GATE EC Engineering Mathematics Previous Year Questions with Solutions

Full question on page 12 of the official paper (PDF)

GATE 2026, Q.12

Engineering Mathematics1 markMCQ

$\vec{n}$ and $-\vec{n}$ are the unit normals to the surface $z^2 = 2x^2 - y^2$ at the point $\hat{i} + \sqrt{2}\,\hat{k}$ . Which vector can $\vec{n}$ be?

Option A: $\hat{i} - \sqrt{2}\,\hat{k}$
Option B: $\tfrac{2}{3}\hat{i} - \tfrac{1}{3}\hat{k}$
Option C: $\sqrt{2}\,\hat{i} - \sqrt{3}\,\hat{k}$
Option D: $\dfrac{\sqrt{2}\,\hat{i} - \hat{k}}{\sqrt{3}}$

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Key idea: The gradient of the surface function is normal to the surface.

Official answer: D

Write the surface as $f = 2x^2 - y^2 - z^2 = 0$ and take its gradient at $(1, 0, \sqrt{2})$ :

\nabla f = (4x,\,-2y,\,-2z) \;\to\; (4,\,0,\,-2\sqrt{2})

\begin{aligned} \nabla f &= (4x,\,-2y,\,-2z) \; \\ &\to\; (4,\,0,\,-2\sqrt{2}) \end{aligned}

This is a multiple of $(\sqrt{2}, 0, -1)$ , whose length is $\sqrt{3}$ , so a unit normal is

\vec{n} = \frac{\sqrt{2}\,\hat{i} - \hat{k}}{\sqrt{3}}

Option (D). Options (A) and (C) aren't unit vectors, and (B) points the wrong way.

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Full question on page 26 of the official paper (PDF)

GATE 2026, Q.26

Engineering Mathematics1 markMSQ

For $M = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 3 & 0 \\ -1 & a & b \end{bmatrix}$ , which values of $a$ and $b$ make $\det M \neq 0$ ?

One or more options are correct.

Option A: $a = -\tfrac{1}{2}$ , $b = -\tfrac{1}{2}$
Option B: $a = \tfrac{1}{2}$ , $b = \tfrac{1}{2}$
Option C: $a = -3$ , $b = 0$
Option D: $a = \tfrac{1}{2}$ , $b = -3$

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Key idea: Write $\det M$ in terms of $a$ and $b$ , then keep the options where it isn't zero.

Official answer: B and D

Expand along the first row:

\det M = 2(3b - 0) - 1(b - 0) + 1(a + 3) = a + 5b + 3

\begin{aligned} \det M &= 2(3b - 0) \\ &\quad - 1(b - 0) \\ &\quad + 1(a + 3) \\ &= a + 5b + 3 \end{aligned}

(A) $a = b = -\tfrac{1}{2}$ : $-\tfrac{1}{2} - \tfrac{5}{2} + 3 = 0$ .
(B) $a = b = \tfrac{1}{2}$ : $6 \neq 0$ .
(C) $a = -3, b = 0$ : $0$ .
(D) $a = \tfrac{1}{2}, b = -3$ : $-11.5 \neq 0$ .

Options (B) and (D).

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Full question on page 33 of the official paper (PDF)

GATE 2026, Q.36

Engineering Mathematics2 marksMCQ

Do $S_A = \displaystyle\sum_{n=1}^{\infty} \frac{n^2}{2^n}$ and $S_B = 1 + \tfrac{1}{2} + \tfrac{1}{8} + \tfrac{1}{16} + \tfrac{1}{64} + \tfrac{1}{128} + \tfrac{1}{512} + \cdots$ converge?

Option A: Both $S_A$ and $S_B$ converge.
Option B: Neither converges.
Option C: $S_A$ converges, $S_B$ does not.
Option D: $S_B$ converges, $S_A$ does not.

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Key idea: Ratio test for the first; compare the second with a geometric series.

Official answer: A

For $S_A$ , use the ratio test:

\frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{2n^2} \to \frac{1}{2} < 1

\begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{(n+1)^2}{2n^2} \\ &\to \frac{1}{2} \\ &< 1 \end{aligned}

So $S_A$ converges (its sum is in fact 6).

Every term of $S_B$ is a different power of $\tfrac{1}{2}$ (the exponents run 0, 1, 3, 4, 6, 7, 9, and so on). Its partial sums are therefore bounded by the full geometric series:

S_B \le \sum_{k=0}^{\infty} \left(\tfrac{1}{2}\right)^k = 2

A series of positive terms with bounded partial sums converges (its sum is 12/7). Both converge. Option (A).

Full question on page 35 of the official paper (PDF)

GATE 2026, Q.38

Engineering Mathematics2 marksMCQ

$X$ is uniform on $(-1, 1)$ and $N$ is an independent standard normal. With $Y = X + N$ and $Z = X^2 + N$ , find the correlations $E[XY]$ and $E[XZ]$ .

