GATE EC 2026 Set 1

The word nocturnal refers to organisms or activities that occur during the night. Its direct antonym is diurnal, which refers to organisms or activities that occur during the day.

• Normal means standard or usual.
• Abnormal means deviating from what is normal.
• Exceptional means unusual or outstanding.

Let the scores of the students in descending order be $x_1, x_2, x_3, x_4, \dots, x_n$.Therefore, (S1) and (S2) together imply (S3). Option (A) is correct.

By observing the line segments entering the missing rectangular portion $WXYZ$ in the main figure:To form three straight line segments, we must connect:

• The top-left corner to the bottom-right corner.
• The top-right corner to the bottom-left corner.
• The middle-left edge to the middle-right edge.

Option S is the only one that correctly connects these points while maintaining the slopes of the segments and showing the three lines intersecting each other as required.

Given the equation: $(\log_{p^{1/n}} y) (\log_{y^{1/n}} p) = 16$We use the change of base formula for logarithms: $\log_b a = \frac{\log_c a}{\log_c b}$.
Let's convert the given logarithms to a common base, say $e$ (natural logarithm).$\log_{p^{1/n}} y = \frac{\ln y}{\ln (p^{1/n})} = \frac{\ln y}{\frac{1}{n} \ln p} = n \frac{\ln y}{\ln p}$$\log_{y^{1/n}} p = \frac{\ln p}{\ln (y^{1/n})} = \frac{\ln p}{\frac{1}{n} \ln y} = n \frac{\ln p}{\ln y}$Substitute these back into the given equation:$\left( n \frac{\ln y}{\ln p} \right) \left( n \frac{\ln p}{\ln y} \right) = 16$As $y > 1$ and $p > 1$, $\ln y \neq 0$ and $\ln p \neq 0$. Thus, $\frac{\ln y}{\ln p}$ and $\frac{\ln p}{\ln y}$ are reciprocals and their product is 1.$n^2 \left( \frac{\ln y}{\ln p} \cdot \frac{\ln p}{\ln y} \right) = 16$$n^2 \cdot 1 = 16$$n^2 = 16$Since $n$ is a real number and greater than 1, we take the positive square root:
$n = \sqrt{16}$
$n = 4$Thus, the value of $n$ is 4.The final answer is $\boxed{4}$

Let the original statement be P:
"For each student (S1), there exists another student (S2) in the class such that their scores are at most ten marks away."
This can be written in logical form as: $\forall S_1 \exists S_2 \text{ such that } |\text{score}(S_1) - \text{score}(S_2)| \le 10$.We are told that statement P is false. To find the negation of P, we apply De Morgan's laws to the quantifiers:
$\neg (\forall S_1 \exists S_2 \text{ such that } |\text{score}(S_1) - \text{score}(S_2)| \le 10)$
$\equiv \exists S_1 \neg (\exists S_2 \text{ such that } |\text{score}(S_1) - \text{score}(S_2)| \le 10)$
$\equiv \exists S_1 \forall S_2 \neg ( |\text{score}(S_1) - \text{score}(S_2)| \le 10)$
$\equiv \exists S_1 \forall S_2 \text{ such that } |\text{score}(S_1) - \text{score}(S_2)| > 10$In words, the negation means:
"There exists at least one student (S1) in the class for whom, for all other students (S2), their scores are more than 10 marks away."Let's compare this with the given options:
(A) "For each student, the scores of all the other students are more than 10 marks away." This would be $\forall S_1 \forall S_2 \text{ such that } |\text{score}(S_1) - \text{score}(S_2)| > 10$, which is stronger than the negation and not necessarily true.
(B) "There exists at least one student in the class for whom the scores of all the other students are more than 10 marks away." This matches our derived negation exactly.
(C) "There is exactly one student in the class for whom the scores of some students are more than 10 marks away." This is not the correct negation. The negation requires all other students to be more than 10 marks away from that one student.
(D) "For each student, the score of exactly one other student is more than 10 marks away." This is also not the correct negation.Therefore, option (B) is necessarily true if the original statement is false.The final answer is $\boxed{B}$

