\(\Delta =\left| \begin{matrix} 1 & 4 & 3 \\ 2 & -6 & 6 \\ 5 & -2 & 3 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -6 & 6 \\ -2 & 3 \end{matrix} \right| -4\left| \begin{matrix} 2 & 6 \\ 5 & 3 \end{matrix} \right| +3\left| \begin{matrix} 2 & -6 \\ 5 & -2 \end{matrix} \right| \)
= 1(-18 + 12) - 4(6 - 30) +3 (- 4 +30)
= 1(- 6) - 4(- 24) + 3(26)
= - 6 + 96 + 78 = 168 \(\neq \) 0
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied
\(\Delta x=\left| \begin{matrix} 2 & 4 & 3 \\ -3 & -6 & 6 \\ -5 & -2 & 3 \end{matrix} \right| \)
= \(2\left| \begin{matrix} -6 & 6 \\ -2 & 3 \end{matrix} \right| -4\left| \begin{matrix} -3 & 6 \\ -5 & 3 \end{matrix} \right| +3\left| \begin{matrix} -3 & -6 \\ -5 & -2 \end{matrix} \right| \)
= 2 (- 18 + 12) - 4(- 9 +30) + 3(6 -30)
= 2(- 6) - 4(21) + 3(- 24)
= -12-84-72 =-168
\(\Delta y=\left| \begin{matrix} 1 & 2 & 3 \\ 2 & - & 6 \\ 5 & -5 & 3 \end{matrix} \right| =1\left| \begin{matrix} -3 & 6 \\ -5 & 3 \end{matrix} \right| -2\left| \begin{matrix} 2 & 6 \\ 5 & 3 \end{matrix} \right| +3\left| \begin{matrix} 2 & -3 \\ 5 & -5 \end{matrix} \right| \)
= 1 (-9+30)-2(6-30)+3(- 10+ 15)
= 1(21) - 2(- 24) + 3(5)
= 21 + 48 + 15 = 84
\(\Delta z=\left| \begin{matrix} 1 & 4 & 2 \\ 2 & -6 & -3 \\ 5 & -2 & -5 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -6 & -3 \\ -2 & -5 \end{matrix} \right| -4\left| \begin{matrix} 2 & -3 \\ 5 & -5 \end{matrix} \right| +2\left| \begin{matrix} 2 & -6 \\ 5 & -2 \end{matrix} \right| \)
= 1(30-6)-4(-10+ 15)+2(-4+30)
= 24 - 4(5) + 2(26)
= 24 - 20 + 52 = 56


Solution set is \(\left\{ -1,\frac { 1 }{ 2 } ,\frac { 1, }{ 3 } \right\} \)

Given

x	0	1	3	4
y	-12	0	6	12

Here the intervals are unequal
∴ By Lagranges interpolation formula, we have
x₀ = 0, x₁ = 1, x₂ = 3, x₃ = 4
y₀ = -12, y₁ = 0, y₂ = 6, y₃ = 12 and x = x.
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (x-1)(x-3)(x-4) }{ (0-1)(0-3)(0-4) } (-12)+\frac { (x-0)(x-3)(x-4) }{ (1-0)(1-3)(1-4) } (0)+\frac { (x-0)(x-1)(x-4) }{ (3-0)(3-1)(3-4) } (6)+\frac { (x-1)(x-3)(x-4) }{ (4-0)(4-1)(4-3) } (12)\)
= \(\frac { (x-1)(x-3)(x-4) }{ (-1)(-3)(-4) } (-12)+0+\frac { x(x-1)(x-4) }{ (3)(2)(-1) } (6)+\frac { x(x-1)(x-3) }{ (4)(3)(1) } (12)\)
= +[(x - 1)(x - 3)(x - 4)] - x (x - 1)(x -  4) + x(x - 1)(x - 3)
= +[(x³-4x+3)(x-4)] -x(x²-5x+4) + x(x²-4x + 3)
= - (x³ - 8x²+ 19x - 12) - 4x² + 3x
= (x - 4)(x² - 4x + 3) - x (x² - 5x + 4) + x(x² - 4x + 3)

= x³ - 7x² + 19x - 12.

Given

Year	1974	1978	1982	1990
Output in 1000 tones	25	60	80	170

 Here the intervals are unequal.
∴ By Lagranges interpolation formula, we have
x₀ = 1974, x₁ = 1978, x₂ = 1982, x₃ = 1990
y₀ = 25, y₁ = 60, y₂ = 80, y₃ = 170 and x = 1986.
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
\(\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times25+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times60+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times80+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times 70+\frac { (1986-1974)(1986-1982)(1986-1990) }{ (1990-1974)(1990-1978)(1990-1982) } \times 170\)

= 6.25 - 60 + 120 + 42.5
y = 108.75

Since the pressure required is at the end of the table, we apply Backward interpolation formula. Let temperature be x and the pressure be y.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
To find y at x = 175
\(\therefore\) x_n + nh = 175 , x_n = 180, h = 10 \(\Rightarrow\) n = −0.5

x	y	\(\Delta y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 4 }y\)
140	3.685
		1.169
150	4.854		0.279
		1.448		02.047
160	6.032		0.326		0.002
		1.774		0.049
170	8.076		0.375
		2.149
180	10.225

\({ y }_{ (x=1750 }=10.225+\left( -0.5 \right) (2.149)+\frac { (-0.5)(0-5) }{ 2! } (0.375)+\frac { (-0.5)(0-5)(1.5) }{ 3! } (0.049)+\frac { (-0-5)(0.5)(1.5)(2.5) }{ 4! } (0.002)\)

= 10.225−1.0745−0.046875−0.0030625 − 0.000078125
= 9.10048438
= 9.1

To find the population for the year 1905 (i.e) the value of y at x = 1905
Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\cfrac { n }{ n! } \Delta { y }_{ 0 }+\cfrac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\cfrac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 } + ...\)
To find y at x = 1905
\(\therefore\) x₀+nh = 1905 , x₀ = 1891, h = 10
1891+n(10) = 1905 \(\Rightarrow\) n = 1.4

x	y	\(\Delta y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 3 }y\)	\(\Delta ^{ 4 }y\)
1891	98,752
		33,533
1901	1,32,285		2,258
		35,791		–10,435
1911	1,68,076		-8,177		41,376
		27,614
1921	1,95,690			30,941
			22,764
		50,360
1931	2,46,050

y_(x=1905) = \(98,752+(1.4)(33533)+\frac { (1.4)(0.4) }{ 2 } (2258)+\frac { (1.4)(0.4)(-0.6) }{ 6 } (-10435)+\frac { (1.4)(0.6)(-0.6)(-1.6) }{ 24 } (41358)\)
=  98,752 + 46946.2 + 632.4 + 584.36 + 1389.63
= 1,48,304.43
= 1,48,304

Here the number of rows and columns are equal
∴ The given assignment problem is balanced.
Step 1 : Select a minimum element in each row and subtract this from all the elements in its row.
∴ The cost matrix of the given assignment problem is

Column 2 has no zero. Go to Step 2.
Step 2 : Select the minimum element in each column and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3 : Examine the rows with exactly one zero, mark it by ׈ and draw a vertical line.
After examining the rows, examine the columns with exactly one zero.
Mark it by ׈ and draw a horizontal line.

