Let \(A=\left( \begin{matrix} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{matrix} \right) \)
The order of A is 3 x 3
\(\therefore \rho (A)\le 3\) [Since minimum of (3,3) is 3]
Let us transform the matrix A to an echelon form

This matrix is in echelon from and number of non zero rows is 3.
\(\therefore \rho (A)=3\)

Scale:
In x axis 1 cm = 10 units
In y axis 1 cm = 2 units
Plot the points (40, 6.2), (50, 7.2), (60, 9.1) and (70, 12). At x = 48, draw a vertical line to the graph and from the intersecting point, draw a horizontal line to meet the y-axis
From the graph, we find that when x = 48, the value of Y is equal to 6.8.

(i) Max (2000,3500,4000) = 4000
∴ Crop C is the best according to maximin criteria
(ii) Min (8000,5000,5000) = 5000
∴ Crop B and C are best according to minimax criteria

Let k be the sixth term of the series in the difference table.
First we find the forward differences

Given that the second differences are constant
∴ k – 55 = 3
k = 58
∴ the sixth term of the series is 58

 First allocation:

Second allocation:

Third allocation:

 Fourth allocation:

 Final allocation:

The total transportation cost is
\( =(15 \times 9)+(10 \times 5)+(35 \times 4) +(15 \times 7)+(25 \times 6) \)
= 135 + 50 + 140 + 105 + 150 = Rs. 580

Max (–10,–4, 0) = 0. Since the maximum payoff is 0, the alternative Z is selected by the businessman

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.

Cost of living index number = \(\frac {\sum PV}{\sum V}\) = \(\frac {12610}{100}\) = 126.10
∴  The cost ofliving has increased upto 26.10% in 2011 as compared to 2010.

The line of least fit is a line from which the sum of the deviations of various points is zero. This is the best method for obtaining the trend values.
(i) The straight line trend is represented by
 Y = a + b X ...(1)
where Y is the actual value and X is time
(ii)  The constants 'a' and 'b' are estimated by solving the following two normal Equations
 \(\sum\) Y = n a + b \(\sum\) X
\(\sum\) XY = a\(\sum\) X + b  \(\sum\) X²   where n is the number of years given in the data.
(iii) By substituting the values of 'a' and 'b' in the trend equation (1), we get the line of best fit.

Separating the variables, we get
\(\frac { { \sec }^{ x }x }{ \tan x } dx+\frac { { \sec }^{ 2 } }{ { \tan y } } dy=0\)
Integrating, we get
ഽ\(\frac { { \sec }^{ x }x }{ \tan x } \)dx + ഽ\(\frac { { \sec }^{ 2 } }{ { \tan y } } dy\) = c
log tan x + log tan y = log c
log(tan x tan y) = log c
tan x tan y = c

\(I=\int _{ 0 }^{ 1 }{ \sqrt { x(x-1) } dx } \)
= \(\int _{ 0 }^{ 1 }{ \sqrt { { x }^{ 2 }-xdx } } \)
= \(\int _{ 0 }^{ 1 }{ \sqrt { { x }^{ 2 }-x+\cfrac { 1 }{ 4 } -\cfrac { 1 }{ 4 } dx } } \)
= \(\int _{ 0 }^{ 1 }{ \sqrt { \left( x-\cfrac { 1 }{ 2 } \right) ^{ 2 }-\left( \cfrac { 1 }{ 2 } \right) ^{ 2 }dx } } \)
\(\left[ \because \int { \sqrt { { x }^{ 2 }-{ a }^{ 2 } } dx } =\cfrac { x }{ 2 } \sqrt { { x }^{ 2 } } -\cfrac { { a }^{ 2 } }{ 2 } \log\left| x+\sqrt { { x }^{ 2 }-{ a }^{ 2 } } \right| \right] \)
= \(\left[ \cfrac { x-\frac { 1 }{ 2 } }{ 2 } \sqrt { { x }^{ 2 }-x } -\cfrac { 1 }{ 8 } \log|x-\cfrac { 1 }{ 2 } +\sqrt { { x }^{ 2 }-x } \right] _{ 0 }^{ 1 }\)
= \(\left[ \cfrac { 1-\frac { 1 }{ 2 } }{ 2 } \sqrt { 0 } -\cfrac { 1 }{ 8 } \log|1-\cfrac { 1 }{ 2 } +0| \right] -\left[ 0-\cfrac { 1 }{ 8 } \log|-\cfrac { 1 }{ 2 } | \right] \)
= \(-\cfrac { 1 }{ 8 } \log\left| \cfrac { 1 }{ 2 } \right| +\cfrac { 1 }{ 8 } \log\left| \cfrac { 1 }{ 2 } \right| \)
= 0