Let's solve each clue to find the keywords and their first letters:Clue 1: Synonym of recognize (8 letters): _ D _NT_FY
The word is IDENTIFY. The first letter is 'I'.Clue 2: A story long enough to fill a book (5 letters): __ EL
The word is NOVEL. The first letter is 'N'.Clue 3: Two of something (6 letters): _ PLE
The word is COUPLE. The first letter is 'C'.Clue 4: A fraction of something, split equally into two parts (4 letters): _ F
The word is HALF. The first letter is 'H'.The first letters of the keywords are I, N, C, H.Now, we need to rearrange these letters to form a four-letter word and check the options.
The letters I, N, C, H can be rearranged to form the word CHIN.Let's check the options:
(A) CHIN - This is a possible word formed by rearranging I, N, C, H.
(B) COIN - Requires 'O' instead of 'H'.
(C) ITCH - Requires 'T' instead of 'N'.
(D) NOSE - Requires 'O', 'S', 'E' which are not in the set of letters.Therefore, CHIN is the possible choice for the four-letter word.The final answer is $\boxed{A}$

Let the parents be $X$ and $Y$. They are parents to two distinct children from the set $\{P, Q, R\}$. One child has no parent playing.Rule (i): Parent and child cannot be on the same team.
Rule (ii): At most one parent-child pair per match.From MATCH 1: $\{P, X\}$ vs $\{Q, R\}$. Since $P$ and $X$ are on the same team, $X$ cannot be the parent of $P$ (Rule i).
From MATCH 3: $\{R, X\}$ vs $\{Q, Y\}$. Since $R$ and $X$ are on the same team, $X$ cannot be the parent of $R$ (Rule i). Also, since $Q$ and $Y$ are on the same team, $Y$ cannot be the parent of $Q$ (Rule i).Since $X$ is a parent of one of the children and it's not $P$ or $R$, $X$ must be the parent of $Q$.Now, $Y$ must be the parent of either $P$ or $R$.
In MATCH 3: $\{R, X\}$ vs $\{Q, Y\}$. We know $X$ is the parent of $Q$. This forms one parent-child pair $(X, Q)$ in the match. Rule (ii) states there can be at most one such pair. If $Y$ were the parent of $R$, then $(Y, R)$ would be a second parent-child pair in this match. Thus, $Y$ cannot be the parent of $R$.Therefore, $Y$ must be the parent of $P$.Conclusion:

• $X$ is the parent of $Q$.
• $Y$ is the parent of $P$.
• $R$ has no parent playing in the tournament.

Thus, option (C) is correct.

The perimeter of a square with side length $k$ is given by the formula:$$P_k = 4k$$We need to find the value of the sum:$$S = P_1 + P_2 + P_3 + \dots + P_{10}$$Substituting the formula for $P_k$:$$S = 4(1) + 4(2) + 4(3) + \dots + 4(10)$$Factoring out the constant 4:$$S = 4(1 + 2 + 3 + \dots + 10)$$The sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$. For $n = 10$:$$\sum_{i=1}^{10} i = \frac{10(10+1)}{2} = \frac{110}{2} = 55$$Therefore, the total sum is:$$S = 4 \times 55 = 220$$Thus, the correct option is (C).

Let $r_0$ be the radius of the Inner Island (land). $r_0 = 2.5$ stades.
Width of water ring 1 ($AB$) = 1 stade. Radius $r_1 = 2.5 + 1 = 3.5$ stades.
Width of land ring 1 ($BC$) = 2 stades. Radius $r_2 = 3.5 + 2 = 5.5$ stades.
Width of water ring 2 ($CD$) = 2 stades. Radius $r_3 = 5.5 + 2 = 7.5$ stades.
Width of land ring 2 ($DE$) = 3 stades. Radius $r_4 = 7.5 + 3 = 10.5$ stades.
Width of water ring 3 ($EF$) = 3 stades. Radius $r_5 = 10.5 + 3 = 13.5$ stades.Total Land Area ($A_L$):
$A_L = \pi r_0^2 + \pi(r_2^2 - r_1^2) + \pi(r_4^2 - r_3^2)$
$A_L = \pi [2.5^2 + (5.5^2 - 3.5^2) + (10.5^2 - 7.5^2)]$
$A_L = \pi [6.25 + (30.25 - 12.25) + (110.25 - 56.25)]$
$A_L = \pi [6.25 + 18 + 54] = 78.25\pi$Total Water Area ($A_W$):
$A_W = \pi(r_1^2 - r_0^2) + \pi(r_3^2 - r_2^2) + \pi(r_5^2 - r_4^2)$
$A_W = \pi [(3.5^2 - 2.5^2) + (7.5^2 - 5.5^2) + (13.5^2 - 10.5^2)]$
$A_W = \pi [(12.25 - 6.25) + (56.25 - 30.25) + (182.25 - 110.25)]$
$A_W = \pi [6 + 26 + 72] = 104\pi$Ratio of Land Area to Water Area:
Ratio $= \frac{A_L}{A_W} = \frac{78.25\pi}{104\pi} = \frac{78.25}{104} \approx 0.7524$Rounding off to two decimal places, the ratio is 0.75.

