What is the exam pattern and passing mark for GTU Mathematics - 1 (3110014)?

The GTU theory examination carries 70 total marks with a duration of 2.5 Hours. Minimum passing criteria is 23 marks (33%). The paper consists of 5 main questions with official marking breakdown and internal choices.

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Gujarat Technological UniversitySummer 2025 examination

GTU 3110014 Mathematics - 1 (Maths 1) Summer 2025 Solved Paper & Solutions

B.E. · Common Engineering · Semesters 1 and 2 · Subject code 3110014

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Total: 70 marks
To pass: 23 marks (33%)
Duration: 2.5 Hours 10:30 AM – 1:30 PM
Questions: 5 with internal choice

5 of 25 questions on this paper were also asked in other years. 11 more are types of question GTU sets again with new data. Each is marked below with the years it appeared.See all 35 repeated 3110014 questions, most asked first

Q.1

14 Marks

(a)

3 marksAsked in 2 of 7 papers

Evaluate:

\lim_{x \to 0} [x^{-2} - \csc^2 x]

Also asked in Winter 2024 Q1b

Step 1 Identify Indeterminate Form

We need to evaluate:

L = \lim_{x \to 0} \left[ \frac{1}{x^2} - \csc^2 x \right] = \lim_{x \to 0} \left[ \frac{1}{x^2} - \frac{1}{\sin^2 x} \right]

Direct substitution of $x = 0$ yields the indeterminate form $(\infty - \infty)$ .

Step 2 Combine and Express in Form $\frac{0}{0}$

L = \lim_{x \to 0} \frac{\sin^2 x - x^2}{x^2 \sin^2 x} = \lim_{x \to 0} \left( \frac{\sin^2 x - x^2}{x^4} \cdot \frac{x^2}{\sin^2 x} \right)

Since $\lim_{x \to 0} \frac{\sin x}{x} = 1$ , we have $\lim_{x \to 0} \frac{x^2}{\sin^2 x} = 1$ . The limit simplifies to:

L = \lim_{x \to 0} \frac{\sin^2 x - x^2}{x^4}

Step 3 Apply Taylor Series Expansion

Using the Maclaurin expansion of $\sin x$ :

\sin x = x - \frac{x^3}{6} + \mathcal{O}(x^5)

\sin^2 x = \left(x - \frac{x^3}{6}\right)^2 = x^2 - \frac{x^4}{3} + \mathcal{O}(x^6)

Substitute this into the numerator:

\sin^2 x - x^2 = -\frac{x^4}{3} + \mathcal{O}(x^6)

Step 4 Compute the Final Limit

L = \lim_{x \to 0} \frac{-\frac{1}{3}x^4 + \mathcal{O}(x^6)}{x^4} = -\frac{1}{3}

Final Result

-\frac{1}{3}

(b)

4 marks

Check the consistency of the system of linear equations. Solve it if consistent:

\begin{aligned} 3x + y - 3z &= 13 \\ 2x - 3y + 7z &= 5 \\ 2x + 19y - 47z &= 32 \end{aligned}

Step 1 Write Augmented Matrix $[A \mid B]$

Given the linear system:

\begin{aligned} 3x + y - 3z &= 13 \\ 2x - 3y + 7z &= 5 \\ 2x + 19y - 47z &= 32 \end{aligned}

The augmented matrix is:

[A \mid B] = \begin{pmatrix} 3 & 1 & -3 & \bigm| & 13 \\ 2 & -3 & 7 & \bigm| & 5 \\ 2 & 19 & -47 & \bigm| & 32 \end{pmatrix}

Step 2 Row Echelon Reduction

Apply $R_1 \to R_1 - R_2$ :

\begin{pmatrix} 1 & 4 & -10 & \bigm| & 8 \\ 2 & -3 & 7 & \bigm| & 5 \\ 2 & 19 & -47 & \bigm| & 32 \end{pmatrix}

Apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 2R_1$ :

\begin{pmatrix} 1 & 4 & -10 & \bigm| & 8 \\ 0 & -11 & 27 & \bigm| & -11 \\ 0 & 11 & -27 & \bigm| & 16 - 0 = 16 \end{pmatrix}

(Wait, checking $R_3$ : $32 - 2(8) = 16$ . But let us verify $R_3 + R_2$ :)

R_3 \to R_3 + R_2: \quad 0 \quad 0 \quad 0 \quad \bigm| \quad 16 + (-11) = 5

Notice:

$\text{Rank}(A) = 2$
$\text{Rank}([A \mid B]) = 3$
Since $\text{Rank}(A) \ne \text{Rank}([A \mid B])$ ( $2 \ne 3$ ):

By Rouché-Capelli Theorem, the system is Inconsistent and has No Solution.

(Double checking original equations:)
Multiply (1) by 2 and (2) by 1: $2(3x+y-3z) - (2x-3y+7z) = 4x + 5y - 13z$ .
For row 3: $4(2x-3y+7z) - 3(2x+19y-47z) = 8x - 12y + 28z - 6x - 57y + 141z = 2x - 69y + 169z$ .
In row reduced form: row 3 yields $0x + 0y + 0z = 5$ , a contradiction ( $0 = 5$ ).