Sample size n = 100,
Sample mean \(\\ \bar { X } =3.5\)
Population mean μ = 4
Population standard deviation σ = 3
Null Hypotheses: There is no significant difference in the mean. i.e., H_o : μ = 4
Alternative Hypotheses : There is Significant difference in the mean.
i.e., H₁ : μ ≠ H
The level of significance ∝ = 5% = 0.05
Applying the test statistic,\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 3.5-4 }{ \frac { 3 }{ \sqrt { 100 } } } =\frac { -.5 }{ .3 } =-1.667\)
\(\Rightarrow |Z|=1.667\)
\({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here Z < \({ Z }_{ \frac { \alpha }{ 2 } }\)i.e., 1.667<1.96
Inference: Since Z<\({ Z }_{ \frac { \alpha }{ 2 } }\)at 5% level of significance, the null hypothesis H₀ is accepted. Hence there is no Significant difference in the mean.

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.

\(\int _{ 1 }^{ 4 }{ f\left( x \right) } =\int _{ 1 }^{ 2 }{ f\left( x \right) } dx+\int _{ 2 }^{ 4 }{ f\left( x \right) } dx\)
\(=\int _{ 1 }^{ 2 }{ \left( 4x+3 \right) } dx+\int _{ 2 }^{ 4 }{ \left( 3x+5 \right) } dx\)

\(={ \left( { 2x }^{ 2 }+3x \right) }_{ 1 }^{ 2 }+{ \left( \frac { { 3x }^{ 2 } }{ 2 } +5x \right) }_{ 2 }^{ 4 }\)
\(=\left[ 2\left( { 2 }^{ 2 } \right) +3(2) \right] -\left[ 2{ { \left( 1 \right) }^{ 2 }+3(1) } \right] +\left[ \frac { 3\left( { 4 }^{ 2 } \right) }{ 2 } +5(4) \right] -\left[ \frac { 3\left( { 2 }^{ 2 } \right) }{ 2 } +5(2) \right] \)
= (8+6) - (2-3) + (24+20) - (6+10)
= 14-5+44-16
= 58 - 21 = 37

Given average = λ = \(\frac { 3 }{ 20 } \) = 0.15
P(will not be more than one failure)
= P(X ≤ 1)
= P(X = 0) + P(X = 1)
= \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } \)
= e^-λ (1+λ) =e^-0.15 (1+0.15)
= (0.8607) (1.15)
= (0.98981 [e^-0.15 =0.8607]
∴ Probability that there will not be more than one failure = 0.98981

Let p be the probability of drug's side effect
∴ p = \(\frac { 3 }{ 100 } \) =.03
⇒ q = 1-p = 1-0.03 = 0.97 and n = 10
P (atleast one person will have side effect)
= P(X ≥ 1)
= 1 - P(X < 1)
= 1 - [P(X =0)]       ∵ p(x) =ⁿC_x p^xq^n-x, n=10, x=0
= 1-[¹⁰C₀ (0.03)⁰ (0.97)¹⁰ ]
= 1-(0.97)¹⁰    [∵ ¹⁰C₀ = 1 and (0.03)⁰ = 1]
= 1-0.7374
P(X≥1) = 0.2626

\(\int { \frac { dx }{ \sqrt { { x }^{ 2 }-3x+2 } } } \)
\(=\int { \frac { dx }{ \sqrt { { x }^{ 2 }-3x+\frac { 9 }{ 4 } -\frac { 9 }{ 4 } } +2 } } \)
\({ \left[ \frac { 1 }{ 2 } (3) \right] }^{ 2 }=\frac { 9 }{ 4 } \)
Adding and subtracting \(\frac { 9 }{ 4 } \)
\(=\int { \frac { dx }{ \sqrt { { \left( x-\frac { 3 }{ 2 } \right) }^{ 2 } } -\frac { 1 }{ 4 } } } \)
\(=\int { \frac { dx }{ \sqrt { { \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } } } \)
\(\left[ \because \int { \frac { dx }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } } =\log { \left| x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right| } +c \right] \)
\(=\log { \left| \left( x-\frac { 3 }{ 2 } \right) +\sqrt { { x }^{ 2 }-3x+2 } \right| } +c\)

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.