Only 3 assignments have been made.
The numbers not lying on the line are and min. is 1. Subtract 1 from all these numbers and add 1 to 23 which lies on the intersecting lines. Other numbers remain the same.
A new cost matrix is formed and repeat step 3.
∴ The new cost matrix is

Thus, all the 4 assignments have been made.
∴The optimal assignment schedule and total cost is

Subordinates	tasks	cost
P	1	8
Q	3	4
R	2	19
S	4	10
Total Cost		Rs. 41

(x² +y²) dx = -2xy dy
⇒ \(\frac { dy }{ dx } =-\frac { ({ x }^{ 2 }+{ { y }^{ 2 }) } }{ 2xy } \)
Since the numerator and denominator is a homogenous function of degree 2,
put y = vx and \(\frac { dy }{ dx } =v+x\frac { dy }{ dx } \)
x\(\frac { dv }{ dx } =-\left( \frac { { x }^{ 2 }+{ v }^{ 2 }{ x }^{ 2 } }{ 2xvx } \right) \)

x\(\frac { dv }{ dx } =-\left( \frac { 1+{ v }^{ 2 } }{ 2v } \right) -v\)
= \(\frac { -1-{ v }^{ 2 }-2{ v }^{ 2 } }{ 2v } -\frac { -1-3{ v }^{ 2 } }{ 2v } =-\left( \frac { 1+3{ v }^{ 2 } }{ 2v } \right) \)
\(\frac { 2v.dv }{ 1+3{ v }^{ 2 } } =-\frac { dx }{ x } \)
Multiplying by 3,
put 1 + 3v² = t
6v dv = dt
\(\int { \frac { 6vdv }{ 1+3v^{ 2 } } } dv=-\int { 3 } \frac { dx }{ x } \)
\(\int { \frac { dt }{ t } } =-3\int { \frac { dx }{ x } } \)
⇒ log t = - 3 log x + log C
⇒ log (1 + 3v²) + 3 log x = log C
⇒ log (1 + 3v²) .x³ = log C
⇒ (1 + 3v²)x³= C
Replace v by \(\frac { y }{ x } \) we get,
\(\left( 1+\frac { { 3y }^{ 2 } }{ { x }^{ 2 } } \right) \).x³= C
⇒ \(\left( 1+\frac { { 3y }^{ 2 } }{ { x }^{ 2 } } \right) ^{ \frac { 1 }{ 3 } }\)= C

The auxiliary equation is 3m² + m -14 = 0
(m-2)\(\frac { 7 }{ 3 } \) = 0
⇒ m = 2, -\(\frac { 7 }{ 3 } \)
The roots are real and different
∴ CF is Ae^2x+\(Be^{ -\frac { 7 }{ 3 } x }\)
Particular Integral PI = \(\frac { 1 }{ \phi (D) } \).f(x)
PI = \(\frac { 1 }{ (3D^{ 2 }+D-14) } \).13 e2x
= \(\frac { 13.e^{ 2x } }{ 3D^{ 2 }+D-14 } \)
= \(\frac { 13e^{ 2x } }{ (D-2)(3D+7) } =\frac { 13e^{ 2x } }{ 3(D-2)\left( D+\frac { 7 }{ 3 } \right) } \)
= \(\frac { 13x{ e }^{ 2x } }{ 3\left( 2+\frac { 7 }{ 3 } \right) } \)

∴ y = CF + PI
∴ The general solution is
y = Ae^2x+\({ Be }^{ -\frac { 7 }{ 3 } x }\) + xe^2x

Auxiliary equation is m² - 3m + 2 = 0
∴ (m - 1)(m - 2) = 0
⇒ m = 1,2
∴ Complementary function CF is A^ex + Be^2x
PI = \(\frac { 1 }{ \phi (D) } \)(x)
= \(\frac { 1 }{ { D }^{ 2 }-3D+2 } \).e3x
= \(\frac { 1 }{ (D-1)(D-2) } \).e3x
= \(\frac { e^{ 3x } }{ (3-1)(3-2) } \).e3x
= \(\frac { { e }^{ 3x } }{ 2 } \)
General solution is y = CF + PI
y = A^ex+Be^2x+\(\frac { 1 }{ 2 } \)e^3x ...(1)
Given when x = 0, y = 0
∴ 0 = Ae⁰+Be₀+\(\frac { 1 }{ 2 } \)e₀ ⇒ 0 = A+B-\(\frac { 1 }{ 2 } \) ..(2)
Also, when x = log2, y = 0
⇒ 0 = Ae^log2+Be^2log2+e^3log3
⇒ 0 = A(2)+\(Be^{ log2^{ 2 }+ }+\frac { 1 }{ 2 } elog2^{ 3 }\)
⇒ 0 = 2A+4B+\(\frac{8}{2}\) ⇒ 2A+4B=-4
⇒ A+2B = -2
(2)-(3) ⇒ A+B = -\(\frac { 1 }{ 2 } \)

Substituting B = -\(\frac { 3 }{ 2 } \) in  (2) we get
\(A-\frac { 3 }{ 2 } =-\frac { 1 }{ 2 } \Rightarrow A=-\frac { 1 }{ 2 } +\frac { 3 }{ 2 } \Rightarrow \frac { 2 }{ 2 } \)=1
∴ y = \(1.e^{ x }-\frac { 3 }{ 2 } e^{ 2x }+\frac { e^{ 3x } }{ 2 } \).

\(\overline {\overline{X}} = \frac {1080+1390+1460+1380+1230+1370+1310+1630+1580+1510+1270+1200}{12}\)
\(\overline {\overline{X}}\) = \(\frac {16410}{12}\) = 1367.5
\(\overline {\overline{R}} = \frac {410+670+180+320+690+450+380+350+270+660+440+310}{12}\)
\(\overline {R}\) =  \(\frac {5130}{12}\) = 427.5
The control limits of mean chart are
UCL = \(\overline {\overline{X}} + A_{2} \overline{R}\)
= 1367.5 + 0.577(427.5)
= 1367.5 + 246.67 = 1614.17
CL =  \(\overline {\overline{X}}\) = 1367.5
LCL  =  \(\overline {\overline{X}} - A_{2} \overline{R}\)
= 1367.5 - 246.67 = 1120.83
The control limits of range chart are
UCL = D₄\(\overline {R}\) = 2.114 (427.5) = 903.74
CL = \(\overline {R}\)  = 427.5
[For n=5, A₂  = 0.577, D₃  = 0, D₄ = 2.114]
LCL = D₃\(\overline {R}\)  = 0
\(\overline {X}\) - chart

 
R-chart


Conclusion: The above diagram shows all the control lines with the data points plotted.
Since one point in mean chart, lie outside the control limits, we can say that the process is out of control.

