BEU CSE Math 2023 PYQ with Solutions | 1st Sem Math Solved Papers ” – your ultimate guide to mastering BEU mathematics with confidence, clarity, and complete solutions at your fingertips.
Q.1 Choose the correct answer of the following:
(a) The value of $\int_0^1 x^3\left(1-x^2\right)^{5 / 2} d x$ is
(i) $\frac{\pi}{2}$
(ii) $\frac{2}{63}$
(iii) 1
(iv) None of the above
Answer: (ii) $\frac{2}{63}$
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BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
(b) The area $\iint_R \sin (x+y) d x d y$ over $\mathrm{R}{(x, y) ; 0 \leq x \leq \pi / 2,0 \leq x \leq \pi / 2}$
(i) 0
(ii) 1
(iii) 2
(iv) -1
Answer: (iii) 2
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(c) The matrix is
$$
\begin{bmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{bmatrix}
$$
(i) Invertible but not orthogonal
(ii) Invertible and orthogonal
(iii) Neither invertible nor orthogonal
(iv) None of the above
Answer: (ii) Invertible and orthogonal
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BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
(d) The sequence ${ a_n }$, where $a_n = \sqrt[n]{n}$, is:
(i) Not convergent
(ii) Convergent and $\lim_{n \to \infty} a_n = 0$
(iii) Convergent and $\lim_{n \to \infty} a_n = 1$
(iv) None of the above
Answer: (iii) Convergent and $\lim_{n \to \infty} a_n = 1$
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(e) Let $f(x)=\cos \left(x^2\right), x \in R$ then
(i) $f$ is uniformly continuous
(ii) $f$ is continuous. but not uniformly continuous
(iii) $f$ continuous but unbounded
(iv) $f$ is Lipschitz continuous
Answer: (ii) $f$ is continuous. but not uniformly continuous
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(f) $\lim _{(x, y) \rightarrow(1,2)} \frac{x^2+7 y}{x+y^2}$
(i) Doés not exists
(ii) Exists and the value is 2
(iii) Exists and the value is 3
(iv) None of the above
Answer: (iii) Exists and the value is 3
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(g) if $A=2 \hat{\imath}+\alpha \hat{\jmath}+\hat{k}, B=\hat{\imath}+3 \hat{\jmath}-8 \hat{k}$. Then $A$ and $B$ are orthogonal if
(i) $\alpha=-2$
(ii) $\alpha=2$
(iii) $\alpha=-1$
(iv) $a=1$
Answer: (ii) $\alpha=2$
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(h) Let $x+2 y+z=0$,$3 x+4 y+z=0$, $x-z=0$be a system of equations. Then
(i) It is inconsistent
(ii) It has only the trivial solution $x=0, y=0, z=0$
(iii) Determinant of the matrix of coefficient is zero
(iv) None of these
Answer: (iii) Determinant of the matrix of coefficient is zero
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(i) The value at which the Rolle’s theorem is applicable for $f(x)=\cos \frac{2}{2}$ in $|\pi, 3 \pi|$ is
(i) $\frac{5 \pi}{2}$
(ii) $\frac{3 \pi}{2}$
(iii) $2 \pi$
(iv) none of the above
Answer: (iii) $2 \pi$
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BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
(j) The rank of the matrix is
$$
\begin{bmatrix}
1 & 2 & 3 \\
2 & 4 & 7 \\
3 & 6 & 10
\end{bmatrix}
$$
(i) 3
(ii) 2
(iii) 1
(iv) none of these
Answer: (ii) 2
See Solution

BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
Q. 2 (a) Show that:
$$
\int_0^{\frac{\pi}{2}} \log (\tan x+\cot x) \,dx = \pi \log 2.
$$
Solution:
First simplify:
$$
\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}
$$
$$
= \frac{\sin^2 x + \cos^2 x}{\sin x \cos x}
= \frac{1}{\sin x \cos x}
= \frac{2}{\sin 2x}
$$
So,
$$
\int_0^{\frac{\pi}{2}} \log \left( \frac{2}{\sin 2x} \right) \, dx
= \int_0^{\frac{\pi}{2}} \left( \log 2 – \log(\sin 2x) \right) \, dx
$$
Split the integral:
$$
= \int_0^{\frac{\pi}{2}} \log 2 \, dx – \int_0^{\frac{\pi}{2}} \log(\sin 2x) \, dx
$$Evaluate first part:
$$
\int_0^{\frac{\pi}{2}} \log 2 \, dx = \log 2 \cdot \frac{\pi}{2} = \frac{\pi}{2} \log 2
$$
Substitute for second part:
$$
\int_0^{\frac{\pi}{2}} \log(\sin 2x) \, dx
= \frac{1}{2} \int_0^{\pi} \log(\sin u) \, du
$$
$$\text{Let } u = 2x \Rightarrow dx = \frac{1}{2} \, du \quad \text{When } x = 0,\, u = 0 \quad \text{and} \quad x = \frac{\pi}{2},\, u = \pi$$
$$
= \frac{1}{2} \cdot (-\pi \log 2)
= -\frac{\pi}{2} \log 2
$$
Then:
$$
\int_0^{\frac{\pi}{2}} \log(\sin 2x) \, dx
= \frac{1}{2} \int_0^{\pi} \log(\sin u) \, du
= \frac{1}{2} (-\pi \log 2) = -\frac{\pi}{2} \log 2
$$
Final result:
$$
\int_0^{\frac{\pi}{2}} \log (\tan x + \cot x) \, dx
= \frac{\pi}{2} \log 2 + \frac{\pi}{2} \log 2 = \pi \log 2
$$
Answer:
$$
\boxed{ \int_0^{\frac{\pi}{2}} \log (\tan x+\cot x) \,dx = \pi \log 2 }
$$
(b) Find the following limit.