Option A: $\tfrac{1}{3}$ and 0
Option B: $\tfrac{1}{3}$ and $\tfrac{1}{9}$
Option C: $\tfrac{1}{3}$ and $\tfrac{1}{3}$
Option D: 1 and 0

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Key idea: Independence and $E[N] = 0$ remove the noise terms, leaving $E[X^2]$ and $E[X^3]$ .

Official answer: A

Here correlation means $E[XY]$ (the correlation coefficient of X and Y is 1/2, which isn't an option). Independence and $E[N] = 0$ remove the noise terms:

E[XY] = E[X^2] + E[X]\,E[N] = \int_{-1}^{1} \frac{x^2}{2}\,dx = \frac{1}{3}

\begin{aligned} E[XY] &= E[X^2] \\ &\quad + E[X]\,E[N] \\ &= \int_{-1}^{1} \frac{x^2}{2}\,dx \\ &= \frac{1}{3} \end{aligned}

E[XZ] = E[X^3] + E[X]\,E[N] = \int_{-1}^{1} \frac{x^3}{2}\,dx = 0

\begin{aligned} E[XZ] &= E[X^3] \\ &\quad + E[X]\,E[N] \\ &= \int_{-1}^{1} \frac{x^3}{2}\,dx \\ &= 0 \end{aligned}

The second is zero because $x^3$ is odd and X is symmetric about 0. Option (A).

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Full question on page 57 of the official paper (PDF)

GATE 2026, Q.61

Engineering Mathematics2 marksNAT

Evaluate $\displaystyle\iint_R (x^2 + y^2 - 1)\,dx\,dy$ over the square $R$ with corners $(0, 0)$ , $(1, 1)$ , $(2, 0)$ and $(1, -1)$ .

The X-Y plane with the square region R shaded: a diamond with corners at (0, 0), (1, 1), (2, 0) and (1, -1), and a dashed vertical line through x = 1. — Figure from the official GATE 2026 EC paper, IIT Guwahati

Answer format: A number, rounded off to two decimal places

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Key idea: Use the square's polar moment about its centre, then shift the centre to $(1, 0)$ .

Official answer: 0.60 to 0.70

The square has diagonals of length 2, so its area is 2 and its side is $\sqrt{2}$ . Its centre is (1, 0).

For a square of area $A$ and side $s$ , $\iint (u^2 + v^2)$ about its centre is $A s^2/6$ , whatever its orientation. Shifting the centre to (1, 0) adds $A \times 1^2$ :

\iint_R (x^2 + y^2)\,dA = \frac{2 \times 2}{6} + 2 \times 1 = \frac{8}{3}

\begin{aligned} &\iint_R (x^2 + y^2)\,dA \\ &\quad = \frac{2 \times 2}{6} + 2 \times 1 \\ &\quad = \frac{8}{3} \end{aligned}

\iint_R (x^2 + y^2 - 1)\,dA = \frac{8}{3} - 2 = \frac{2}{3} \approx 0.67

\begin{aligned} &\iint_R (x^2 + y^2 - 1)\,dA \\ &\quad = \frac{8}{3} - 2 \\ &\quad = \frac{2}{3} \\ &\quad \approx 0.67 \end{aligned}

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Full question on page 10 of the official paper (PDF)

GATE 2025, Q.11

Engineering Mathematics1 markMCQ

$A = \begin{bmatrix} 2 & 3 & 4 & 5 \\ 0 & 6 & 7 & 8 \\ 0 & 0 & \alpha & \beta \\ 0 & 0 & 0 & \gamma \end{bmatrix}$ . For which of these is the rank of $A$ at least three? (i) $\alpha = 0$ and $\beta = \gamma \neq 0$ ; (ii) $\alpha = \beta = \gamma = 0$ ; (iii) $\beta = \gamma = 0$ and $\alpha \neq 0$ ; (iv) $\alpha = \beta = \gamma \neq 0$ .

Option A: Only (i), (iii) and (iv)
Option B: Only (iv)
Option C: Only (ii)
Option D: Only (i) and (iii)

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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).

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Full question on page 11 of the official paper (PDF)

GATE 2025, Q.12

Engineering Mathematics1 markMCQ

Which of these series converge? (i) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ , (ii) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ , (iii) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n!}$

Option A: Only (ii) converges
Option B: Only (ii) and (iii) converge
Option C: Only (iii) converges
Option D: All three converge

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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).

Full question on page 12 of the official paper (PDF)

GATE 2025, Q.13

Engineering Mathematics1 markMCQ

A pot holds two red balls and two blue balls. Two balls are drawn at random without replacement. What is the probability that they are different colours?

Option A: $\tfrac{2}{3}$
Option B: $\tfrac{1}{3}$
Option C: $\tfrac{1}{2}$
Option D: $1$

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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.