Year X	Sales Y	X = \(\frac {x-1983.5}{0.5}\)	XY	X2
1980	50.3	-7	-352.1	49
1981	52.7	-5	-263.5	25
1982	49.3	-3	-147.9	9
1983	57.3	-1	-57.3	1
1984	56.8	1	56.8	1
1985	60.7	3	182.1	9
1986	62.1	5	310.5	25
1987	58.7	7	410.9	49
	447.9	0	139.5	168

Since \(\sum X = 0, a = \frac {\sum Y}{n} = \frac {447.9}{8} = 55.9875\)
b  = \(\frac {\sum XY}{\sum X^{2}} = \frac{139.5}{168}\)  = 0.83035
∴  The straight line trend is obtained by
Y = a + bX  ⇒ Y = 55.9875 + 0.830 \((\frac {x-1983.5}{0.5})\)
= 55.9875 + 0.8304 (-5) 
= 55.9875 - 4.152  
= 51.8355

The given differential equation is of this form
\(\frac { dy }{ dx } \)+Py = Q Where
P = \(\frac { 3x^{ 2 } }{ 1+{ x }^{ 3 } } \); Q=\(\frac { 1+x^{ 2 } }{ 1+{ x }^{ 3 } } \)
∴ \(\int { p } dx=\int { \frac { 3x^{ 2 } }{ 1+{ x }^{ 3 } } } \)dx
put t = 1+x³ ⇒ dt = 3x² dx
=\(\int { \frac { dt }{ t } } \) = log t =log(1+x³) [∵ t = 1+x³]
∴ Integrating factor (I. F) = \(e^{ \int { p } dx }=e^{ log(1+x^{ 3 }) }\)
= 1 + x³
Hence the solution is
\(ye^{ \int { p } dx }=\int { Q } .e^{ \int { p } dx }dx+c\)

=\(\int { (1+{ x }^{ 2 }) } dx\) + c
⇒ y(1+x³) = x + \(\frac { { x }^{ 3 } }{ 3 } \) + c

Sample No	Observations			Total	\(\overline {X}\)	R = x max - xmin
	1	2	3
1	32	36	42	110	\(\frac {100}{3} = 36.7\)	42-32 = 10
2	28	32	40	100	\(\frac {100}{3} = 33.3\)	40-28 = 12
3	39	52	28	119	\(\frac {119}{3} = 39.7\)	52 -28 = 24
4	50	42	31	123	\(\frac {123}{3} = 41\)	50-31 = 19
5	42	45	34	121	\(\frac {123}{3} = 40.3\)	45-34 = 29
6	50	29	21	100	\(\frac {100}{3} = 33.3\)	50-21 = 29
7	44	52	35	131	\(\frac {131}{3} = 43.7\)	52-35 = 17
8	22	35	44	101	\(\frac {101}{3} = 33.7\)	44-22 = 22
					301.7	144

\(\overline {\overline{X}}\)   =  \(\frac {301.7}{8}\) = 37.71
\(\overline{R}\) = \(\frac {144}{8}\) = 18
Control limits for  \(\overline {X}\) - chart
UCl =  \(\overline {\overline{X}} + A_{2} \overline {R}\)
= 37.71 + 0.58 (18)
= 37.71 + 18.44
[when n = 3, A₂ = 1.023]
UCL = 56.12
CL =  \(\overline {\overline{X}}\) = 37.71
LCL =  \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 37.71 - 18.44
LCL = 19.28
Control limits for R-chart
UCL = D₄ \(\overline {R}\) = 2.2574(18) = 46.33
CL = \(\overline {R}\) = 18
LC = D3 \(\overline {R}\) = 0
when n = 3, D₄ = 2.574]
\(\overline {X}\) - chart

.\(\overline {R}\) - chart


Conclusion: The above diagram shows all the control lines with the data points plotted. Since all the points lie within the control limits, we can say that the process is in control.

The given equation can be reduced to
\(\frac { dy }{ dx } +\frac { 2x }{ { x }^{ 2 }+1 } y=\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)
It is of the form \(\frac { dy }{ dx } \) + Py = Q
Here P \(=\frac { 2x }{ { x }^{ 2 }+1 } ,Q=\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)
ഽPdx = ഽ\(\frac { 2x }{ { x }^{ 2 }+1 } \)dx = log(x² +1)
I.F = e^ഽpdx = e^log(x2+1) = x² + 1
The required solution is y(IF) = ഽQ(I.F)dx + c
y(x² +1) = ഽ\(\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)(x² + 1)dx + c
y(x² +1) = \(\frac { 4{ x }^{ 3 } }{ 3 } \) + c

Commodity	1995		2005
	Price(p0)	(q0)	Price	Quantity
A	5	60	15	70
B	4	20	8	35
C	3	15	6	20

p0q0	p0q1	p1q0	p1q1
300	350	900	1050
80	140	160	280
45	60	90	120
425	550	1150	1450

Laspeyre's price index number
\(P^{L}_{01}\) = \({\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}}} \times 100\) 
= \(\frac {1150}{425} \times {100}\) = 270.58
Paasche's index number
\(P^{L}_{01}\) =  \({\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{1}}} \times 100\)
=  \(\frac {1450}{550} \times {100}\) = 263.63

\(x\frac { dy }{ dx } =y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { \left[ y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right] }{ x } \)
Since the numerator and denominator is a homogeneous function of degree 1,
put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ v + x\(\frac { dv }{ dx } =\frac { vx+\sqrt { { x }^{ 2 }+{ v }^{ 2 }{ x }^{ 2 } } }{ x } =\frac { vx+x\sqrt { 1+{ v }^{ 2 } } }{ x } \)

⇒ x\(\frac { dv }{ dx } =v+\sqrt { 1+{ v }^{ 2 } } -v=\sqrt { 1+v^{ 2 } } \)
Separating the variables we get,
\(\frac { dx }{ \sqrt { 1+{ v }^{ 2 } } } =\frac { dx }{ x } \)
Integrating both sides we get,
\(\int { \frac { dv }{ \sqrt { 1+{ v }^{ 2 } } } } =\int { \frac { dx }{ x } } \)
\(\left[ \therefore \int { \frac { dx }{ \sqrt { { x }^{ 2 }+a^{ 2 } } } } =log\left| x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } +c \right| \right] \)
⇒ \(log\left| v+\sqrt { { v }^{ 2 }+1 } \right| \) =log x + log c
⇒ \(log(v+\sqrt { { v }^{ 2 }+1 } )\) =log xc
⇒ v+\(\sqrt { { v }^{ 2 }+1 } \)= xc
Replace v by \(\frac { y }{ x } \) we get,
\(\frac { y }{ x } +\sqrt { \frac { { y }^{ 2 } }{ { x }^{ 2 } } +1 } \)= xc
⇒ \(\frac { y }{ x } +\sqrt { \frac { { x }^{ 2 }+{ y }^{ 2 } }{ x } } \)= xc
⇒ y+\(\\ \frac { y+\sqrt { \frac { { x }^{ 2 }+{ y }^{ 2 } }{ x } } }{ x } \)= xc
⇒ y+\(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \) = x²c

Year 2010	Sales (in tones)
Jan	280
Feb	240
Mar	270
Apr	300
May	280
June	290

Average
\(\frac {280 + 240 + 270 + 300 + 280 + 290}{6}\) = \(\frac {1660}{6}\) = 276.6666

Year 2010	Sales (in tones)
Jul	210
Aug	200
Sep	230
Oct	200
Nov	230
Dec	210

\(\frac {210+ 200+ 230+ 200+ 230+ 210}{6}\) = \(\frac {1280}{6}\) = 213.333

Since the number of years is even (12), we can equally divide the given data into two equal parts and obtain the averages of first 6 months and last 6 months.

