(i) Max (3000, 1000, 6000) = 6000
∴ Debentures is the best using maximin principle
(ii) Max (10000, 8000, 6000) = 6000
∴ Debentures is the best using minimax principle.

y = f(x) = x³+2x+1 for x = 1,2,3,4,5

Total Supply = Total Demand = 24
\(\therefore\)The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
Given Transportation Problem is:

The least cost is 1 corresponds to the cells (O₁, D₁) and (O3, D₄)
Take the Cell (O₁, D₁) arbitrarily.
Allocatemin (6,4) = 4 units to this cell.

The reduced table is

The least cost corresponds to the cell (O₃, D₄). Allocate min (10,6) = 6 units to this cell.

The reduced table is

The least costis 2 corresponds to the cells (O₁, D₂), (O₂, D₃), (O₃, D₂), (O₃, D₃)
Allocate min (2,6) = 2 units to this cell.

The reduced table is

The least cost is 2 corresponds to the cells (O₂, D₃), (O₃, D₂), (O₃, D₃)
Allocate min ( 8,8) = 8 units to this cell.

The reduced table is

Here allocate 4 units in the cell (O₃, D₂)

Thus we have the following allocations:

Transportation schedule :
O₁⟶D₁, O₁⟶D₂,O₂⟶D₃,O₃⟶D₂,O₃⟶D₄
Total transportation cost
= (4×1)+ (2×2)+(8×2)+(4×2)+(6×1)
= 4+4+16+8+6
= Rs. 38.

Given \(\sigma\) = 10, S.E. \(\bar { X } \) = 3
We know that S.E = \(\frac { \sigma }{ \sqrt { n } } \)
Therefore, \(3=\frac { 10 }{ \sqrt { n } } \Rightarrow \sqrt { n } =\frac { 10 }{ 3 } \)
Taking Squaring on both sides we get
\(n=\left(\frac{10}{3}\right)^{2}=\frac{100}{9}=11.11 \cong 11\), 
The required sample size is 11.

Here p = q = 1/2
(a) probability of A beating B in exactly 3 games out of 4
\(\left( \begin{matrix} 4 \\ 3 \end{matrix} \right) { (\frac { 1 }{ 2 } ) }^{ 3 }\left( \frac { 1 }{ 2 } \right) ^{ 4-3 }\)
= 1/4 = 25%
(b) probability of A beating B in exactly 5 games out of 8
\(\left( \begin{matrix} 8 \\ 5 \end{matrix} \right) { (\frac { 1 }{ 2 } ) }^{ 5 }\left( \frac { 1 }{ 2 } \right) ^{ 8-5 }\)
\(=\frac { 7 }{ 32 } \) = 21.875%
Clearly, the first event is more probable.

Given probability mass function is

∴  The cumulative distribution function F_x(x) is
F_x(0) = 0 if x < 3
F_x(3) = P(X = 3) = 0.3, for 3 ≤x<5
F_x(5) = P(X = 3)+P(X = 5) = 0.3+0.2 = 0.5, for 5≤X<8
F_x(8) = P(X = 3)+P(X+5)+P(X = 8)
= 0.3+0.2+0.3
= 0.8,8≤x<10
F_x(10) = P(X = 3)+P(X = 5)+P(X = 8)+P(X = 10)
= 0.3+0.2+0.3+0.2
= 1 for x ≥10
\(F_{x}(x)= \begin{cases}0, & \text { if } x<3 \\ 0.3, & \text { if } 3 \leq x < 1 \\ 0.5, & \text { if } 5 \leq x < 8 \\ 0.8, & \text { if } 8 \leq x < 10 \\ 1, & \text { if } x ≥ 10 \\ \end{cases}\)

