B.Tech 1st Year Mathematics-I (BT-102) Solved Question Paper June 2026

Question 114 marks

Q1(a)Unit 1 · Rolle's & Mean Value theorems7 marks

Verify Rolle's theorem for the function

f (x) = e^{x} sin x

[0, 2 π]

Solution

$e^{x}$ and $sin x$ are continuous and differentiable everywhere, so their product is continuous on $[0, 2 π]$ and differentiable in $(0, 2 π)$ .

f (0) = e^{0} sin 0 = 0, f (2 π) = e^{2 π} sin 2 π = 0.

f^{'} (x) = e^{x} sin x + e^{x} cos x = e^{x} (sin x + cos x) .

Since $e^{x} \neq = 0$ , $f^{'} (c) = 0 \Rightarrow sin c + cos c = 0 \Rightarrow tan c = - 1$ .

c = \frac{3 π}{4} or c = \frac{7 π}{4} .

c = \frac{3 π}{4}, \frac{7 π}{4} \in (0, 2 π)

; Rolle's theorem is verified.

Q1(b)Unit 1 · Taylor & Maclaurin (one variable)7 marks

Expand

f (x) = sin x

as Taylor series about

x = \frac{π}{3}

Also asked in: Dec 2024 Q1(b)

Solution

Taylor's series about $x = a$ : $f (x) = f (a) + (x - a) f^{'} (a) + \frac{( x - a ) ^{2}}{2 !} f^{''} (a) + \frac{( x - a ) ^{3}}{3 !} f^{'''} (a) + \dots$

Derivatives of $sin x$ repeat: $sin x, cos x, - sin x, - cos x, sin x, \dots$

(ii) About $a = π /3$ : values are $\frac{3}{2}, \frac{1}{2}, - \frac{3}{2}, - \frac{1}{2}, \frac{3}{2}, \dots$

sin x = \frac{3}{2} + \frac{1}{2} (x - \frac{π}{3}) - \frac{3}{4} (x - \frac{π}{3})^{2} - \frac{1}{12} (x - \frac{π}{3})^{3} + \frac{3}{48} (x - \frac{π}{3})^{4} + \dots

Question 214 marks

Q2(a)Unit 1 · Partial differentiation7 marks

u = lo g (x^{3} + y^{3} + z^{3} - 3 x y z)

, show that

(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z})^{2} u = - \frac{9}{( x + y + z ) ^{2}} .

Also asked in: Dec 2025 Q1(b)

Solution

$u_{x} = \frac{3 x ^{2} - 3 y z}{x ^{3} + y ^{3} + z ^{3} - 3 x y z}$ , and similarly for $u_{y}, u_{z}$ . Adding:

u_{x} + u_{y} + u_{z} = \frac{3 ( x ^{2} + y ^{2} + z ^{2} - x y - y z - z x )}{x ^{3} + y ^{3} + z ^{3} - 3 x y z} .

Use the identity $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$ :

(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}) u = \frac{3}{x + y + z} .

Apply the operator once more:

(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}) \frac{3}{x + y + z} = - \frac{3}{( x + y + z ) ^{2}} - \frac{3}{( x + y + z ) ^{2}} - \frac{3}{( x + y + z ) ^{2}} .

(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z})^{2} u = - \frac{9}{( x + y + z ) ^{2}}

. Hence proved.

Q2(b)Unit 1 · Lagrange's multipliers7 marks

A rectangular box open at the top is to have a volume of 32 cubic ft. Find the dimensions of the box requiring least material for its construction.

Solution

Method: for extremum of $f$ subject to $ϕ = 0$ , form $F = f + λ ϕ$ and solve $F_{x} = F_{y} = F_{z} = 0$ together with $ϕ = 0$ .

Let the length, breadth and height be $x, y, z$ . Surface (no lid) $S = x y + 2 y z + 2 z x$ , with $x y z = 32$ .