Given marginal cost function is (x² + xy)dy + (3xy + y²)dx = 0 
\(\frac { dy }{ dx } =\frac { -(3xy+{ y }^{ 2 }) }{ { x }^{ 2 }+xy } \)    (1)
Put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \) in (1)
\(v+x\frac { dv }{ dx } =\frac { -(3xvx+{ v }^{ 2 }{ x }^{ 2 }) }{ { x }^{ 2 }+xvx } \)
\(=\frac { -(3v+{ v }^{ 2 }) }{ 1+v } \)
Now, \(x\frac { dv }{ dx } =\frac { -3v-{ v }^{ 2 } }{ 1+v } -v\)
\(=\frac { -3v-{ v }^{ 2 }-v-{ v }^{ 2 } }{ 1+v } \)
\(x\frac { dv }{ dx } =\frac { -4v-{ 2v }^{ 2 } }{ 1+v } \)
\(\frac { 1+v }{ 4v+2{ v }^{ 2 } } dv=\frac { -dx }{ x } \)
On Integration
\(\int { \frac { 1+v }{ 4v+2{ v }^{ 2 } } } =-\int { \frac { dx }{ x } } \)
Now, multiply 4 on both sides
\(\int { \frac { 1+v }{ 4v+2{ v }^{ 2 } } } =-4\int { \frac { dx }{ x } } \)
log (4v+2v²) = −4 logx+logc
4v + 2v² = \(\frac { c }{ { x }^{ 4 } } \)
x⁴(4v + 2v²) = c
Replace \(v=\frac { y }{ x } \)
\({ x }^{ 4 }\left( 4\frac { y }{ x } +2\frac { { y }^{ 2 } }{ { x }^{ 2 } } \right) =c\)
\({ x }^{ 4 }\left[ \frac { 4xy+2{ y }^{ 2 } }{ { x }^{ 2 } } \right] \) = c
c = 2x²(2xy + y²)   (2)
Cost of producing a pair of shoes = Rs. 12
(i.e) y = 12 when x = 2
c = 8[48 + 144]= 1536
∴ The cost function is x²(2xy + y²) = 768

Commodities Price Weights
(V) \(P=\frac { { P }_{ 1 } }{ { P }_{ 0 } } \times 100\) PV 2007
(P₀) 2011
(P₁) A 350 400 40 114.285 4571.4 B 175 250 35 142.857 4999.99 C 100 115 15 115 1725 D 75 105 20 114 2280 E 60 80 25 133.333 3333.33     Total 135   17429 76 Value of Cost of Living Index Number\(\frac { \sum { PV } }{ \sum { V } } =\frac { 17429 76 }{ 135 } =129.1093\)
Hence, the Cost of Living Index Number for a particular class of people for the year 2011 is increased by 129.1093 % as compared to the year 2007.

Commodities	Base Year		Current Year		p0q0	p0q1	p1q0	p1q1
	Price (p0)	Quantity (q1)	Price ((p0))	Quantity (q1)
Rice	40	5	48	4	200	160	240	192
Wheat	45	2	42	3	90	135	84	126
Rent	90	4	95	6	360	540	380	570
Fuel	85	3	80	2	255	170	240	160
Transport	50	5	65	8	250	400	325	520
Miscellaneous	65	1	72	3	65	195	72	216
				Total	1220	1600	1341	1784

Fisher’s price index number
\({ P }_{ 01 }^{ F }=\left( \sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 10 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \right) \times 100=\left( \sqrt { \frac { 1341\times 1784 }{ 1220\times 1600 } } \right) \times 100=110.706\)
Time Reversal Test:P₀₁x P₁₀=1
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { 1341\times 1784\times 1600\times 1220 }{ 1220\times 1600\times 1784\times 1341 } \right) } \)
P₀₁x P₁₀ = 1
Factor Reversal Test
\(P_{01} \times Q_{01}=\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{0}}\)
\(P_{01} \times Q_{01}=\sqrt{\left(\frac{\sum p_{1} q_{0} \times \sum p_{1} q_{1} \times \sum q_{1} p_{0} \times \sum q_{1} p_{1}}{\sum p_{0} q_{0} \times \sum p_{0} q_{1} \times \sum q_{0} p_{0} \times \sum q_{0} p_{1}}\right)}\)
\(P_{01} \times Q_{01}=\sqrt{\left(\frac{1341 \times 1784 \times 1600 \times 1784}{1220 \times 1600 \times 1220 \times 1341}\right)}\)
\(P_{01} \times Q_{01}=\sqrt{\left(\frac{1784 \times 1784}{1220 \times 1220}\right)}=\frac{1784}{1220}\)
\(\Rightarrow P_{01} \times Q_{01}=\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{0}}\)

Given x - y \(\frac { dx }{ dy } =a\left( { x }^{ 2 }+\frac { dx }{ dy } \right) \)
\(x-y\frac { dx }{ dy } ={ ax }^{ 2 }+a\frac { dx }{ dy } \)
\(x-{ ax }^{ 2 }=a\frac { dx }{ dy } +y\frac { dx }{ dy } \)
x(1 − ax) = (a + y)\(\frac { dx }{ dy } \)
By separating the variables, we get
\(\frac { dx }{ x(1-ax) } =\frac { dy }{ a+y } \)
\(\left( \frac { a }{ 1-ax } +\frac { 1 }{ x } \right) dx=\frac { dy }{ a+y } \)
Integrating, ഽ\(\left( \frac { a }{ 1-ax } +\frac { 1 }{ x } \right) dx=\frac { dy }{ a+y } \)
−log(1 − ax) + log  x = log(a + y) + logc
\(log\left( \frac { x }{ 1-ax } \right) \) = log(c(a + y))
\(\left( \frac { x }{ 1-ax } \right) \) = c(a + y)
x = (1 − ax)(a + y)c which is the required solution

Computation of seasonal Indices by method of simple averages.
Months Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec Year 2001 15 41 25 31 29 47 41 19 35 38 40 30 2002 20 21 27 19 17 25 29 31 35 39 30 44 2003 18 16 20 28 24 25 30 34 30 38 37 39 Monthly Total 53  78 72 78 70 97 100 84 100 115 107 113 Monthly Averages 17.67 26 24 26 23.33 32.33 33.33 28 33.33 38.33 35.67 37.67 S.I for Jan = \(\frac{Monthly\ Average\ (for\ Jan)}{Grand\ average}\times100\)
Grand Average = \(\frac{355.582}{12}=29.63\)
S.I for Jan = \(\frac{17.666}{29.361}\times100=59.62; \) 
S.I for Feb = \(\frac{26}{29.361}\times 100=87.72\)
Similarly other seasonal index values can be obtained.
 
