\({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) =\Delta \left( \Delta \left( \frac { 1 }{ x } \right) \right) \)
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 \({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) =\frac { 2 }{ x(x+1)(x+2) } \)

The objective function is minimize Z = \(\overset { m }{ \underset { i=1 }{ \Sigma } } \overset { n }{ \underset { j=1 }{ \Sigma } } { { C }_{ ij } }{ x }_{ ij }\) subject to the constraints
\(\overset { n }{ \underset { j=1 }{ \Sigma } }{ x }_{ ij } = a_i, i=1,2,.....m\) (Supply constraints)
\(\overset { m }{ \underset { i=1 }{ \Sigma } }{ x }_{ ij } = b_j, j=1,2,.....n\) (demand constraints)
x_ij ≥, 0 for all i, j (non-negative restrictions)

The components of time series are
(i) Secular trend
(ii) Seasonal variations
(iii) Cyclic variations
(iv) Irregular variations

There many ways to select 15 random samples from the given Tippet’s random number table. Since the population size is 83(two-digit number). Here the door numbers are assigned from 1 to 83. Assume that at random we first choose 2^nd column. So the first sample is 66 and other 14 samples are 74, 52, 39, 15, 34, 11, 14, 13, 27, 61, 79, 72, 35, and 60. If the numbers are above 83, choose the next number ranging from 1 to 83.

2952	6641	3992	9792	7969	5911	3170	5624
4167	9524	1545	1396	7203	5356	1300	2693
2670	7483	3408	2762	3563	1089	6913	7991
0560	5246	1112	6107	6008	8125	4233	8776
2754	9143	1405	9025	7002	6111	8816	6446

Let X be a random variable follows binomial distribution with probability value p = 1/2 and q = 1/2
probability that exactcy 2 heads occur are as follows
P(X = 2) = \(\left( \begin{matrix} 6 \\ x \end{matrix} \right) { p }^{ x }{ q }^{ n-x }\)
\(=\left( \begin{matrix} 6 \\ 2 \end{matrix} \right) { \left( \frac { 1 }{ 2 } \right) }^{ 2 }{ { \left( \frac { 1 }{ 2 } \right) } }^{ 6-2 }\)
\(=\frac { 15 }{ 64 } \)

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

\(y=4{ - }x^{ 2 }\)
Required area = \(\int _{ 0 }^{ 2 }{ ydx } =\int _{ 0 }^{ 2 }{ (4- } { x }^{ 2 })dx\)
= \({ \left[ 4x-\frac { { x }^{ 2 } }{ 3 } \right] }_{ 0 }^{ 2 }=8-\frac { 8 }{ 3 } \)
= \(\frac { 16 }{ 3 } \) sq.units

\(\int { \sqrt { 3x+5 } dx } \)
\(=\int { { \left( 3x+5 \right) }^{ 1/2 } } dx\)
\(=\frac { { \left( 3x+5 \right) }^{ 1/2+1 } }{ 3\left( \frac { 1 }{ 2 } +1 \right) } +c\)
\(\left[ \because \int { { \left( ax+b \right) }^{ n }dx=\frac { { \left( ax+b \right) }^{ n+1 } }{ a\left( n+1 \right) } } +c \right] \)
\(=\frac { { \left( 3x+5 \right) }^{ 3/2 } }{ 3\left( \frac { 3 }{ 2 } \right) } +c=\frac { { \left( 3x+5 \right) }^{ 3/2 } }{ \frac { 9 }{ 2 } } +c\)
\(=\frac { 2 }{ 9 } { \left( 3x+5 \right) }^{ 3/2 }+c\)

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

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