$$
\lim _{n \rightarrow \infty}\left[t \ln \left(1+\frac{3}{t}\right)\right]
$$
Solution:
$$
\lim_{n \to \infty} \left[ t \cdot \ln \left(1 + \frac{3}{t} \right) \right]
$$
Now let:
$$
\lim_{t \to \infty} \left[ \frac{ \ln \left(1 + \frac{3}{t} \right)}{\frac{1}{t}} \right]
$$
$$
\Rightarrow \lim_{t \to \infty} \left[ \frac{3 \ln \left(1 + \frac{3}{t} \right)}{\frac{3}{t}} \right]
$$
$$
\text{Let } A = \frac{3}{t}, \quad \text{then as } t \to \infty, \quad A \to 0 \Rightarrow \lim_{t \to \infty} = \lim_{A \to 0}
$$So the limit becomes:$$
\lim_{A \to 0} \left[ \frac{3 \ln(1 + A)}{A} \right]
$$
$$
\text{This is an indeterminate form of type } \frac{0}{0}, \text{ so we apply L’Hospital’s Rule:}
$$$$
\lim_{A \to 0} \left[ 3 \cdot \frac{1}{1 + A} \right] = 3
$$
BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
Q. 3 (a) Test the convergence of: : ***miss printing
$$
x^2+\frac{2^2}{3.4} x^4+\frac{2^2 4^2}{3.4 \cdot 5 \cdot 6.7} x^8+\cdots.
$$
Solution:
The series is: ***correct question
$$
x^2 + \frac{2^2}{3 \cdot 4} x^4 + \frac{2^2 \cdot 4^2}{3 \cdot 4 \cdot 5 \cdot 6} x^6 + \frac{2^2 \cdot 4^2 \cdot 6^2}{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} x^8 + \cdots
$$
The general term is:
$$
a_n = \frac{(2 \cdot 4 \cdot 6 \cdots (2n))^2}{(2n+1)!} x^{2n+2}, \quad \text{for } n \ge 0
$$
Next term:
$$
a_{n+1} = \frac{(2 \cdot 4 \cdot 6 \cdots (2n) \cdot (2n+2))^2}{(2n+3)!} x^{2n+4}
$$
Now take the ratio:
$$
\frac{a_{n+1}}{a_n} = \frac{ \left[(2 \cdot 4 \cdot 6 \cdots 2n)(2n+2)\right]^2 }{(2n+3)!} \cdot \frac{(2n+1)!}{(2 \cdot 4 \cdot 6 \cdots 2n)^2} \cdot \frac{x^{2n+4}}{x^{2n+2}}
$$
Simplify:
$$
= \frac{(2n+2)^2 x^2}{(2n+3)(2n+2)} = \frac{(2n+2) x^2}{2n+3}
$$
Now take the limit:
$$
\lim_{n \to \infty} \left| \frac{(2n+2)x^2}{2n+3} \right|
= |x^2| \cdot \lim_{n \to \infty} \frac{2n+2}{2n+3}
= |x^2| \cdot 1 = |x^2|
$$
Therefore, the series converges when:
$$
|x^2| < 1 \Rightarrow |x| < 1
$$
Final conclusion:
$$
\text{The series converges for } -1 < x < 1
$$
(b) Examine the convergence of the series with the general term:
$$
\sum_{n=1}^{\infty} \frac{1}{n(n+1)}.
