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.

Read off the six stubs that stop at the edges of the missing rectangle $WXYZ$, and the direction each one is heading:Since there are exactly three straight segments, the stubs must pair up with matching slopes:

• stub 1 (down-right) continues to stub 6, leaving the rectangle at the right edge near $Y$;
• stub 2 (down-left) continues to stub 5, leaving through the bottom edge near $Z$;
• stub 3 continues to stub 4 as a gently rising, nearly horizontal line.

So the missing portion must show two lines that start on the top edge and cross each other just below it, one running to the corner region at $Y$ and the other to the bottom edge near $Z$, plus a nearly horizontal line that is lower on the left than on the right.

• P: the nearly horizontal line falls from left to right, and it enters high on the left edge. It does not match stubs 3 and 4.
• Q: both steep lines rise to the right; there is no line heading down-right from the top edge to $Y$.
• R: the line rising to the right ends at the right edge near $X$, not on the top edge, and it starts in the middle of the bottom edge rather than near $Z$.
• S: the down-right line runs from the top edge to $Y$, the down-left line runs from the top edge to the bottom near $Z$, and the third line rises gently from the lower left to the right. All six stubs continue correctly.

Answer: (D) S

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.

Find the rule from the reference figure.The reference block has Ref Fig on the first line and 12 34 on the second. After the arrow:Apply the rule to the new block. GATE - 2026 is on top and Aptitude Test is below. So the result must have:

• Aptitude Test on top and GATE - 2026 underneath;
• both lines flipped upside down, with the word order unchanged (Aptitude before Test, GATE before 2026).

Check the panels.

• P: the top line reads Test Aptitude, so the word order has been reversed.
• Q: the bottom line reads 2026 - GATE, so the word order has been reversed.
• R: Aptitude Test on top, GATE - 2026 below, both flipped upside down, with the word order kept. This matches the rule.
• S: both lines have their word order reversed.

Answer: (C) R

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).

Write the surface as the level set $F(x,y,z) = 2x^2 - y^2 - z^2 = 0$. A normal is $\nabla F = 4x\,\hat{\mathbf{i}} - 2y\,\hat{\mathbf{j}} - 2z\,\hat{\mathbf{k}}$.At $P = (1, 0, \sqrt{2})$ (which lies on the surface since $2 = 2\cdot 1 - 0$):$$\nabla F\big|_P = 4\hat{\mathbf{i}} - 2\sqrt{2}\,\hat{\mathbf{k}} = 2\sqrt{2}\left(\sqrt{2}\,\hat{\mathbf{i}} - \hat{\mathbf{k}}\right)$$Its magnitude is $\sqrt{16 + 8} = 2\sqrt{6}$, so the unit normals are$$\pm\vec{n} = \pm\frac{\sqrt{2}\,\hat{\mathbf{i}} - \hat{\mathbf{k}}}{\sqrt{3}}$$Checking the options:

• (A) $\hat{\mathbf{i}} - \sqrt{2}\hat{\mathbf{k}}$: magnitude $\sqrt{3}$ (not unit) and direction ratio $1:-\sqrt{2}$, not $\sqrt{2}:-1$. Wrong.
• (B) $\frac{2}{3}\hat{\mathbf{i}} - \frac{1}{3}\hat{\mathbf{k}}$: magnitude $\sqrt{5}/3 \ne 1$ and ratio $2:-1 \ne \sqrt{2}:-1$. Wrong.
• (C) $\sqrt{2}\hat{\mathbf{i}} - \sqrt{3}\hat{\mathbf{k}}$: ratio $\sqrt{2}:-\sqrt{3}$, not parallel; magnitude $\sqrt{5}$. Wrong.
• (D) $\frac{\sqrt{2}\hat{\mathbf{i}} - \hat{\mathbf{k}}}{\sqrt{3}}$: parallel to $\nabla F$ and of unit length. Correct.

Answer: (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.

The circuit is a non-inverting amplifier. The feedback resistance $R_f$ is the sum of the $2\text{ k}\Omega$ and $1\text{ k}\Omega$ resistors in series: $R_f = 2\text{ k}\Omega + 1\text{ k}\Omega = 3\text{ k}\Omega$. The resistance from the inverting terminal to ground is $R_1 = 1\text{ k}\Omega$.
The gain is given by:
$$A_{v1} = 1 + \frac{R_f}{R_1} = 1 + \frac{3\text{ k}\Omega}{1\text{ k}\Omega} = 4$$ Given $V_i = 1\text{ V}$, the output voltage is:
$$V_o = A_{v1} V_i = 4 \times 1 = 4\text{ V}$$ Thus, $x = 4$. When S is closed, the $1\text{ k}\Omega$ feedback resistor is short-circuited. The feedback resistance becomes $R_f' = 2\text{ k}\Omega$.
The new gain is:
$$A_{v2} = 1 + \frac{R_f'}{R_1} = 1 + \frac{2\text{ k}\Omega}{1\text{ k}\Omega} = 3$$ The new output voltage is:
$$V_o' = A_{v2} V_i = 3 \times 1 = 3\text{ V}$$ We have $V_o' = 3$ and $x = 4$.
Therefore, $V_o' = \frac{3}{4}x$.Hence, the correct option is (C).