\(\begin{array}{|c|c|} \hline \text { Months } & \text { Seasonal Index } \\ \hline \text { Jan } & 59.62 \\ \hline \text { Feb } & 87.72 \\ \hline \text { Mar } & 80.97 \\ \hline \text { Apr } & 87.72 \\ \hline \text { May } & 78.71 \\ \hline \text { June } & 109.08 \\ \hline \text { July } & 112.45 \\ \hline \text { Aug } & 94.47 \\ \hline \text { Sep } & 112.45 \\ \hline \text { Oct } & 129.32 \\ \hline \text { Nov } & 120.34 \\ \hline \text { Dec } & 127.09 \\ \hline \end{array}\)

Sample size n=100
Sample mean \(\bar { X } \) = 67.45
Sample variance s² = 9
∴ Sample standard deviation s=√9=3
Let μ the population mean.
Standard error \(S.E=\frac { s }{ \sqrt { n } } =\frac { 3 }{ \sqrt { 100 } } =\frac { 3 }{ 10 } =0.3\)
a) The 95% confidence limits for μ are given by
\(\bar { X } -{ Z }_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { X } +{ Z }_{ \frac { \alpha }{ 2 } }(S.E)\)
As the level of significance is α = 0.05,
\({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
∴ (1) becomes,
67.45-(1.96)(0.3) ≤ μ ≤ 67.45+(1.96)(0.3)
⇒ 67.45-0.588 ≤ μ ≤ 67.45+0.588
⇒ 66.862 ≤ μ ≤ 68.038
Thus, the 95% confidence intervals for estimating μ is given by (66.86, 68.04).
b) The 99% confidence limits for estimating μ are given by
\(\bar { X } -{ Z }_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { X } +{ Z }_{ \frac { \alpha }{ 2 } }(S.E)\)
As the level of significance is α = 0.01,
\({ Z }_{ \frac { \alpha }{ 2 } }=2.58\)
∴ 67.45-2.58(0.3) ≤ μ ≤ 67.45+2.58(0.3)
⇒ 67.45-0.774 ≤ μ ≤ 67.45+0.774
⇒ 66.676 ≤ μ ≤ 68.224
Thus, the 99% confidence intervals for estimating μ is given by (66.67, 68.22).

Let I = \(\int _{ 0 }^{ 3 }{ \frac { xdx }{ \sqrt { x+1 } +\sqrt { 5x+1 } } } \)
\(I=\int _{ 0 }^{ 3 }{ \cfrac { x\left( \sqrt { x+1 } -\sqrt { 5x+1 } \right) }{ \left( \sqrt { x+1 } +\sqrt { 5x+1 } \right) \left( \sqrt { x+1 } -\sqrt { 5x+1 } \right) } } \)
= \(\int _{ 0 }^{ 3 }{ \cfrac { x\left( \sqrt { x+1 } -\sqrt { 5x+1 } \right) dx }{ \left( x+1 \right) -\left( 5x+1 \right) } } \)
\(\left[ \because \left( a+b \right) \left( a-b \right) ={ a }^{ 2 }-{ b }^{ 2 } \right] \)

= \(-\cfrac { 1 }{ 4 } \int _{ 0 }^{ 3 }{ \left( \sqrt { x+1 } -\sqrt { 5x+1 } \right) dx } \)
=  \(-\cfrac { 1 }{ 4 } \left[ \cfrac { \left( x+1 \right) ^{ \cfrac { 3 }{ 2 } } }{ \cfrac { 3 }{ 2 } } -\cfrac { \left( 5x+1 \right) ^{ \cfrac { 3 }{ 2 } } }{ 5\left( \cfrac { 3 }{ 2 } \right) } \right] _{ 0 }^{ 3 }\)
= \(-\cfrac { 1 }{ 4 } \left[ \cfrac { 2 }{ 3 } \left( x+1 \right) ^{ \frac { 3 }{ 2 } }-\cfrac { 1 }{ 15 } \left( 5x+1 \right) ^{ \frac { 3 }{ 2 } } \right] _{ 0 }^{ 3 }\)
= \(-\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ 3 } \left( x+1 \right) ^{ \cfrac { 3 }{ 2 } }-\cfrac { 1 }{ 15 } \left( 5x+1 \right) ^{ \cfrac { 3 }{ 2 } } \right] _{ 0 }^{ 3 }\)
= \(-\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ 3 } (4)^{ \cfrac { 3 }{ 2 } }-\cfrac { 1 }{ 15 } \left( 16 \right) ^{ \cfrac { 3 }{ 2 } } \right] -\left[ \cfrac { 1 }{ 3 } \left( 1 \right) ^{ \cfrac { 3 }{ 2 } }-\cfrac { 1 }{ 15 } \left( 1 \right) ^{ \cfrac { 3 }{ 2 } } \right] \)
= \(-\cfrac { 1 }{ 2 } \left[ \left( \cfrac { 1 }{ 3 } \left( 4 \right) \sqrt { 4 } -\cfrac { 1 }{ 15 } 16\sqrt { 16 } \right) -\left( \cfrac { 1 }{ 3 } -\cfrac { 1 }{ 15 } \right) \right] \)
= \(\cfrac { 1 }{ 2 } \left[ \left( \cfrac { 8 }{ 3 } -\cfrac { 64 }{ 15 } \right) -\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 5 } \right] \)
= \(-\cfrac { 1 }{ 2 } \left[ \cfrac { 8 }{ 3 } -\cfrac { 64 }{ 15 } -\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 15 } \right] \)
= \(-\cfrac { 1 }{ 2 } \left[ \cfrac { 8 }{ 3 } -\cfrac { 1 }{ 3 } -\cfrac { 64 }{ 15 } +\cfrac { 1 }{ 15 } \right] \)
= \(-\cfrac { 1 }{ 2 } \left[ \cfrac { 7 }{ 3 } -\cfrac { 63 }{ 15 } \right] =-\cfrac { 1 }{ 2 } \left[ \cfrac { 105-189 }{ 45 } \right] \)
= \(-\cfrac { 1 }{ 2 } \left[ \cfrac { -84 }{ 45 } \right] =\cfrac { 42 }{ 45 } =\cfrac { 14 }{ 15 } \)
\(\therefore\) \(I=\cfrac { 14 }{ 15 } \)

In simple random sampling the samples are selected in such a way that each and every unit in the population has an equal and independent chance of being selected as a sample. It can be done with or without replacement of the samples selected. If the sampling is With replacement, there is a possibility of selecting the same sample any number of times. So, in simple random sampling without replacement is followed. Thus in simple random sampling from a population of N units, the probability is 1/ N the probability of drawing any unit in the second draw from among the available (N - 1) units is 1/(N - 1) and so on.
Simple random sampling without replacement is followed. The following two methods are generally used.
(A) Lottery method:
This is the most popular and simplest method when the population is finite. In this method, all the items of the population are numbered on separate slips of paper of same size, shape and colour. They are folded and placed in a container and shuffled thoroughly. Then the required numbers of slips are selected.
(B) Table of Random number:
The random number table has been so constructed that each of the digits 0,1,2, ... ,9 will appear approximately with the same frequency and independently of each other.
The various random number tables available are
a. L.H.C. Tippet random number series
b. Fisher and Yates random number series
c. Kendall and Smith random number series
d. Rand Corporation random number series.
Example: Tippett's table of random numbers is 20 items out of 6000.
Here we consider row wise selection of random numbers.

