Given values are

The difference table is

x₀ + nh = x ⇒ 1 + n = x ⇒ n = x - 1
∴ By Newton's forward interpolation formula,
y= y₀\(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)

[∵ Δ³y₀ = 0]
⇒ y = -1 + 0 + x² - 3x + 2
⇒ y = x² - 3x + 1
Hence, the required second degree polynomial is  y = x² - 3x + 1

Δ(log ax) = log (ax + h) -log (ax)
= log\(\left( \frac { ax+h }{ ax } \right) \)
=log \(\left( \frac { ax }{ ax } +\frac { h }{ ax } \right) \)
= log \(\left( 1+\frac { h }{ ax } \right) \)
∴ ∆ (log ax) = log \(\left( 1+\frac { h }{ ax } \right) \)

(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

 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

Since the number of columns is less than the number of rows, given assignment problem is unbalanced one. To balance it , introduce a dummy column with all the entries zero. The revised assignment problem is

Here only 3 tasks can be assigned to 3 men.
Step 1: Its not necessary, since each row contains zero entry. Go to Step 2.
Step 2:

Step 3 (Assignment) :

Since each row and each columncontains exactly one assignment,all the three men have been assigned a task. But task S is not assigned to any Man. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = Rs. 35

\(\frac { dy }{ dx } +\frac { 2 }{ y } y\) = x³
The given differential equation is of this form [Divided by x]
\(\frac { dy }{ dx } \)+Py = Q where
P =\(\int { \frac { 2 }{ x } } \) and Q = x³
∴ \(\int { P } dx=\int { \frac { 2 }{ x } dx } \) = 2 log x = log x²
∴ Integrating factor (I. F) =\(e^{ \int { p } dx }=e^{ logx^{ 2 } }\)= x²
Hence the solution is
\(ye^{ \int { p } dx }=\int { Q } .e^{ \int { p.dx } }dx+c\)
⇒ y.x²=\(\int { { x }^{ 3 }.{ x }^{ 2 } } dx+c\)
⇒ x²y=\(\\ \int { { x }^{ 5 }dx } +c\)
⇒ x²y = \(\frac { { x }^{ 6 } }{ 6 } \) + c

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.

Given y = ae^4x + be^−x.    (1)
Here a and b are arbitrary constants
From (1), \(\frac { dy }{ dx } \) = 4ae^4x− be^−x     (2)
and \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 16ae^4x+ be^−x          (3)
(1) + (2) ⇒ \(y+\frac { dy }{ dx } \) = 5ae^4x      (4)
= (2) + (3) ⇒ \(\frac { dy }{ dx } +\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 20ae^4x
= 4(5ae^4x)
= \(4\left( y+\frac { dy }{ dx } \right) \)
\(\frac { dy }{ dx } +\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =4y+4\frac { dy }{ dx } \)
⇒ \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -3\frac { dy }{ dx } -4y=0\) which is the required differential equation

Computation of three- yearly moving averages.

Let I = ഽ(x +1)² log x dx
Let u = logx dv = (x+1)²dx
\(du=\cfrac { 1 }{ x } dx\quad v=\cfrac { \left( x+1 \right) ^{ 3 } }{ 3 } \)
\(\therefore\) Using integration by parts,
\(I=\int { udv } =uv-\int { vdu } \)
\(I=\int { \left( x+1 \right) ^{ 2 }\log xdx } \)
= \(\cfrac { (x+1)^{ 3 } }{ 3 } \log x-\int { \cfrac { \left( x+1 \right) ^{ 3 } }{ 3 } .\cfrac { 1 }{ x } dx } \)
= \(\cfrac { \left( x+1 \right) ^{ 3 } }{ 3 } \log x-\cfrac { 1 }{ 3 } \int { \cfrac { { x }^{ 3 }+{ 3x }^{ 2 }+3x+1 }{ x } } \)
= \(\cfrac { \left( x+1 \right) ^{ 3 } }{ 3 } \log x-\cfrac { 1 }{ 3 } \int { \left( { x }^{ 2 }+3x+3+\cfrac { 1 }{ x } \right) dx } \)
= \(\cfrac { \left( x+1 \right) ^{ 3 } }{ 3 } \log x-\cfrac { 1 }{ 3 } \left[ \cfrac { { x }^{ 3 } }{ 3 } +\cfrac { { 3x }^{ 2 } }{ 2 } +3x+\log|x| \right] +c\)

= \(\left[ \left( x+1 \right) ^{ 3 }\log x-\cfrac { { x }^{ 3 } }{ 3 } -\cfrac { { 3x }^{ 2 } }{ 2 } -3x-log|x| \right] +c\)

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.

