\(\frac { { d }^{ 2 }y }{ { dx }^{ 3 } } -3{ \left( \frac { dy }{ dx } \right) }^{ 6 }+2y={ x }^{ 2 }\)
∴ order = 3, 
∴ Degree = 1

\(\Gamma \left( \frac { 9 }{ 2 } \right) =\frac { 7 }{ 2 } \Gamma \left( \frac { 7 }{ 2 } \right) \)
\(=\frac { 7 }{ 2 } \Gamma \left( \frac { 5 }{ 2 } \right) \)
\(=\frac { 7 }{ 2 } \times \frac { 5 }{ 2 } \times \Gamma \left( \frac { 3 }{ 2 } \right) \)
\(=\frac { 7 }{ 2 } \times \frac { 5 }{ 2 } \times \left( \frac { 3 }{ 2 } \right) \times \Gamma \left( \frac { 1 }{ 2 } \right) \)
\(=\frac { 7 }{ 2 } \times \frac { 5 }{ 2 } \times \frac { 3 }{ 2 } \times \frac { 1 }{ 2 } \sqrt { \pi } \)
\(=\frac { 105 }{ 16 } \sqrt { \pi } \)

LHS = (1 + Δ) (1 - ∇)

= (1 + Δ) (1 - ∇)
= (1 + E - 1) \((1-\frac { E-1 }{ E } )\)
= E\((1-\frac { E-1 }{ E } )\)
= E - E\((\frac { E-1 }{ E } )\)
= E - (E - 1)
= E - E + 1 = 1
= RHS.

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

The auxiliary equation is m² - 4m + 4 = 0
⇒ (m - 2)² = 0
⇒ m 2,2
The roots are real and equal
∴ Complementary function CF is (Ax + B)e^2x
∴ The general solution is y = (Ax + B)e^2x

The statistical tool applied in process control is the Control Chart. Control Charts are the devices to describe the patterns of variation. It is an instrument to be used in specitication, production and inspection and is the core of statistical quality control. It is essentially a graphic device, simple to construct and easy to interpret.

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 = \(\sqrt {\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0} }}{\times}{\frac {\sum p_{1}q_{1}}{\sum p_{0}q_{1}} \times {100}}\)

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.

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.

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.

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

Given that \(\int _{ 1 }^{ a }{ { 3 }x^{ 2 } } dx=-1\)
\({ \left[ { x }^{ 3 } \right] }_{ 1 }^{ a }=-1\)
a³ −1 = –1
a³ = 0 ⇒ a = 0

A random experiment whose outcomes are of two types namely success S and failure P, occurring with probabilities p and q is called a Bernoulli trial.

ഽ\(\frac { dx }{ \sqrt { { 4x }^{ 2 }-9 } } \)=ഽ\(\frac { dx }{ \sqrt { 4\left[ { x }^{ 2 }-\frac { 9 }{ 4 } \right] } } \)
= \(\frac { 1 }{ 2 } \) ഽ\(\frac { dx }{ \sqrt { { x }^{ 2 }-{ \left( \frac { 3 }{ 2 } \right) }^{ 2 } } } \)
= \(\frac { 1 }{ 2 } \log\left| x+\sqrt { { x }^{ 2 }-{ \left( \frac { 3 }{ 2 } \right) }^{ 2 } } \right| +c\)
= \(\frac { 1 }{ 2 } \log\left| x\sqrt { { x }^{ 2 }-\frac { 9 }{ 4 } } \right| +c\)
= \(\frac { 1 }{ 2 } \log\left| 2x+\sqrt { 4{ x }^{ 2 }-9 } \right| +c\)

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

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

Given MR = 20-5x+3x²
\(\Rightarrow \frac { dR }{ dx } =20-5x+3{ x }^{ 2 }\)
⇒ dR = (20 - 5x + 3x²)dx
⇒ഽdR = ഽ(20-5x+3x²)dx
\(\Rightarrow R=20x-\frac { { 5x }^{ 2 } }{ 2 } +\frac { { 3x }^{ 3 } }{ 3 } +k\)
When x = 0, R = 0 ⇒ k = 0
∴ R = 20x - \(\frac { 5{ x }^{ 2 } }{ x } +{ x }^{ 3 }\)

When two coins are tossed,
Sample space S = {HH, HT, TH, TT}
⇒n(S) = 4
Since getting a head is termed as success,
X takes the values 2, 1, 1,0
∴ P(X = 2) = \(\frac{1}{4}\)[∵ only one (HH) favourable event]
P(X = 1) = \(\frac{1}{4}\)+\(\frac{1}{4}\) = \(\frac{1}{2}\)[∵ favourable events are HT, TH]
P(X = 0) = \(\frac{1}{4}\)[∵only one favourable event]
∴ Probability distribution function is

Here each ρ_i > 0 and Σρ_i = \(\frac{1}{4}\)+\(\frac{1}{2}+\frac{1}{4}=1\)

∫ (2 cos x - 3 sin x + 4 sec² x - 5cosec² x)dx
= 2 ∫cos x dx - 3∫sin x dx +4∫sec² x dx - 5 ∫cosec² x dx
= 2 (sin x) - 3 (-cos x) +4 tan x - 5 (-cot x) + c
= 2sin x + 3cos x + 4 tan x + 5cot x + c

Given A =\(\left( \begin{matrix} 1 & -3 \\ 9 & 1 \end{matrix}\begin{matrix} 4 & 7 \\ 2 & 0 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & -3 \\ 0 & 28 \end{matrix}\begin{matrix} 4 & 0 \\ -34 & -63 \end{matrix} \right) { R }_{ 2 }\rightarrow { R }_{ 2 }-9{ R }_{ 1 }\)
\(-\left( \begin{matrix} 1 & -3 \\ 0 & 0 \end{matrix}\begin{matrix} 4 & 0 \\ \frac { 10 }{ 3 } & -63 \end{matrix} \right) { R }_{ 2 }\rightarrow { R }_{ 2 }+\frac { 28 }{ 3 } .{ R }_{ 1 }\)
The last equivalent matrix is in echelon form and there are 2 non-zero rows
\(\therefore \rho (A)=2\)

Given f'(x) = 8x³-2x, f(2) = 8
f'(x) = 8x³-2x
\(\Rightarrow \int { f'(x)dx=\int { \left( { 8x }^{ 3 }-2x \right) } } dx\)

⇒ f(x) = 2x⁴-x²+c...(1)
Given f(2) = 8
⇒ 8 = 2(2⁴)-2²+c
⇒ 8 = 32 - 4+c
⇒ 8 = 32 - 4+c
⇒ 8 - 28 = c
⇒ c = -20
Substituting c = -20 in (1) we get,
f(x) = 2x⁴-x²-20