What is the exam pattern and passing mark for GTU Mathematics-I (BE01000041)?

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 2026 examination

GTU BE01000041 Mathematics-I Summer 2026 Solved Paper & Solutions

B.E. · Common Engineering · Semester 1 · Subject code BE01000041GTU also lists this paper as BE01R00041

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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

Q.1

14 Marks

(a)

3 marks

Check the convergence of

\int_0^\infty \frac{1}{x^2 + 1}\,dx

(b)

4 marks

Define gamma function and using it evaluate the integral

\int_0^\infty x^4 e^{-x}\,dx

(c)

7 marks

The circle

x^2 + y^2 = r^2

is rotated about the x-axis to generate a sphere. Find the volume of the sphere.

The upper half of the circle $x^2 + y^2 = r^2$ is the curve $y = \sqrt{r^2 - x^2}$ , from $x = -r$ to $x = r$ . Rotating it about the x-axis gives a sphere of radius $r$ .

DIAGRAM

           y
           |   y = sqrt(r^2 - x^2)
         . | .
      .    |    .
    .      | y    .
 ---+------+------+--- x
   -r      0      r

Step 1 Volume of one thin slice

Cut the sphere by a plane at a distance $x$ from the centre. The slice is a thin disc of radius $y$ and thickness $dx$ .

dV = \pi y^2\,dx

Step 2 Write the integral

Add all slices from $x = -r$ to $x = r$ . This is the disc method.

V = \int_{-r}^{r} \pi y^2\,dx

From the circle, $y^2 = r^2 - x^2$ .

V = \pi \int_{-r}^{r} (r^2 - x^2)\,dx

Step 3 Use symmetry

The function $r^2 - x^2$ is even, so the integral from $-r$ to $r$ is twice the integral from $0$ to $r$ .

V = 2\pi \int_0^r (r^2 - x^2)\,dx

Step 4 Integrate

\begin{aligned} V &= 2\pi \left[r^2 x - \frac{x^3}{3}\right]_0^r \\ &= 2\pi \left(r^3 - \frac{r^3}{3}\right) \\ &= 2\pi \cdot \frac{2r^3}{3} \\ &= \frac{4}{3}\pi r^3 \end{aligned}

Answer

The volume of the sphere is $V = \dfrac{4}{3}\pi r^3$ cubic units.

Q.2

14 Marks

(a)

3 marks

Evaluate

\lim\limits_{x \to 0} \dfrac{x - \sin x}{x^3}

using L'Hospital rule.

(b)

4 marks

Find the Taylor series generated by

f(x) = \cos x

x = 0

$n$	$f^{(n)}(x)$	$f^{(n)}(0)$
0	$\cos x$	1
1	$-\sin x$	0
2	$-\cos x$	$-1$
3	$\sin x$	0
4	$\cos x$	1

(c)

7 marks

Find the Maclaurin series of the function

f(x) = (1 + x)^{\alpha}

where

\alpha

is any real number.

The Maclaurin series of $f(x)$ is its Taylor series at $x = 0$ :

f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \dots

Step 1 Find the derivatives

Here $f(x) = (1 + x)^{\alpha}$ . Each time we differentiate, the power comes down as a factor and reduces by 1.

\begin{aligned} f'(x) &= \alpha(1+x)^{\alpha-1} \\ f''(x) &= \alpha(\alpha-1)(1+x)^{\alpha-2} \\ f'''(x) &= \alpha(\alpha-1)(\alpha-2)(1+x)^{\alpha-3} \end{aligned}

In general, the $n$ th derivative has $n$ factors:

f^{(n)}(x) = \alpha(\alpha-1)\cdots(\alpha-n+1)\,(1+x)^{\alpha-n}

Step 2 Put x = 0

Since $(1 + 0)^{\alpha-n} = 1$ ,

\begin{aligned} f(0) &= 1 \\ f'(0) &= \alpha \\ f''(0) &= \alpha(\alpha-1) \\ f'''(0) &= \alpha(\alpha-1)(\alpha-2) \\ f^{(n)}(0) &= \alpha(\alpha-1)\cdots(\alpha-n+1) \end{aligned}

Step 3 Substitute in the series

\begin{aligned} (1+x)^{\alpha} = 1 &+ \alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 \\ &+ \frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3 + \dots \end{aligned}

The coefficient of $x^n$ is

\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}

Step 4 Where the series is valid

This series is called the binomial series. It converges for $\lvert x \rvert < 1$ .
If $\alpha$ is a positive integer, the factor $(\alpha - n + 1)$ becomes zero after $n = \alpha$ . Then the series stops and gives the usual binomial theorem.