Final Result

Since $\text{Rank}(A) = 2$ while $\text{Rank}([A \mid B]) = 3$ , $\text{Rank}(A) < \text{Rank}([A \mid B])$ . Therefore, the system of linear equations is inconsistent and possesses no solution.

(c)

7 marksAsked in 2 of 7 papers

Find the Fourier Series of the function

f(x) = x^2

in the interval

(-\pi, \pi)

Also asked in Summer 2026 Q2c OR

Step 1 Check Parity of Function

$f(x) = x^2$ in the interval $(-\pi, \pi)$ .

f(-x) = (-x)^2 = x^2 = f(x)

Since $f(x)$ is an even function, the Fourier sine coefficients vanish:

b_n = 0 \quad \text{for all } n \ge 1

The Fourier series contains only cosine terms:

f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(nx)

Step 2 Calculate $a_0$

a_0 = \frac{2}{\pi} \int_0^\pi f(x) \, dx = \frac{2}{\pi} \int_0^\pi x^2 \, dx = \frac{2}{\pi} \left[ \frac{x^3}{3} \right]_0^\pi = \frac{2}{\pi} \cdot \frac{\pi^3}{3} = \frac{2\pi^2}{3}

\implies \frac{a_0}{2} = \frac{\pi^2}{3}

Step 3 Calculate $a_n$

a_n = \frac{2}{\pi} \int_0^\pi x^2 \cos(nx) \, dx

Integrating by parts using Bernoulli's rule ( $u = x^2, v = \cos nx$ ):

\begin{aligned} a_n &= \frac{2}{\pi} \left[ x^2 \left(\frac{\sin nx}{n}\right) - 2x \left(-\frac{\cos nx}{n^2}\right) + 2 \left(-\frac{\sin nx}{n^3}\right) \right]_0^\pi \\ &= \frac{2}{\pi} \left[ 0 + 2\pi \frac{\cos(n\pi)}{n^2} - 0 \right] \\ &= \frac{4}{n^2} \cos(n\pi) = \frac{4(-1)^n}{n^2} \end{aligned}

Step 4 Write the Complete Fourier Series

x^2 = \frac{\pi^2}{3} + 4 \sum_{n=1}^\infty \frac{(-1)^n}{n^2} \cos(nx) = \frac{\pi^2}{3} - 4 \left( \cos x - \frac{\cos 2x}{4} + \frac{\cos 3x}{9} - \frac{\cos 4x}{16} + \dots \right)

Final Result

f(x) = \frac{\pi^2}{3} + 4 \sum_{n=1}^\infty \frac{(-1)^n}{n^2} \cos(nx)

Q.2

14 Marks

(a)

3 marks

Define the improper integrals of the first kind and the second kind. State the relation between Beta and Gamma function.

(b)

4 marksSame type asked in 2 of 7 papers

Find the area of the surface of revolution generated by revolving the curve

x = y^3

from

y = 0

y = 2

Also asked in Summer 2024 Q2c

(c)

7 marksSame type asked in 5 of 7 papers

Find eigen values and corresponding eigen vectors of the matrix:

\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}

Also asked in Summer 2026 Q5c OR, Summer 2024 Q5c OR, Winter 2023 Q2c, Summer 2023 Q1c

Step 1 Characteristic Equation

Given matrix $A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}$ .
The characteristic equation is $|A - \lambda I| = 0$ :

\begin{vmatrix} -\lambda & 1 & 1 \\ 1 & -\lambda & 1 \\ 1 & 1 & -\lambda \end{vmatrix} = 0

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

Factoring: $(\lambda - 2)(\lambda + 1)^2 = 0$ .
The eigenvalues are:

\lambda_1 = 2, \quad \lambda_2 = -1, \quad \lambda_3 = -1

Step 2 Eigenvector for $\lambda = 2$

(A - 2I)X = 0 \implies \begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

From $R_1: -2x_1 + x_2 + x_3 = 0$ and $R_2: x_1 - 2x_2 + x_3 = 0$ :
Subtracting gives $-3x_1 + 3x_2 = 0 \implies x_1 = x_2$ .
Substitute into $R_1$ : $-2x_1 + x_1 + x_3 = 0 \implies x_3 = x_1$ .
Choosing $x_1 = 1$ :

X_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}

Step 3 Eigenvectors for $\lambda = -1$ (Algebraic Multiplicity = 2)

(A - (-1)I)X = 0 \implies (A + I)X = 0 \implies \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

This reduces to a single equation:

x_1 + x_2 + x_3 = 0 \implies x_1 = -x_2 - x_3

Since there are 2 free variables, we choose linearly independent vectors:

Set $x_2 = 1, x_3 = 0 \implies x_1 = -1$ :

X_2 = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}

Set $x_2 = 0, x_3 = 1 \implies x_1 = -1$ :

X_3 = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}

Final Result

Eigenvalues: $\lambda = 2, -1, -1$
Eigenvectors: For $\lambda = 2$ : $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ ; For $\lambda = -1$ : $\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$ .

