The order of A is 3 \(\times\) 3.
\(\therefore \rho (A)=\le 3.\)
Let us transform the matrix A to an echelon form by using elementary transformations.

The number of non zero rows is 2
∴ Rank of A is 2.
\(\rho (A)=2.\)

(2 - y'')² = y''² + 2y'

⇒ 4-y'' = 2y
The highest derivative is of second order and its poweris 1.
∴ Order is 2 and degree is 1.

(i) Δe^ax = e^a(x+h)−e^x
= e^aX.e^h-e^ax [∵ a^m+n = a^m.aⁿ]
= e^ax[e^h-1]
 (ii) Δ²e^x = Δ.[Δe^x]
= Δ[e^x+h - e^x]
= Δ[e^xe^h - e^x]
= Δe^x [e^h - 1]
= (e^h - 1)Δe^x 
= (e^h−1).(e^h−1).e^x
= (e^h−1)².e^x
(iii) Δ log x = log(x+h) − log x
= log \(\frac{x + h}{x}\)
= log \(\left( \frac { x }{ x } +\frac { h }{ x } \right) \)
= log \(\left( 1+\frac { h }{ x } \right) \)

To assign the different jobs to the different machines (one job per machine) to minimize the overall cost is known as assignment problem.

Given (D²−3D−4)y = 0
The auxiliary equations is
m²− 3m − 4 = 0
⇒ (m − 4)(m + 1) = 0
m = −1, 4
Roots are real and different
∴ The complementary function is Ae^−x+ Be^4x
The general solution is y = Ae^−x + Be^4x

The control charts of variables are
(i) Charts for mean (\(\overline { X } \))
(ii) Charts for Range (R)

An index numberis a device which shows by its variations the changes in a magnitude which is not capable of accurate measurements in itself or of direct valuation in practice

A time series consists of a set of observations arranged in chronological order (either ascending or descending). It is a statistical data which relates to successive intervals or point of time.

When a single value is used as an estimate, the estimate is called a point estimate of the population parameter In other words, an estimate of a population parameter given by a single numberis called as point estimation.

The group of individuals considered under study is called as population. It refers not only to people but to all items that have been chosen for the study.

\(\int _{ 0 }^{ 1 }{ { e }^{ 2x } } dx\)
\(={ { \left[ \frac { { e }^{ 2x } }{ 2 } \right] }_{ 0 } }^{ 1 }=\frac { 1 }{ 2 } \left[ { e }^{ 2(1) }-{ e }^{ 2(0) } \right] \)
\(=\frac { 1 }{ 2 } \left[ { e }^{ 2 }-{ e }^{ 0 } \right] \)  =\(\frac { 1 }{ 2 } \) [e² - 1]

A random variable X is said to follow a normal distribution with parameters mean μ and variance σ², if its probability density function is given by

Let p he the probability of getting a girls
∴ p = \(\frac { 1 }{ 2 } \) [∵ one favourable event and total no of events is 2]
⇒ q = 1-p = 1-\(\frac { 1 }{ 2 } \) = \(\frac { 1 }{ 2 } \) and n = 3
∴ (getting exactly 2 girls)
= P(X = 2)
= ³C₂\(\left( \frac { 1 }{ 2 } \right) ^{ 2 }\left( \frac { 1 }{ 2 } \right) ^{ 1 }\)   [∵ p(x) = ⁿC_x p^xq^n-x, n = 3 and x = 2]
= \(3\left( \frac { 1 }{ 2 } \right) ^{ 3 }=3\times \left( \frac { 1 }{ 2 } \right) ^{ 3 }=3\times \frac { 1 }{ 8 } \) = 0.375
P (getting exactly 2 girls) = 0.375.

ഽ\(\frac { dx }{ \sqrt { { x }^{ 2 }+25 } } \) = ഽ\(\frac { dx }{ \sqrt { { x }^{ 2 }+5^2 } } \)
= \(\log\left| x+\sqrt { { x }^{ 2 }+{ 5 }^{ 2 } } \right| +c\) 
= \(\log\left| x+\sqrt { { x }^{ 2 }+{ 25 }^{ } } \right| +c\)

Let X be a discrete random variable with probability mass function p(x), then its expected value is defined by
E(X) =\(\sum _{ x }^{ }{ x.p(x) } \)

Continuous random variable :
A random variable X which can take on any value (integral as well as fraction) in the interval is called continuous random variable. For eg., height of students in a school.

\(F(x)=\int _{ -\infty }^{ x }{ f(t)dt } \)
\(=\int _{ 100 }^{ x }{ \frac { 100 }{ { t }^{ 2 } } dt,\quad x\ge 100 } \)
\(={ \left[ \frac { 100 }{ -t } \right] }_{ 100 }^{ x },\quad x\ge 100\)
\(F(x)=\left[ 1-\frac { 100 }{ x } \right] ,\ge 100\)

x - 2y - 12 = 0
x = 2y + 12
Required Area
= \(\int _{ 2 }^{ 5 }{ xdy } \)
= \(\int _{ 2 }^{ 5 }{ (2y+12)dy= } [{ { y }^{ 2 }+12y] }_{ 2 }^{ 5 }\)
= (25 + 60)−(4 + 24) = 57 sq.units

\(\int { \sqrt { 1+\sin2x \ dx } } =\int { \sqrt { { \left( \sin x+\cos x \right) }^{ 2 }dx } } \)
=∫(sinx + cosx)dx
=−cos x + sin x + c
[ Change into simple integrands 
1+ sin2x = sin² x + cos² x + 2sin x cos x
= (sin x + cos x)² ]

\(\int { \left( { x }^{ 3 }+7 \right) \left( x-4 \right) dx } \)
= \(\int { \left( { x }^{ 4 }-{ 4x }^{ 3 }+7x-28 \right) dx } \)
\(= \frac { { x }^{ 5 } }{ 5 } -{ x }^{ 4 }+\frac { { 7x }^{ 2 } }{ 2 } -28x+c\)