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

⇒ \(\frac { dy }{ dx } \) = -e^x(1+y)
Separating the variables we get,
⇒ \(\frac { dy }{ 1+y } \) = -e^x dx
⇒ log(1+y) = -e^x+c

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.

If the total supply = total demand, then the given problem is a balanced transportation 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 main objective in any product process is to control and maintain a satisfactory quality level of the manufactured product. This is done by Process Control. In process control the proportion of defective items in the production process is to be minimized and it is achieved through the technique of control charts.

(i) It is an important tool for the formulating decision and management policies
(ii) It helps in studying the trends and tendencies
(iii) It determines the inflation and deflation in an economy

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

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

Poisson distribution is the only distribution in which the mean and variance are equal.

\(\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \sin x } dx={ \left[ -\cos x \right] }_{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }\)
\(=-\left( \cos\frac { \pi }{ 3 } -\cos\frac { \pi }{ 6 } \right) \)
\(=\frac { 1 }{ 2 } (\sqrt { 3 } -1)\)

The average number of typographical errors per page in the book is given by \(\lambda\) = (390/520) = 0.75.
Hence using Poisson probability law, the probability of x errors per page is given by
\(P(X=x)=\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } ={ e }^{ -0.75 }=\frac{(0.75)^x}{x!}\) x = 0,1,2,3……
The required probability that a random sample of 5 pages will contain no error is given by :
[P(X = 0)]⁵ = (e-0.75)⁵ = e^-3.75

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

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

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.

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 - \(\frac { 5{ x }^{ 2 } }{ x } +{ x }^{ 3 }\)

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 \(\bar { X } \) = 0532., σ = 0.002, n = 5
The control limits for \(\overset{-} {X}\) chart is 
\(UCL=\overset { = }{ X } +3\frac { \sigma }{ \sqrt { n } } =0.532+3\frac { 0.002 }{ \sqrt { 5 } } =0.5346\)
\(CL=\overset { = }{ X } =0.532\)
\(UCL=\overset { = }{ X } -3\frac { \sigma }{ \sqrt { n } } =0.532-3\frac { 0.002 }{ \sqrt { 5 } } =0.5293\)

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