(c)

7 marksSame type asked in 3 of 7 papers

Find the inverse of the matrix using Gauss-Jordan elimination method:

A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 0 \\ 0 & 1 & 2 \end{pmatrix}

Also asked in Winter 2024 Q5b OR, Summer 2024 Q5b OR

Step 1 Set up Augmented Matrix $[A \mid I]$

A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 0 \\ 0 & 1 & 2 \end{pmatrix}

[A \mid I] = \begin{pmatrix} 1 & 2 & 3 & \bigm| & 1 & 0 & 0 \\ 2 & 3 & 0 & \bigm| & 0 & 1 & 0 \\ 0 & 1 & 2 & \bigm| & 0 & 0 & 1 \end{pmatrix}

Step 2 Perform Elementary Row Operations

$R_2 \to R_2 - 2R_1$ :

\begin{pmatrix} 1 & 2 & 3 & \bigm| & 1 & 0 & 0 \\ 0 & -1 & -6 & \bigm| & -2 & 1 & 0 \\ 0 & 1 & 2 & \bigm| & 0 & 0 & 1 \end{pmatrix}

$R_2 \to -R_2$ :

\begin{pmatrix} 1 & 2 & 3 & \bigm| & 1 & 0 & 0 \\ 0 & 1 & 6 & \bigm| & 2 & -1 & 0 \\ 0 & 1 & 2 & \bigm| & 0 & 0 & 1 \end{pmatrix}

$R_1 \to R_1 - 2R_2$ and $R_3 \to R_3 - R_2$ :

\begin{pmatrix} 1 & 0 & -9 & \bigm| & -3 & 2 & 0 \\ 0 & 1 & 6 & \bigm| & 2 & -1 & 0 \\ 0 & 0 & -4 & \bigm| & -2 & 1 & 1 \end{pmatrix}

$R_3 \to -\frac{1}{4}R_3$ :

\begin{pmatrix} 1 & 0 & -9 & \bigm| & -3 & 2 & 0 \\ 0 & 1 & 6 & \bigm| & 2 & -1 & 0 \\ 0 & 0 & 1 & \bigm| & 1/2 & -1/4 & -1/4 \end{pmatrix}

$R_1 \to R_1 + 9R_3$ and $R_2 \to R_2 - 6R_3$ :

For $R_1$ : $[-3 + 9/2, \ 2 - 9/4, \ 0 - 9/4] = [3/2, \ -1/4, \ -9/4]$
For $R_2$ : $[2 - 3, \ -1 + 6/4, \ 0 + 6/4] = [-1, \ 1/2, \ 3/2]$

\begin{pmatrix} 1 & 0 & 0 & \bigm| & 3/2 & -1/4 & -9/4 \\ 0 & 1 & 0 & \bigm| & -1 & 1/2 & 3/2 \\ 0 & 0 & 1 & \bigm| & 1/2 & -1/4 & -1/4 \end{pmatrix}

Final Result

A^{-1} = \frac{1}{4} \begin{pmatrix} 6 & -1 & -9 \\ -4 & 2 & 6 \\ 2 & -1 & -1 \end{pmatrix} = \begin{pmatrix} 1.5 & -0.25 & -2.25 \\ -1 & 0.5 & 1.5 \\ 0.5 & -0.25 & -0.25 \end{pmatrix}

Q.3

14 Marks

(a)

3 marks

Determine whether

\lim_{(x, y) \to (0, 0)} \frac{x^2 - y^2}{x^2 + y^2}

exists and find it if exists.

(b)

4 marksSame type asked in 5 of 7 papers

Find the equation of tangent plane and normal line to the surface

2xz^2 - 3xy - 4x = 7

at the point

(1, -1, 2)

Also asked in Winter 2025 Q3b, Winter 2024 Q1a, Winter 2023 Q3b OR, Summer 2023 Q2b

Step 1 Define the Implicit Surface

F(x, y, z) = 2xz^2 - 3xy - 4x - 7 = 0

The given point of tangency is $P(1, -1, 2)$ .

Step 2 Compute Gradient Vector $\nabla F$

\nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)

$\frac{\partial F}{\partial x} = 2z^2 - 3y - 4$
$\frac{\partial F}{\partial y} = -3x$
$\frac{\partial F}{\partial z} = 4xz$

Step 3 Evaluate Gradient at $P(1, -1, 2)$

$F_x(1, -1, 2) = 2(2^2) - 3(-1) - 4 = 8 + 3 - 4 = 7$
$F_y(1, -1, 2) = -3(1) = -3$
$F_z(1, -1, 2) = 4(1)(2) = 8$

The normal vector is $\mathbf{n} = \langle 7, -3, 8 \rangle$ .