$$
Solution:
$$
\text{Given: } a_n = \sum_{n=1}^{\infty} \frac{1}{n(n+1)}
$$
$$
\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}
$$
$$
1 = A(n+1) + Bn
$$
$$
\text{If } n = 0, \text{ then } 1 = A(1) + B(0) \Rightarrow A = 1
$$
$$
n = -1 \Rightarrow 1 = A(0) + B(-1) \Rightarrow B = -1
$$
$$
\frac{1}{n(n+1)} = \frac{1}{n} – \frac{1}{n+1}
$$
$$
\text{Let } S_n \text{ be the } n^\text{th} \text{ partial sum of the series}
$$
$$
S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \left( \frac{1}{k} – \frac{1}{k+1} \right)
$$
$$
S_n = \left( \frac{1}{1} – \frac{1}{2} \right) + \left( \frac{1}{2} – \frac{1}{3} \right) + \left( \frac{1}{3} – \frac{1}{4} \right) + \cdots + \left( \frac{1}{n} – \frac{1}{n+1} \right)
$$
$$
\text{This is a telescoping series, so the intermediate terms cancel out.}
$$
$$
S_n = 1 – \frac{1}{n+1}
$$
$$
\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 1 – \frac{1}{n+1} \right) = 1 – \lim_{n \to \infty} \frac{1}{n+1} = 1 – 0 = 1
$$
$$
\text{The series converges.}
$$
BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
Q. 4 (a) Prove that
$$
\log (1+x) = \frac{x}{1+\theta x}, \quad \text{where } 0<\theta<1.
$$
Hence deduce:
$$
\frac{x}{1+x}<\log (1+x)<x.
$$
Solution: $Let F(t) = \log(1 + t)$
Then:
$$
F'(t) = \frac{1}{1 + t}
$$
Apply the Mean Value Theorem on the interval ([0, x]):
$$
F'(c) = \frac{F(x) – F(0)}{x – 0}
$$
We know:
$$
F(x) = \log(1 + x), \quad F(0) = \log(1 + 0) = 0
$$
So:
$$
F'(c) = \frac{\log(1 + x) – 0}{x} = \frac{\log(1 + x)}{x}
$$
Also, by MVT:
$$
F'(c) = F'(\theta x) = \frac{1}{1 + \theta x}
$$
Thus:
$$
\frac{1}{1 + \theta x} = \frac{\log(1 + x)}{x}
$$
Rewriting:
$$
\log(1 + x) = \frac{x}{1 + \theta x} \quad \text{proved}
$$
From this, we deduce:
$$
\frac{x}{1 + x} < \log(1 + x) < x
$$
(b) It is given that Rolle’s theorem holds for the function:
$$
f(x) = x^3 + b x^2 + c x, \quad 1 \leq x \leq 2.
$$$$
\text{At the point } x = \frac{4}{3}, \text{ find the values of } b \text{ and } c.
$$
Solution:
$$
\text{Let } f(x) = x^3 + bx^2 + cx, \quad 1 \leq x \leq 2
$$
$$
\text{Given: At the point } x = \frac{4}{3}, \text{ Rolle’s Theorem holds.}
$$
So, we must have:
$$
f(1) = f(2)
$$
Compute ( f(1) ) and ( f(2) ):
$$
f(1) = 1^3 + b(1)^2 + c(1) = 1 + b + c
$$
$$
f(2) = 2^3 + b(2)^2 + c(2) = 8 + 4b + 2c
$$
Set ( f(1) = f(2) ):
$$
1 + b + c = 8 + 4b + 2c
$$
Bring terms together:
$$
1 + b + c – 8 – 4b – 2c = 0 \Rightarrow -7 – 3b – c = 0
$$
Rewriting:
$$
3b + c = -7 \quad \text{(Equation 1)}
$$
Now compute the derivative:
$$
f'(x) = 3x^2 + 2bx + c
$$
Apply Rolle’s Theorem at $$ ( x = \frac{4}{3} \Rightarrow f’\left(\frac{4}{3}\right) = 0 )$$
$$
f’\left(\frac{4}{3}\right) = 3\left(\frac{4}{3}\right)^2 + 2b\left(\frac{4}{3}\right) + c = 0
$$
Simplify:
$$
3 \cdot \frac{16}{9} + \frac{8b}{3} + c = 0 \Rightarrow \frac{16}{3} + \frac{8b}{3} + c = 0
$$
Multiply entire equation by 3:
$$
16 + 8b + 3c = 0 \quad \text{(Equation 2)}
$$
Now subtract Equation (1) multiplied by 3 from Equation (2):
\begin{aligned}
\text{Equation (1) × 3:} \quad 9b + 3c = -21
\end{aligned}
\begin{aligned}
\text{Equation (2):} \quad 8b + 3c = -16
\end{aligned}
\begin{aligned}
\text{Subtract:} \quad (8b + 3c) – (9b + 3c) = -16 – (-21)
\end{aligned}
\begin{aligned}
-b = 5 \Rightarrow b = -5
\end{aligned}
Substitute ( b = -5 ) into Equation (1):$$
3(-5) + c = -7 \Rightarrow -15 + c = -7 \Rightarrow c = 8
$$
Final Answer:
$$
\boxed{b = -5}, \quad \boxed{c = 8}
$$
BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
Q. 5 (a) Evaluate:
$$
\int_0^{\pi / 2} \frac{1}{a \cos^2 x + b \sin^2 x} \,dx.