F = x y + 2 y z + 2 z x + λ (x y z - 32)

x F_{x} : x y + 2 z x + λ x y z = 0 (1) y F_{y} : x y + 2 y z + λ x y z = 0 (2) z F_{z} : 2 y z + 2 z x + λ x y z = 0 (3)

(1) − (2): $2 z (x - y) = 0 \Rightarrow x = y$ . (2) − (3): $y (x - 2 z) = 0 \Rightarrow x = 2 z$ .

So $x = y = 2 z$ ; then $4 z^{3} = 32 \Rightarrow z = 2, x = y = 4$ .

Length

= 4

ft, breadth

= 4

ft, height

= 2

ft; least surface

= 16 + 16 + 16 = 48

sq ft.

Question 314 marks

Q3(a)Unit 2 · Beta & Gamma functions7 marks

Prove that

B (m, n) = \frac{Γ ( m ) Γ ( n )}{Γ ( m + n )}

where

m > 0, n > 0

Also asked in: Dec 2023 Q6(a), Dec 2024 Q6(a)

Solution

Definitions.

B (m, n) = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} d x, Γ (n) = \int_{0}^{\infty} e^{- x} x^{n - 1} d x (m, n > 0) .

Step 1. In $Γ (m)$ put $x = t^{2}$ , $d x = 2 t d t$ :

Γ (m) = 2 \int_{0}^{\infty} e^{- t^{2}} t^{2 m - 1} d t .

Step 2. Multiply $Γ (m)$ and $Γ (n)$ (dummy variables $x$ and $y$ ):

Γ (m) Γ (n) = 4 \int_{0}^{\infty} \int_{0}^{\infty} e^{- (x^{2} + y^{2})} x^{2 m - 1} y^{2 n - 1} d x d y .

Step 3. Change to polar coordinates $x = r cos θ, y = r sin θ, d x d y = r d r d θ$ ; the first quadrant is $0 \leq r < \infty, 0 \leq θ \leq \frac{π}{2}$ :

Γ (m) Γ (n) = [2 \int_{0}^{\infty} e^{- r^{2}} r^{2 (m + n) - 1} d r] [2 \int_{0}^{π /2} cos^{2 m - 1} θ sin^{2 n - 1} θ d θ] .

Step 4. The first bracket is $Γ (m + n)$ (Step 1 with $m + n$ ). The second is $B (m, n)$ : put $x = cos^{2} θ$ in $B (m, n)$ .

Γ (m) Γ (n) = Γ (m + n) B (m, n) .

B (m, n) = \frac{Γ ( m ) Γ ( n )}{Γ ( m + n )}

. Hence proved.

Q3(b)Unit 2 · Beta & Gamma functions7 marks

Prove that

\int_{0}^{π /2} sin^{2} θ cos^{4} θ d θ = \frac{π}{32}

Solution

$\int_{0}^{π /2} sin^{p} θ cos^{q} θ d θ = \frac{Γ ( \frac{p + 1}{2} ) Γ ( \frac{q + 1}{2} )}{2Γ ( \frac{p + q + 2}{2} )}$

Here $p = 2, q = 4$ :

I = \frac{Γ ( \frac{3}{2} ) Γ ( \frac{5}{2} )}{2Γ ( 4 )} = \frac{\frac{π}{2} \cdot \frac{3 π}{4}}{2 \cdot 6} = \frac{3 π /8}{12} = \frac{π}{32} .

Wallis check: $\frac{( 1 ) ( 3 \cdot 1 )}{6 \cdot 4 \cdot 2} \cdot \frac{π}{2} = \frac{3}{48} \cdot \frac{π}{2} = \frac{π}{32}$ ✓ (both powers even, so factor $π /2$ ).

Question 414 marks

Q4(a)Unit 2 · Change of order of integration7 marks

Change the order of integration

\int_{0}^{4 a} \int_{x^{2} /4 a}^{2 a x} d y d x

Solution

The given limits describe exactly the region between $x^{2} = 4 a y$ (lower) and $y^{2} = 4 a x$ (upper); the curves meet at $(0, 0)$ and $(4 a, 4 a)$ .

(a) For a horizontal strip at height $y$ , $x$ runs from the parabola $x = \frac{y ^{2}}{4 a}$ to $x = 2 a y$ , and $y$ from 0 to $4 a$ :

\int_{0}^{4 a} \int_{x^{2} /4 a}^{2 a x} d y d x = \int_{0}^{4 a} \int_{y^{2} /4 a}^{2 a y} d x d y .