Step 4 Equation of the Tangent Plane

F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0

7(x - 1) - 3(y - (-1)) + 8(z - 2) = 0

7x - 7 - 3y - 3 + 8z - 16 = 0 \implies 7x - 3y + 8z = 26

Step 5 Equation of the Normal Line

\frac{x - 1}{7} = \frac{y + 1}{-3} = \frac{z - 2}{8}

Final Result

Tangent Plane: $7x - 3y + 8z = 26$
Normal Line: $\frac{x - 1}{7} = \frac{y + 1}{-3} = \frac{z - 2}{8}$

(c)

7 marksSame type asked in 7 of 7 papers

Find the extreme values of the function

x^3 + y^3 - 63(x + y) + 12xy

Also asked in Summer 2026 Q3b OR, Winter 2025 Q3c OR, Winter 2024 Q3c OR, Summer 2024 Q4c, Winter 2023 Q3c, Summer 2023 Q4c

Step 1 Find Critical Points

f(x, y) = x^3 + y^3 - 63(x + y) + 12xy

Compute first partial derivatives:

f_x = 3x^2 - 63 + 12y = 0 \implies x^2 + 4y = 21 \quad \text{--- (1)}

f_y = 3y^2 - 63 + 12x = 0 \implies y^2 + 4x = 21 \quad \text{--- (2)}

Subtract (2) from (1):

(x^2 - y^2) - 4(x - y) = 0 \implies (x - y)(x + y - 4) = 0

Case 1: $x = y$

Substitute into (1): $x^2 + 4x - 21 = 0 \implies (x + 7)(x - 3) = 0 \implies x = 3, -7$ .
Critical points: $(3, 3)$ and $(-7, -7)$ .

Case 2: $x + y = 4 \implies y = 4 - x$

Substitute into (1): $x^2 + 4(4 - x) = 21 \implies x^2 - 4x - 5 = 0 \implies (x - 5)(x + 1) = 0$ .

If $x = 5 \implies y = -1 \implies (5, -1)$ .
If $x = -1 \implies y = 5 \implies (-1, 5)$ .

Step 2 Second Derivative Test

r = f_{xx} = 6x, \quad s = f_{xy} = 12, \quad t = f_{yy} = 6y

D = rt - s^2 = 36xy - 144

At $(3, 3)$ :

$D = 36(9) - 144 = 324 - 144 = 180 > 0$
$r = 6(3) = 18 > 0$
$\implies$ Local Minimum at $(3, 3)$ .
$f(3, 3) = 27 + 27 - 63(6) + 12(9) = 54 - 378 + 108 = -216$ .

At $(-7, -7)$ :

$D = 36(49) - 144 = 1764 - 144 = 1620 > 0$
$r = 6(-7) = -42 < 0$
$\implies$ Local Maximum at $(-7, -7)$ .
$f(-7, -7) = -343 - 343 - 63(-14) + 12(49) = -686 + 882 + 588 = 784$ .

At $(5, -1)$ and $(-1, 5)$ :

$D = 36(-5) - 144 = -180 - 144 = -324 < 0$
$\implies$ Saddle Points at $(5, -1)$ and $(-1, 5)$ .

Final Result

Local Maximum: Value $784$ at point $(-7, -7)$ .
Local Minimum: Value $-216$ at point $(3, 3)$ .
Saddle Points: At $(5, -1)$ and $(-1, 5)$ .

(a)

3 marksSame type asked in 2 of 7 papers

State chain rule for

\frac{\partial u}{\partial x}

for

u = f(v, w)

, where

v = g(x, y), w = h(x, y)

. If

u = f(x - 2y, 2y - 3z, 3z - x)

then show that

\frac{\partial u}{\partial x} + \frac{1}{2}\frac{\partial u}{\partial y} + \frac{1}{3}\frac{\partial u}{\partial z} = 0

Also asked in Winter 2025 Q3a

Chain Rule for Partial Derivatives

Let $u = f(v, w)$ be a differentiable function of two intermediate variables $v$ and $w$ , where $v = v(x, y)$ and $w = w(x, y)$ are themselves differentiable functions of independent variables $x$ and $y$ .

The partial derivative of $u$ with respect to $x$ is given by:

\mathbf{\frac{\partial u}{\partial x} = \frac{\partial u}{\partial v} \frac{\partial v}{\partial x} + \frac{\partial u}{\partial w} \frac{\partial w}{\partial x}}

Dependency Tree Diagram:

ALGORITHM

u
       /   \
      v     w
     / \   / \
    x   y x   y

To find $\frac{\partial u}{\partial x}$ , sum the products of partial derivatives along all branches leading from $u$ to $x$ :

Branch 1 (via $v$ ): $\frac{\partial u}{\partial v} \cdot \frac{\partial v}{\partial x}$
Branch 2 (via $w$ ): $\frac{\partial u}{\partial w} \cdot \frac{\partial w}{\partial x}$

(b)

4 marksSame type asked in 5 of 7 papers

Find the directional derivative of

f(x, y, z) = xy^2 + yz^2

at the point

(2, -1, 1)

in the direction of the vector

\hat{i} + 2\hat{j} + 2\hat{k}

Also asked in Summer 2026 Q3b, Winter 2024 Q2a, Summer 2024 Q4a OR, Winter 2023 Q3b

Step 1 Formula for Directional Derivative

The directional derivative of $f(x, y, z)$ at point $P$ along unit vector $\hat{\mathbf{u}}$ is:

D_{\mathbf{u}} f = \nabla f \cdot \hat{\mathbf{u}}

Step 2 Find Gradient $\nabla f$

Given $f(x, y, z) = xy^2 + yz^2$ :

\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle = \langle y^2, \ 2xy + z^2, \ 2yz \rangle

At point $P(2, -1, 1)$ :

$f_x(2, -1, 1) = (-1)^2 = 1$
$f_y(2, -1, 1) = 2(2)(-1) + (1)^2 = -4 + 1 = -3$
$f_z(2, -1, 1) = 2(-1)(1) = -2$

\nabla f(2, -1, 1) = \langle 1, -3, -2 \rangle

Step 3 Find Unit Vector $\hat{\mathbf{u}}$

Given direction vector $\mathbf{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ :

|\mathbf{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3

\hat{\mathbf{u}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{1}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k}

Step 4 Compute Dot Product

D_{\mathbf{u}} f = \nabla f \cdot \hat{\mathbf{u}} = (1)\left(\frac{1}{3}\right) + (-3)\left(\frac{2}{3}\right) + (-2)\left(\frac{2}{3}\right) = \frac{1 - 6 - 4}{3} = -\frac{9}{3} = -3

Final Result

The directional derivative is $-3$ .

(c)

7 marksSame type asked in 4 of 7 papers

The temperature

T(x, y, z)

at any point in space is

T = 400xyz^2

. Find the highest temperature on surface of the sphere

x^2 + y^2 + z^2 = 1

Also asked in Summer 2026 Q3c, Winter 2025 Q3c, Winter 2023 Q3c OR

Step 1 Lagrange Multipliers Setup

We wish to maximize temperature $T = 400xyz^2$ subject to constraint $g(x, y, z) = x^2 + y^2 + z^2 - 1 = 0$ .
By Lagrange's method: $\nabla T = \lambda \nabla g$ :

$400yz^2 = 2\lambda x \implies 200yz^2 = \lambda x$
$400xz^2 = 2\lambda y \implies 200xz^2 = \lambda y$
$800xyz = 2\lambda z \implies 400xyz = \lambda z$

Step 2 Solve the System for Non-zero Coordinates

For maximum temperature $T > 0$ , we require $x \ne 0, y \ne 0, z \ne 0$ .
Multiply (1) by $x$ and (2) by $y$ :

200xyz^2 = \lambda x^2 = \lambda y^2 \implies x^2 = y^2

From (3), since $z \ne 0$ , divide by $z$ :

400xy = \lambda

Multiply by $z^2$ :

\lambda z^2 = 400xyz^2 = 2(200xyz^2) = 2(\lambda x^2) \implies z^2 = 2x^2

Step 3 Substitute into Constraint $x^2 + y^2 + z^2 = 1$

x^2 + x^2 + 2x^2 = 1 \implies 4x^2 = 1 \implies x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2}

Then:

$y^2 = x^2 = \frac{1}{4} \implies y = \pm \frac{1}{2}$
$z^2 = 2x^2 = 2\left(\frac{1}{4}\right) = \frac{1}{2} \implies z = \pm \frac{1}{\sqrt{2}}$

Step 4 Calculate Highest Temperature

To maximize $T = 400xyz^2$ , choose $x$ and $y$ with identical signs ( $x = 1/2, y = 1/2$ ):

T_{\max} = 400 \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = 400 \left(\frac{1}{8}\right) = 50

Final Result

The highest temperature on the surface of the sphere is $50$ units (occurring at points $(\pm 1/2, \pm 1/2, \pm 1/\sqrt{2})$ with $xy > 0$ ).

Q.4

14 Marks

(a)

3 marks

Evaluate

\int_{-1}^1 \int_0^2 (1 - 6x^2 y) \, dx \, dy

(b)

4 marks

Find the value of

\iint_R (2x - y^2) \, dA

over the triangular region

R

enclosed between the lines

y = -x + 1

y = x + 1

and

y = 3

Step 1 Find Vertices of the Triangular Region

The region $R$ is bounded by:

$y = -x + 1 \implies x = 1 - y$
$y = x + 1 \implies x = y - 1$
$y = 3$

Intersection points:

Lines 1 & 2: $-x + 1 = x + 1 \implies x = 0, y = 1 \implies (0, 1)$ .
Lines 1 & 3: $y = 3 \implies 3 = -x + 1 \implies x = -2 \implies (-2, 3)$ .
Lines 2 & 3: $y = 3 \implies 3 = x + 1 \implies x = 2 \implies (2, 3)$ .

Step 2 Set up Horizontal Strips

For a fixed $y \in [1, 3]$ , $x$ ranges from the left boundary to the right boundary:

x \in [1 - y, \ y - 1]

I = \int_1^3 \int_{1-y}^{y-1} (2x - y^2) \, dx \, dy

Step 3 Evaluate Inner Integral

\int_{1-y}^{y-1} (2x - y^2) \, dx = \left[ x^2 - y^2 x \right]_{1-y}^{y-1}

Notice that $1 - y = -(y - 1)$ .

Since $x^2$ is an even function: $(y - 1)^2 - [-(y - 1)]^2 = 0$ .
For the $-y^2 x$ term: $-y^2 [(y - 1) - (1 - y)] = -y^2 [2(y - 1)] = -2y^3 + 2y^2$ .

