Separating the variables, we get,
\(\frac { dy }{ x } \)= sin 2x dx
Integrating both sides we get,
\(\int { \frac { dy }{ y } } =\int { \sin 2x } \)
⇒ log y = \(\frac { -\cos 2x }{ 1 }\)+c

Let A= \(\left( \begin{matrix} 5 & 3 & 0 \\ 1 & 2 & -4 \\ -2 & -4 & 8 \end{matrix} \right) \)
Order of A is 3 \(\times\) 3.
∴\(\rho \)(A)\(\le \)3
Consider the third order minor \(\left| \begin{matrix} 5 & 3 & 0 \\ 1 & 2 & -4 \\ -2 & -4 & 8 \end{matrix} \right| =0\)
Since the third order minor vanishes, therefore \(\rho (A)\neq 3\)
Consider a second order minor \(\left| \begin{matrix} 5 & 3 \\ 1 & 2 \end{matrix} \right| =7\neq 0\)
There is a minor of order 2, which is not zero.
\(\therefore \rho (A)=2\)

Grand average = \(\frac{67 + 61 + 63.33 + 63.37}{4}\)
= \(\frac{255}{4}=63.75\)
Seasonal index (S.I) = \(\frac{Quarterly average}{Grand average}\times100\)
Hence, S.I for I quarter = \(\frac{67}{63.75}\times100\) = 105.01
S.I for II quarter = \(\frac{61}{63.75}\times100\) = 95.68
S.I for III quarter = \(\frac{63.33}{63.75}\times100\) = 99.35
S.I for IV quarter = \(\frac{63.67}{63.75}\times100\) = 99.87

Let the straight line of best fit be y = ax + b.
The normal equations are
Σy = a Σx + nb
Σxy = a Σx² + bΣx
⇒ 10a + 7b    = 16.9 ....(1)
30a + 10b  = 47.4.... (2)

Substituting b = 0.3 in (2) we get
30a + 3 = 47.4 ⇒ 30a = 44.4
\(a=\frac{44.4}{30}=1.48\)
∴ The straight line trend is y = 1.48x + 0.3

Given sample size n = 400
Sample mean \(\bar { x } \) = 171.38
Population mean μ = 171.17
Population Standard deviation σ = 3.3
Null hypotheses: H₀ : μ = 171.17
Alternative hypotheses: H₁ : μ ≠ 171.17
The test statistic, z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } =\frac { 171.38-171.17 }{ \frac { 3.3 }{ \sqrt { 400 } } } \)
=\(\frac { 0.21 }{ 0.165 } \) = 1.273
As the level of significance is α = 0.005, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
Here z < \(Z_{ \frac { \alpha }{ 2 } }\) as 1.273 < 1.96
Inference: since z < \(Z_{ \frac { \alpha }{ 2 } }\), we accept the null hypotheses at 5% level of significance.
Hence, we can conclude that the sample of 400 has taken from the population with mean height of 171.17 cm.

Given population size N = 1500
Sample size n = 150
Sample proportion p =\(\frac { 10 }{ 150 } \)=0.07
∴ q = 1 - P = 1 - 0.07 = 0.93
Standard error of sample proportion =\(\sqrt { \frac { pq }{ n } } \)
=\(\sqrt { \frac { (0.07)(0.93) }{ 150 } } \)
S.E_(p) = 0.02 

Given mean μ = 60 and S.D. σ = 5

To find P(X > 60)
When X = 60, Z =\(\frac { X-\mu }{ \sigma } =\frac { 60-60 }{ 5 } \) = 0
∴ P(X > 60) = P(Z > 0) = P (0 < Z < ∞)
= 0.5
∴ 50% of students scored more than 60 marks

Given n = 9 and mean = 6 ⇒ np = 6
9p = 6 ⇒ \(\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)
∴ q=1-p = \(1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
Variance = npq = \(9\times \frac { 2 }{ 3 } \times \frac { 1 }{ 3 } \) = 2

Mean = E(X)=\(\int _{ -\infty }^{ \infty }{ x.f(x)dx=\int _{ -12 }^{ 12 }{ x.\left( \frac { 1 }{ 24 } \right) } dx } \)
\(=\frac { 1 }{ 24 } \int _{ -12 }^{ 12 }{ x.dx } \)
\(=0[\because \int _{ -a }^{ a }{ f(x)dx=0 } when\ f(x)\ is\ an\ odd\ function]\)
\(\therefore E(X)=0\)

Since the random variable. X is the probability mass function, Σp_i = 1
\(\Rightarrow \frac { k }{ 6 } +\frac { k }{ 4 } +\frac { k }{ 12 } =1\Rightarrow \frac { 2k+3k+k }{ 12 } =1\)
\(\Rightarrow \frac { 6k }{ 12 } =1\Rightarrow k=\frac { 12 }{ 6 } =2\quad \therefore k=2\)

Min (45, 70, 53) = 45
∴ Choosing Bonds is the best decision under minimax principle.