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

\(\int _{ a }^{ b }{ f\left( x \right) } dx=\lim _{ x\rightarrow \infty \\ h\rightarrow 0 }{ \sum _{ r=1 }^{ n }{ h\ f(a+rh) } } \)
Here a = 0, b = 0
and f(x) = x + 4
f (a + rh) = \(f\left( 0+\frac { r }{ n } \right) =f\left( \frac { r }{ n } \right) \)
\(=\frac { r }{ n } +4\)
\(\therefore \int _{ 0 }^{ 1 }{ \left( x+4 \right) } dx=\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { 1 }{ n } } } \left( \frac { r }{ n } +4 \right) \)
\(=\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \left( \frac { r }{ { n }^{ 2 } } +\frac { 4 }{ n } \right) } } \)
\(=\lim _{ n\rightarrow \infty }{ \left( \frac { 1 }{ { n }^{ 2 } } \sum _{ r=1 }^{ n }{ r } +\frac { 4 }{ n } \sum _{ r=1 }^{ n }{ 1 } \right) } \)

\(\left( \sum _{ r=1 }^{ n }{ 1=n)\sum _{ r=1 }^{ n }{ r } =\frac { n(n+2) }{ 2 } } \right) \)

\(=\lim _{ n\rightarrow \infty }{ \frac { 1+\frac { 1 }{ n } }{ 2 } } +4\)
\(=\frac { 1+0 }{ 2 } 4\)
[∵ when \(n\rightarrow \infty ,\frac { 1 }{ n } \rightarrow 0\)]
\(=\frac { 1 }{ 2 } +4=\frac { 1+8 }{ 2 } =\frac { 9 }{ 2 } \)

Sample size n = 50 Sample mean \(\bar x\) = 10 km sample standard deviation s = 3.5 km 
Population mean \(\mu\) = 9.5km
Since population SD is unknown we consider \(\sigma\) = s
The sample is a large sample and so we apply Z-test
Null Hypothesis :
There is no significant difference between the sample average and the company’s claim, i.e., \(H_0 : \mu=9.5\)
Alternative Hypothesis :
There is significant difference between the sample average and the company’s claim, i.e., H₁ : \(\mu\)\(\neq \) 9.5 (two tailed test)
The level of significance \(\alpha\) = 5% = 0.05
Applying the test statistic
\(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1);\ \)
\(Z=\frac { 10-9.5 }{ \frac { 3.5 }{ \sqrt { 50 } } } \sim N(0,1)=\frac { 0.5 }{ 0.495 } =1.01\)
Thus the calculated value 1.01 and the significant value or table value \({ Z }_{ \frac { \sigma }{ 2 } }=1.96\)
Comparing the calculated and table value ,Here Z<\({ Z }_{ \frac { \sigma }{ 2 } }\)i.e., 1.01<1.96.
Inference : Since the calculated value is less than table value i.e.,  Z < \({ Z }_{ \frac { \sigma }{ 2 } }\) at 5% level of sinificance, the null hypothesis H₀ is accepted. Hence we conclude that the company’s claim that the new car petrol consumption is 9.5 km per litre is acceptable.

Let I = \(\int _{ 2 }^{ 5 }{ \frac { \sqrt { x } }{ \sqrt { x } +\sqrt { 7-x } } } \)  dx ...(1)
I = \(\int _{ 2 }^{ 5 }{ \frac { \sqrt { 2+5-x } }{ \sqrt { 2+5-x } +\sqrt { 7-(2+5-x) } } } \) dx  \(​​\left[∵ \int _{ 0 }^{ a }{ f(x) } dx=\int _{ 0 }^{ a }{ f(a+b-x)dx } \right] \)
I = \(\int _{ 2 }^{ 5 }{ \frac { \sqrt { 7-x } }{ \sqrt { 7-x } +\sqrt { x } } } \)dx ..... (2)
(1) + (2) ⇒
2I = \(\int _{ 2 }^{ 5 }{ \left[ \frac { \sqrt { x } }{ \sqrt { x } +\sqrt { 7-x } } +\frac { \sqrt { 7+x } }{ \sqrt { 7-x } +\sqrt { x } } \right] } \) dx
= \(\int _{ 2 }^{ 5 }{ \left[ \frac { \sqrt { x } +\sqrt { 7-x } }{ \sqrt { x } +\sqrt { 7-x } } \right] } \) dx
= \(\int _{ 2 }^{ 5 }{ dx } \) = \({ \left[ x \right] }_{ 2 }^{ 5 }\) = 3
ஃ I = \(\frac { 3 }{ 2 } \)

The first thing is to number the population (8585 children). The numbering has already been provided by the frequency table. There are 2 children with height of 105 cm, therefore we assign number 1 and 2 to the children those in the group 105 cm, number 3 to 6 is assigned to those in the group 107 cm and similarly all other children are assigned the numbers. In the last group 145 cms there are two children with assigned number 8584 and 8585.

Height (cm)	Number of children	Cumulative Frequency
105	2	2
107	4	6
109	14	20
111	41	61
113	83	144
115	169	313
117	394	707
119	669	1376
121	990	2366
123	1223	3589
125	1329	4918
127	1230	6148
129	1063	7211
131	646	7857
133	392	8249
135	202	8451
137	79	8530
139	32	8562
141	16	8578
143	5	8583
145	2	8585
Total	8585

Now we take 10 samples from the tables, since the population size is in 4 digits we can use the given random number table. Select the10 random numbers from 1 to 8585 in the table, Here, we consider column wise selection of random numbers, starting from first column.

Tippet’s random number table
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

The children with assigned number 2952 is selected and then see the cumulative frequency table where 2952 is present, now select the corresponding row height which is 123 cm, similarly all the selected random numbers are considered for the selection of the child with their corresponding height. The following table shows all the selected 10 children with their heights.

Child with assigned Number	2952	4167	2670	0560	2754
Corresponding Height (cms)	123	125	123	117	123
Child with assigned Number	6641	7483	5246	3996	1545
Corresponding Height (cms)	129	131	127	125	121

Given μ = 400, σ = 100
(i) Company pays a penalty of atleast Rs. 2,00,000
Penalty per day = Rs.10,000
Number of days = \(\frac { 2,00,000 }{ 10,000 } \) = 20
Hence, the company has taken excess of 20 days
∴ P(atleast 470 days)     [∵ 450 + 20 = 470]
= P(X ≥ 470)
When X = 470, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 470-400 }{ 100 } =\frac { 70 }{ 100 } \) = 0.7
∴ P(X ≥ 470) = P(Z ≥ 0.7)
= P(0.7 = 0.5-0.2580
P(X ≥ 470) = 0.2420

(ii) P (atmost 500 days) = P(X≤500)
When X = 500, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 500-400 }{ 100 } =\frac { 100 }{ 100 } \)=1
∴ P(X ≤ 500) = P(Z≤1)
= P(-∞ = 0.5+0.3413
= 0.8413
P(X ≤ 500) = 0.8413

Plot the variable X = 50 on the left side and
X = 86 on the right side of the curve.
Given P( -∞< Z₁ < -Z₁) = 0.3
⇒ P(-Z₁ < Z < Z₁) = 0.2       [∵ 0.5 - 0.3 = 0.2]
⇒ P(0 < Z < Z₁) = 0.2 [By symmetry]
⇒ Z₁ = -0.52 [From the normal distribution table 'and it lies on the negative side]
⇒ -0.52 = \(\frac { X-\mu }{ \sigma } \) ⇒ -0.52 σ = 50-μ
⇒ 50-μ = -0.52 σ ....(1)
Also given P(Z₂ ∴ P(Z₂2) = 0.4    [∵ 0.5-0.1 = 0.4]
⇒ Z₂ = 1.28 (from the table)
⇒ 1.28 =\(\frac { 86-\mu }{ \sigma } \)
⇒ 86-μ = 1.28 σ ...(2)

Substituting σ = 20 in (2) we get,
86-μ = (1.28)(20)
86-μ = 25.6
μ = 86-25.6
μ = 60.4

Hence, the mean is 60.4 and standard deviation is 20.