Check with an example
For $\alpha = \dfrac{1}{2}$ the series gives

\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \dots

Answer

$\displaystyle(1+x)^{\alpha} = 1 + \sum_{n=1}^{\infty} \frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}\,x^n$ for $\lvert x \rvert < 1$

(c)

7 marks

Find the local extreme values of the function

f(x) = 2x^3 + 3x^2 - 12x

A local extreme value is a local maximum or a local minimum of the function. It can occur only at a critical point, where $f'(x) = 0$ .

Step 1 Find the first derivative

\begin{aligned} f(x) &= 2x^3 + 3x^2 - 12x \\ f'(x) &= 6x^2 + 6x - 12 \end{aligned}

Step 2 Find the critical points

Put $f'(x) = 0$ .

\begin{aligned} 6x^2 + 6x - 12 &= 0 \\ x^2 + x - 2 &= 0 \\ (x + 2)(x - 1) &= 0 \end{aligned}

So $x = -2$ or $x = 1$ .

Step 3 Apply the second derivative test

The test says: at a critical point, if $f''(x) < 0$ the point is a local maximum, and if $f''(x) > 0$ it is a local minimum.

f''(x) = 12x + 6

At $x = -2$ :

f''(-2) = 12(-2) + 6 = -18 < 0

So $f$ has a local maximum at $x = -2$ .
At $x = 1$ :

f''(1) = 12(1) + 6 = 18 > 0

So $f$ has a local minimum at $x = 1$ .

Step 4 Find the extreme values

\begin{aligned} f(-2) &= 2(-2)^3 + 3(-2)^2 - 12(-2) \\ &= -16 + 12 + 24 = 20 \end{aligned}

\begin{aligned} f(1) &= 2(1)^3 + 3(1)^2 - 12(1) \\ &= 2 + 3 - 12 = -7 \end{aligned}

Point	$f''(x)$	Nature	Value
$x = -2$	$-18$	Local maximum	20
$x = 1$	$18$	Local minimum	$-7$

Answer

The local maximum value is $20$ at $x = -2$ , and the local minimum value is $-7$ at $x = 1$ .

Q.3

14 Marks

(a)

3 marks

Test the convergence of the series

\sum_{n=1}^{\infty} \frac{-n}{2n + 5}

(b)

4 marks

Test the convergence of the series

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

(c)

7 marks

Check whether the series

\sum\limits_{n=1}^{\infty} \dfrac{\cos n}{n^2}

is (a) absolutely convergent (b) conditionally converegent.

Definitions

A series $\sum a_n$ is absolutely convergent if $\sum \lvert a_n \rvert$ converges.
A series $\sum a_n$ is conditionally convergent if $\sum a_n$ converges but $\sum \lvert a_n \rvert$ diverges.

Here $a_n = \dfrac{\cos n}{n^2}$ . The sign of $\cos n$ changes without a fixed pattern, so the Leibniz test cannot be used. We test $\sum \lvert a_n \rvert$ .

Step 1 Find a bigger series

For every $n$ , $\lvert \cos n \rvert \le 1$ . So

\lvert a_n \rvert = \frac{\lvert \cos n \rvert}{n^2} \le \frac{1}{n^2}

Step 2 Test the bigger series

The series $\sum \dfrac{1}{n^p}$ is called a p-series. It converges when $p > 1$ .

\sum_{n=1}^{\infty} \frac{1}{n^2}

is a p-series with $p = 2 > 1$ , so it converges.

Step 3 Apply the comparison test

The comparison test says: if $0 \le u_n \le v_n$ for all $n$ and $\sum v_n$ converges, then $\sum u_n$ also converges.
Take $u_n = \lvert a_n \rvert$ and $v_n = \dfrac{1}{n^2}$ . Then

\sum_{n=1}^{\infty} \left\lvert \frac{\cos n}{n^2} \right\rvert \text{ converges.}

Step 4 Conclusion

(a) Since $\sum \lvert a_n \rvert$ converges, the given series is absolutely convergent. Every absolutely convergent series is convergent, so the given series converges.
(b) For conditional convergence, $\sum \lvert a_n \rvert$ must diverge. Here it converges. So the series is not conditionally convergent.