$$
Solution:
$$
= \int_0^{\frac{\pi}{2}} \frac{\frac{1}{\cos^2 x}}{\frac{a \cos^2 x + b \sin^2 x}{\cos^2 x}} \, dx
$$
$$
= \int_0^{\frac{\pi}{2}} \frac{\sec^2 x}{a + b \tan^2 x} \, dx
$$
$$
\text{Let } u = \tan x \Rightarrow du = \sec^2 x \, dx
$$
$$
x = 0 \Rightarrow u = \tan 0 = 0, \quad x = \frac{\pi}{2} \Rightarrow u = \tan\left( \frac{\pi}{2} \right) = \infty
$$
$$
\Rightarrow \int_0^{\infty} \frac{1}{a + b u^2} \, du
$$
$$
= \frac{1}{b} \int_0^{\infty} \frac{1}{\frac{a}{b} + u^2} \, du
$$
$$
= \frac{1}{b} \cdot \frac{\pi}{2 \sqrt{a/b}} = \frac{\pi}{2b} \cdot \sqrt{\frac{b}{a}}
$$
$$
= \frac{\pi}{2} \cdot \frac{\sqrt{b}}{b \sqrt{a}} = \frac{\pi}{2 \sqrt{ab}}
$$
$$
\boxed{ \int_0^{\frac{\pi}{2}} \frac{1}{a \cos^2 x + b \sin^2 x} \, dx = \frac{\pi}{2 \sqrt{ab}} }
$$
(b) Discuss the convergence of the sequence whose ( n )-th term is:
$$
a_n = \frac{(-1)^n}{n} + 1.
$$
Solution:To test for convergence, take the limit of the sequence as ( $n \to \infty $):
$$
\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left( \frac{(-1)^n}{n} + 1 \right)
$$
Since $( \frac{(-1)^n}{n} \to 0 ) as ( n \to \infty )$, we have:
$$
\lim_{n \to \infty} a_n = 0 + 1 = 1
$$
So the sequence ($a_n$) converges to:
$$
\boxed{1}
$$
$\textbf{Conclusion:}$ The sequence is convergent.
BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
Q. 6 (a) Show that the equation of the evolute of the curve:
$$
x^{2/3} + y^{2/3} = a^{2/3}
$$
is:
$$
(x+y)^{2/3} + (x-y)^{2/3} = 2 a^{2/3}.
$$
Solution:
Solution for Curve Coordinates and Derivatives
1. Parametric Equations
$$ x = a \cos^3 \theta $$ $$ y = a \sin^3 \theta $$
2. First Derivatives
$$ \frac{dy}{d\theta} = 3a \sin^2 \theta \cos \theta $$ $$ \frac{dx}{d\theta} = -3a \sin \theta \cos^2 \theta $$
3. First Derivative dy/dx
$$ \frac{dy}{dx} = \frac{3a \sin^2 \theta \cos \theta}{-3a \sin \theta \cos^2 \theta} = -\tan \theta $$
4. Second Derivative
$$ \frac{d^2y}{dx^2} = -\sec^2 \theta \cdot \frac{d\theta}{dx} = \frac{\sec^2 \theta}{3a \sin \theta \cos^2 \theta} = \frac{1}{3a \sin \theta \cos^4 \theta} $$
5. Calculation of α
$$ \alpha = x – \frac{y'(1 + y’^2)}{y”} $$ $$ = a \cos^3 \theta – \frac{(-\tan \theta)(1 + \tan^2 \theta)}{\frac{1}{3a \sin \theta \cos^4 \theta}} $$ $$ = a \cos^3 \theta + \frac{\tan \theta \sec^2 \theta}{\frac{1}{3a \sin \theta \cos^4 \theta}} $$ $$ = a \cos^3 \theta + 3a \sin^2 \theta \cos \theta $$
$$ \boxed{\alpha = a \cos \theta (\cos^2 \theta + 3 \sin^2 \theta)} $$
6. Calculation of β
$$ \beta = y + \frac{1 + y’^2}{y”} $$ $$ = a \sin^3 \theta + \frac{1 + \tan^2 \theta}{\frac{1}{3a \sin \theta \cos^4 \theta}} $$ $$ = a \sin^3 \theta + 3a \sin \theta \cos^2 \theta $$
$$ \boxed{\beta = a \sin \theta (\sin^2 \theta + 3 \cos^2 \theta)} $$
Final Relationship Proof
$$ \alpha + \beta = a (\sin \theta + \cos \theta)^3 $$ $$ \alpha – \beta = a (\cos \theta – \sin \theta)^3 $$ $$ (\alpha + \beta)^{2/3} = a^{2/3} (\sin \theta + \cos \theta)^2 $$ $$ (\alpha – \beta)^{2/3} = a^{2/3} (\cos \theta – \sin \theta)^2 $$
Combining both expressions: $$ (\alpha + \beta)^{2/3} + (\alpha – \beta)^{2/3} = a^{2/3} \left[ (\sin \theta + \cos \theta)^2 + (\cos \theta – \sin \theta)^2 \right] $$
Expanding the squares: $$ = a^{2/3} \left[ (1 + 2 \sin \theta \cos \theta) + (1 – 2 \sin \theta \cos \theta) \right] $$
Simplifying: $$ = a^{2/3} \cdot 2 $$
$$ \boxed{(\alpha + \beta)^{2/3} + (\alpha – \beta)^{2/3} = 2a^{2/3}} $$
(b) Find the areas of the regions that lie inside the circle $( r=a \cos \theta )$ and outside the cardioid $( r=a(1-\cos \theta) )$.
