Let A =\(\left( \begin{matrix} 1 & 2 & -1 \\ 2 & 4 & 1 \\ 3 & 6 & 3 \end{matrix}\begin{matrix} 3 \\ -2 \\ -7 \end{matrix} \right) \)
The order of A is 3 x 4
\(\therefore \rho \left( A \right) \le 3\) [Since minimum of (3, 3) is 3]
Let us transform the matrix to an echelon form.

The matrix is in echelon form and the number of non- zero rows is 2.
\(\therefore \rho (A)=2\)

The Forward difference table is given below

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

First Allocation:

Second Allocation:

Third Allocation:

Fourth Allocation:

Fifth Allocation:

Sixth Allocation:

Transportation schedule :
T⟶H,T⟶P,B⟶C,B⟶H,M⟶H,M⟶K
The total Transportation cost = ( 5×8) + (25×5)+ (35×5) + (5×11)+ (18×9) + (32×7)
= 40+125+175+55+162+224
= Rs.781

The given differential equation is of the follows
\(\frac { dy }{ dx } \)+ Py =Q where
P = cos x, Q = sinx cosx
∴ \(\int { P } dx=\int { cosx } dx\)
∴ Integrating factor (I.F) =\(e^{ \int { p.dx } }=e^{ sinx }\)
Hence, the solution is
\(ye^{ \int { p.dx } }=\int { Qe^{ \int { p.dx } } } dx+c\)
⇒ \(y.e^{ sinx }=\int { sinx } cosx.e^{ sinx }dx+c\)
put t = sin x ⇒ dt =cos x dx
⇒ \(ye^{ sinx }=\int { t{ e }^{ t } } dt\)  ....(1)

put u = t; dv = et dt
du = dt; v = et
Using integration by parts
\(\int { u } dv=uv-\int { v } du\)
⇒ \(\int { t } e^{ t }dt=te^{ t }-\int { e^{ t }dt } \) = te^t-e^t....(2)
Substituting (2) in (1) we get,
ye^sinx = t e^t-e^t + c
⇒ y e^sinx = e^t(t-1)+c
⇒ y e^sinx = e^sinx (sin x-1)+c   [∵ t = sinx]

Equation of the family of paraboles with latus rectum 4a and whose axes are parallel to the x-axis is (y- k)² = 4a(x- h)
[Where (h, k) is the centre of the parabola]
Differentiating w.r.t. 'x' we get,
2(y-k)\(\left( \frac { dy }{ dx } \right) \) = 4a(1)
⇒ 2(y-k)\(\left( \frac { dy }{ dx } \right) \) = 4a           (1)
Differentiating again w.r.t x we get,
2(y-k)\(\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) (2)\frac { dy }{ dx } \) = 0  (Product rule)
(y-k)\(\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) ^{ 2 }\) = 0 [Divided by 2]
y-k= \(\frac { -\left( \frac { dy }{ dx } \right) ^{ 2 } }{ \frac { { d }^{ 2 }y }{ dx^{ 2 } } } \)    (2)
Substituting (2) in (1) we get,
2\(\frac { -\left( \frac { dy }{ dx } \right) ^{ 2 } }{ \frac { { d }^{ 2 }y }{ dx^{ 2 } } } \left( \frac { dy }{ dx } \right) \)= 4a
⇒ \(-2\left( \frac { dy }{ dx } \right) ^{ 3 }=4a\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) \)
⇒ \(4a\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) ^{ 3 }\)= 0
\(2a\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 3 }=0\).

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.

The errors that arise due to human factors which always vary from one investigator to another in selecting, estimating or using measuring instruments are called Non-Sampling errors.
It may arise in the following ways:
a) Due to negligence and carelessness of the part of either investigator or respondents.
b) Due to lack of trained and qualified investigators.
c) Due to framing of a wrong questionnaire.
d) Due to applying wrong statistical measure.
e) Due to incomplete investigation and sample survey.

Let t = x² + 3x + 7
⇒ dt = (2x + 3)
When x = -1, t = (-1)² +3(-1)+7 = 1-3 + 7= 5
When x = 1, t = 1² +3(1)+7 = 11
∴ I = \(\int _{ 5 }^{ 11 }{ \frac { dt }{ t } } ={ \left( \log { t } \right) }_{ 5 }^{ 11 }\)
\(=\log { 11 } -\log { 5 } \)
\(=\log { \left( \frac { 11 }{ 5 } \right) } \)

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

We are given mean μ = 800 and standard deviation σ = 100
Probability that the procurement of milk between 800 litres to 1000 litres per day is
P(800 < X < 1000)
\(P(\frac { 800-800 }{ 100 })\) < z < \((\frac { 1000-800 }{ 100 })\)
P(0 < Z < 2) = 0.4772 (table value)
Therefore 47.75 percent of societies procure milk between 800 litres to 1000 litres per day.

