\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -2\frac { dy }{ dx } +3y=0\)
Highest order derivative is \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }\)
∴ order = 2
Power of the highest order derivative \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }\) is 1
∴ Degree = 1

Let I = \(\int _{ 0 }^{ \infty }{ { e }^{ -mx } } { x }^{ 6 }dx\)
Gamma Integral \(\int _{ 0 }^{ \infty }{ { x }^{ n } } { e }^{ -ax }dx=\frac { n! }{ { a }^{ n+1 } } \)
Here n = 6, and a = m
\(\therefore I=\int _{ 0 }^{ \infty }{ { e }^{ -mx } } { x }^{ 6 }dx\)
\(=\frac { 6! }{ { m }^{ 6+1 } } =\frac { 6! }{ { m }^{ 7 } } \)

The assignment problem is a special case of transportation problem where the number of sources and destinations are equal. Here, jobs represent sources and machines represent destinations.

The mathematical form of assignment problem is minimize Z = \(\overset { n }{ \underset { i=1 }{ \Sigma } } \overset { n }{ \underset { j=1 }{ \Sigma } } { { C }_{ ij } }{ x }_{ ij }\) subject to the
Constraints \(\overset { n }{ \underset { i=1 }{ \Sigma } }{ x }_{ ij } = 1, j=1,2,.....n\)
\(\overset { n }{ \underset { j=1 }{ \Sigma } }{ x }_{ ij } = 1, i=1,2,.....n\) and x_ij = 0 or 1 for all i, j

A feasible solution to a transportation problem is a set of non negative values x_ij (i = 1, 2, m, j = 1, 2, ....... n) that satisfies the constraints.
If a basic feasible solution to a transportation problem contains exactly m + n - l allocations. in independent positions, it is called a non degenerate basic feasible solution. Here m is the number of rows and n is the number of columns in a transportation problem.

I. It is a powerful technique used to diagnose the lack of quality in any of the raw materials, process, machines etc.
II. It is essential that the end products should possess the qualities that the consumer expects from the manufacturer.

The R chart is to show the variability or dispersion of the samples taken from the given process.

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

In this method, the weights are calculated by multiplying prices and quantity of the base year.
Cost of living index number = \(\frac {\sum pV}{\sum V}\) where
P = \(\frac {p_{1}}{p_{0}} \times 100 \) is the price relative and
V = \(\sum p_{0}q_{0}\) is the value relative

The ratio between the total value of current period and total value of the base period is known as true value ratio.
(\( \frac{\sum P_{1} q_{1}}{\sum p_{0} q_{0}}\) is a true value tario)

Time reversal test is an important test for testing the consistency of a good index number. This test maintains time consistency by working both forward and backward with respect to time (here time refers to base year and current year). Symbolically the following Relationship should be satisfied, P₀₁ \(\times\) P₁₀ = 1 Fisher's index number formula satisfies the above relationship
\(\text { i.e. } P_{01}=\frac{1}{P_{10}} \text { or } P_{01} \times P_{10}=1(\text { Except the factor } 100 \text { ) }\)

Index numbers are studied to know the relative changes in price and quantity for any two years compared. There are two tests which are used to test the adequacy for an index number. The two tests are as follows.
(i) Time reversal test
(ii) Factor reversal test
The criterion for a good index number is to satisfy the above two tests.

Fisher's price index number is the geometric mean of Laspeyre's and Paasche's price index number. Hence it is weighted index number.
Fisher's price index number = \(\sqrt {\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0} }}{\times}{\frac {\sum p_{1}q_{1}}{\sum p_{0}q_{1}} \times {100}}\)

The weighted aggregate index number using base period weights is called Laspeyre’s price index number.
\(P_{01}^{L}=\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}} \times 100\)
Where p₁ is current year price 
p₀ is base year price 
q₀ is base year quantity

Index number can be classified as follows
(i) Price index number
It measures the general changes in the retail or wholesale price level of a particular or group of commodities.
(ii) Quantity index number
These are indices to measure the changes in the quantity of goods manufactured in a factory.
(iii) Cost of living index number
These are intended to study the effect of change in the price level on the cost of living of diferent classes of people.