Solution:
Given: $$ \text{Circle: } r = a \cos \theta $$ $$ \text{Cardioid: } r = a(1 – \cos \theta) $$
Step 1: Find Points of Intersection Set the equations equal: $$ a \cos \theta = a(1 – \cos \theta) $$ $$ \cos \theta = 1 – \cos \theta $$ $$ 2 \cos \theta = 1 $$ $$ \cos \theta = \frac{1}{2} $$ $$ \theta = \pm \frac{\pi}{3} $$
Step 2: Area Calculation The area is given by: $$ A = \frac{1}{2} \int_{-\pi/3}^{\pi/3} \left[ (a \cos \theta)^2 – (a(1 – \cos \theta))^2 \right] d\theta $$
Simplify the integral: $$ = \frac{a^2}{2} \int_{-\pi/3}^{\pi/3} \left[ \cos^2 \theta – (1 – 2 \cos \theta + \cos^2 \theta) \right] d\theta $$ $$ = \frac{a^2}{2} \int_{-\pi/3}^{\pi/3} (2 \cos \theta – 1) d\theta $$
Step 3: Evaluate the Integral $$ A = \frac{a^2}{2} \left[ 2 \sin \theta – \theta \right]_{-\pi/3}^{\pi/3} $$ $$ = \frac{a^2}{2} \left[ \left(2 \cdot \frac{\sqrt{3}}{2} – \frac{\pi}{3}\right) – \left(2 \cdot \left(-\frac{\sqrt{3}}{2}\right) – \left(-\frac{\pi}{3}\right)\right) \right] $$ $$ = \frac{a^2}{2} \left[ \sqrt{3} – \frac{\pi}{3} + \sqrt{3} – \frac{\pi}{3} \right] $$ $$ = \frac{a^2}{2} \left[ 2\sqrt{3} – \frac{2\pi}{3} \right] $$
$$ \boxed{A = a^2 \left( \sqrt{3} – \frac{\pi}{3} \right)} $$
BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
Q. 7 (a) Find the solution of the following system of equations:
$$ x_1 + 2x_2 + 2x_3 = 2 $$
$$ x_1 + 8x_3 + 5x_4 = -6 $$
$$ x_1 + x_2 + 5x_3 + 5x_4 = 3 $$
Also, find the basis and dimension of the solution space.
Solution:
Step 1: Construct Augmented Matrix $$ \left[\begin{array}{cccc|c} 1 & 2 & 2 & 0 & 2 \\ 1 & 0 & 8 & 5 & -6 \\ 1 & 1 & 5 & 5 & 3 \end{array}\right] $$
Step 2: Row Reduction (RREF)
$( R_2 \rightarrow R_2 – R_1 ), ( R_3 \rightarrow R_3 – R_1 )$: $$ \left[\begin{array}{cccc|c} 1 & 2 & 2 & 0 & 2 \\ 0 & -2 & 6 & 5 & -8 \\ 0 & -1 & 3 & 5 & 1 \end{array}\right] $$
$( R_2 \rightarrow \frac{-1}{2} R_2 )$: $$ \left[\begin{array}{cccc|c} 1 & 2 & 2 & 0 & 2 \\ 0 & 1 & -3 & -2.5 & 4 \\ 0 & -1 & 3 & 5 & 1 \end{array}\right] $$
$( R_3 \rightarrow R_3 + R_2 )$: $$ \left[\begin{array}{cccc|c} 1 & 2 & 2 & 0 & 2 \\ 0 & 1 & -3 & -2.5 & 4 \\ 0 & 0 & 0 & 2.5 & 5 \end{array}\right] $$
Step 3: Back Substitution From$ ( R_3 )$: $$ 2.5x_4 = 5 \Rightarrow x_4 = 2 $$ $From ( R_2 )$: $$ x_2 – 3x_3 – 2.5(2) = 4 \Rightarrow x_2 = 9 + 3x_3 $$ $From ( R_1 )$: $$ x_1 + 2(9 + 3x_3) + 2x_3 = 2 \Rightarrow x_1 = -16 – 8x_3 $$
Final Parametric Solution: $$ \begin{cases} x_1 = -16 – 8t \\ x_2 = 9 + 3t \\ x_3 = t \\ x_4 = 2 \end{cases} $$
Solution Space Properties:
Particular solution when $( t = 0 )$: $$ \begin{bmatrix} -16 \\ 9 \\ 0 \\ 2 \end{bmatrix} $$
Dimension: 1 (one free variable $( t )$
Basis vector: $$ \begin{bmatrix} -8 \\ 3 \\ 1 \\ 0 \end{bmatrix} $$
(b) Verify the Cayley-Hamilton theorem for the matrix:
$$
A = \begin{bmatrix} 2 & 4\\ 2 & 5 \end{bmatrix}
$$
and find its inverse. Also, express:
$$
A^5 – 4A^4 – 7A^3 + 11A^2 – A – 10I
$$
as a linear polynomial in ( A ).