Step 4 Evaluate Outer Integral

I = \int_1^3 (-2y^3 + 2y^2) \, dy = \left[ -\frac{y^4}{2} + \frac{2y^3}{3} \right]_1^3

I = \left( -\frac{81}{2} + \frac{54}{3} \right) - \left( -\frac{1}{2} + \frac{2}{3} \right) = \left( -\frac{81}{2} + 18 \right) - \left( \frac{1}{6} \right) = -\frac{45}{2} - \frac{1}{6} = -\frac{136}{6} = -\frac{68}{3}

Final Result

-\frac{68}{3}

(c)

7 marksSame type asked in 7 of 7 papers

Change the order of integral and evaluate

\int_0^{4a} \int_{x^2/4a}^{2\sqrt{ax}} dy \, dx, \quad (a > 0)

Also asked in Summer 2026 Q4b, Winter 2025 Q4c, Winter 2024 Q4c OR, Summer 2024 Q4c OR, Winter 2023 Q4b, Summer 2023 Q5c

Step 1 Identify the Region of Integration

Given integral:

I = \int_0^{4a} \int_{x^2/4a}^{2\sqrt{ax}} dy \, dx

Region $R$ is bounded by the parabolas:

Bottom curve: $y = \frac{x^2}{4a} \implies x^2 = 4ay$
Top curve: $y = 2\sqrt{ax} \implies y^2 = 4ax$
Points of intersection: $(x^2/4a)^2 = 4ax \implies x^4 = 64a^3 x \implies x = 0$ and $x = 4a$ .

Corresponding $y$ values: $y = 0$ to $y = 4a$ .

Step 2 Change the Order of Integration

Integrating with horizontal strips first:

$y$ varies from $0$ to $4a$ .
For a given $y$ , $x$ varies from the parabola $y^2 = 4ax \implies x = \frac{y^2}{4a}$ to the parabola $x^2 = 4ay \implies x = 2\sqrt{ay}$ .

I = \int_0^{4a} \int_{y^2/4a}^{2\sqrt{ay}} dx \, dy

Step 3 Evaluate the Integral

I = \int_0^{4a} [x]_{y^2/4a}^{2\sqrt{ay}} \, dy = \int_0^{4a} \left( 2\sqrt{a} y^{1/2} - \frac{y^2}{4a} \right) \, dy

I = \left[ 2\sqrt{a} \left(\frac{2}{3} y^{3/2}\right) - \frac{y^3}{12a} \right]_0^{4a}

Substitute $y = 4a$ :

$y^{3/2} = (4a)^{3/2} = 8a\sqrt{a} \implies 2\sqrt{a} \cdot \frac{2}{3} (8a\sqrt{a}) = \frac{32a^2}{3}$
$\frac{y^3}{12a} = \frac{64a^3}{12a} = \frac{16a^2}{3}$

I = \frac{32a^2}{3} - \frac{16a^2}{3} = \frac{16a^2}{3}

Final Result

\frac{16a^2}{3}

(a)

3 marksAsked in 2 of 7 papers

Calculate

\int_0^2 \int_1^z \int_0^{yz} xyz \, dx \, dy \, dz

Also asked in Winter 2024 Q4a OR

(b)

4 marksSame type asked in 7 of 7 papers

Evaluate

\int_0^1 \int_x^1 \sin y^2 \, dy \, dx

Also asked in Summer 2026 Q4b, Winter 2025 Q4c, Winter 2024 Q4c OR, Summer 2024 Q4c OR, Winter 2023 Q4b, Summer 2023 Q5c

(c)

7 marksAsked in 2 of 7 papers

Compute

\int_0^2 \int_0^{\sqrt{2x-x^2}} \frac{x}{x^2+y^2} \, dy \, dx

by transforming into polar coordinates.

Also asked in Summer 2024 Q5c

Step 1 Identify the Region in Cartesian Coordinates

I = \int_0^2 \int_0^{\sqrt{2x-x^2}} \frac{x}{x^2+y^2} \, dy \, dx

The curve $y = \sqrt{2x - x^2} \implies y^2 = 2x - x^2 \implies x^2 + y^2 = 2x$ , which is a circle of radius $1$ centered at $(1, 0)$ in the upper half plane ( $y \ge 0$ ).

Step 2 Convert to Polar Coordinates

$x = r \cos\theta, \quad y = r \sin\theta, \quad x^2 + y^2 = r^2, \quad dy \, dx = r \, dr \, d\theta$
Circle equation: $r^2 = 2r \cos\theta \implies r = 2\cos\theta$ .
Since $y \ge 0$ and $x \in [0, 2]$ , $\theta$ sweeps from $0$ to $\frac{\pi}{2}$ .
For a given $\theta$ , $r$ goes from $0$ to $2\cos\theta$ .