Given mean = λ = 1.5
X follows poisson distribution with
p(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
(i) ∴ P(neither car is used)
= P(X = 0) = \(\frac { e^{ -1.5 }.(1.5)^{ 0 } }{ 0! } \) =e^-1.5
= 0.2231    [∵ e^-1.5 =0.2231]
(ii) P (Some demand is refused)
The demand may be either 0 car, 1 car or 2 cars
∴ P (Some demand is refused) = 1 - P(X ≤ 2)
= 1 - [P(X = 0) + P (X = 1) + P(X = 2)]
= 1-\(\left[ \frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { e^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } +\frac { e^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } \right] \)
= 1-e^-λ (1+λ+\(\frac { { \lambda }^{ 2 } }{ 2 } \))
= 1-e^-1.5 (1+1.5+\(\frac { (1.5)^{ 2 } }{ 2 } \))
= 1 - 0.2231 (3.625) = 1 - 0.8087 = 0.1912
∴ Probability of some demand is refused = 0.1912.

Let the probability for boy or girl is p =\(\frac { 1 }{ 2 } \) and q = \(\frac { 1 }{ 2 } \), n = 4
(i) P (atleast one boy) = P(X≥1)
= 1-P(X<1)
= 1-[P(X = 0)]
= 1-\(\left[ { { 4 }_{ C } }_{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 4 } \right] \)  p(x) = ⁿC_x p^xq^n-x, n = 4, x = 0
= \(\left( \frac { 1 }{ 2 } \right) ^{ 4 }=1-\frac { 1 }{ 16 } \) = 1-.0625
= 0.9375
∴ Probability of one family having atleast one boy = 0.9375.
Hence, probability of 750 families having atleast one boy
= 0.9375 \(\times\) 750 = 703
(ii) P (atmost 2 girls) = P(X ≤ 2)
= P(X = 0) + P(X = 1) + P(X = 2)
= ⁴C₀ \(\left( \frac { 1 }{ 2 } \right) ^{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 4 }\) + ⁴C₁ \(\left( \frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 2 } \right) ^{ 3 }\) + ⁴C₂ \(\left( \frac { 1 }{ 2 } \right) ^{ 2 }\left( \frac { 1 }{ 2 } \right) ^{ 2 }\)
= \(\left( \frac { 1 }{ 2 } \right) ^{ 4 }\) [⁴C₀ + ⁴C₁ + ⁴C₁]
= \(\frac { 1 }{ 16 } \left[ 1+4+\frac { 4\times 3 }{ 2\times 1 } \right] \) = 0.0625 (5+6)
= 0.0625 \(\times\) 11 = 0.6815
∴ Probability of one family having atmost 2 girls = 0.6875.
Hence, probability of 750 families having atmost 2 girls
= 0.6875 \(\times\) 750 = 515.6 = 516.
(iii) P(Chiidren of both sexes) = P (atleast one girl and atleast one boy)
= 1[P(X = 0) + P(X = 4)]
= 1-⁴C₀ \(\left( \frac { 1 }{ 2 } \right) ^{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 4 }-{ C }_{ 4 }\left( -\frac { 1 }{ 2 } \right) ^{ 4 }\left( \frac { 1 }{ 2 } \right) ^{ 0 }\)
= \(1-\frac { 1 }{ 16 } -\frac { 1 }{ 16 } =1-\frac { 2 }{ 16 } =1-\frac { 1 }{ 8 } =\frac { 7 }{ 8 } \)
∴ Probability of 1 family of having children of both sexes = 0.875.
∴ Probability of 750 families having children of both sexes = 0.875 \(\times\) 75 = 656.

Let X be the waiting time of a customer in the queue and it is normally distributed with mean 5 and SD 0.7.

(i) for less than 6 minutes
\(Z=\frac { X-\mu }{ \sigma } =\frac { 6-5 }{ 0.7 } =1.4285\)
P(X < 6) = P(Z < 1.43)
= 0.5 + 0.4236
= 0.9236
(ii) between 3.5 and 6.5 minutes
When X = 3.5
\(Z=\frac { X-\mu }{ \sigma } =\frac { 3.5-5 }{ 0.7 } =2.1429\)
When X = 6.5
\(Z=\frac { X-\mu }{ \sigma } =\frac { 6.5-5 }{ 0.7 } =2.1429\)
P(3.5 < X < 6.5)
= P (–2.1429 < Z < 2.1429)
= P(0 < Z < 2.1429) + P(0 < Z < 2.1429)
= 2 P(0 < Z < 2.1429)
= 2 x .4838
= 0.9676

Here mean μ= 30 and standard deviation σ = 5
(i) When X = 26 Z=(X−μ)/\(\sigma \) = (26 – 30)/5 = –0.8
And when X = 40 , z =\(\frac{40-30}{5}=2\)
Therefore,
P(26 < x < 40)) = P(–0.8 ≤ Z < 2)
= P(–0.8 ≤ Z ≤ 0) + p(0 ≤ Z ≤ 2)
= P(0 ≤ Z ≤ 0.8) + P(0 ≤ Z ≤ 2)
= 0.2881 + 0.4772 (By tables)
= 0.7653

(ii) The probability that X≥45
When X = 45
\(z=\frac { X-\mu }{ \sigma } =\frac { 45-30 }{ 5 } =3\)
P( X ≥ 45) = P(Z ≥ 3)
= 0.5 – 0.49865
= 0.00135

Probability of raining on a particular day is given by p = 9/30 = 3/10 and q = 1–p = 7/10.
The binomial distribution is P(X = x) = ⁿC_x p^x q^n–x
There are 7 days in a week
P(X = x) =\(\left( \begin{matrix} 7 \\ x \end{matrix} \right) { \left( \frac { 3 }{ 10 } \right) }^{ x }{ \left( \frac { 7 }{ 10 } \right) }^{ 7-x }\)
The probability of raining for atleast 2 days is given by
P(X\(\ge\)2) = 1-P(X<2)
= 1-[P(X = 0)+P(X = 1)]
Here, P(X = 0)\(\left( \begin{matrix} 7 \\ 0 \end{matrix} \right) { \left( \frac { 3 }{ 10 } \right) }^{ 0 }{ \left( \frac { 7 }{ 10 } \right) }^{ 7-0 }\)
= 0.0823
and \(P(X=1)=\left( \begin{matrix} 7 \\ 1 \end{matrix} \right) \left( \frac { 3 }{ 10 } \right) { \left( \frac { 7 }{ 10 } \right) }^{ 7-1 }\)
= 0.2471
Therefore the required probability = 1– [P(x = 0) +P(x = 1)]
= 1– [0.082+ 0.247]
= 0.6706

Given p.d.f is = \(f(x)=\left\{\begin{array}{l} a+b x^{2}, 0 \leq x \leq 1 \\ 0, \text { otherwise } \end{array}\right.\)
Since f(x) is a p.d.f.\(\\ \int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 0 }^{ 1 }{ (a+{ bx }^{ 2 })dx=1 } \)
\(\Rightarrow { \left[ ax+\frac { { bx }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1 }=1\Rightarrow a+\frac { b }{ 3 } =1\)
3a+b = 3  [multiplied by 3] ...(1)
Also it is given that E(X) = \(\frac{3}{5}\)
\(\Rightarrow \int _{ 0 }^{ 1 }{ x.f(x)dx=\frac { 3 }{ 5 } } \)
\(\Rightarrow \int _{ 0 }^{ 1 }{ (a+{ bx }^{ 2 })dx=\frac { 3 }{ 5 } } \)
\(\Rightarrow \int _{ 0 }^{ 1 }{ (ax+{ bx }^{ 3 })dx=\frac { 3 }{ 5 } } \)
\(\Rightarrow { \left[ \frac { { ax }^{ 2 } }{ 2 } +\frac { { bx }^{ 4 } }{ 4 } \right] }_{ 0 }^{ 1 }=\frac { 3 }{ 5 } \Rightarrow \frac { a }{ 2 } +\frac { b }{ 4 } =\frac { 3 }{ 5 } \)
\(\Rightarrow 2a+b=\frac { 12 }{ 5 } \)  [Multiplies by 4] ...(2)