Solution:Given Matrix:
$$
A = \left[\begin{array}{cc}
2 & 4 \\
2 & 5
\end{array}\right]
$$
Part 1: Verify the Cayley-Hamilton Theorem
Step 1: Characteristic Polynomial
$$ \left|\begin{array}{cc} 2 – \lambda & 4 \\ 2 & 5 – \lambda \end{array}\right| = (2 – \lambda)(5 – \lambda) – (2)(4) $$ $$ = \lambda^2 – 7\lambda + 2 $$
Step 2: Compute ( $A^2 $)
$$ A^2 = \left[\begin{array}{cc} 2 & 4 \\ 2 & 5 \end{array}\right] \left[\begin{array}{cc} 2 & 4 \\ 2 & 5 \end{array}\right] = \left[\begin{array}{cc} 12 & 28 \\ 14 & 33 \end{array}\right] $$
Step 3: Evaluate $( A^2 – 7A + 2I )$
$$ -7A = \left[\begin{array}{cc} -14 & -28 \\ -14 & -35 \end{array}\right], \quad 2I = \left[\begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array}\right] $$ $$ A^2 – 7A + 2I = \left[\begin{array}{cc} 12 & 28 \\ 14 & 33 \end{array}\right] + \left[\begin{array}{cc} -14 & -28 \\ -14 & -35 \end{array}\right] + \left[\begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array}\right] = \left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right] $$
Part 2: Find the Inverse of ( A )
Step 1: Determinant and Adjugate
$$ \det(A) = (2)(5) – (4)(2) = 10 – 8 = 2 $$ $$ \text{Adj}(A) = \left[\begin{array}{cc} 5 & -4 \\ -2 & 2 \end{array}\right] $$
Step 2: Inverse
$$ A^{-1} = \frac{1}{2} \left[\begin{array}{cc} 5 & -4 \\ -2 & 2 \end{array}\right] = \left[\begin{array}{cc} \frac{5}{2} & -2 \\ -1 & 1 \end{array}\right] $$
Part 3: Simplify the Matrix Polynomial
Use the identity $( A^2 = 7A – 2I )$ repeatedly:
Step 1: Reduce Higher Powers
$$ A^3 = A \cdot A^2 = A(7A – 2I) = 7A^2 – 2A = 47A – 14I $$ $$ A^4 = A \cdot A^3 = A(47A – 14I) = 315A – 94I $$ $$ A^5 = A \cdot A^4 = A(315A – 94I) = 2111A – 630I $$
Step 2: Combine Like Terms
$$ A^5 – 4A^4 – 7A^3 + 11A^2 – A – 10I $$ $$ = (2111A – 630I) – 4(315A – 94I) – 7(47A – 14I) + 11(7A – 2I) – A – 10I $$
Combine coefficients: $$ A\text{-term: } 2111 – 1260 – 329 + 77 – 1 = 598 $$ $$ I\text{-term: } -630 + 376 + 98 – 22 – 10 = -188 $$ $$ \Rightarrow 598A – 188I $$
Final Answer
- Cayley-Hamilton: $( A^2 – 7A + 2I = 0 )$, verified.
- Inverse: $$ A^{-1} = \left[\begin{array}{cc} \frac{5}{2} & -2 \\ -1 & 1 \end{array}\right] $$
- Polynomial: $$ A^5 – 4A^4 – 7A^3 + 11A^2 – A – 10I = 598A – 188I $$
BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
Q. 8 (a) Expand ( $sin x$ ) in a Taylor series about ( $x=\frac{\pi}{2} $).
Solution:We need to expand sin x in a Taylor series about ($ x = \frac{\pi}{2} $). The Taylor series of a function f(x) about a point a is:
$$
f(x) = f(a) + f'(a)(x – a) + \frac{f”(a)}{2!}(x – a)^2 + \frac{f”'(a)}{3!}(x – a)^3 + \cdots
$$
Step 1: Compute the Function and Derivatives
For $$ ( f(x) = sin x ) at ( a = \frac{\pi}{2} ):$$
- $$( f(x) = sin x ) → ( f(\frac{\pi}{2}) = 1 )$$
- $$( f'(x) = cos x ) → ( f'(\frac{\pi}{2}) = 0 )$$
- $$( f”(x) = -sin x ) → ( f”(\frac{\pi}{2}) = -1 )$$
- $$( f”'(x) = -cos x ) → ( f”'(\frac{\pi}{2}) = 0 )$$
- $$( f^{(4)}(x) = sin x ) → ( f^{(4)}(\frac{\pi}{2}) = 1 )$$
- $$( f^{(5)}(x) = cos x ) → ( f^{(5)}(\frac{\pi}{2}) = 0 )$$
- $$( f^{(6)}(x) = -sin x ) → ( f^{(6)}(\frac{\pi}{2}) = -1 )$$
Pattern repeats every 4 steps: ( 1, 0, -1, 0 ).