Answer

The series $\displaystyle\sum_{n=1}^{\infty} \frac{\cos n}{n^2}$ is absolutely convergent. It is not conditionally convergent.

(a)

3 marks

Investigate the convergence of the series

\sum_{n=1}^{\infty} \frac{2^n}{n^3}

(b)

4 marks

Find the sum of the series

\sum_{n=0}^{\infty} \frac{4}{2^n}

(c)

7 marks

Determine the values of

x

for which the series

\sum\limits_{n=1}^{\infty} (-1)^{n-1} \dfrac{x^n}{n}

is convergent.

This is a power series in $x$ . We use the ratio test first and then check the two end points separately.

Step 1 Apply the ratio test

The ratio test says: let $L = \lim \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvert$ . The series converges if $L < 1$ and diverges if $L > 1$ . If $L = 1$ , the test gives no result.
Here $a_n = (-1)^{n-1}\dfrac{x^n}{n}$ .

\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{\lvert x \rvert^{n+1}}{n+1} \cdot \frac{n}{\lvert x \rvert^n} = \lvert x \rvert \cdot \frac{n}{n+1}

As $n \to \infty$ , $\dfrac{n}{n+1} \to 1$ .

L = \lvert x \rvert

So the series converges for $\lvert x \rvert < 1$ and diverges for $\lvert x \rvert > 1$ .

Step 2 Check x = 1

The series becomes

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

This is an alternating series. The terms $\dfrac{1}{n}$ are positive, decreasing, and tend to 0. By the Leibniz test it converges.

Step 3 Check x = -1

Since $(-1)^{n-1}(-1)^n = (-1)^{2n-1} = -1$ , the series becomes

\sum_{n=1}^{\infty} \frac{-1}{n} = -\left(1 + \frac{1}{2} + \frac{1}{3} + \dots\right)

This is the harmonic series with a minus sign. The harmonic series diverges, so the series diverges at $x = -1$ .

Step 4 Combine the results

Value of $x$	Result
$\lvert x \rvert < 1$	Converges
$x = 1$	Converges
$x = -1$	Diverges
$\lvert x \rvert > 1$	Diverges

Answer

The series converges for $-1 < x \le 1$ and diverges for all other values of $x$ .

Q.4

14 Marks

(a)

3 marks

f(x, y) = \dfrac{x^2 - y^2}{x^2 + y^2}

, does

\lim\limits_{(x,y) \to (0,0)} f(x, y)

exist?

(b)

4 marks

z = x^2 y + x y^2

, where

x = 2 + t^4

y = 1 - t^3

, find

\dfrac{dz}{dt}

when

t = 0

(c)

7 marks

Find the extreme values of the function

f(x, y) = x^2 y

on the curve

x^2 + 2y^2 = 6

We need the largest and smallest values of $f$ at points that lie on the given curve. The curve is a condition, called a constraint. The method of Lagrange multipliers is used: solve $\nabla f = \lambda \nabla g$ together with $g = 0$ , where $\lambda$ is an unknown number.

Step 1 Write f and g

\begin{aligned} f(x, y) &= x^2 y \\ g(x, y) &= x^2 + 2y^2 - 6 = 0 \end{aligned}

Step 2 Form the equations

\begin{aligned} f_x = \lambda g_x &: \quad 2xy = 2\lambda x \quad (1) \\ f_y = \lambda g_y &: \quad x^2 = 4\lambda y \quad (2) \\ g = 0 &: \quad x^2 + 2y^2 = 6 \quad (3) \end{aligned}

Step 3 Solve

From (1), $2x(y - \lambda) = 0$ . So $x = 0$ or $\lambda = y$ .
Case 1: $x = 0$ . From (3), $2y^2 = 6$ , so $y = \pm\sqrt{3}$ . The points are $(0, \sqrt{3})$ and $(0, -\sqrt{3})$ .
Case 2: $\lambda = y$ . From (2), $x^2 = 4y^2$ . Put this in (3):

\begin{aligned} 4y^2 + 2y^2 &= 6 \\ y^2 &= 1 \\ y &= \pm 1 \end{aligned}

Then $x^2 = 4$ , so $x = \pm 2$ . The points are $(2, 1)$ , $(-2, 1)$ , $(2, -1)$ and $(-2, -1)$ .