Given  \(\begin{cases} cx, \\ 0, \end{cases}\begin{matrix} 0 < x < 1 \\ \text{otherwise} \end{matrix}\)
Also, Given \(\int _{ 0 }^{ 1 }{ f\left( x \right) } dx=2\)
\(\Rightarrow \int _{ 0 }^{ 1 }{ cx } dx=2\)
\(\Rightarrow c{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }=2\)
\(\Rightarrow \frac { c }{ 2 } \left[ { 1 }^{ 2 }-{ 0 }^{ 2 } \right] =2\)
\(\Rightarrow \frac { c }{ 2 } \left( 1-0 \right) =2\)
\(\Rightarrow \frac { c }{ 2 } =2\)
⇒ c = 2

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

\(\int _{ 0 }^{ 1 }{ [{ e }^{ a \log x }+{ e }^{ x \log a }] } dx=\int _{ 0 }^{ 1 }{ ({ x }^{ a }+{ a }^{ x }) } dx\)
\(={ \left[ \frac { { x }^{ a+1 } }{ a+1 } +\frac { { a }^{ x } }{ \log a } \right] }_{ 0 }^{ 1 }\)
\(=\left( \frac { 1 }{ a+1 } +\frac { a }{ \log a } \right) -\left( 0+\frac { 1 }{ \log a } \right) \)
\(=\frac { 1 }{ a+1 } +\frac { a }{ \log a } -\frac { 1 }{ \log a } \)
\(=\frac { 1 }{ a+1 } +\frac { (a-1) }{ \log a } \)

\(\int { \frac { { x }^{ 3 }dx }{ \sqrt { { x }^{ 8 }-1 } } } \)
\(=\int { \frac { { x }^{ 3 }dx }{ \sqrt { { \left( { x }^{ 4 } \right) }^{ 2 }-1 } } } \)
\(Put\quad { x }^{ 4 }=t\Rightarrow { 4x }^{ 3 }dx=dt\Rightarrow { x }^{ 3 }dx=\frac { dt }{ 4 } \)
\(=\frac { 1 }{ 4 } \int { \frac { dt }{ \sqrt { { t }^{ 2 } } -1 } } \)
\(=\frac { 1 }{ 4 } \int { \frac { dt }{ \sqrt { { t }^{ 2 }-{ 1 }^{ 2 } } } } \)
\(=\frac { 1 }{ 4 } \log { \left| t+\sqrt { { t }^{ 2 }-1 } \right| } +c\)
\(\left[ \because \int { \frac { dx }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } =\log { \left| x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right| +c } } \right] \)
\(=\frac { 1 }{ 4 } \log { \left| { x }^{ 4 }+\sqrt { { x }^{ 8 }-1 } \right| } +c\)

ഽ\(\frac { { x }^{ 3 }dx }{ \sqrt { x^{ 8 }+1 } } \) = \(\frac { 1 }{ 4 } \)ഽ\(\frac { { 4x }^{ 3 } }{ \sqrt { { \left( { x }^{ 4 } \right) }^{ 2 }+{ 1 }^{ 2 } } } \)
= \(\frac { 1 }{ 4 } \log\left| { x }^{ 4 }+\sqrt { { \left( { x }^{ 4 } \right) }^{ 2 }+{ 1 }^{ 2 } } \right| +c\)
= \(\frac { 1 }{ 4 } \log\left| { x }^{ 4 }+\sqrt { { x }^{ 8 }+1 } \right| +c\)

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,
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(=\int _{ 0 }^{ \infty }{ x{ e }^{ -3x } } dx\)
\(=3\int _{ 0 }^{ \infty }{ { xe }^{ -3x }dx } \)
\(=3\left\{ { \left[ \frac { { xe }^{ -3x } }{ -3 } \right] }_{ 0 }^{ \infty }\int _{ 0 }^{ \infty }{ \frac { { e }^{ -3x } }{ -3 } dx } \right\} \)( ∵ ∫udv = uv - ∫ vdu)
\(=\int _{ 0 }^{ \infty }{ { e }^{ -3x }dx } \)
\(=\frac{1}{3}\)
Therefore, the expected life of the piece of equipment is \(=\frac{1}{3}\)hrs (in thousands).

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

Annual consumption at timet = 0 (In the year 2000) = p₀ = 2547 metric ton.
Total production of Gold from 2000 to 2013 = \(\int _{ 0 }^{ 13 }{ 2547e^{ 0.006t } } dt\)
= \(\frac { 2547 }{ 0.006 } \left[ e^{ 0.006t } \right] _{ 0 }^{ 13 }\)
= 424500 (e^0.078 −1)
= 34,426.95 metric tons approximately.