P(\text{different colours}) = \frac{4}{6} = \frac{2}{3}

\begin{aligned} &P(\text{different colours}) \\ &\quad = \frac{4}{6} \\ &\quad = \frac{2}{3} \end{aligned}

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

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Full question on page 24 of the official paper (PDF)

GATE 2025, Q.25

Engineering Mathematics1 markMSQ

For $f(x) = 2x^3 - 3x^2 - 12x + 1$ on all real $x$ , which statements are correct?

One or more options are correct.

Option A: $f$ has no global maximizer
Option B: $f$ has no global minimizer
Option C: $x = -1$ is a local minimizer of $f$
Option D: $x = 2$ is a local maximizer of $f$

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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'(x) = 6x^2 - 6x - 12 = 6(x - 2)(x + 1), \qquad f''(x) = 12x - 6

\begin{aligned} f'(x) &= 6x^2 - 6x - 12 \\ &= 6(x - 2)(x + 1), \\ f''(x) &= 12x - 6 \end{aligned}

$f''(-1) = -18 < 0$ , 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).

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Full question on page 29 of the official paper (PDF)

GATE 2025, Q.32

Engineering Mathematics1 markNAT

$y(t)$ satisfies $t^2 y''(t) - 2t\,y'(t) + 2y(t) = 0$ , with $y'(0) = 1$ and $y'(1) = -1$ . What is the maximum value of $y(t)$ on $[0, 1]$ ?

Answer format: Rounded off to two decimal places

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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

m(m-1) - 2m + 2 = m^2 - 3m + 2 = (m-1)(m-2) = 0

\begin{aligned} &m(m-1) - 2m + 2 \\ &\quad = m^2 - 3m + 2 \\ &\quad = (m-1)(m-2) \\ &\quad = 0 \end{aligned}

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 $c_2 = -1$ :

y(t) = t - t^2

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

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Full question on page 45 of the official paper (PDF)

GATE 2025, Q.51

Engineering Mathematics2 marksMSQ

$f(x)$ is non-negative, continuous and bounded on $[2, 8]$ , with maximum $M$ and minimum $m$ there. For which pairs is $\beta \le \displaystyle\int_2^8 f(x)\,dx \le \alpha$ guaranteed?

One or more options are correct.

Option A: $\beta = 5m$ , $\alpha = 7M$
Option B: $\beta = 6m$ , $\alpha = 5M$
Option C: $\beta = 7m$ , $\alpha = 6M$
Option D: $\beta = 7m$ , $\alpha = 5M$

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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

6m \le \int_2^8 f(x)\,dx \le 6M

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.

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Full question on page 46 of the official paper (PDF)

GATE 2025, Q.52

Engineering Mathematics2 marksMSQ

Which of these statements about contour integrals, taken counter-clockwise round the unit circle $C$ , are TRUE?

One or more options are correct.

Option A: $\oint_C e^z\,dz = 0$
Option B: $\oint_C z^n\,dz = 0$ , where $n$ is an even integer
Option C: $\oint_C \cos z\,dz \neq 0$
Option D: $\oint_C \sec z\,dz \neq 0$

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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).

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Full question on page 50 of the official paper (PDF)

GATE 2025, Q.58

Engineering Mathematics2 marksNAT

Two fair six-sided dice are rolled, and $X$ is the sum of the outcomes. What is the expectation of $X$ ?

Answer format: Rounded off to two decimal places

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Key idea: Expectation is linear: the expected sum is the sum of the expected values.

Official answer: 6.95 to 7.05

E[\text{one die}] = \frac{1 + 2 + \cdots + 6}{6} = 3.5, \qquad E[X] = 3.5 + 3.5 = 7

\begin{aligned} E[\text{one die}] &= \frac{1 + 2 + \cdots + 6}{6} \\ &= 3.5, \\ E[X] &= 3.5 + 3.5 \\ &= 7 \end{aligned}

So $E[X] = 7.00$ . Listing all 36 outcomes gives the same answer.

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Full question on page 50 of the official paper (PDF)

GATE 2025, Q.59

Engineering Mathematics2 marksNAT

With $\mathbf{a} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix} 0 \\ 3\sqrt{2} \end{bmatrix}$ , what is $\displaystyle\min_x \|\mathbf{a}x - \mathbf{b}\|_2$ over real $x$ , where $\|\cdot\|_2$ is the Euclidean norm?

Answer format: Rounded off to two decimal places

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Key idea: The minimum is the distance from $\mathbf{b}$ to the line through $\mathbf{a}$ : the part of $\mathbf{b}$ perpendicular to $\mathbf{a}$ .