The transformation rule demonstrated in the example involves two steps:Applying this to the target text:

• Original Top Line: 'GATE - 2026' $\rightarrow$ New Bottom Line: 'GATE - 2026' (mirrored horizontally).
• Original Bottom Line: 'Aptitude Test' $\rightarrow$ New Top Line: 'Aptitude Test' (mirrored horizontally).

Comparing this result with the provided options:

• Option P: Mirrors the text but does not swap the lines.
• Option Q: Swaps the lines and mirrors them, but also reverses the word order within the lines (e.g., '2026 - GATE').
• Option R: Correctly swaps the lines and mirrors the text while maintaining the relative word order within each line.
• Option S: Swaps the lines and mirrors them, but with a different word-order reversal pattern.

Therefore, Option R (which corresponds to choice C) is the correct answer.

Given the solution vector $\mathbf{w}(t) = e^t \hat{\mathbf{u}}_x + e^{-2t} \hat{\mathbf{u}}_y$. We can write this as a column vector:
$\mathbf{w}(t) = \begin{bmatrix} e^t \\ e^{-2t} \end{bmatrix}$First, we find the derivative of $\mathbf{w}(t)$ with respect to $t$:
$\dot{\mathbf{w}}(t) = \frac{d}{dt} \begin{bmatrix} e^t \\ e^{-2t} \end{bmatrix} = \begin{bmatrix} \frac{d}{dt}(e^t) \\ \frac{d}{dt}(e^{-2t}) \end{bmatrix} = \begin{bmatrix} e^t \\ -2e^{-2t} \end{bmatrix}$The given differential equation is $\dot{\mathbf{w}} = \mathbf{A}\mathbf{w}$. Substituting the expressions for $\dot{\mathbf{w}}$ and $\mathbf{w}$:
$\begin{bmatrix} e^t \\ -2e^{-2t} \end{bmatrix} = \mathbf{A} \begin{bmatrix} e^t \\ e^{-2t} \end{bmatrix}$Let $\mathbf{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Then:
$\begin{bmatrix} e^t \\ -2e^{-2t} \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} e^t \\ e^{-2t} \end{bmatrix} = \begin{bmatrix} ae^t + be^{-2t} \\ ce^t + de^{-2t} \end{bmatrix}$Comparing the components:
1) $ae^t + be^{-2t} = e^t$
2) $ce^t + de^{-2t} = -2e^{-2t}$For these equations to hold for all $t$, we must equate the coefficients of $e^t$ and $e^{-2t}$ on both sides.
From equation (1):
Coefficient of $e^t$: $a = 1$
Coefficient of $e^{-2t}$: $b = 0$From equation (2):
Coefficient of $e^t$: $c = 0$
Coefficient of $e^{-2t}$: $d = -2$Therefore, the matrix $\mathbf{A}$ is:
$\mathbf{A} = \begin{bmatrix} 1 & 0 \\ 0 & -2 \end{bmatrix}$We can also verify the initial condition $\mathbf{w}(t=0) = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$:
$\mathbf{w}(0) = \begin{bmatrix} e^0 \\ e^{-2(0)} \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$, which matches the given condition.This corresponds to option (A).