Step 3 Transform and Evaluate the Integral

I = \int_0^{\pi/2} \int_0^{2\cos\theta} \frac{r \cos\theta}{r^2} \cdot r \, dr \, d\theta = \int_0^{\pi/2} \int_0^{2\cos\theta} \cos\theta \, dr \, d\theta

\int_0^{2\cos\theta} \cos\theta \, dr = \cos\theta [r]_0^{2\cos\theta} = 2\cos^2\theta

I = \int_0^{\pi/2} 2\cos^2\theta \, d\theta = \int_0^{\pi/2} (1 + \cos 2\theta) \, d\theta = \left[ \theta + \frac{\sin 2\theta}{2} \right]_0^{\pi/2} = \frac{\pi}{2}

Final Result

\frac{\pi}{2}

Q.5

14 Marks

(a)

3 marks

Define Monotonic sequence. Test the convergence of the sequence

\{2 - (-1)^n\}

(b)

4 marksSame type asked in 4 of 7 papers

Express the function

f(x) = \log(1 + x)

in power series using the formula of Maclaurin’s series.

Also asked in Summer 2026 Q5b OR, Winter 2025 Q5b, Winter 2023 Q5a

Step 1 Formula of Maclaurin's Series

f(x) = f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f'''(0) + \dots + \frac{x^n}{n!} f^{(n)}(0) + \dots

Step 2 Compute Successive Derivatives of $f(x) = \log(1 + x)$

$f(x) = \log(1 + x) \implies f(0) = \log 1 = 0$
$f'(x) = \frac{1}{1 + x} \implies f'(0) = 1$
$f''(x) = -\frac{1}{(1 + x)^2} \implies f''(0) = -1$
$f'''(x) = \frac{2}{(1 + x)^3} \implies f'''(0) = 2$
$f^{(4)}(x) = -\frac{6}{(1 + x)^4} \implies f^{(4)}(0) = -6$
In general, $f^{(n)}(0) = (-1)^{n-1} (n - 1)!$

Step 3 Form the Maclaurin Series

f(x) = 0 + x(1) + \frac{x^2}{2!}(-1) + \frac{x^3}{3!}(2) + \frac{x^4}{4!}(-6) + \dots

\log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots = \sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n} \quad \text{for } -1 < x \le 1

Final Result

\log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots

(c)

7 marks

iTest the convergence of

\sum_{n=1}^\infty \frac{n!}{(2n+1)!}

iiState Cauchy’s root test and discuss the convergence of

\sum_{n=1}^\infty \frac{1}{(\log n)^n}

Part (i): Test Convergence of $\sum_{n=1}^\infty \frac{n!}{(2n+1)!}$

Let $u_n = \frac{n!}{(2n+1)!}$
Then $u_{n+1} = \frac{(n+1)!}{(2(n+1)+1)!} = \frac{(n+1)!}{(2n+3)!}$
Apply D'Alembert's Ratio Test:

\lim_{n \to \infty} \frac{u_n}{u_{n+1}} = \lim_{n \to \infty} \frac{n!}{(2n+1)!} \cdot \frac{(2n+3)(2n+2)(2n+1)!}{(n+1) n!} = \lim_{n \to \infty} \frac{(2n+3)(2n+2)}{n+1}

= \lim_{n \to \infty} 2(2n + 3) = \infty > 1

By D'Alembert's Ratio Test, since $\lim \frac{u_n}{u_{n+1}} = \infty > 1$ , the series is convergent.

Part (ii): Cauchy's Root Test and $\sum_{n=1}^\infty \frac{1}{(\log n)^n}$

Statement of Cauchy's Root Test: For a positive-term series $\sum u_n$ , let $L = \lim_{n \to \infty} (u_n)^{1/n}$ .
If $L < 1$ , the series converges.
If $L > 1$ , the series diverges.
If $L = 1$ , the test is inconclusive.
Application:
$u_n = \frac{1}{(\log n)^n}$
Compute $(u_n)^{1/n}$ :

(u_n)^{1/n} = \left( \frac{1}{(\log n)^n} \right)^{1/n} = \frac{1}{\log n}

Take limit as $n \to \infty$ :

L = \lim_{n \to \infty} \frac{1}{\log n} = 0

Since $L = 0 < 1$ , by Cauchy's Root Test, the series $\sum_{n=1}^\infty \frac{1}{(\log n)^n}$ is convergent.

Final Result

(i) Convergent by D'Alembert's Ratio Test. (ii) Convergent by Cauchy's Root Test ( $L = 0 < 1$ ).

(a)

3 marksAsked in 3 of 7 papers

State sandwich theorem for sequence. Show that the sequence

u_n = \frac{\sin n}{n}

converges to zero.

Also asked in Winter 2024 Q3a, Summer 2024 Q2a

(b)

4 marks

Using Taylor’s theorem find the approximate value of

\sqrt{10}

up to three decimal places.

Step 1 Taylor's Expansion Formula

f(a + h) = f(a) + h f'(a) + \frac{h^2}{2!} f''(a) + \frac{h^3}{3!} f'''(a) + \dots

Step 2 Choose Function and Base Point

We want to evaluate $\sqrt{10}$ .

Let $f(x) = \sqrt{x}$ , base point $a = 9$ , and step $h = 1$ (since $9 + 1 = 10$ ).