Official answer: 2.95 to 3.05

The best $x$ projects $\mathbf{b}$ onto $\mathbf{a}$ :

x^* = \frac{\mathbf{a}\cdot\mathbf{b}}{\mathbf{a}\cdot\mathbf{a}} = \frac{3\sqrt{2}}{2}, \qquad \mathbf{a}x^* - \mathbf{b} = \begin{bmatrix} 3\sqrt{2}/2 \\ -3\sqrt{2}/2 \end{bmatrix}

\begin{aligned} x^* &= \frac{\mathbf{a}\cdot\mathbf{b}}{\mathbf{a}\cdot\mathbf{a}} \\ &= \frac{3\sqrt{2}}{2}, \\ \mathbf{a}x^* - \mathbf{b} &= \begin{bmatrix} 3\sqrt{2}/2 \\ -3\sqrt{2}/2 \end{bmatrix} \end{aligned}

\|\mathbf{a}x^* - \mathbf{b}\|_2 = \sqrt{\frac{18}{4} + \frac{18}{4}} = \sqrt{9} = 3

\begin{aligned} &\|\mathbf{a}x^* - \mathbf{b}\|_2 \\ &\quad = \sqrt{\frac{18}{4} + \frac{18}{4}} \\ &\quad = \sqrt{9} \\ &\quad = 3 \end{aligned}

The minimum is 3.00.

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Full question on page 10 of the official paper (PDF)

GATE 2024, Q.11

Engineering Mathematics1 markMCQ

The general form of the complementary function of a differential equation is $y(t) = (At + B)e^{-2t}$ , where $A$ and $B$ are real constants set by the initial conditions. Which is the differential equation?

Option A: $\dfrac{d^2y}{dt^2} + 4\dfrac{dy}{dt} + 4y = f(t)$
Option B: $\dfrac{d^2y}{dt^2} + 4y = f(t)$
Option C: $\dfrac{d^2y}{dt^2} + 3\dfrac{dy}{dt} + 2y = f(t)$
Option D: $\dfrac{d^2y}{dt^2} + 5\dfrac{dy}{dt} + 6y = f(t)$

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Key idea: A factor of $t$ in front of an exponential means a repeated root of the characteristic equation.

Official answer: A

$(At + B)e^{-2t}$ is what a double root at $m = -2$ produces, so the characteristic polynomial is

(m + 2)^2 = m^2 + 4m + 4

That is option (A). Options (C) and (D) have distinct roots ( $-1, -2$ and $-2, -3$ ), and (B) has roots $\pm 2j$ .

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Full question on page 22 of the official paper (PDF)

GATE 2024, Q.25

Engineering Mathematics1 markMSQ

$\rho(x, y, z, t)$ and $\mathbf{u}(x, y, z, t)$ are density and velocity, with $\partial\rho/\partial t$ continuous. $V$ is any volume with closed surface $S$ and outward unit normal $\hat{n}$ . Which equations are equivalent to $\dfrac{\partial\rho}{\partial t} + \nabla\cdot(\rho\mathbf{u}) = 0$ ?

One or more options are correct.

Option A: $\displaystyle\int_V \frac{\partial\rho}{\partial t}\,dv = -\oint_S \rho\,\mathbf{u}\cdot\hat{n}\,ds$
Option B: $\displaystyle\int_V \frac{\partial\rho}{\partial t}\,dv = \oint_S \rho\,\mathbf{u}\cdot\hat{n}\,ds$
Option C: $\displaystyle\int_V \frac{\partial\rho}{\partial t}\,dv = -\int_V \nabla\cdot(\rho\mathbf{u})\,dv$
Option D: $\displaystyle\int_V \frac{\partial\rho}{\partial t}\,dv = \int_V \nabla\cdot(\rho\mathbf{u})\,dv$

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Key idea: Integrate the equation over any volume, then use the divergence theorem to turn the volume integral of the divergence into a flux.

Official answer: A and C

Integrating over $V$ gives (C) directly. The divergence theorem, $\int_V \nabla\cdot(\rho\mathbf{u})\,dv = \oint_S \rho\,\mathbf{u}\cdot\hat{n}\,ds$ , turns it into (A): mass inside falls at the rate it flows out.

Because $V$ is arbitrary and the integrands are continuous, either integral form also gives back the differential equation. (B) and (D) have the wrong sign. Options (A) and (C).

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Full question on page 24 of the official paper (PDF)

GATE 2024, Q.30

Engineering Mathematics1 markNAT

For which $\alpha$ do the vectors $(2, -3, \alpha)$ , $(3, -1, 3)$ and $(1, -5, 7)$ fail to form a basis of $\mathbb{R}^3$ ?

Answer format: A number

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Key idea: Three vectors in three dimensions form a basis exactly when the determinant of the matrix they make is non-zero.

Official answer: 5

\det\begin{bmatrix} 2 & -3 & \alpha \\ 3 & -1 & 3 \\ 1 & -5 & 7 \end{bmatrix} = 2(-7 + 15) + 3(21 - 3) + \alpha(-15 + 1) = 70 - 14\alpha

\begin{aligned} &\det\begin{bmatrix} 2 & -3 & \alpha \\ 3 & -1 & 3 \\ 1 & -5 & 7 \end{bmatrix} \\ &\quad = 2(-7 + 15) \\ &\qquad + 3(21 - 3) \\ &\qquad + \alpha(-15 + 1) \\ &\quad = 70 - 14\alpha \end{aligned}

It is zero at $\alpha = 5$ , so then the vectors are dependent and not a basis.