The surface is given by the equation $z^2 = 2x^2 - y^2$. We can rewrite this as a level surface of a function $F(x, y, z) = 2x^2 - y^2 - z^2 = 0$.The normal vector to the surface at any point $(x, y, z)$ is given by the gradient of $F$, $\nabla F$.
$\nabla F = \frac{\partial F}{\partial x} \hat{\mathbf{i}} + \frac{\partial F}{\partial y} \hat{\mathbf{j}} + \frac{\partial F}{\partial z} \hat{\mathbf{k}}$Let's compute the partial derivatives:
$\frac{\partial F}{\partial x} = \frac{\partial}{\partial x}(2x^2 - y^2 - z^2) = 4x$
$\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(2x^2 - y^2 - z^2) = -2y$
$\frac{\partial F}{\partial z} = \frac{\partial}{\partial z}(2x^2 - y^2 - z^2) = -2z$So, the gradient vector is $\nabla F = 4x \hat{\mathbf{i}} - 2y \hat{\mathbf{j}} - 2z \hat{\mathbf{k}}$.We need to find the normal vector at the point $P = \hat{\mathbf{i}} + \sqrt{2} \hat{\mathbf{k}}$. This corresponds to $(x, y, z) = (1, 0, \sqrt{2})$.
First, let's verify that this point lies on the surface:
$z^2 = (\sqrt{2})^2 = 2$
$2x^2 - y^2 = 2(1)^2 - (0)^2 = 2 - 0 = 2$
Since $2 = 2$, the point $P$ is indeed on the surface.Now, substitute the coordinates of $P$ into the gradient vector:
$\nabla F \Big|_{P} = 4(1) \hat{\mathbf{i}} - 2(0) \hat{\mathbf{j}} - 2(\sqrt{2}) \hat{\mathbf{k}}$
$\nabla F \Big|_{P} = 4 \hat{\mathbf{i}} - 2\sqrt{2} \hat{\mathbf{k}}$This is a normal vector to the surface at $P$. To find a unit normal vector $\hat{\mathbf{n}}$, we normalize this vector:
Magnitude of the normal vector: $|\nabla F \Big|_{P}| = \sqrt{(4)^2 + (0)^2 + (-2\sqrt{2})^2} = \sqrt{16 + 0 + 8} = \sqrt{24} = 2\sqrt{6}$.So, a unit normal vector is $\hat{\mathbf{n}} = \frac{4 \hat{\mathbf{i}} - 2\sqrt{2} \hat{\mathbf{k}}}{2\sqrt{6}} = \frac{2 \hat{\mathbf{i}} - \sqrt{2} \hat{\mathbf{k}}}{\sqrt{6}}$.Now, let's compare this with the given options:
(A) $\hat{\mathbf{i}} - \sqrt{2} \hat{\mathbf{k}}$: This vector is not a unit vector (magnitude is $\sqrt{1^2 + (-\sqrt{2})^2} = \sqrt{1+2} = \sqrt{3}$) and its direction is not proportional to $2 \hat{\mathbf{i}} - \sqrt{2} \hat{\mathbf{k}}$.
(B) $\frac{2}{3} \hat{\mathbf{j}} - \frac{1}{3} \hat{\mathbf{k}}$: This vector has a $\hat{\mathbf{j}}$ component and no $\hat{\mathbf{i}}$ component, which is incorrect.
(C) $\sqrt{2} \hat{\mathbf{i}} - \sqrt{3} \hat{\mathbf{k}}$: This vector is not proportional to $2 \hat{\mathbf{i}} - \sqrt{2} \hat{\mathbf{k}}$.
(D) $\frac{\sqrt{2} \hat{\mathbf{i}} - \hat{\mathbf{k}}}{\sqrt{3}}$: Let's check its direction and magnitude.
The direction of our calculated normal vector $4 \hat{\mathbf{i}} - 2\sqrt{2} \hat{\mathbf{k}}$ can be simplified by factoring out $2\sqrt{2}$: $2\sqrt{2}(\sqrt{2} \hat{\mathbf{i}} - \hat{\mathbf{k}})$.
So, the direction is indeed $(\sqrt{2} \hat{\mathbf{i}} - \hat{\mathbf{k}})$.
Now, let's check the magnitude of the vector in option (D):
Magnitude $= \left|\frac{\sqrt{2} \hat{\mathbf{i}} - \hat{\mathbf{k}}}{\sqrt{3}}\right| = \frac{1}{\sqrt{3}} \sqrt{(\sqrt{2})^2 + (-1)^2} = \frac{1}{\sqrt{3}} \sqrt{2 + 1} = \frac{1}{\sqrt{3}} \sqrt{3} = 1$.Thus, option (D) represents a unit vector in the correct direction of the normal to the surface at point $P$.The final answer is $\boxed{\text{D}}$.

The Laplace transform of $u(t-2)$ is $e^{-2s}/s$.
The Laplace transform of $tu(t)$ is $1/s^2$.
For convolution in the time domain, $x(t) = f(t) * g(t)$, the Laplace transform is $X(s) = F(s)G(s)$.
Therefore, $X(s) = \mathcal{L}\{u(t-2)\} \cdot \mathcal{L}\{tu(t)\} = \frac{e^{-2s}}{s} \cdot \frac{1}{s^2} = \frac{e^{-2s}}{s^3}$.
Comparing this with the given options, option (D) is the correct expression.

In a Zener diode, when it is operated in the breakdown region, the dominant mechanism for current flow is Zener breakdown, which is a quantum mechanical tunneling phenomenon. This occurs when a high electric field across the p-n junction causes electrons to tunnel from the valence band to the conduction band, leading to a sudden increase in reverse current.