y-1 = x

Given line is y - 1 = x ⇒ y = x + 1
Given limits are from x = - 2 to 3.
In the diagram, the area from x = - 2 to x = -1 lies below the X-axis and the area from x = -1 to x = 3 lies above the X-axis.
∴ Required Area
\(=\int _{ 2 }^{ -1 }{ -ydx+ } \int _{ -1 }^{ 3 }{ ydx } \)
\(=-\int _{ -2 }^{ -1 }{ ydx } +\int _{ -1 }^{ 3 }{ ydx } \)
\(=\int _{ -1 }^{ -2 }{ ydx } +\int _{ -1 }^{ 3 }{ ydx } \left[ \because \int _{ a }^{ b }{ f(x)dx=-\int _{ b }^{ a }{ f(x)dx } } \right] \)
\(=\int _{ -1 }^{ -2 }{ (x+1)dx+\int _{ -1 }^{ 3 }{ (x+1)dx } } \)
\(={ \left( \frac { { x }^{ 2 } }{ 2 } +x \right) }_{ -1 }^{ -2 }+{ \left( \frac { { x }^{ 2 } }{ 2 } +x \right) }_{ -1 }^{ 3 }\)
\(=\left( \frac { 4 }{ 2 } -2 \right) -\left( \frac { 1 }{ 2 } -1 \right) +\left( \frac { 9 }{ 2 } +3 \right) -\left( \frac { 1 }{ 2 } -1 \right) \)
\(=(2-2)-\left( -1\frac { 1 }{ 2 } \right) +\left( \frac { 15 }{ 2 } \right) -\left( -\frac { 1 }{ 2 } \right) \)
\(=0+\frac { 1 }{ 2 } +\frac { 15 }{ 2 } +\frac { 1 }{ 2 } =\frac { 17 }{ 2 } \) sq.units.

Split into simple integrands
\({ \frac { x+2 }{ \sqrt { 2x+3 } } } = { \frac { 1/2( {2x+4) }}{ {( 2x+3 )^{1/2}} } } \)
\(=\int { \frac { 1 }{ 2 } \left\{ { \left( 2x+3 \right) }^{ \frac { 1 }{ 2 } }+{ \left( 2x+3 \right) }^{ -\frac { 1 }{ 2 } } \right\} } dx\)
\(=\frac { 1 }{ 2 } \left\{ \frac { { \left( 2x+3 \right) }^{ \frac { 1 }{ 2 } } }{ (2x+3) + 1} \right\} \)
\(\int { \frac { x+2 }{ \sqrt { 2x+3 } } } dx=\int { \frac { 1 }{ 2 } \left\{ { \left( 2x+3 \right) }^{ \frac { 1 }{ 2 } }+{ \left( 2x+3 \right) }^{ -\frac { 1 }{ 2 } } \right\} } dx\)
\(=\frac { 1 }{ 2 } \left\{ \frac { { \left( 2x+3 \right) }^{ \frac { 3 }{ 2 } } }{ 3 } +{ \left( 2x+3 \right) }^{ -\frac { 1 }{ 2 } } \right\} +c\)

Let A = \(\left( \begin{matrix} 1 & 2 & -1 \\ 2 & 4 & 1 \\ 3 & 6 & 3 \end{matrix}\begin{matrix} 3 \\ -2 \\ -7 \end{matrix} \right) \)
Order of A is 3 \(\times\) 4
∴\(\rho \)(A)\(\le \)3
Consider the third order minors
\(\left| \begin{matrix} 1 & 2 & -1 \\ 2 & 4 & 1 \\ 3 & 6 & 3 \end{matrix} \right| =\) 0, \(\left| \begin{matrix} 1 & -1 & 3 \\ 2 & 1 & -2 \\ 3 & 3 & -7 \end{matrix} \right| =0\) 
\(\left| \begin{matrix} 1 & 2 & 3 \\ 2 & 4 & -2 \\ 3 & 6 & -7 \end{matrix} \right| =0,\) \(\left| \begin{matrix} 2 & -1 & 3 \\ 4 & 1 & -2 \\ 6 & 3 & -7 \end{matrix} \right| =0\)
Since all third order minors vanishes, \(\rho (A)\neq 3\)
Now, let us consider the second order minors,
Consider one of the second order minors \(\\ \\ \left| \begin{matrix} 2 & -1 \\ 4 & 1 \end{matrix} \right| =6\neq 0\)
There is a minor of order 2 which is not zero.
\(\therefore \rho (A)=2\)

Given equation is y = cx + c - c³ ....(1)
Differentiating w.r.t 'x' we get,
\(\frac { dy }{ dx } \) = c(1) + 0-0
⇒ \(\frac { dy }{ dx } \) = c ....(2)
Substituting (2) in (1) we get,
y=\(x\left( \frac { dy }{ dx } \right) +\left( \frac { dy }{ dx } \right) -\left( \frac { dy }{ dx } \right) ^{ 3 }\) which is the required differential equation.