\int_{0}^{4 a} \int_{y^{2} /4 a}^{2 a y} d x d y

Q4(b)Unit 2 · Area by double integration7 marks

Find the area of the region bounded by the parabolas

y^{2} = 4 a x

and

x^{2} = 4 a y

Solution

The parabolas $x^{2} = 4 a y$ and $y^{2} = 4 a x$ meet at $(0, 0)$ and $(4 a, 4 a)$ . For a vertical strip, $y$ runs from $\frac{x ^{2}}{4 a}$ to $2 a x$ , and $x$ from 0 to $4 a$ .

(b) Area $= \iint d y d x$ :

A = \int_{0}^{4 a} (2 a x^{1/2} - \frac{x ^{2}}{4 a}) d x = 2 a \cdot \frac{2}{3} (4 a)^{3/2} - \frac{( 4 a ) ^{3}}{12 a} = \frac{32 a ^{2}}{3} - \frac{16 a ^{2}}{3} .

Area

= \frac{16}{3} a^{2}

square units

Question 514 marks

Q5(a)Unit 3 · Convergence of series7 marks

Test the convergence of the series

1 + \frac{x}{2} + \frac{x ^{2}}{5} + \frac{x ^{3}}{10} + \dots + \frac{x ^{n}}{n + 1} + \dots (x > 0)

Also asked in: Jun 2023 Q5(a), Dec 2023 Q5(a)

Solution

Note on the printed paper: the denominators $2, 5, 10, \dots$ are $n^{2} + 1$ (for $n = 1, 2, 3$ ), so the general term is $\frac{x ^{n}}{n ^{2} + 1}$ ; the " $n + 1$ " in the paper is a misprint. Both readings are handled below.

Leaving out the first term, $u_{n} = \frac{x ^{n}}{n ^{2} + 1}$ . By D'Alembert's ratio test:

\frac{u _{n + 1}}{u _{n}} = \frac{x ^{n + 1}}{( n + 1 ) ^{2} + 1} \cdot \frac{n ^{2} + 1}{x ^{n}} = \frac{1 + \frac{1}{n ^{2}}}{( 1 + \frac{1}{n} ) ^{2} + \frac{1}{n ^{2}}} x ⟶ x .

$x < 1$ : convergent; $x > 1$ : divergent.
$x = 1$ : the test fails. Then $u_{n} = \frac{1}{n ^{2} + 1}$ . Compare with $v_{n} = \frac{1}{n ^{2}}$ : $\frac{u _{n}}{v _{n}} = \frac{n ^{2}}{n ^{2} + 1} \to 1$ , and $\sum \frac{1}{n ^{2}}$ converges ( $p = 2 > 1$ ). So the series converges.

The series converges for

0 < x \leq 1

and diverges for

x > 1

If the general term is read literally as $\frac{x ^{n}}{n + 1}$ , the ratio limit is again $x$ , but at $x = 1$ the series $\sum \frac{1}{n + 1}$ diverges; it then converges only for $x < 1$ .

Q5(b)Unit 3 · Fourier series7 marks

Find a Fourier series to represent

f (x) = x^{2}

in the interval

(0, 2 π)

Solution

Euler's formulae on $(c, c + 2 π)$ : $a_{0} = \frac{1}{π} \int f d x, a_{n} = \frac{1}{π} \int f cos n x d x, b_{n} = \frac{1}{π} \int f sin n x d x$ ; $f (x) = \frac{a _{0}}{2} + \sum (a_{n} cos n x + b_{n} sin n x)$ .

a_{0} = \frac{1}{π} \int_{0}^{2 π} x^{2} d x = \frac{8 π ^{2}}{3} .

a_{n} = \frac{1}{π} [x^{2} \frac{sin n x}{n} + 2 x \frac{cos n x}{n ^{2}} - 2 \frac{sin n x}{n ^{3}}]_{0}^{2 π} = \frac{1}{π} \cdot \frac{4 π}{n ^{2}} = \frac{4}{n ^{2}} .