Step 3 Compute Derivatives at $a = 9$

$f(9) = \sqrt{9} = 3$
$f'(x) = \frac{1}{2\sqrt{x}} \implies f'(9) = \frac{1}{2(3)} = \frac{1}{6} \approx 0.16667$
$f''(x) = -\frac{1}{4x^{3/2}} \implies f''(9) = -\frac{1}{4(27)} = -\frac{1}{108} \approx -0.00926$
$f'''(x) = \frac{3}{8x^{5/2}} \implies f'''(9) = \frac{3}{8(243)} = \frac{1}{648} \approx 0.00154$

Step 4 Substitute into Taylor's Series

\begin{aligned} \sqrt{10} &= f(9) + (1) f'(9) + \frac{1}{2} f''(9) + \frac{1}{6} f'''(9) \\ &= 3 + \frac{1}{6} - \frac{1}{216} + \frac{1}{3888} \\ &= 3 + 0.16667 - 0.00463 + 0.00026 \\ &= 3.1623 \end{aligned}

Rounding to 3 decimal places:

\sqrt{10} \approx 3.162

Final Result

\sqrt{10} \approx 3.162

(c)

7 marks

iExamine the convergence of the series

\frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} - \frac{1}{5 \cdot 6} + \frac{1}{7 \cdot 8} - \dots

iiState Cauchy’s Integral test for convergence of series and test the convergence of the series

\sum_{n=1}^\infty n^2 e^{-n^3}

Part (i): Examine Convergence of the Alternating Series

S = \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} - \dots = \sum_{n=1}^\infty (-1)^{n-1} \frac{1}{n(n+1)}

Let $u_n = \frac{1}{n(n+1)}$
Absolute Convergence Test:

\sum |a_n| = \sum_{n=1}^\infty \frac{1}{n(n+1)} < \sum_{n=1}^\infty \frac{1}{n^2}

Since $\sum \frac{1}{n^2}$ is a $p$ -series with $p = 2 > 1$ , it converges.
Therefore, the alternating series converges absolutely (and is hence convergent).

Part (ii): Cauchy's Integral Test and $\sum_{n=1}^\infty n^2 e^{-n^3}$

Statement of Cauchy's Integral Test: If $f(x)$ is a positive, continuous, and monotonically decreasing function for $x \ge 1$ such that $f(n) = u_n$ , then the series $\sum u_n$ and the improper integral $\int_1^\infty f(x) \, dx$ either both converge or both diverge.
Evaluation:

I = \int_1^\infty x^2 e^{-x^3} \, dx

Substitute $t = x^3 \implies dt = 3x^2 \, dx \implies x^2 \, dx = \frac{dt}{3}$ :

I = \frac{1}{3} \int_1^\infty e^{-t} \, dt = \frac{1}{3} [-e^{-t}]_1^\infty = \frac{1}{3} (0 - (-e^{-1})) = \frac{1}{3e}

Since the integral converges to a finite positive value ( $\frac{1}{3e}$ ), by Cauchy's Integral Test, the series $\sum_{n=1}^\infty n^2 e^{-n^3}$ is convergent.

Final Result

(i) The series is absolutely convergent. (ii) Convergent by Cauchy's Integral Test ( $I = \frac{1}{3e}$ ).

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Official Gujarat Technological University (GTU) examination paper and step-by-step solutions for Mathematics - 1 (Maths 1) (Summer 2025, B.E. · Common Engineering, Sem 1). Features complete 70-mark regular & remedial examination pattern, official marking distribution across all 5 questions, and direct 1-click official PDF download.

Transcribed for student exam preparation from Gujarat Technological University official examination archives. Questions, syllabus guidelines, and curriculum marking schemes remain the intellectual property of Gujarat Technological University.

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GTU 3110014 Mathematics - 1 (Maths 1) Summer 2025 Solved Paper & Solutions

Q.1

Q.2

1. Improper Integral of the First Kind

2. Improper Integral of the Second Kind

3. Comparison Test for Convergence

Q.3

Chain Rule for Partial Derivatives

Dependency Tree Diagram:

Q.4

Q.5

1. Definition of Monotonic Sequence

2. Convergence Test for Sequence $a_n = 2 - \frac{1}{2^n}$

1. Statement of Sandwich (Squeeze) Theorem

2. Convergence of $u_n = \frac{\cos n}{n}$

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About this paper

GTU 3110014 Mathematics - 1 (Maths 1) Summer 2025 Solved Paper & Solutions

1. Improper Integral of the First Kind

2. Improper Integral of the Second Kind

3. Comparison Test for Convergence

Chain Rule for Partial Derivatives

Dependency Tree Diagram:

1. Definition of Monotonic Sequence

2. Convergence Test for Sequence an=2−12na_n = 2 - \frac{1}{2^n}an​=2−2n1​

1. Statement of Sandwich (Squeeze) Theorem

2. Convergence of un=cos⁡nnu_n = \frac{\cos n}{n}un​=ncosn​

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2. Convergence Test for Sequence $a_n = 2 - \frac{1}{2^n}$

2. Convergence of $u_n = \frac{\cos n}{n}$