Step 2: Taylor Series Terms
$$ \sin x = f(\frac{\pi}{2}) + f'(\frac{\pi}{2})(x – \frac{\pi}{2}) + \frac{f”(\frac{\pi}{2})}{2!}(x – \frac{\pi}{2})^2 + \frac{f”'(\frac{\pi}{2})}{3!}(x – \frac{\pi}{2})^3 + \cdots $$
Step 3: Write the Series
$$ \sin x = 1 – \frac{1}{2}(x – \frac{\pi}{2})^2 + \frac{1}{24}(x – \frac{\pi}{2})^4 – \frac{1}{720}(x – \frac{\pi}{2})^6 + \cdots $$
General Form: $$ \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(x – \frac{\pi}{2})^{2n} $$
(b) By using the Beta function, evaluate:
$$
\int_0^1 x^4(1-\sqrt{x})^5 \,dx.
$$
Solution:
Given integral: $$ I = \int_{0}^{1} x^4 (1 – \sqrt{x})^5 \, dx $$
Step 1: Substitution
Let ($ u = \sqrt{x} $), then: $$ x = u^2 \quad \Rightarrow \quad dx = 2u \, du $$
Change of limits: $$ x = 0 \Rightarrow u = 0, \quad x = 1 \Rightarrow u = 1 $$
Step 2: Transform the Integral $$ I = \int_{0}^{1} (u^2)^4 (1 – u)^5 \cdot 2u \, du = 2 \int_{0}^{1} u^9 (1 – u)^5 \, du $$
Step 3: Express as Beta Function
The Beta function is defined as: $$ B(m,n) = \int_{0}^{1} t^{m-1} (1 – t)^{n-1} \, dt = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} $$
Comparing with the standard form: $$ 2 \int_{0}^{1} u^{10 – 1} (1 – u)^{6 – 1} \, du = 2B(10,6) $$
Step 4: Evaluate Beta Function $$ 2B(10,6) = 2 \cdot \frac{\Gamma(10)\Gamma(6)}{\Gamma(16)} = 2 \cdot \frac{9! \cdot 5!}{15!} $$ $$ = 2 \cdot \frac{362880 \cdot 120}{1307674368000} = \frac{87091200}{1307674368000} = \frac{1}{15015} $$
$$ \boxed{I = \frac{1}{15015}} $$
BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
Q. 9 (a) Develop the Fourier series on ( $-\pi < x < \pi $) to represent the function defined as:
$$
f(x) =
\begin{cases}
0, & -\pi < x < 0 \\
\pi, & 0 < x < \pi
\end{cases}
$$
Solution:
Fourier series representation: $$ f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) $$
Step 1: Compute $a_0$ $$ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\, dx = \frac{1}{\pi} \left( \int_{-\pi}^{0} 0\, dx + \int_{0}^{\pi} \pi\, dx \right) = \frac{1}{\pi} (\pi^2 – 0) = \pi $$
Step 2: Compute $ a_n$ $$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx)\, dx = \frac{1}{\pi} \int_{0}^{\pi} \pi \cos(nx)\, dx $$\\$$= \int_{0}^{\pi} \cos(nx)\, dx = \left. \frac{\sin(nx)}{n} \right|_0^{\pi} = 0 $$
Step 3: Compute $ b_n$ $$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx)\, dx = \frac{1}{\pi} \int_{0}^{\pi} \pi \sin(nx)\, dx $$\\$$= \int_{0}^{\pi} \sin(nx)\, dx = \left. \frac{-\cos(nx)}{n} \right|_0^{\pi} = \frac{1 – \cos(n\pi)}{n} $$ $$ = \frac{1 – (-1)^n}{n} = \left\{ \begin{array}{ll} \frac{2}{n}, & \text{if } n \text{ is odd} \\ 0, & \text{if } n \text{ is even} \end{array} \right. $$
Step 4: Final Fourier Series $$ f(x) \sim \frac{\pi}{2} + \sum_{n=1}^{\infty} \frac{1 – (-1)^n}{n} \sin(nx) $$
Only odd ( n ) contribute. So: $$ f(x) \sim \frac{\pi}{2} + 2 \left( \frac{\sin x}{1} + \frac{\sin(3x)}{3} + \frac{\sin(5x)}{5} + \cdots \right) $$
(b) (i) Evaluate $( \operatorname{div} \vec{F} )$ and $( \operatorname{Curl} \vec{F} )$, where:
$$
\vec{F} = \operatorname{grad} \left(x^2+y^2+z^2\right) e^{-\sqrt{x^2+y^2+z^2}}.