b_{n} = \frac{1}{π} [- x^{2} \frac{cos n x}{n} + 2 x \frac{sin n x}{n ^{2}} + 2 \frac{cos n x}{n ^{3}}]_{0}^{2 π} = \frac{1}{π} (- \frac{4 π ^{2}}{n}) = - \frac{4 π}{n} .

x^{2} = \frac{4 π ^{2}}{3} + 4 n = 1 \sum \infty (\frac{cos n x}{n ^{2}} - \frac{π sin n x}{n}), 0 < x < 2 π

Question 614 marks

Q6(a)Unit 3 · Half range series7 marks

Find the half range sine series for

f (x) = x (π - x)

0 < x < π

Solution

$b_{n} = \frac{2}{π} \int_{0}^{π} (π x - x^{2}) sin n x d x$ . By Bernoulli's rule with $u = π x - x^{2}, u^{'} = π - 2 x, u^{''} = - 2$ :

\int_{0}^{π} u sin n x d x = [- u \frac{cos n x}{n} + u^{'} \frac{sin n x}{n ^{2}} + u^{''} \frac{cos n x}{n ^{3}}]_{0}^{π} = - \frac{2}{n ^{3}} (cos nπ - 1) = \frac{2 ( 1 - ( - 1 ) ^{n} )}{n ^{3}} .

So $b_{n} = \frac{4 ( 1 - ( - 1 ) ^{n} )}{π n ^{3}}$ : $b_{n} = \frac{8}{π n ^{3}}$ for odd $n$ , 0 for even $n$ .

x (π - x) = \frac{8}{π} [\frac{sin x}{1 ^{3}} + \frac{sin 3 x}{3 ^{3}} + \frac{sin 5 x}{5 ^{3}} + \dots]

Q6(b)Unit 3 · Fourier series7 marks

Expand

f (x) = x sin x, 0 < x < 2 π

as a Fourier series.

Also asked in: Jun 2023 Q6(a), Dec 2023 Q5(b)

Solution

Useful results for an integer $k \neq = 0$ : $\int_{0}^{2 π} x sin k x d x = - \frac{2 π}{k}$ , $\int_{0}^{2 π} x cos k x d x = 0$ .

$a_{0}$ : $a_{0} = \frac{1}{π} \int_{0}^{2 π} x sin x d x = \frac{1}{π} (- 2 π) = - 2$ .

$a_{n}$ , $n \neq = 1$ : $x sin x cos n x = \frac{x}{2} [sin (1 + n) x + sin (1 - n) x]$ :

a_{n} = \frac{1}{2 π} [- \frac{2 π}{1 + n} - \frac{2 π}{1 - n}] = - [\frac{1}{1 + n} + \frac{1}{1 - n}] = \frac{2}{n ^{2} - 1} .

$a_{1}$ : $a_{1} = \frac{1}{2 π} \int_{0}^{2 π} x sin 2 x d x = \frac{1}{2 π} (- \frac{2 π}{2}) = - \frac{1}{2}$ .

$b_{n}$ : $x sin x sin n x = \frac{x}{2} [cos (n - 1) x - cos (n + 1) x]$ . For $n \geq 2$ both integrals are 0, so $b_{n} = 0$ . For $n = 1$ : $b_{1} = \frac{1}{2 π} \int_{0}^{2 π} x (1 - cos 2 x) d x = \frac{1}{2 π} \cdot 2 π^{2} = π$ .

x sin x = - 1 + π sin x - \frac{1}{2} cos x + 2 n = 2 \sum \infty \frac{cos n x}{n ^{2} - 1} = - 1 + π sin x - \frac{1}{2} cos x + \frac{2}{3} cos 2 x + \frac{2}{8} cos 3 x + \dots

Question 714 marks

Q7(a)Unit 5 · Rank & normal form7 marks

Reduce the following matrix to normal form and hence find its rank:

1236242833170235

Solution

A R_{2} - 2 R_{1}, R_{3} - 3 R_{1} R_{4} - 6 R_{1} 1000 20 - 4 - 4 3 - 3 - 8 - 11 0235 R_{2} \leftrightarrow R_{3} R_{4} - R_{2} 1000 2 - 4 00 3 - 8 - 3 - 3 0322 R_{4} - R_{3} 1000 2 - 4 00 3 - 8 - 3 0 0320

Dividing the rows by their pivots and using column operations to clear everything to the right of each pivot gives

A \sim [I_{3} O O O] .