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Full question on page 25 of the official paper (PDF)

GATE 2024, Q.33

Engineering Mathematics1 markNAT

$X$ and $Y$ are independent and identically distributed, uniform on $[0, 1]$ . What is $P(X \ge Y)$ ?

Answer format: Rounded off to two decimal places

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Key idea: By symmetry $P(X \ge Y) = P(Y \ge X)$ , and $X = Y$ has probability zero.

Official answer: 0.5

The two events $X \ge Y$ and $Y \ge X$ are mirror images and together cover everything, overlapping only on the line $X = Y$ , which has zero probability. So each has probability $\tfrac{1}{2}$ : the area of the triangle below the diagonal of the unit square.

P(X \ge Y) = \int_0^1 \int_0^x dy\,dx = \frac{1}{2} = 0.50

\begin{aligned} P(X \ge Y) &= \int_0^1 \int_0^x dy\,dx \\ &= \frac{1}{2} \\ &= 0.50 \end{aligned}

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Full question on page 27 of the official paper (PDF)

GATE 2024, Q.36

Engineering Mathematics2 marksMCQ

Taking the Earth as a sphere of radius $R$ , what is the surface area of the region inside the 60°N latitude circle that contains the north pole?

Option A: $(2 - \sqrt{3})\pi R^2$
Option B: $\dfrac{(\sqrt{2} - 1)\pi R^2}{2}$
Option C: $\dfrac{2\pi R^2}{3}$
Option D: $\dfrac{(2 + \sqrt{3})\pi R^2}{8\sqrt{2}}$

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Key idea: A spherical cap of height $h$ has area $2\pi R h$ , the same as the band of a cylinder round it.

Official answer: A

The 60°N circle sits at height $R\sin 60° = \tfrac{\sqrt{3}}{2}R$ above the equator, so the cap above it has height $h = R(1 - \tfrac{\sqrt{3}}{2})$ :

A = 2\pi R h = 2\pi R^2\left(1 - \frac{\sqrt{3}}{2}\right) = (2 - \sqrt{3})\,\pi R^2

\begin{aligned} A &= 2\pi R h \\ &= 2\pi R^2\left(1 - \frac{\sqrt{3}}{2}\right) \\ &= (2 - \sqrt{3})\,\pi R^2 \end{aligned}

Option (A). About 6.7% of the sphere.

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Full question on page 33 of the official paper (PDF)

GATE 2024, Q.43

Engineering Mathematics2 marksMCQ

For $f(z) = \dfrac{\sin(\pi z)}{z^2(z - 2)}$ and $C$ the circle $|z| = 3$ , what is $\oint_C f(z)\,dz$ ?

Option A: $\pi^2 j$
Option B: $j\pi\left(\tfrac{1}{2} - \pi\right)$
Option C: $j\pi\left(\tfrac{1}{2} + \pi\right)$
Option D: $-\pi^2 j$

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Key idea: Both singularities, $z = 0$ and $z = 2$ , are inside the circle; add their residues.

Official answer: D

Near $z = 0$ , $\sin(\pi z) \approx \pi z$ , so the double pole is really simple: $f(z) \approx \dfrac{\pi}{z(z - 2)}$ , with residue $\dfrac{\pi}{-2}$ . At $z = 2$ the residue is $\dfrac{\sin 2\pi}{4} = 0$ .

\oint_C f(z)\,dz = 2\pi j\left(-\frac{\pi}{2} + 0\right) = -\pi^2 j

\begin{aligned} \oint_C f(z)\,dz &= 2\pi j\left(-\frac{\pi}{2} + 0\right) \\ &= -\pi^2 j \end{aligned}

Option (D).

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Full question on page 42 of the official paper (PDF)

GATE 2024, Q.54

Engineering Mathematics2 marksMSQ

$F_1$ , $F_2$ and $F_3$ are functions of $(x, y, z)$ , and for every pair of points $A$ and $B$ the line integral $\int_C (F_1\,dx + F_2\,dy + F_3\,dz)$ is the same along any path $C$ from $A$ to $B$ . Which statements are true?

One or more options are correct.

Option A: Every closed path $\gamma$ has $\oint_\gamma (F_1\,dx + F_2\,dy + F_3\,dz) = 0$ .
Option B: There is a differentiable scalar $f(x, y, z)$ with $F_1 = \frac{\partial f}{\partial x}$ , $F_2 = \frac{\partial f}{\partial y}$ , $F_3 = \frac{\partial f}{\partial z}$ .
Option C: $\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} = 0$ .
Option D: $\frac{\partial F_3}{\partial y} = \frac{\partial F_2}{\partial z}$ , $\frac{\partial F_1}{\partial z} = \frac{\partial F_3}{\partial x}$ , $\frac{\partial F_2}{\partial x} = \frac{\partial F_1}{\partial y}$ .