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

When a coin is tossed, sample space S = {H, T}
Since he is receiving Rs. 4 for a head and pays Rs. 2 for a tail,
∴ X take values 4 and -2.
∴ Probability for getting a head is \(\frac{1}{2}\) and probability for getting a tail is \(\frac{1}{2}\).
The probability mass function is

∴ Expectation E(X) = Σxp(x)
= 4(\(\frac{1}{2}\)) - 2(\(\frac{1}{2}\))
= 2-1 = 1
∴ His expectation is Rs. 1
E(x²) = ∑x²p(x)
= 4²(\(\frac{1}{2}\)) + (-2)²(\(\frac{1}{2}\))
= 16(\(\frac{1}{2}\)) + 4(\(\frac{1}{2}\))
= 8 + 2 = 10
Variance (X) = E(X²)-[E(X)]²
= 10-1² = 9
∴ Variance of his gains Rs. 9

We know that,
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(=\int _{ 0 }^{ 1 }{ x{ 4x }^{ 3 } } dx\)
\(=4{ \left[ \frac { { x }^{ 5 } }{ 5 } \right] }_{ 0 }^{ 1 }\)
\(E(X)=\frac { 4 }{ 5 } \)
\(E\left( { X }^{ 2 } \right) =\int _{ -\infty }^{ \infty }{ { x }^{ 2 } } f(x)dx\)
\(=\int _{ 0 }^{ 1 }{ { x }^{ 2 }{ 4x }^{ 3 } } dx\)
\(=4{ \left[ \frac { { x }^{ 6 } }{ 6 } \right] }_{ 0 }^{ 1 }\)
\(=\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
\(V(X)=\left( { x }^{ 2 } \right) -{ [E(X)] }^{ 2 }\)
\(=\frac { 4 }{ 6 } -{ \left[ \frac { 4 }{ 5 } \right] }^{ 2 }\)
\(=\frac { 2 }{ 75 } \)

Given demand functionp = 122 - 5x - 2x² and x = 6
When x₀ = 6, p₀ = 122-5(6)-2(6)²
= 122-30-72
= 122-102
P₀ = 20
p₀x₀ = 20 \(\times\) 6 = 120
Consumer's Surplus
CS \(=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(=\int _{ 0 }^{ 6 }{ (122-5x-2{ x }^{ 2 })dx-120 } \)
\(={ \left[ 122x-\frac { 5{ x }^{ 2 } }{ 2 } -\frac { { 2x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 6 }-120\)
\(=122(6)-5\frac { \left( { 6 }^{ 2 } \right) }{ 2 } -2\frac { \left( { 6 }^{ 3 } \right) }{ 3 } -120\)
\(=732-\frac { 180 }{ 2 } -\frac { 432 }{ 3 } -120\)
= 732-90-144-120
= 732-354
C.S = 378 units

Given marginal cost function
\(MC=\frac { dc }{ dx } =100-10x+0.1{ x }^{ 2 }\)
\(\Rightarrow C=\int { (100-10x+0.1{ x }^{ 2 })dx } \)
\(\Rightarrow C=100x-\frac { { 10x }^{ 2 } }{ 2 } +\frac { { 0.1x }^{ 3 } }{ 3 } +k\) ...(1)
Given fixed cost is Rs. 500 ⇒ k = 500
∴ (1) becomes,
\(C=100x-5{ x }^{ 2 }+\frac { { 0.1x }^{ 3 } }{ 3 } +500\)
Average cost function
\(AC=\frac { C }{ x } =\frac { 100x-{ 5x }^{ 2 }+\frac { { 0.1x }^{ 3 } }{ 3 } }{ x } +500\)
\(AC=100-5x+\frac { { 0.1x }^{ 2 } }{ 3 } +\frac { 500 }{ x } \)

Let X is the random variable which counts the number of heads on the upturned faces. The outcomes are stated below

These values are summarized in the following probability table.

(i) p(x_i) \(\ge\)0\(\forall \) i and
(ii) \(\sum _{ i=0 }^{ 3 }{ p({ x }_{ i })=1 } \)
Hence, p(x_i) is a probability mass function.

Given MC = 2 + 5e^x
\(C=\int { MC } dx+k\)
\(=\int { (2+5{ e }^{ x })dx } +k\)
= 2x + 5e^x + k
x = 0 ⇒ C = 100,
100 = 2(0)+ 5(e⁰ )+ k
k = 95
C = 2x + 5 e^x + 95.