$$
Solution:
Given vector field: $$ \vec{F} = \nabla (x^2 + y^2 + z^2)\, e^{-\sqrt{x^2 + y^2 + z^2}} $$
Step 1: Simplify the expression
Let $r = \sqrt{x^2 + y^2 + z^2}$, then: $$ \vec{F} = \nabla (r^2)\, e^{-r} = 2r\, e^{-r} \, \hat{r} $$
where $\hat{r} = \dfrac{x\hat{i} + y\hat{j} + z\hat{k}}{r}$ is the unit radial vector.
Step 2: Compute Divergence $\nabla \cdot \vec{F}$
In spherical coordinates, for a radial vector field $\vec{F} = f(r)\hat{r}$: $$ \nabla \cdot \vec{F} = \frac{1}{r^2} \frac{d}{dr} \left( r^2 f(r) \right) $$
Here $f(r) = 2r e^{-r}$, so: $$ \nabla \cdot \vec{F} = \frac{1}{r^2} \frac{d}{dr} \left( 2r^3 e^{-r} \right) = \frac{1}{r^2} \left( 6r^2 e^{-r} – 2r^3 e^{-r} \right) $$ $$ = (6 – 2r) e^{-r} $$
Step 3: Compute Curl $\nabla \times \vec{F}$
For any radial vector field $\vec{F} = f(r)\hat{r}$: $$ \nabla \times \vec{F} = 0 $$
This is because: $$ \nabla \times (f(r)\hat{r}) = \nabla f(r) \times \hat{r} + f(r)(\nabla \times \hat{r}) = 0 $$
since $\nabla f(r)$ is parallel to $\hat{r}$ and $\nabla \times \hat{r} = 0$.
Final Results: $$ \boxed{\nabla \cdot \vec{F} = (6 – 2r)\, e^{-r}}, \quad \boxed{\nabla \times \vec{F} = \vec{0}} $$
(ii) Let:
$$
\vec{F} = (-4x – 3y + az) \hat{\imath} + (bx + 3y + 5z) \hat{\jmath} + (4x + cy + 3z) \hat{k}.
$$
Find the values of ( a, b ), and ( c ) such that ($ \vec{F} $) is irrotational.
Solution:
Condition for Irrotational Field:
A vector field $\vec{F}$ is irrotational if its curl is zero: $$ \nabla \times \vec{F} = \vec{0} $$
Step 1: Compute the Curl
The curl in Cartesian coordinates is: $$ \nabla \times \vec{F} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{vmatrix} $$
Which expands to: $$ \nabla \times \vec{F} = \left( \frac{\partial F_z}{\partial y} – \frac{\partial F_y}{\partial z} \right)\! \hat{\imath} – \left( \frac{\partial F_z}{\partial x} – \frac{\partial F_x}{\partial z} \right)\! \hat{\jmath} + \left( \frac{\partial F_y}{\partial x} – \frac{\partial F_x}{\partial y} \right)\! \hat{k} $$
Step 2: Calculate Partial Derivatives
Compute each component of the curl:
1. $\hat{\imath}$-component:
$$ \frac{\partial F_z}{\partial y} – \frac{\partial F_y}{\partial z} = c – 5 $$
2. $\hat{\jmath}$-component:
$$ \frac{\partial F_z}{\partial x} – \frac{\partial F_x}{\partial z} = 4 – a $$
3. $\hat{k}$-component:
$$ \frac{\partial F_y}{\partial x} – \frac{\partial F_x}{\partial y} = b – (-3) = b + 3 $$
Step 3: Set Curl to Zero
For $\nabla \times \vec{F} = \vec{0}$, each component must vanish: $$ \begin{cases} c – 5 = 0 \\ 4 – a = 0 \\ b + 3 = 0 \end{cases} $$
Solving gives: $$ a = 4,\quad b = -3,\quad c = 5 $$
Final Answer
The vector field $\vec{F}$ is irrotational when: $$ \boxed{a = 4,\quad b = -3,\quad c = 5} $$
Verification
Substituting these values back into the curl components: $$ \nabla \times \vec{F} = (5 – 5)\hat{\imath} – (4 – 4)\hat{\jmath} + (-3 + 3)\hat{k} = \vec{0} $$
BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams
Conclusion:
In this post, we provided detailed solutions for BEU Maths PYQs for CSE / BEU B.Tech CSE Math Solved Papers previous year questions.BEU CSE Math 2023 PYQ with Solutions and 1st Semester Math Solved Papers are now available to help students prepare effectively for Bihar Engineering University exams Understanding these solutions will help you strengthen your core concepts and improve your exam performance. Keep practicing and exploring related topics to enhance your knowledge.
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