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Key idea: Path independence, zero circulation, being a gradient and having zero curl are four faces of the same property; zero divergence is a different one.

Official answer: A, B and D

(A) True: going out along one path and back along another gives the difference of two equal integrals.
(B) True: define $f$ as the integral from a fixed point; path independence makes it well defined, and $\nabla f = \mathbf{F}$ .
(D) True: $\mathbf{F} = \nabla f$ has zero curl, and these are its three components.
(C) False: that is zero divergence. $\mathbf{F} = (x, 0, 0) = \nabla(x^2/2)$ is path independent but has divergence 1.

Options (A), (B) and (D).

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Full question on page 43 of the official paper (PDF)

GATE 2024, Q.55

Engineering Mathematics2 marksMSQ

For the matrix $\begin{bmatrix} 1 & k \\ 2 & 1 \end{bmatrix}$ , with $k$ a positive real number, which vectors are eigenvectors?

One or more options are correct.

Option A: $\begin{bmatrix} 1 \\ -\sqrt{2/k} \end{bmatrix}$
Option B: $\begin{bmatrix} 1 \\ \sqrt{2/k} \end{bmatrix}$
Option C: $\begin{bmatrix} \sqrt{2k} \\ 1 \end{bmatrix}$
Option D: $\begin{bmatrix} \sqrt{2k} \\ -1 \end{bmatrix}$

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Key idea: Find the eigenvalues, then read the ratio of components off one row of $A - \lambda I$ .

Official answer: A and B

(1 - \lambda)^2 = 2k \;\Rightarrow\; \lambda = 1 \pm \sqrt{2k}

\begin{aligned} (1 - \lambda)^2 &= 2k \\ \Rightarrow\; \lambda &= 1 \pm \sqrt{2k} \end{aligned}

The first row of $A - \lambda I$ gives $\mp\sqrt{2k}\,v_1 + k\,v_2 = 0$ , so $v_2 = \pm\frac{\sqrt{2k}}{k}v_1 = \pm\sqrt{2/k}\,v_1$ . Taking $v_1 = 1$ gives (A) and (B).

(C) and (D) have $v_2/v_1 = \pm 1/\sqrt{2k}$ , which equals $\pm\sqrt{2/k}$ for no positive $k$ . Options (A) and (B).

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Full question on page 9 of the official paper (PDF)

GATE 2023, Q.11

Engineering Mathematics1 markMCQ

Let $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$ . What value of the coefficient $\alpha$ in $\mathbf{v}_1 = \alpha \mathbf{v}_2 + \mathbf{e}$ minimises the length of the error vector $\mathbf{e}$ ?

Option A: $\dfrac{7}{2}$
Option B: $-\dfrac{2}{7}$
Option C: $\dfrac{2}{7}$
Option D: $-\dfrac{7}{2}$

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Key idea: The error is shortest when it is perpendicular to $\mathbf{v}_2$ , which makes $\alpha \mathbf{v}_2$ the projection of $\mathbf{v}_1$ .

Official answer: C

Setting $\mathbf{e} \cdot \mathbf{v}_2 = 0$ gives the projection coefficient:

\alpha = \frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2} = \frac{2 + 2 + 0}{4 + 1 + 9} = \frac{4}{14} = \frac{2}{7}

\begin{aligned} \alpha &= \frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2} \\ &= \frac{2 + 2 + 0}{4 + 1 + 9} \\ &= \frac{4}{14} \\ &= \frac{2}{7} \end{aligned}

Option (C).

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Full question on page 9 of the official paper (PDF)

GATE 2023, Q.12

Engineering Mathematics1 markMCQ

What is the rate of increase of the scalar field $f(x, y, z) = xyz$ in the direction $\mathbf{v} = (2, 1, 2)$ at the point $(0, 2, 1)$ ?

Option A: $\dfrac{2}{3}$
Option B: $\dfrac{4}{3}$
Option C: 2
Option D: 4

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Key idea: The rate of increase along a direction is the gradient dotted with the unit vector in that direction.

Official answer: B

\nabla f = (yz,\ xz,\ xy) = (2,\ 0,\ 0) \ \text{at}\ (0, 2, 1), \qquad \hat{\mathbf{v}} = \frac{(2, 1, 2)}{3}

\begin{aligned} \nabla f &= (yz,\ xz,\ xy) \\ &= (2,\ 0,\ 0) \ \text{at}\ (0, 2, 1), \\ \hat{\mathbf{v}} &= \frac{(2, 1, 2)}{3} \end{aligned}

\nabla f \cdot \hat{\mathbf{v}} = \frac{2 \times 2}{3} = \frac{4}{3}

Option (B). Forgetting to normalise $\mathbf{v}$ gives 4, option (D).