Here total availability = 150 + 100 + 250 = 500
total requirement = 50 + 150 + 300 = 500
∴ Total availability = total requirement
∴ The given problem is a balanced transportation problem
Hence, there exists a feasible solution to the given problem
I - allocation:

[∵ least cost is 4 & min (50,150) = 50]
II - allocation:

[∵ least cost is 6 & min (150, 250) = 150]
III - allocation:

[∵ least cost is 8 & min (300, 100) = 100]
IV - allocation:

[∵ least cost is 9 & min (200,100) = 100]
V - allocation:

[∵ min (100, 100) = 100]
Thus, the allocations are

∴ The transportation schedule is
O₁ → D₁, O₁ → D₃, O₂ → D₃, O₃ → D₂, O₃ → D₃
Hence, the total transportation cost is
= 50(4) + 100(8) + 100(11) + 150(6) + 100(9)
= 200 + 800 + 1100 + 900 + 900
= Rs. 3900

Given
 ⇒ \(\int _{ 0 }^{ 1 }{ \left( { 3x }^{ 2 }+2x+k \right) } dx=0\)
⇒ \({ \left[ \frac { { 3x }^{ 3 } }{ 3 } +\frac { { 2x }^{ 2 } }{ 2 } +kx \right] }_{ 0 }^{ 1 }=0\)
⇒ \(\therefore { \left[ { x }^{ 3 }+{ x }^{ 2 }+kx \right] }_{ 0 }^{ 1 }=0\)
⇒ [1 + 1 + k(1)] - (0) = 0
⇒ 2 + k = 0
⇒ k = -2

Given f'(x) = 8x³ -2x²
∴ ∫ f'(x) dx = ∫ (8x³ - 2x²) dx
⇒ f(x) = \(\frac { { 8x }^{ 4 } }{ 4 } -\frac { { { 2x }^{ 3 } } }{ 3 } +c\)
⇒ f (x) = 2x⁴ - \(\frac { { { 2x }^{ 3 } } }{ 3 } +c\) ...(1)
Also, f(2) = 1
⇒ 1 = 2(2⁴) - \(\frac { { 2\left( { { 2 }^{ 3 } } \right) } }{ 3 } +c\)
⇒ 1 = 32 - \(\frac { { 16 } }{ 3 } +c\)
⇒ 1 - 32 + \(\frac { { 16 } }{ 3 } \) = c ⇒ -31 + \(\frac { { 16 } }{ 3 } \) =c
⇒ \(\frac { { -93+16 } }{ 3 } =c\)
⇒ c = \(\frac { { -77 } }{ 3 } \)
∴ (1) ⟶ f(x) = 2x⁴ - \(\frac { { 2x }^{ 3 } }{ 3 } -\frac { { -77 } }{ 3 } \)

Equation of family of rectangular hyperbolas whose asymptotes are the Co-ordinate axis is
xy - c²
Differentiating w.r.t. 'x' we get,
x.\(\frac { dy }{ dx } \)+y(1) = 0
⇒ x\(\left( \frac { dy }{ dx } \right) \)+y(1) = 0 which is the required differential equation.

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

By data we have,
x₀ = 0, x₁ = 1, x₂ = 3
y₀ = 5, y₁ = 6, y₂ = 50 and x = 2
Using Lagrange's formula, we get
\(y=\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 }) } { y }_{ 0 }+\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 }) } { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 0 }-{ x }_{ 1 }) } { y }_{ 2 }\)
= \(5\times \frac { (2-0)(2-3) }{ (0-1)(0-3) } +6\times \frac { (2-0)(2-3) }{ (1-0)(1-3) } +50\times \frac { (2-0)(2-1) }{ (3-0)(3-1) } \)

= \(\frac { 10 }{ 3 } -6+\frac { 50 }{ 3 } =\frac { 60 }{ 3 } -6=20-6=14\)
∴ f(2) = 14

Since only two values of yare given, the polynomial which fits the data is of degree 1.
Hence 2nd differences are zeros
∴ Δ²(y₀) = 0
⇒ (E-1)²y_o=0
⇒(E² - 2E + 1) y_o = 0
⇒y₂ - 2y₁ +y_o = 0
⇒34 - 2y₁ + 51 = 0
​​​​​​​⇒85 - 2y₁ =0
​​​​​​​⇒2y₁ + 85 ​​​​​​​⇒ y₁ = \(\frac{85}{2}\)
​​​​​​​⇒y₁ = 42.5 