Probability of getting a degree p = 0.4
∴ q = 1– p
= 1 - 0.4
= 0.6
(i) P (one will be a graduate) = P(X = 1) = 5C₁ (0.4)(0.6)⁴
= 0.2592
(ii) P ( atleast one will be a graduate) = 1–P (none will be a graduate)
= 1-5C₀(P⁰)(Q)5-0
= 1-5C₀(0.4)⁰(0.6)5
= 1-0.0777
= 0.9222

\(Let\ I=\int { \frac { { x }^{ e-1 }+{ e }^{ x-1 } }{ { x }^{ e }+{ e }^{ x } } } dx\)
\(put\ t={ x }^{ e }+{ e }^{ x }\)
\(\Rightarrow dt=\left( e{ x }^{ e-1 }+{ e }^{ x } \right) dx\)
\(dt=e\left( { x }^{ e-1 }+{ e }^{ x-1 } \right) dx\)
\(\Rightarrow \frac { dt }{ e } =\left( { x }^{ e-1 }+{ e }^{ x-1 } \right) dx\)
\(\therefore I=\int { \frac { dt }{ e(t) } } =\frac { 1 }{ e } \int { \frac { dt }{ t } } \)
\(=\frac { 1 }{ e } \log { \left| t \right| } +c\)
\(=\frac { 1 }{ e } \log { \left| { x }^{ e }+{ e }^{ x } \right| } +c\quad \left[ \because t={ x }^{ e }+{ e }^{ x } \right] \)

Given MR = 275 - x - 0.3x²
ഽMR = f(275 - x - 0.3x²)dx
To find the total revenue, when it is increased from 10 to 20 units
\(R=\int _{ 10 }^{ 20 }{ (275-x-0.3{ x }^{ 2 })dx } \)
\(={ \left( 275x-\frac { { x }^{ 2 } }{ 2 } -0.3\frac { { x }^{ 3 } }{ 3 } \right) }_{ 10 }^{ 20 }\)
\(={ \left( 275x-\frac { { x }^{ 2 } }{ 2 } -0.1{ x }^{ 3 } \right) }_{ 10 }^{ 20 }\)
\(=\left[ 275(20)-\frac { { 10 }^{ 2 } }{ 2 } -0.1({ 10 }^{ 3 }) \right] \)
= [5500 - 200 - 800] - [2750 - 50 - 100]
= [5500 - 1000] - [2750 - 150]
= 4500-2600 = 1,900
R = Rs. 1,900

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 function P_d = 85 - 5x and 
Supply function p₅ = 3x - 35
At equilibrium prices,P_d = P_s
⇒ 85 - 5x = 3x - 35
⇒ 85 + 35 = 3x + 5x
⇒ 120 = 8x
⇒ x = \(\frac{120}{8}\) = 15
When x₀ = 15, p₀ = 85-5(15)
= 85 - 75 = 10
p₀ = 10
∴ p₀x₀ = 15\(\times\)10 = 150
Consumer's Surplus
\(Cs=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(=\int _{ 0 }^{ 15 }{ (85-5x)dx-150 } \)
\(={ \left[ 85x-\frac { 5{ x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 15 }-150\)
\(=85(15)-5\frac { { (15) }^{ 2 } }{ 2 } -150\)
\(=1275-\frac { 1125 }{ 2 } -150\)
= 1275 - 562.5 - 150
= 1275 - 712.5
Cs = 562.5

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
\(F_{x}(x)= \begin{cases}0, & \text { if } x<3 \\ 0.3, & \text { if } 3 \leq x < 1 \\ 0.5, & \text { if } 5 \leq x < 8 \\ 0.8, & \text { if } 8 \leq x < 10 \\ 1, & \text { if } x ≥ 10 \\ \end{cases}\)

Given,
Marginal cost MC = 6 +10x − 6x²
C = \(\int { MC } dx+k\)
= \(\int { (6+10x-{ 6x }^{ 2 })dx+k } \)
= 6x + 5x² − 2x³ + k (1)
when x = 2, C = 12 (given)
12 = 12 + 20 −16 + k
k = -4
C = 6x + 5x² − 2x³ − 4
Average cost = \(\frac { C }{ x } =\frac { 6x+{ 5x }^{ 2 }-2{ x }^{ 3 }+{ 4 } }{ x } \)
= 6 + 5x − 2x² − \(\frac { 4 }{ x } \)