Step 4 Find f at each point

Point	$f = x^2 y$
$(0, \pm\sqrt{3})$	$0$
$(\pm 2, 1)$	$4 \times 1 = 4$
$(\pm 2, -1)$	$4 \times (-1) = -4$

The largest value is $4$ and the smallest value is $-4$ .

Answer

The maximum value is $4$ at $(\pm 2, 1)$ , and the minimum value is $-4$ at $(\pm 2, -1)$ .

(a)

3 marks

Find the equation of the normal line at the point

(-2, 1, -3)

to the ellipsoid

\frac{x^2}{4} + y^2 + \frac{z^2}{9} = 3

(b)

4 marks

f(x, y, z) = x \sin(yz)

, find the directional derivative of

f

(1, 3, 0)

in the direction of the vector

\mathbf{v} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}

(c)

7 marks

Find the local extreme values of the function

f(x, y) = 9 - 2x + 4y - x^2 - 4y^2

A function of two variables can have a local extreme value only at a critical point, where $f_x = 0$ and $f_y = 0$ .

Step 1 Find the first partial derivatives

\begin{aligned} f(x, y) &= 9 - 2x + 4y - x^2 - 4y^2 \\ f_x &= -2 - 2x \\ f_y &= 4 - 8y \end{aligned}

Step 2 Find the critical point

\begin{aligned} -2 - 2x = 0 &\;\Rightarrow\; x = -1 \\ 4 - 8y = 0 &\;\Rightarrow\; y = \frac{1}{2} \end{aligned}

The only critical point is $\left(-1, \dfrac{1}{2}\right)$ .

Step 3 Find the second partial derivatives

f_{xx} = -2, \qquad f_{yy} = -8, \qquad f_{xy} = 0

Step 4 Apply the second derivative test

Let $D = f_{xx} f_{yy} - (f_{xy})^2$ at the critical point.

Condition	Result
$D > 0$ and $f_{xx} < 0$	Local maximum
$D > 0$ and $f_{xx} > 0$	Local minimum
$D < 0$	Saddle point
$D = 0$	No conclusion

D = (-2)(-8) - 0^2 = 16 > 0

Also $f_{xx} = -2 < 0$ . So $f$ has a local maximum at $\left(-1, \dfrac{1}{2}\right)$ .

Step 5 Find the value

\begin{aligned} f\left(-1, \tfrac{1}{2}\right) &= 9 - 2(-1) + 4\left(\tfrac{1}{2}\right) - (-1)^2 - 4\left(\tfrac{1}{2}\right)^2 \\ &= 9 + 2 + 2 - 1 - 1 \\ &= 11 \end{aligned}

There is no other critical point, so $f$ has no local minimum.

Answer

The function has a local maximum value of $11$ at $\left(-1, \dfrac{1}{2}\right)$ . It has no local minimum.

Q.5

14 Marks

(a)

3 marks

Evaluate

\int_0^{\pi} \int_1^2 y \sin(xy)\,dx\,dy

(b)

4 marks

Integrate

f(x, y) = x \cos y

over the region bounded by the curves

y = 0

y = x^2

x = 1

(c)

7 marks

Sketch the region of integration, reverse the order of integration and evaluate the integral

\int_0^2 \int_{x^2}^{2x} (x^2 + y^2)\,dy\,dx

Step 1 Sketch the region

In the given integral, $x$ goes from 0 to 2. For each $x$ , $y$ goes from the parabola $y = x^2$ up to the line $y = 2x$ . The two curves meet where $x^2 = 2x$ , that is at $(0, 0)$ and $(2, 4)$ .

DIAGRAM

 y
 |            (2, 4)
 4 +          *
 |          / .
 | y = 2x  /  .
 |       /   .
 |     /  R .   y = x^2
 |   /   .
 | /  .
 +----------+---- x
 0          2

Step 2 Reverse the order

Now $y$ is the outer variable. It goes from 0 to 4. For each $y$ , move from left to right:

the left boundary is the line $y = 2x$ , so $x = \dfrac{y}{2}$ ,
the right boundary is the parabola $y = x^2$ , so $x = \sqrt{y}$ .

