $A_v = v_{out}/v_{in}$ for each single-stage MOSFET amplifier.$$A_v = -g_m (r_o \parallel R_D)$$The negative sign means a $180^\circ$ phase shift, so the CS stage is inverting.$$A_v \approx +g_m R_D \quad (\text{neglecting } r_o \text{ and body effect})$$The positive sign means the output is in phase with the input, so the CG stage is non-inverting.(For completeness, the common-drain / source follower has a gain slightly below $+1$ and is also non-inverting.)Hence the CS amplifier is inverting and the CG amplifier is non-inverting, option (C).

Label the bottom op-amp as A1 and the top op-amp as A2.$$V_{A2} = V_{out}$$$$V_{-,1} = V_{A2} = V_{out}$$$$V_{in} = V_{out}$$The output follows the input with unity gain, and the input $V_{in}$ drives only the (ideal, infinite-impedance) non-inverting terminal of A1. No resistors are present, so there is no summing, differencing or logarithmic action.The circuit is a buffer, option (D).

Applying Kirchhoff's Current Law (KCL) at node X:
The current $I_{XY}$ entering node X must equal the sum of the currents leaving node X through the parallel branches.$$I_{XY} = I_{source} + I_{resistor}$$$$I_{XY} = 1\text{ A} + \frac{V_{XY}}{1\text{ k}\Omega}$$$$I_{XY} = 1 + 0.001 V_{XY}$$This is a linear equation of the form $y = mx + c$ where:

• $y = I_{XY}$
• $x = V_{XY}$
• Intercept $c = 1\text{ A}$ (at $V_{XY} = 0$)
• Slope $m = 1/1000$ (positive slope)

Graph (B) correctly shows a straight line starting at $1\text{ A}$ on the vertical axis and increasing linearly with $V_{XY}$.

A nullator is a theoretical two-terminal network element defined by having zero voltage across its terminals ($V = 0$) and zero current flowing through it ($I = 0$). When a nullator is connected in series with a resistor of value $R$:Since the total voltage across the combination is zero and the total current through it is zero, the combination behaves exactly like a nullator.

The impulse response $h[n]$ of the system is defined as the output when the input is a unit impulse $\delta[n]$.
Given: $h[n] = \boldsymbol{S}\{\delta[n]\} = \begin{cases} 1, & n = 0, 1, 2 \\ 0, & \text{otherwise} \end{cases}$
This can be written as: $h[n] = \delta[n] + \delta[n-1] + \delta[n-2]$For any LTI system, the output $y[n]$ for an input $x[n]$ is given by the convolution sum:
$y[n] = x[n] * h[n]$
$y[n] = x[n] * (\delta[n] + \delta[n-1] + \delta[n-2])$
Using the property $x[n] * \delta[n-k] = x[n-k]$:
$y[n] = x[n] + x[n-1] + x[n-2]$

The sifting property of the unit impulse function states that:$$\int_{-\infty}^{\infty} f(t)\delta(t - t_0)dt = f(t_0)$$In this problem, we have $\delta(t + 3)$, which means $t_0 = -3$.
So, the integral is equal to $x(-3)$.Given $x(t) = -t^2 \{u(t + 4) - u(t - 4)\}$:
$x(-3) = -(-3)^2 \{u(-3 + 4) - u(-3 - 4)\}$
$x(-3) = -9 \{u(1) - u(-7)\}$Since $u(t) = 1$ for $t \geq 0$ and $u(t) = 0$ for $t < 0$:
$u(1) = 1$
$u(-7) = 0$Therefore:
$x(-3) = -9 \{1 - 0\} = -9$

The step response of a stable first-order LTI system is given by $y(t) = K(1 - e^{-t/\tau})$, where $K$ is the steady-state gain and $\tau$ is the time constant.
From the table, as $t \to \infty$, $y(t) \to 3$, so $K = 3$.
Using the data point at $t = 0.6$ s:
$y(0.6) = 3(1 - e^{-0.6/\tau}) = 0.78$
$1 - e^{-0.6/\tau} = \frac{0.78}{3} = 0.26$
$e^{-0.6/\tau} = 1 - 0.26 = 0.74$
Taking the natural logarithm on both sides:
$-\frac{0.6}{\tau} = \ln(0.74) \approx -0.301$
$\tau = \frac{0.6}{0.301} \approx 1.993 \approx 2$ s.
Checking with $t = 1.6$ s:
$y(1.6) = 3(1 - e^{-1.6/2}) = 3(1 - e^{-0.8}) \approx 3(1 - 0.449) = 3(0.551) = 1.653$, which matches the table value of 1.65.

The closed-loop characteristic equation of the unity negative-feedback loop is $1 + G(s) = 0$, and the Nyquist criterion relates the encirclements of the critical point $-1 + j0$ by the Nyquist plot of $G(s)$ to

• $P$ = number of open-loop poles of $G(s)$ in the right half $s$-plane, and
• $Z$ = number of closed-loop poles (roots of $1 + G(s) = 0$) in the right half $s$-plane.

Step 1 - find $P$. $G(s)$ is strictly stable, so all its poles lie in the left half plane and $P = 0$. The factor $(s - 3)$ is in the numerator: it is a right-half-plane zero of $G(s)$ and does not change $P$.Step 2 - condition for closed-loop stability. A stable closed loop needs $Z = 0$. With $P = 0$, this happens only if the Nyquist plot makes no net encirclement of $-1$. So any net encirclement of $-1$, in either direction, means the closed loop is not stable.Step 3 - count the unstable closed-loop poles. The sense of the encirclement depends on the direction in which the Nyquist contour is traversed. Taking the contour anti-clockwise around the right half plane, each right-half-plane zero of $1 + G(s)$ (closed-loop pole) produces one anti-clockwise encirclement and each right-half-plane pole of $G(s)$ one clockwise encirclement:$$N_{\text{anti-clockwise}} = Z - P$$With $N_{\text{anti-clockwise}} = 1$ and $P = 0$,$$Z = N + P = 1 + 0 = 1$$so one closed-loop pole lies in the right half plane.Step 4 - rule out the other options. The encirclement count is definite, so stability can be decided (not (A)). Marginal stability would require closed-loop poles on the $j\omega$-axis, i.e. the plot passing exactly through $-1$, which is not the case here (not (B)). A non-zero encirclement with $P = 0$ excludes a stable loop (not (C)).Answer: (D) The system is unstable.