The sampling rate is $F_s = 40$ Hz. The sampling interval is $T_s = 1/F_s = 1/40$ seconds.
The samples are taken at $t_k = k T_s = k/40$ for $k = 0, 1, \dots, 9$ (first ten samples).For $x_1(t) = \cos(20\pi t)$, the samples are $x_1[k] = x_1(k/40) = \cos(20\pi k/40) = \cos(\pi k/2)$.
For $x_2(t) = \cos(100\pi t)$, the samples are $x_2[k] = x_2(k/40) = \cos(100\pi k/40) = \cos(5\pi k/2)$.Let's evaluate the samples for $k = 0, 1, \dots, 9$:$k=0$: $x_1[0] = \cos(0) = 1$, $x_2[0] = \cos(0) = 1$
$k=1$: $x_1[1] = \cos(\pi/2) = 0$, $x_2[1] = \cos(5\pi/2) = \cos(2\pi + \pi/2) = \cos(\pi/2) = 0$
$k=2$: $x_1[2] = \cos(\pi) = -1$, $x_2[2] = \cos(10\pi/2) = \cos(5\pi) = -1$
$k=3$: $x_1[3] = \cos(3\pi/2) = 0$, $x_2[3] = \cos(15\pi/2) = \cos(6\pi + 3\pi/2) = \cos(3\pi/2) = 0$
$k=4$: $x_1[4] = \cos(2\pi) = 1$, $x_2[4] = \cos(20\pi/2) = \cos(10\pi) = 1$
$k=5$: $x_1[5] = \cos(5\pi/2) = 0$, $x_2[5] = \cos(25\pi/2) = \cos(12\pi + \pi/2) = \cos(\pi/2) = 0$
$k=6$: $x_1[6] = \cos(3\pi) = -1$, $x_2[6] = \cos(30\pi/2) = \cos(15\pi) = -1$
$k=7$: $x_1[7] = \cos(7\pi/2) = 0$, $x_2[7] = \cos(35\pi/2) = \cos(16\pi + 3\pi/2) = \cos(3\pi/2) = 0$
$k=8$: $x_1[8] = \cos(4\pi) = 1$, $x_2[8] = \cos(40\pi/2) = \cos(20\pi) = 1$
$k=9$: $x_1[9] = \cos(9\pi/2) = 0$, $x_2[9] = \cos(45\pi/2) = \cos(22\pi + \pi/2) = \cos(\pi/2) = 0$From the calculations, we observe that $x_1[k] = x_2[k]$ for all $k = 0, 1, \dots, 9$.Let's check the given options:
(A) "All of the first three samples of $x_1(t)$ are greater than the corresponding samples of $x_2(t)$." This is false, as $x_1[k] = x_2[k]$.
(B) "All of the last three samples of $x_1(t)$ are greater than the corresponding samples of $x_2(t)$." This is false, as $x_1[k] = x_2[k]$.
(C) "All of the samples of $x_2(t)$ are greater than the corresponding samples of $x_1(t)$." This is false, as $x_1[k] = x_2[k]$.
(D) "All of the fourth to seventh samples of $x_1(t)$ are equal to the corresponding samples of $x_2(t)$." The fourth sample corresponds to $k=3$, and the seventh sample corresponds to $k=6$. For $k=3, 4, 5, 6$, we have:
$x_1[3]=0, x_2[3]=0$ (Equal)
$x_1[4]=1, x_2[4]=1$ (Equal)
$x_1[5]=0, x_2[5]=0$ (Equal)
$x_1[6]=-1, x_2[6]=-1$ (Equal)
This statement is TRUE.The final answer is $\boxed{D}$

Given the difference equation:
$y[n] - \frac{5}{6}y[n - 1] + \frac{1}{6}y[n - 2] = x[n]$Taking the Z-transform on both sides:
$Y(z) \left(1 - \frac{5}{6}z^{-1} + \frac{1}{6}z^{-2}\right) = X(z)$The transfer function $H(z)$ is:
$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1 - \frac{5}{6}z^{-1} + \frac{1}{6}z^{-2}} = \frac{1}{(1 - \frac{1}{2}z^{-1})(1 - \frac{1}{3}z^{-1})}$The input is $x[n] = \delta[n] - \frac{1}{3}\delta[n - 1]$, so its Z-transform is:
$X(z) = 1 - \frac{1}{3}z^{-1}$The Z-transform of the output $y[n]$ is:
$Y(z) = H(z)X(z) = \frac{1 - \frac{1}{3}z^{-1}}{(1 - \frac{1}{2}z^{-1})(1 - \frac{1}{3}z^{-1})} = \frac{1}{1 - \frac{1}{2}z^{-1}}$$Y(z)$ has a single pole at $z = \frac{1}{2}$.Standard systems described by such difference equations are typically assumed to be causal. Therefore, the response is stable and causal.