Given demand function is p = 25 - x - X²
and p₀ = 19
⇒ 19 = 25-x-x²
⇒ x² + x - 6 = 0
⇒ (x + 3) (x - 2) = 0
⇒ x = -3 or x = 2
Since x cannot be negative x_o = 2
p_o x_o = 19(2) = 38
\(CS=\int _{ 0 }^{ 2 }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ 2 }{ \left( 25-x-{ x }^{ 2 } \right) dx-38 } \)
\(={ \left( 25x-\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ 2 }-38\)
\(=25(2)-\frac { 4 }{ 2 } -\frac { 8 }{ 3 } -38\)
\(=50-2-\frac { 8 }{ 3 } -38\)
\(=10-\frac { 8 }{ 3 } =\frac { 30-8 }{ 3 } \)
\(CS=\frac { 22 }{ 3 } \) units

Given \(MC=\frac { 100 }{ x } \)
\(\\ \int { MC } =\int { \frac { 100 }{ x } } \)
⇒ C = 100 log x + k
Given C(16) =100 ⇒ When x = 16,
C = 100
∴ 100 = 100 log 16 + k
⇒ k = 100-100 log 16
C = 100 log x + 100 - 100 log 16
= 100 (log x - log16 + 1)
\(C=100(log\left( \frac { x }{ 16 } \right) +1)\)

Since A and B are non-singular,
|A| \(\neq \) 0, |B|\(\neq \) 0
Consider |AB| |A|·|B|
\(\neq \) 0 since |A|\(\neq \) 0 and |B|\(\neq \) 0.=? |AB| \(\neq \) 0
\(\therefore\) AB is non-singular.

Let A = \(\left( \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right) \)
The order of A is 2 \(\times\) 2
\(\rho (A)\le min(2,2)\)
\(\Rightarrow \rho (A)\le 2\)
\(\left| \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right| =4-4=0\)
Since the second order minor vanishes \(\rho (A)\neq 2\)
 We have to try for atleast one non-zero first order minor.
ie. atleast one non-zero element of A.
This is possible because A has non-zero element
\(\therefore \rho (A)-1\)

Let \(A=\left( \begin{matrix} i & -1 \\ 3 & -6 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\) [Since minimum of (2, 2) is 2]
 Consider the second order minor
\(\left| \begin{matrix} 1 & -1 \\ 3 & -6 \end{matrix} \right| =-6-(-3)\)
= -6 + 3 = -3
\(\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)

\({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) =\Delta \left( \Delta \left( \frac { 1 }{ x } \right) \right) \)
Now \(\Delta \left[ \frac { 1 }{ x } \right] =\frac { 1 }{ 1+x } -\frac { 1 }{ x } \)
\({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) ={ \Delta }\left( \frac { 1 }{ 1+x } -\frac { 1 }{ x } \right) \)
\(=\Delta \left( \frac { 1 }{ 1+x } \right) -\Delta \left( \frac { 1 }{ x } \right) \)
Similarly \({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) =\frac { 2 }{ x(x+1)(x+2) } \)

(i) Δe^ax = e^a(x+h)−e^x
= e^aX.e^h-e^ax [∵ a^m+n = a^m.aⁿ]
= e^ax[e^h-1]
 (ii) Δ²e^x = Δ.[Δe^x]
= Δ[e^x+h - e^x]
= Δ[e^xe^h - e^x]
= Δe^x [e^h - 1]
= (e^h - 1)Δe^x 
= (e^h−1).(e^h−1).e^x
= (e^h−1)².e^x
(iii) Δ log x = log(x+h) − log x
= log \(\frac{x + h}{x}\)
= log \(\left( \frac { x }{ x } +\frac { h }{ x } \right) \)
= log \(\left( 1+\frac { h }{ x } \right) \)

The assignment problem is a special case of transportation problem where the number of sources and destinations are equal. Here, jobs represent sources and machines represent destinations.

A transportation problem is to determine the amount to be transported from each origin to each destinations such that the total transportation cost is minimized.

Given  \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -4\frac { dy }{ dx } +5y\) = 0
(D²−4D+5)y = 0
The auxiliary equation is m²−4m + 5 = 0
⇒ (m−2)²−4 + 5 = 0
(m− 2)² = –1
m - 2 = 土\(\sqrt{-1}\)
m = 2 土 i , it is if the form α 土 iβ
∴ C.F = e^2x[A cos x + B sin x]
The general solution is y = e^2x[A cos x + B sin x]

Index numbers are studied to know the relative changes in price and quantity for any two years compared. There are two tests which are used to test the adequacy for an index number. The two tests are as follows.
(i) Time reversal test
(ii) Factor reversal test
The criterion for a good index number is to satisfy the above two tests.