 Seasonal index is a measure of how a particular season compares with the average season.

The components of time series are
(i) Secular trend
(ii) Seasonal variations
(iii) Cyclic variations
(iv) Irregular variations

Time series has an important objective to identify the variations and try to eliminate the variations and also helps us to estimate or predict the future values.

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.

Given (x²+ x + 1)dx + (y²−y + 3)dy = 0
It is of the form f(x)dx + g(y)dy = 0
Integrating , we get
ഽ(x²+ x + 1)dx + ഽ(y² − y + 3)dy = c
\(\left( \frac { x^{ 3 } }{ 3 } +\frac { { x }^{ 2 } }{ 2 } +x \right) +\left( \frac { { y }^{ 3 } }{ 3 } -\frac { { y }^{ 2 } }{ 2 } +3y \right) =c\)

The probability of type I error is known as level of significance denoted by \(\alpha\). It is always fixed in advance before collecting the sample information.

The value of test statistic which separates the critical (or rejection) region and the acceptance region is called the critical value or significant value.

The interval within which the unknown value of parameter is, expected to lie is called confidence interval. It indicates the probability that the population parameter lies within a specilied range. If o is the population paramcter, then we choose a small value a, known as level of significance (1% or 5%) and determine 2 constants c, and c,such that p(c₁ < 0 < c₂/t) = 1- \(\alpha\). When t is the value of statistic. The quantities c₁ and c₂ are determined as confidence limits and the interval [c₁, c₂]  within which the unknown value of the population parameter is expected to lieis known as confidence interval.

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.

1. Systematic samples are not random samples.
2. If N is not a multiple of n, then the sampling interval (k) cannot be an integer, thus sample selection becomes difficult.

The standard deviation of the sampling distribution of a statistic is known as its Standard Error.

The statistical constants of the population like mean (μ), variance (σ²) are referred as population parameters.

A selection of a group of individuals from a population in such a way that it represents the population is called as sample.

A random variable Z = \(\frac { X-\mu }{ \sigma } \) follows the standard normal distribution is called the standard normal variate with mean 0 and standard deviation 1. i.e. Z~ N(0,1). Its,probability density function is given by:
\(\varphi(Z)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{Z^{2}}{2}},-\infty< Z< \infty \)

(i) Number of lightnings per second.
(ii) Number of printing mistakes per page in a textbook.

Let p be the probability of getting doublet in a pair of dice.
∴ p = \(\frac { 6 }{ 36 } \) [∵ favourable events are (1, 1) (2,2) (3,3) (4,4) (5,5) (6,6) and n(S) = 36]
⇒ p = \(\frac { 1 }{ 6 } \) ∴ q=1-p =\(1-\frac { 1 }{ 6 } =\frac { 5 }{ 6 } \)]
∴ P (getting 2 success) = P(X = 2)
=⁴C₂ \(\left( \frac { 1 }{ 6 } \right) ^{ 2 }\left( \frac { 5 }{ 6 } \right) ^{ 2 }\)     [∵ p(x) =ⁿC_x p^xq^n-x, n = 4, x = 2 ]

∴ P(X = 2) =\(\frac { 25 }{ 216 } \).

The binomial distribution can be used under the following conditions.
(i) The number of trials en' is finite,
(ii) The trials are independent of each other.
(iii) The probability of success 'p' is constant for each trial.
(iv) In every trial there are only two possible outcomes namely success or failure.