Let I = ∫ xⁿ log x dx
Let u = log x; dv = xⁿ dx
\(du=\frac { 1 }{ x } dx;v=\frac { { x }^{ n+1 } }{ n+1 } \)
Using integration by parts we get,
∫ udv = uv - ∫ vdu
⇒ ∫ xⁿ log x dx
\(=\frac { { x }^{ n+1 } }{ n+1 } \log\ x-\int { \frac { { x }^{ n+1 } }{ n+1 } } dx\)
\(=\frac { { x }^{ n+1 } }{ n+1 } \log\ x-\frac { 1 }{ n+1 } \int { { x }^{ n+1-1 }dx } \)
\(=\frac { { x }^{ n+1 } }{ n+1 } \log\ x-\frac { 1 }{ n+1 } \int { { x }^{ n } } dx\)
\(=\frac { { x }^{ n+1 } }{ n+1 } \log\ x-\frac { 1 }{ n+1 } \frac { { x }^{ n+1 } }{ n+1 } +c\)
\(=\frac { { x }^{ n+1 } }{ n+1 } \left( \log\ x-\frac { 1 }{ n+1 } \right) +c\)

\(\int { \cos^{ 3 }xdx } =\frac { 1 }{ 4 } \int { \cos 3xdx+\frac { 3 }{ 4 } } \int { \cos x dx } \)
\(=\frac { \sin 3x }{ 12 } +\frac { 3 \sin\ x }{ 4 } +c\)
[Change into simple integrands
cos3x = 4cos³ x − 3cos x 
\({ \cos }^{ 3 }x=\frac { 1 }{ 4 } [\cos 3x+3 \cos x]\)
\(=\frac { 1 }{ 4 } \cos 3x+\frac { 3 }{ 4 } \cos x\)

\(\int { \frac { 7x-1 }{ { x }^{ 2 }-5x+6 } dx } =\int { \left[ \frac { 20 }{ x-3 } -\frac { 13 }{ x-2 } \right] dx } \)
\(=20\int { \frac { dx }{ x-3 } -13\int { \frac { dx }{ x-2 } } } \)
\(20log\left| x-3 \right| -13 \log\left| x-2 \right| +c\)
[ By partial fractions,
\(\frac { 7x-1 }{ { x }^{ 2 }-5x+6 } =\frac { A }{ x-3 } +\frac { B }{ x-2 } \Rightarrow \frac { 7x-1 }{ { x }^{ 2 }-5x+6 } =\frac { 20 }{ x-3 } -\frac { 13 }{ x-2 } \)]

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

Given  A =\(\left( \begin{matrix} 4 & 5 & 2 \\ 3 & 2 & 1 \\ 4 & 4 & 8 \end{matrix}\begin{matrix} 2 \\ 6 \\ 0 \end{matrix} \right) \)
\(\left( \begin{matrix} 4 & 5 & 2 \\ 3 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}\begin{matrix} 2 \\ 6 \\ 0 \end{matrix} \right) { R }_{ 3 }\rightarrow { R }_{ 3 }\div 4\)
\(\left( \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 4 & 5 & 2 \end{matrix}\begin{matrix} 0 \\ 6 \\ 2 \end{matrix} \right) { R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\left( \begin{matrix} 1 & 1 & 2 \\ 0 & -1 & -5 \\ 0 & 1 & -6 \end{matrix}\begin{matrix} 0 \\ 6 \\ 2 \end{matrix} \right) \)\({ R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\)
\(\left( \begin{matrix} 1 & 1 & 2 \\ 0 & -1 & -5 \\ 0 & 1 & -6 \end{matrix}\begin{matrix} 0 \\ 6 \\ 2 \end{matrix} \right) \begin{matrix} { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } \\ { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 1 } \end{matrix}\)
\(\left( \begin{matrix} 1 & 1 & 2 \\ 0 & -1 & -5 \\ 0 & 0 & -11 \end{matrix}\begin{matrix} 0 \\ 6 \\ 8 \end{matrix} \right) { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 }\)
The last equivalent matrix is in echelon form and there are 3 non-zero rows
\(\therefore \rho (A)=3\)

Let ‘x’ be the cost of a pencil
Let ‘y’ be the cost of an eraser
\(\therefore \) By given data, we get the following equations
11x + 3y = 64
8x + 3y = 49
\(\triangle =\left| \begin{matrix} 11 & 3 \\ 8 & 3 \end{matrix} \right| =9\neq 0,\) It has unique solution.
\({ \triangle }_{ x }\left| \begin{matrix} 64 & 3 \\ 49 & 3 \end{matrix} \right| =45\)
\({ \triangle }_{ y }\left| \begin{matrix} 11 & 64 \\ 8 & 49 \end{matrix} \right| =27\)
\(\therefore \) By Cramer’s rule
\(x={ \frac { \triangle x }{ \triangle } =\frac { 45 }{ 9 } =5 }\)
\(y={ \frac { \triangle y }{ \triangle } =\frac { 27 }{ 9 } =3 }\)
\(\therefore \) The cost of a pencil is Rs. 5 and the cost of an eraser is Rs. 3.

Let A =\(\left( \begin{matrix} 0 & -1 & 5 \\ 2 & 4 & -6 \\ 1 & 1 & 5 \end{matrix} \right) \)
Order of A is 3 \(\times\) 3  
∴ \(\rho \)(A)\(\le \)3
Consider the third order minor \(\left| \begin{matrix} 0 & -1 & 5 \\ 2 & 4 & -6 \\ 1 & 1 & 5 \end{matrix} \right| =6\neq 0\)
There is a minor of order 3, which is not zero 
\(\therefore \rho (A)=3\)