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Full question on page 10 of the official paper (PDF)

GATE 2023, Q.13

Engineering Mathematics1 markMCQ

Let $w^4 = 16j$ . Which of the following cannot be a value of $w$ ?

Option A: $2e^{j2\pi/8}$
Option B: $2e^{j\pi/8}$
Option C: $2e^{j5\pi/8}$
Option D: $2e^{j9\pi/8}$

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Key idea: Write $16j$ in polar form; its four fourth roots have modulus 2 and angles a quarter turn apart.

Official answer: A

16j = 16 e^{j(\pi/2 + 2\pi k)} \quad\Rightarrow\quad w = 2 e^{j(\pi/8 + k\pi/2)},\ k = 0, 1, 2, 3

\begin{aligned} 16j &= 16 e^{j(\pi/2 + 2\pi k)} \\ \Rightarrow\; \\ w &= 2 e^{j(\pi/8 + k\pi/2)},\ k \\ &= 0, 1, 2, 3 \end{aligned}

The angles are $\pi/8$ , $5\pi/8$ , $9\pi/8$ and $13\pi/8$ . $2\pi/8$ is not among them: $(2e^{j\pi/4})^4 = 16e^{j\pi} = -16$ . Option (A).

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Full question on page 10 of the official paper (PDF)

GATE 2023, Q.14

Engineering Mathematics1 markMCQ

What is the value of the contour integral $\oint_C \dfrac{z + 2}{z^2 + 2z + 2}\, dz$ , where $C$ is the circle $\{z : |z + 1 - \tfrac{3}{2} j| = 1\}$ , taken counter-clockwise?

Option A: $-\pi(1 + j)$
Option B: $\pi(1 + j)$
Option C: $\pi(1 - j)$
Option D: $-\pi(1 - j)$

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Key idea: Find which poles lie inside the circle, then use the residue theorem.

Official answer: B

The poles are the roots of $z^2 + 2z + 2$ , $z = -1 \pm j$ . The circle has centre $-1 + 1.5j$ and radius 1, so $-1 + j$ (distance 0.5) is inside and $-1 - j$ (distance 2.5) is not.

\text{Res}_{z = -1 + j} = \frac{z + 2}{z - (-1 - j)}\bigg|_{z = -1 + j} = \frac{1 + j}{2j}

\begin{aligned} \text{Res}_{z = -1 + j} &= \frac{z + 2}{z - (-1 - j)}\bigg|_{z = -1 + j} \\ &= \frac{1 + j}{2j} \end{aligned}

\oint_C = 2\pi j \times \frac{1 + j}{2j} = \pi(1 + j)

\begin{aligned} \oint_C &= 2\pi j \times \frac{1 + j}{2j} \\ &= \pi(1 + j) \end{aligned}

Option (B).

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Full question on page 11 of the official paper (PDF)

GATE 2023, Q.15

Engineering Mathematics1 markMCQ

The eigenvalues and eigenvectors of a matrix $B$ are $\{\lambda_k \mid 1 \le k \le n\}$ and $\{\mathbf{v}_k \mid 1 \le k \le n\}$ . For any invertible matrix $P$ , what are the eigenvalues and eigenvectors of the matrix $A$ , where $B = P^{-1} A P$ ?

Option A: $\{\lambda_k \det(A)\}$ and $\{P\mathbf{v}_k\}$
Option B: $\{\lambda_k\}$ and $\{\mathbf{v}_k\}$
Option C: $\{\lambda_k\}$ and $\{P\mathbf{v}_k\}$
Option D: $\{\lambda_k\}$ and $\{P^{-1}\mathbf{v}_k\}$

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Key idea: Similar matrices share eigenvalues; multiply $B\mathbf{v}_k = \lambda_k \mathbf{v}_k$ by $P$ to see the eigenvectors.

Official answer: C

P^{-1} A P \mathbf{v}_k = \lambda_k \mathbf{v}_k \quad\Rightarrow\quad A (P\mathbf{v}_k) = \lambda_k (P\mathbf{v}_k)

\begin{aligned} P^{-1} A P \mathbf{v}_k &= \lambda_k \mathbf{v}_k \\ \Rightarrow\; \\ A (P\mathbf{v}_k) &= \lambda_k (P\mathbf{v}_k) \end{aligned}

So $A$ has the same eigenvalues $\lambda_k$ , with eigenvectors $P\mathbf{v}_k$ . Option (C).

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Full question on page 24 of the official paper (PDF)

GATE 2023, Q.36

Engineering Mathematics2 marksMCQ

A random variable $X$ , distributed normally as $N(0, 1)$ , undergoes the transformation $Y = h(X)$ in the figure. What is the form of the probability density function of $Y$ ? (In the options, $a$ , $b$ , $c$ are non-zero constants and $g(y)$ is a piecewise continuous function.)