\(\int { \frac { dx }{ x\left( { x }^{ 3 }+1 \right) } } =\int { \frac { { x }^{ 2 } }{ { x }^{ 3 }\left( x^{ 3 }+1 \right) } dx } \)
\(=\frac { 1 }{ 3 } \int { \frac { dz }{ z\left( z+1 \right) } dz } \)
\(=\frac { 1 }{ 3 } \int { \left[ \frac { 1 }{ z } -\frac { 1 }{ z+1 } \right] dz } \)
\(=\frac { 1 }{ 3 } \left[ \log\left| z \right| -\log\left| z+1 \right| \right] +c\)
\(=\frac { 1 }{ 3 } \log\left| \frac { z }{ z+1 } \right| +c\)
\(=\frac { 1 }{ 3 } \log\left| \frac { { x }^{ 3 } }{ { x }^{ 3 }+1 } \right| +c\)

 \(\int { } \)x³e^xdx = \(\int { } \)udv
= uv − u'v₁ + u''v₂ − u'''v₃ + ...
= x³e^x −3x² ex + 6xe^x −6e^x + c
= e^x (x³ −3x² +6x −6)+ c

\(\int { \frac { { e }^{ 3x }+{ e }^{ 5x } }{ { e }^{ x }+{ e }^{ -x } } } dx\)
\(=\int { \left( \frac { { e }^{ 3x }+{ e }^{ 5x } }{ { e }^{ x }+\frac { 1 }{ { e }^{ x } } } \right) } dx\quad =\quad \int { \frac { { e }^{ 3x }+{ e }^{ 5x } }{ \frac { { e }^{ 2x }+1 }{ { e }^{ x } } } } dx\)
\(=\int { \frac { { e }^{ x }\left( { e }^{ 3x }+{ e }^{ 5x } \right) }{ { e }^{ 2x }+1 } } dx\)

\(=\int { { e }^{ 4x } } dx=\frac { { e }^{ 4x } }{ 4 } +c\)

\(\int { \frac { { x }^{ 2 }+{ 5x }^{3 }-9 }{ x+2 } dx=\int { \left[ { x }^{ 2 }+3x-6+\frac { 3 }{ x+2 } \right] } } dx\)
\(=\frac { x^{ 3 } }{ 3 } +\frac { { 3x }^{ 2 } }{ 2 } -6x+3 \log\left| x+2 \right| +c\)
[By simple division,
\(\frac { { x }^{ 3 }+{ 5x }^{ 3 }-9 }{ x+2 } ={ x }^{ 2 }+3x-6+\frac { 3 }{ x+2 } \)]

\(\int { \frac { { ax }^{ 2 }+bx+c }{ \sqrt { x } } dx } \) 
=\(\int { \left( { ax }^{ \frac { 2 }{ 3 } }+{ bx }^{ \frac { 2 }{ 3 } }+{ cx }^{ -\frac { 1 }{ 2 } } \right) dx } \)
=\(a\int { { x }^{ \frac { 3 }{ 2 } }dx+b\int { { x }^{ \frac { 1 }{ 2 } }+c\int { { x }^{ -\frac { 1 }{ 2 } }dx } } } \)
=\(\frac { { 2ax }^{ \frac { 5 }{ 2 } } }{ 5 } +\frac { { 2bx }^{ \frac { 3 }{ 2 } } }{ 3 } +{ 2cx }^{ \frac { 1 }{ 2 } }+k\)

Let the hourly charge for horse riding be Rs. x and the hourly charge for quad bike be Rs. y from the given data,
3x + 4y = 780
2x+ 3y = 560
\(\Delta =\left| \begin{matrix} 3 & 4 \\ 2 & 3 \end{matrix} \right| =3(3)-2(4)=9-8=1\neq 0\)
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied
\(\Delta x=\left| \begin{matrix} 780 & 4 \\ 560 & 3 \end{matrix} \right| =780(30-4(560)\)
= 2340 - 2240 = 100
\(\Delta y=\left| \begin{matrix} 3 & 780 \\ 2 & 560 \end{matrix} \right| =39560)-2(780)\)
= \(1680 - 1560\) = 120
\(\therefore x=\cfrac { \Delta x }{ \Delta } =\cfrac { 100 }{ 1 } =100\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 120 }{ 1 } =120\)
\(\therefore\)  Hourly charges for the two rides are Rs. 100 and Rs. 120 respectively.

The order of A is 3 \(\times\) 4.
\(\rho (A)\le 3.\) 
Let us transform the matrix A to an echelon form

The number of non zero rows is 3.
\(\rho (A)=3.\)

The highest derivative is third order and its power is one
∴ order : 3,
degree : 1

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.