Rank

ρ (A) = 3

Q7(b)Unit 5 · Consistency & solution of equations7 marks

Solve the system of equations

3 x + y + 2 z = 3, 2 x - 3 y - z = - 3, x + 2 y + z = 4

using Gauss elimination method.

Solution

Take $x + 2 y + z = 4$ as the first (pivot) equation.

132 21 - 3 12 - 1 43 - 3 R_{2} - 3 R_{1} R_{3} - 2 R_{1} 100 2 - 5 - 7 1 - 1 - 3 4 - 9 - 11 5 R_{3} - 7 R_{2} 100 2 - 5 0 1 - 1 - 8 4 - 9 8

Back substitution: $- 8 z = 8 \Rightarrow z = - 1$ ; $- 5 y + 1 = - 9 \Rightarrow y = 2$ ; $x + 4 - 1 = 4 \Rightarrow x = 1$ .

x = 1, y = 2, z = - 1

Check: $3 + 2 - 2 = 3$ ✓, $2 - 6 + 1 = - 3$ ✓

Question 814 marks

Q8(a)Unit 5 · Cayley-Hamilton theorem7 marks

A = 25 - 1 130 33 - 2

, then verify Cayley-Hamilton theorem and hence find

A^{- 1}

Solution

Characteristic equation. $tr A = 3$ ; $S_{2} = (- 6 - 0) + (- 4 + 3) + (6 - 5) = - 6$ ; $∣ A ∣ = 2 (- 6) - 1 (- 10 + 3) + 3 (0 + 3) = 4$ .

λ^{3} - 3 λ^{2} - 6 λ - 4 = 0.

Verification of $A^{3} - 3 A^{2} - 6 A - 4 I = O$ :

A^{2} = 6220 514 - 1 3181, A^{3} = 3496 - 6 2164 - 3 2772 - 5

(1,1): $34 - 18 - 12 - 4 = 0$ ; (1,2): $21 - 15 - 6 - 0 = 0$ ; (2,1): $96 - 66 - 30 = 0$ ; (3,3): $- 5 - 3 + 12 - 4 = 0$ . Every entry vanishes ⇒ theorem verified.

Inverse. Multiply by $A^{- 1}$ : $A^{2} - 3 A - 6 I - 4 A^{- 1} = O \Rightarrow A^{- 1} = \frac{1}{4} (A^{2} - 3 A - 6 I)$ .

A^{- 1} = \frac{1}{4} - 6 73 2 - 1 - 1 - 6 91

Q8(b)Unit 5 · Consistency & solution of equations7 marks

Find for what values of

λ

the equations

x + y + z = 1, x + 2 y + 4 z = λ, x + 4 y + 10 z = λ^{2}

have a solution, and solve them completely in each case.

Solution

[A ∣ B] R_{2} - R_{1} R_{3} - R_{1} 100113139 1 λ - 1 λ^{2} - 1 R_{3} - 3 R_{2} 100110130 1 λ - 1 λ^{2} - 3 λ + 2

$ρ (A) = 2$ . For consistency we need $ρ (A ∣ B) = 2$ , i.e. $λ^{2} - 3 λ + 2 = 0 \Rightarrow (λ - 1) (λ - 2) = 0$ .

$λ = 1$ : $y + 3 z = 0$ . Put $z = k$ : $y = - 3 k, x = 1 - y - z = 1 + 2 k$ .

$λ = 2$ : $y + 3 z = 1$ . Put $z = k$ : $y = 1 - 3 k, x = 1 - y - z = 2 k$ .

Solutions exist only for

λ = 1

(1 + 2 k, - 3 k, k)

; and

λ = 2

(2 k, 1 - 3 k, k)