Graph of y = h(x): equal to -1 for x below -2, rising linearly to 0 between x = -2 and x = -1, equal to 0 between -1 and 1, rising linearly to 1 between 1 and 2, and equal to 1 for x above 2. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $a\delta(y - 1) + b\delta(y + 1) + g(y)$
Option B: $a\delta(y + 1) + b\delta(y) + c\delta(y - 1) + g(y)$
Option C: $a\delta(y + 2) + b\delta(y) + c\delta(y - 2) + g(y)$
Option D: $a\delta(y + 2) + b\delta(y - 2) + g(y)$

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Key idea: Wherever $h$ is flat over an interval of $x$ , all that probability lands on one value of $y$ : an impulse in the PDF.

Official answer: B

$h$ is flat at three levels: $y = -1$ for $x < -2$ , $y = 0$ for $-1 < x < 1$ and $y = 1$ for $x > 2$ . Each of these ranges has non-zero probability under $N(0, 1)$ , so the PDF has impulses at $y = -1$ , $0$ and $1$ :

P(Y = 0) = P(|X| < 1) \approx 0.683, \qquad P(Y = \pm 1) = P(X > 2) \approx 0.0228

\begin{aligned} P(Y = 0) &= P(|X| < 1) \\ &\approx 0.683, \\ P(Y = \pm 1) &= P(X > 2) \\ &\approx 0.0228 \end{aligned}

The sloped parts map $1 < |x| < 2$ continuously onto $0 < |y| < 1$ , which gives $g(y)$ . Option (B).

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Full question on page 25 of the official paper (PDF)

GATE 2023, Q.37

Engineering Mathematics2 marksMCQ

What is the value of the line integral $\int_P^Q (z^2\, dx + 3y^2\, dy + 2xz\, dz)$ along the straight line joining $P(1, 1, 2)$ and $Q(2, 3, 1)$ ?

Option A: 20
Option B: 24
Option C: 29
Option D: -5

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Key idea: The integrand is an exact differential, so the integral depends only on the end points.

Official answer: B

z^2\, dx + 2xz\, dz = d(xz^2), \qquad 3y^2\, dy = d(y^3)

\begin{aligned} &z^2\, dx + 2xz\, dz \\ &\quad = d(xz^2), \\ 3y^2\, dy &= d(y^3) \end{aligned}

\int_P^Q d(xz^2 + y^3) = (2 \cdot 1 + 27) - (1 \cdot 4 + 1) = 29 - 5 = 24

\begin{aligned} &\int_P^Q d(xz^2 + y^3) \\ &\quad = (2 \cdot 1 + 27) \\ &\qquad - (1 \cdot 4 + 1) \\ &\quad = 29 - 5 \\ &\quad = 24 \end{aligned}

Option (B). Parametrising the line as $(1 + t, 1 + 2t, 2 - t)$ gives the same 24.

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Full question on page 25 of the official paper (PDF)

GATE 2023, Q.38

Engineering Mathematics2 marksMCQ

Let $\mathbf{x}$ be an $n \times 1$ real column vector with length $l = \sqrt{\mathbf{x}^T\mathbf{x}}$ . What is the trace of the matrix $P = \mathbf{x}\mathbf{x}^T$ ?

Option A: $l^2$
Option B: $\dfrac{l^2}{4}$
Option C: $l$
Option D: $\dfrac{l^2}{2}$

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Key idea: The diagonal of $\mathbf{x}\mathbf{x}^T$ holds the squares $x_i^2$ .

Official answer: A

\operatorname{tr}(\mathbf{x}\mathbf{x}^T) = \sum_i x_i^2 = \mathbf{x}^T\mathbf{x} = l^2

\begin{aligned} \operatorname{tr}(\mathbf{x}\mathbf{x}^T) &= \sum_i x_i^2 \\ &= \mathbf{x}^T\mathbf{x} \\ &= l^2 \end{aligned}

Option (A).

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Full question on page 41 of the official paper (PDF)

GATE 2023, Q.55

Engineering Mathematics2 marksNAT

What is the value of $\iint_R xy\, dx\, dy$ over the region $R$ in the figure, the square bounded by $y = x$ , $y = -x$ , $y = x + 2$ and $y = -x + 2$ ?

The region R: a square with corners at the origin, (-1, 1), (0, 2) and (1, 1), bounded by the lines y = x, y = -x, y = x + 2 and y = -x + 2. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: rounded off to the nearest integer

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Key idea: The region is symmetric about the y-axis and $xy$ is odd in $x$ .

Official answer: 0

Replacing $x$ by $-x$ maps $R$ onto itself and changes the sign of $xy$ , so the two halves cancel and the integral is 0.

Directly, for $0 \le y \le 1$ the region spans $-y \le x \le y$ , and for $1 \le y \le 2$ it spans $-(2 - y) \le x \le 2 - y$ ; each inner integral $\int x\, dx$ over a symmetric range is 0. Answer: 0.

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