Required area = \(\int _{ -1 }^{ 0 }{ -xdx } +\int _{ 0 }^{ 2 }{ xdx } \)
\(=-{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ -1 }^{ 0 }{ +\left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 2 }=-\left[ 0-\frac { 1 }{ 2 } \right] +\left[ \frac { 4 }{ 2 } -0 \right] \)
\(=\frac { 5 }{ 2 } \) sq.units

\(Let\ I=\int { { x }^{ 3 }{ e }^{ 3x } } dx\)
\(Let\ u={ x }^{ 3 };dv= { e }^{ 3x }dx\)
Using Bernoulli's formula
= uv − u'v₁ + u''v2 − u'''v₃ + ...
\(\therefore \int { { x }^{ 3 }{ e }^{ 3x }dx } \)
\(={ x }^{ 3 }\frac { { e }^{ 3x } }{ 3 } -{ 3x }^{ 2 }\left( \frac { { e }^{ 3x } }{ 9 } \right) +6x\left( \frac { { e }^{ 3x } }{ 27 } \right) -6\left( \frac { { e }^{ 3x } }{ 81 } \right) +c\)
\(=\frac { { x }^{ 3 }{ e }^{ 3x } }{ 3 } -\frac { { x }^{ 2 }{ e }^{ 3x } }{ 3 } +\frac { { 2xe }^{ 3x } }{ 9 } -\frac { { 2e }^{ 3x } }{ 27 } +c\)
\(={ e }^{ 3x }\left[ \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 3 } +\frac { 2x }{ 9 } -\frac { 2 }{ 27 } \right] +c\)

Let I  \(=\int { \frac { 1 }{ x{ \left( \log x \right) }^{ 2 } } } dx\)
put log x = t on differentiating we get,
\(\frac { 1 }{ x } dx=dt\)
\(\therefore I=\int { \frac { 1 }{ { t }^{ 2 } } } dt\) [∵ t = log x]
\(I=\int { { t }^{ -2 } } dt\)
\(=\frac { { t }^{ -2+1 } }{ -2+1 } +c\)
\(=\frac { { t }^{ -1 } }{ -1 } +c\Rightarrow \frac { -1 }{ t } +c\)
\(=\frac { -1 }{ \log { \left| x \right| } } +c\) [∵ t = log |x|]

Let the amount invested in the two accounts be Rs. x and Rs. y respectively
By the given data, x + y = 8600   ..(1)
\(4\cfrac { 3 }{ 4 } \times \cfrac { x }{ 100 } +6\cfrac { 1 }{ 2 } \times \cfrac { y }{ 100 } =431.25\) \(\left[ \therefore interest=\cfrac { PNR }{ 100 } \right] \)
\(\Rightarrow \cfrac { 19x }{ 400 } +\cfrac { 13y }{ 3200 } =431.25\)
\(\Rightarrow \cfrac { 19x+26y }{ 400 } =431.25\)
19x + 26y = 172500   ...(2)
\(\Delta =\left| \begin{matrix} 1 & 1 \\ 19 & 26 \end{matrix} \right| =1(26)-1(19)\)
= 26-19 =7
\({ \Delta }x=\left| \begin{matrix} 8600 & 1 \\ 172500 & 26 \end{matrix} \right| =8600(26)-1(172500)\)
= 223600 - 172500 = 51100
\(\Delta y=\left| \begin{matrix} 1 & 8600 \\ 19 & 172500 \end{matrix} \right| =1(172500)-19(8600)\)
= 172500 - 163400 = 9100
\(x=\cfrac { \Delta x }{ \Delta } -\cfrac { 51100 }{ 7 } =7300\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 9100 }{ 7 } =1300\)
\(\therefore\) Investment in the interest of \(4\frac { 3 }{ 4 } \) % account is Rs. 7300 and investment in the rate of \(6\frac { 1 }{ 2 } \) account is Rs.1300. 

The equations are
2x + 3y = 7
3x + 5y = 9
Here \(\triangle =\left| \begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix} \right| =1\)
\(\neq 0\)
\(\therefore \) we can apply Cramer’s Rule
Now \({ \triangle }_{ x }=\left| \begin{matrix} 7 & 3 \\ 9 & 5 \end{matrix} \right| =8\)  \({ \triangle }_{ y }=\left| \begin{matrix} 2 & 7 \\ 3 & 9 \end{matrix} \right| =-3\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }_{ X } }{ \triangle } =\frac { 8 }{ 1 } =8\)   \(y=\frac { { \triangle }_{ y } }{ \triangle } =\frac { -3 }{ 1 } =-3\)
\(\therefore \) Solution is x = 8, y = −3

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

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