Let X be a random variable follows binomial distribution with p = q = 1/2
P (3 heads) = \(5{ C }_{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ 5-x }\)
\(={ 5C }_{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 5-3 }\)
\(=5{ C }_{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }\)
\(=\frac { 5 }{ 16 } \)

Let X be a random variable follows binomial distribution with probability value p = 1/2 and q = 1/2
probability that exactcy 2 heads occur are as follows
P(X = 2) = \(\left( \begin{matrix} 6 \\ x \end{matrix} \right) { p }^{ x }{ q }^{ n-x }\)
\(=\left( \begin{matrix} 6 \\ 2 \end{matrix} \right) { \left( \frac { 1 }{ 2 } \right) }^{ 2 }{ { \left( \frac { 1 }{ 2 } \right) } }^{ 6-2 }\)
\(=\frac { 15 }{ 64 } \)

\(Let\ I=\int { \left( 4x+2 \right) } \sqrt { { x }^{ 2 }+x+1 } \ dx\)
\(put\ t={ x }^{ 2 }+x+1\)
dt = (2x+1) dx
\(\therefore I=2\int { \left( 2x+1 \right) } \sqrt { { x }^{ 2 }+x+1 } \ dx\)
\(=2\int { \sqrt { t } } dt\)
\(=2\int { { t }^{ \frac { 1 }{ 2 } } } dt=2\frac { { t }^{ \frac { 1 }{ 2 } +1 } }{ \frac { 1 }{ 2 } +1 } +c\)
\(=2\frac { { t }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } +c\)
\(=\frac { 4 }{ 3 } { t }^{ \frac { 3 }{ 2 } }+c\)
\(=\frac { 4 }{ 3 } { \left( { x }^{ 2 }+x+1 \right) }^{ \frac { 3 }{ 2 } }+c\quad \left[ \because t={ x }^{ 2 }+x+1 \right] \)

Var (X) = E(X²) - [E(X)]² where
\(E({ X }^{ 2 })=\begin{cases} \sum { { x }^{ 2 }p(x)\text {for discrete random variable} } \\ \int _{ -\infty }^{ \infty }{ { x }^{ 2 }P(x)\text{dx for continuous random variable} } \end{cases}\)

Given that in an investment profit is Rs. 5000 with probability of 0.62 or a loss of Rs.8000 with a probability of 0.38.
Hence, the probability mass function is

X = x	5000	-8000
P(X = x)	0.61	0.38

∴ Expected gain E(X) = 5000(0.62) - 8000 (0.32)
= 3100-3040
= Rs. 60
Hence, the expected gain is = Rs. 60

The discrete cumulative distribution function or distribution function of a real valued discrete random variable X takes the countable number of points x₁,x₂, .... with corresponding probabilities p(x₁)p(x₂).... and the distribution function is defined by
F_x(x) = P(X≤x) for all x∈R
ie.F_x(x) = \(\sum _{ { x }_{ i }\le x }^{ }{ p({ x }_{ i }) } \)
For a continuous random variable with the probability density function f_x(x) then the distribution function F_x(x) is defined by
F_x(x) = P(X≤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.

A random variable is a real valued function defined on a sample space S and taking values in (-∞, ∞) or whose possible values are numerical outcomes of a random experiment.

The given probability function is

X	1	2	3	4
P(X = x)	k	2k	3k	4k

Since the given function is a probability function, each ρ_i>0 and Σρ_i = 1
⇒ k + 2k + 3k + 4k = 1
⇒10k = 1 ⇒ k = \(\frac{1}{10}\)
⇒ k = 0.1