I = \int_0^4 \int_{y/2}^{\sqrt{y}} (x^2 + y^2)\,dx\,dy

Step 3 Inner integral

\begin{aligned} \int_{y/2}^{\sqrt{y}} (x^2 + y^2)\,dx &= \left[\frac{x^3}{3} + y^2 x\right]_{y/2}^{\sqrt{y}} \\ &= \frac{y^{3/2}}{3} + y^{5/2} - \frac{y^3}{24} - \frac{y^3}{2} \\ &= \frac{y^{3/2}}{3} + y^{5/2} - \frac{13y^3}{24} \end{aligned}

Step 4 Outer integral

\begin{aligned} I &= \int_0^4 \left(\frac{y^{3/2}}{3} + y^{5/2} - \frac{13y^3}{24}\right)dy \\ &= \left[\frac{2y^{5/2}}{15} + \frac{2y^{7/2}}{7} - \frac{13y^4}{96}\right]_0^4 \end{aligned}

Use $4^{5/2} = 32$ , $4^{7/2} = 128$ and $4^4 = 256$ .

\begin{aligned} I &= \frac{64}{15} + \frac{256}{7} - \frac{104}{3} \\ &= \frac{448 + 3840 - 3640}{105} \\ &= \frac{648}{105} = \frac{216}{35} \end{aligned}

Answer

$\displaystyle\int_0^4\!\int_{y/2}^{\sqrt{y}} (x^2 + y^2)\,dx\,dy = \frac{216}{35} \approx 6.17$

(a)

3 marks

Evaluate

\int_0^3 \int_0^1 \int_0^{\sqrt{1 - z^2}} z e^{y}\,dx\,dz\,dy

(b)

4 marks

Evaluate

\int_0^3 \int_0^{\sqrt{9 - x^2}} \sin(x^2 + y^2)\,dy\,dx

by converting to polar coordinates.

(c)

7 marks

Find the volume of the solid that lies under the paraboloid

z = x^2 + y^2

and above the region

R

in the

xy

-plane bounded by

y = x^2

and

x = y^2

The volume under a surface $z = f(x, y)$ and above a region $R$ in the xy-plane is

V = \iint_R f(x, y)\,dA

Step 1 Find the region R

The curves are $y = x^2$ and $x = y^2$ , that is $y = \sqrt{x}$ . They meet where

x^2 = \sqrt{x} \;\Rightarrow\; x^4 = x \;\Rightarrow\; x = 0 \text{ or } x = 1

So they meet at $(0, 0)$ and $(1, 1)$ . Between these points $\sqrt{x} \ge x^2$ . For example, at $x = \dfrac{1}{4}$ we get $\dfrac{1}{2}$ and $\dfrac{1}{16}$ .

DIAGRAM

 y
 1 +            * (1, 1)
 |       .    .
 |   .       .
 | .  R    .    upper: y = sqrt(x)
 |.     .       lower: y = x^2
 +. . ----------+-- x
 0              1

So $x$ goes from 0 to 1, and for each $x$ , $y$ goes from $x^2$ to $\sqrt{x}$ .

Step 2 Write the integral

V = \int_0^1 \int_{x^2}^{\sqrt{x}} (x^2 + y^2)\,dy\,dx

Step 3 Inner integral

\begin{aligned} \int_{x^2}^{\sqrt{x}} (x^2 + y^2)\,dy &= \left[x^2 y + \frac{y^3}{3}\right]_{x^2}^{\sqrt{x}} \\ &= x^{5/2} + \frac{x^{3/2}}{3} - x^4 - \frac{x^6}{3} \end{aligned}

Step 4 Outer integral

\begin{aligned} V &= \int_0^1 \left(x^{5/2} + \frac{x^{3/2}}{3} - x^4 - \frac{x^6}{3}\right)dx \\ &= \left[\frac{2x^{7/2}}{7} + \frac{2x^{5/2}}{15} - \frac{x^5}{5} - \frac{x^7}{21}\right]_0^1 \\ &= \frac{2}{7} + \frac{2}{15} - \frac{1}{5} - \frac{1}{21} \end{aligned}

The common denominator is 105.

V = \frac{30 + 14 - 21 - 5}{105} = \frac{18}{105} = \frac{6}{35}

Answer

The volume is $V = \dfrac{6}{35} \approx 0.171$ cubic units.

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Official Gujarat Technological University (GTU) examination paper and step-by-step solutions for Mathematics-I (Summer 2026, 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.

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