(i) It helps in the analysis of the past behavior
(i) It helps in forecasting and for future plans
(ii) It helps in the evaluation of current achievements
(iv) It helps in making comparative studies between one time period and others

When the hypothesis about the population parameter is rejected only for the value of sample statistic falling into one of the tails of the sampling distribution, then it is known as one tailed test.

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

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

Poisson distribution is a limiting case of binomial distribution under the following conditions.
(i) n, the number of trials is indefinitely large ie., n⟶∞.
(ii) p, the constant probability of success in each trial is very small ie., p ⟶0.
(iii) np = λ is finite.Thus p = λ/n and q = 1 -(λ/n) where λ is a positive real number.

The discrete cumulative distribution function or distribution function of a real valued discrete random variable X takes the countable number of points x₁,x₂, .... with corresponding probabilities p(x₁)p(x₂).... and the distribution function is defined by
F_x(x) = P(X≤x) for all x∈R
ie.F_x(x) = \(\sum _{ { x }_{ i }\le x }^{ }{ p({ x }_{ i }) } \)
For a continuous random variable with the probability density function f_x(x) then the distribution function F_x(x) is defined by
F_x(x) = P(X≤x)

We know Fx (x) = P(X ≤ x) for all x ∈ R
∴ F(0) = P(X ≤ 0) = P(0) = 0.3
F(1) = P(X ≤ 1) = P(0)+P(l)
= 0.3 + 0.2 = 0.5
F(2) = P(X ≤ 2) = P(0) + P(1) + P(2)
= 0.3 + 0.2 + 0.4 = 0.9
F(3) = P(X ≤ 3) = P(0) + P(1) + P(2) + P(3)
= 0.3 + 0.2 + 0.4 + 0.1 = 1
∴ Cumulative distribution function for the given probability distribution is 1

2y + x = 8

Given 2y + x = 8
2y = 8-x
y = \(\frac{1}{2}\) (8-x)
Given limits are x = 2 and x = 4
Area of the shaded region between the given limits
\(A=\int _{ a }^{ b }{ y\quad dx } =\int _{ 2 }^{ 4 }{ \frac { 1 }{ 2 } (8-x)dx } \)
\(\frac { 1 }{ 2 } \int _{ 2 }^{ 4 }{ (8-x)dx } =\frac { 1 }{ 2 } { \left[ 8x-\frac { { x }^{ 2 } }{ x } \right] }_{ 2 }^{ 4 }\)
\(=\frac { 1 }{ 2 } \left[ \left( 8(4)-\frac { { 4 }^{ 2 } }{ 2 } \right) \left( 8(2)-\frac { { 2 }^{ 2 } }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } [(32-8)-(16-21)]\)
\(=\frac { 1 }{ 2 } [24-14]\)

A = 5 sq. units.

x - 2y - 12 = 0
x = 2y + 12
Required Area
= \(\int _{ 2 }^{ 5 }{ xdy } \)
= \(\int _{ 2 }^{ 5 }{ (2y+12)dy= } [{ { y }^{ 2 }+12y] }_{ 2 }^{ 5 }\)
= (25 + 60)−(4 + 24) = 57 sq.units

\(\int { x } \sqrt { { x }^{ 2 }+1 }\ dx=\frac { 1 }{ 2 } \int { { \left( { x }^{ 2 }+1 \right) }^{ \frac { 1 }{ 2 } } } \left( 2x \right) dx\)
\(=\frac { 1 }{ 2 } \int { [f(x){ ] }^{ \frac { 1 }{ 2 } }f'(x)dx } \)
\(=\frac { 1 }{ 2 } \frac { [f(x){ ] }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } +c\)
\(=\frac { 1 }{ 3 } ({ x }^{ 2 }+1{ ) }^{ \frac { 3 }{ 2 } }+c\)
[ Take f (x) = x² +1
 \(\therefore \) f '(x) = 2x ]

\(\int { \frac { x }{ \sqrt { { x }^{ 2 }+1 } } dx } =\frac { 1 }{ 2 } \int { \frac { x }{ \sqrt { { x }^{ 2 }+1 } } } dx\)
\(=\frac { 1 }{ 2 } \int { \frac { f'\left( x \right) }{ \sqrt { f\left( x \right) } } dx } \)
\(=\frac { 1 }{ 2 } [2\sqrt { f(x) } ]+c\)
\(=\sqrt { { x }^{ 2 }+1+ } c\)
[ Take f (x) = x² +1
 \(\therefore \) f '(x) = 2x ]