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

y - 2x - 4 = 0

x	0	-2
y	4	0

Given y - 2.x - 4 = 0
⇒ y-4 = 2x
⇒
Since the area lies to the left of Y-axis, with the limits y = 1 &y = 3.
Area =\(\int _{ 1 }^{ 3 }{ -xdy } \)
\(=\int _{ 1 }^{ 3 }{ -\left( \frac { 1 }{ 2 } \right) (y-4)dy } \)
\(=\frac { 1 }{ 2 } \int _{ 1 }^{ 3 }{ (4-y)dy } =\frac { 1 }{ 2 } { \left[ 4y-\frac { { y }^{ 2 } }{ 2 } \right] }_{ 1 }^{ 3 }\)
\(=\frac { 1 }{ 2 } \left[ \left( 4(3)-\frac { { 3 }^{ 2 } }{ 2 } \right) -\left( 4(1)-\frac { { 1 }^{ 2 } }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ \left( 12-\frac { 9 }{ 2 } \right) -\left( 4-\frac { 1 }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ \left( \frac { 24-9 }{ 2 } \right) -\left( \frac { 8-1 }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ \frac { 15 }{ 2 } -\frac { 7 }{ 2 } \right] =\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } \right] \)

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

\(\int { \left( 3x+x \right) \left( 2-5x \right) } dx\)
\(=\int { \left( 6-15x+2x-{ 5x }^{ 2 } \right) } dx\)
\(=\int { \left( 6-13x-{ 5x }^{ 2 } \right) } dx\)
\(=6x-\frac { { 13x }^{ 2 } }{ 2 } -\frac { { 5x }^{ 3 } }{ 3 } +c\)

\( \int \sqrt{2 x+1} \ d x=\int(2 x+1)^{\frac{1}{2}} d x\)
\(=\frac { { \left( 2x+1 \right) }^{ \frac { 3 }{ 2 } } }{ 3 } +c\)

Let \(A=\left( \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\)  [Since minimum of (2, 2) is 2]
Consider the second order minor
\(\left| \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right| =40-42\)
= \(-2\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)

Given MR = 6 − 3x² − x³
⇒ ഽMR =ഽ(6 − 3x² − x³)dx
\(\Rightarrow \mathrm{R}=6 x-\frac{\not{3} x^{3}}{\not3}-\frac{x^{4}}{4}+k\)
\(\Rightarrow 6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } +k\)
When x = 0, R = 0 ⇒ k = 0
\(\Rightarrow R=6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } \)
Demand function \(P=\frac { R }{ x } =6-{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 4 } \)

Given MR = 14 - x + 9x²
⇒ \(\frac{dR}{dx}\) = 14 - 6x + 9x²
⇒ dR = (14 - 6x + 9x²)dx
⇒ ഽdR = ഽ(14 - 6x + 9x²)dx
\(\Rightarrow \ R=14x-\frac { 6{ x }^{ 2 } }{ 2 } +\frac { { 9x }^{ 3 } }{ 3 } +k\)
When x = 0, R = 0 ⇒ k = 0
R = 14x - 3x²+ 3x³
Demand function \(P=\frac { R }{ x } =\frac { 14x-3{ x }^{ 2 }+{ 3x }^{ 3 } }{ x } \)
⇒ P = 14 − 3x + 3x²

Given m²\(\frac { dc }{ dm } \)+2mc = 2
Dividing by m², we get,
\(\frac { dC }{ dm } +\frac { 2c }{ m } =\frac { 2 }{ { m }^{ 2 } } \)
This is of form \(\frac { dy }{ dx } \)+Py = Q
where P = \(\frac { 2 }{ m } \) and Q = \(\frac { 2 }{ m } \)
\(\int { p } dm=\int { \frac { 2 }{ m } } \)dm = 2 log m = log m²
∴ Integrating Factor (LF.) = \(e^{ \int { P } dm }=e^{ logm^{ 2 } }\)
= m²
∴ The solution is
\(ce^{ \int { P } dm }=\int { Q } e^{ \int { P } dm }m+c.K\)
cm²=\(\int { \frac { 2 }{ { m }^{ 2 } } { m }^{ 2 } } dm+K=\int { 2 } dm+cK\)
cm² = 2m+K
Given that c = 4, when m = 2
4(2²) = 2(2) + K ⇒ 16 - 4 = K
⇒ K=12
∴ (1) becomes
cm² = 2m+ 12
⇒ cm² = 2 (m + 6)