\(a=3-\frac { 12 }{ 5 } =\frac { 15-12 }{ 5 } =\frac { 3 }{ 5 } \)
Substituting a=\(\frac{3}{5}\) in(2) we get,
\(2(\frac { 3 }{ 5 } )+b=\frac { 12 }{ 5 } \Rightarrow \frac { 6 }{ 5 } +b=\frac { 12 }{ 5 } \)
\(\Rightarrow b=\frac { 12 }{ 5 } -\frac { 6 }{ 5 } =\frac { 6 }{ 5 } \)
\(\therefore a=\frac { 3 }{ 5 } ,b=\frac { 6 }{ 5 } \)
ii) \(E({ X }^{ 2 })=\int _{ 0 }^{ 1 }{ { x }^{ 2 }f(x)dx=\int _{ 0 }^{ 1 }{ { x }^{ 2 }\left( \frac { 3 }{ 5 } +\frac { 6 }{ 5 } { x }^{ 2 } \right) dx } } \)
\(\left[ \because a=\frac { 3 }{ 5 } ,b=\frac { 6 }{ 5 } \right] \)
\(=\int _{ 0 }^{ 1 }{ \left( \frac { 3 }{ 5 } { x }^{ 2 }+\frac { 6 }{ 5 } { x }^{ 4 } \right) dx } \)

\(=\frac { 1 }{ 5 } (1-0)+\frac { 6 }{ 25 } (1-0)\)
\(=\frac { 1 }{ 5 } +\frac { 6 }{ 25 } =\frac { 5+6 }{ 25 } =\frac { 11 }{ 25 } \)
\(\therefore\) Var (X) = E(X²)-[E(X)]²
\(=\frac { 11 }{ 25 } -{ \left( \frac { 3 }{ 5 } \right) }^{ 2 }\)
\(=\frac { 11 }{ 25 } -\frac { 9 }{ 25 } =\frac { 2 }{ 25 } \)
\(\therefore\) Var (X) =\(\frac{2}{25}\)

Let I = ∫ e^x(1+x) log(xe^x)dx
Put t = x e^x
⇒ dt = (x.e^x + e^x(1))dx
= e^x(x+1)dx
∴ I = ∫ log t.dt
Let u = log t; dv = dt
\(du=\frac { 1 }{ t } dt;v=t\)
∴Using integration by parts we get,
I = ∫udv = vu - ∫vdu
= t log t - ∫dt = t logt - t + c
= t log t - t + c
= xe^x log(xe^x) - (xe^x) + c  [∵ t = xe^x]
= xe^x (log(xe^x)-1)+c

Given demand function P_d = 20 - 5x and
Supply function P_s = 4x + 8
Under market equilibrium p_s = P_d
⇒ 20-5x = 4x+8
⇒ 20-8 = 4x+5x
⇒ 12 = 9x
\(\Rightarrow x=\frac{\not 12}{\not 9}=\frac{4}{3}\)
When \({ x }_{ 0 }=\frac { 4 }{ 3 } ,{ p }_{ 0 }=20-5\left( \frac { 4 }{ 3 } \right) =20-\frac { 20 }{ 3 } \)
\(=\frac { 60-20 }{ 3 } =\frac { 40 }{ 3 } \)
\(\therefore { p }_{ 0 }{ x }_{ 0 }=\frac { 40 }{ 3 } \times \frac { 4 }{ 3 } =\frac { 160 }{ 9 } \)
Consumer Surplus (CS)
\(=\int _{ 0 }^{ x }{ f(x)dx } -{ p }_{ 0 }{ x }_{ 0 }\)
\(=\int _{ 0 }^{ \frac { 4 }{ 3 } }{ (20-5x)dx } \)
\(={ \left[ 20x-\frac { { 5x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ \frac { 4 }{ 3 } }-\frac { 160 }{ 9 } \)
\(=20\left( \frac { 4 }{ 3 } \right) -\frac { 5 }{ 2 } \left( \frac { 16 }{ 9 } \right) -\frac { 160 }{ 9 } \)
\(=\frac { 80 }{ 3 } -\frac { 40 }{ 9 } -\frac { 160 }{ 9 } \)
\(=\frac { 240-40-160 }{ 9 } =\frac { 40 }{ 9 } \)
\(\therefore CS=\frac { 40 }{ 9 }\)units
Producer's Surplus 
\((PS)={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x }{ g(x)dx } \)
\(=\frac { 160 }{ 9 } -\int _{ 0 }^{ \frac { 4 }{ 3 } }{ (4x+8)dx } \)
\(=\frac { 160 }{ 9 } -{ \left[ \frac { { 4x }^{ 2 } }{ 2 } +8x \right] }_{ 0 }^{ \frac { 4 }{ 3 } }\)
\(=\frac { 160 }{ 9 } -\left( 2\left( \frac { 16 }{ 9 } \right) +8\left( \frac { 4 }{ 3 } \right) \right) \)
\(=\frac { 160 }{ 9 } -\left( \frac { 32 }{ 9 } +\frac { 32 }{ 3 } \right) \)
\(=\frac { 160 }{ 9 } -\frac { 32 }{ 9 } -\frac { 32 }{ 3 } \)
\(PS=\frac { 160-32-96 }{ 9 } =\frac { 32 }{ 9 } \)units

Mean of the random variableX = E(X) = \({ \sum_x { x{ P }_{ x }(x) } } \)
= (1 × 0.15) + (2 × 0.10) + (3 × 0.10) + (4 × 0.01) + (5 × 0.08) + (6 × 0.01) +(7 × 0.05) + (8 × 0.02) + (9 × 0.28) + (10 × 0.20)
E(X) = 6.18
E(X²) = \(\sum _{ x }^{ }{ { x }^{ 2 } } { P }_{ X }(x)\)
= (1² × 0.15) + (2² × 0.10) + (3² × 0.10) + (4² × 0.01) + (5² × 0.08) + (6² × 0.01) + (7² × 0.05) + (8² × 0.02) + (9² × 0.28) + (10² × 0.20).
= 50.38
Variance of the Random Variagble X = V(X) = E(X²)-[E(X)]²
=50.38-(6.56)²
= 12.19
Therefore, the mean and variance of the given discrete distribution are 6.18 and 12.19 respectively.

Given C'(x) = 5 + 0.13x
R'(x) = 18
Fixed cost is Rs. 120
C'(x) = 5 + 0.13x
⇒ ∫C'(x) = ∫(5+0.13x)dx
⇒ C(x) = 5x \(+\frac { 0.13{ x }^{ 2 } }{ 2 } +{ k }_{ 1 }\)
Since fixed cost is Rs.120 ⇒ k₁ = 120
\(\therefore C(x)=5x+\frac { 0.13{ x }^{ 2 } }{ 2 } +120\) ...(1)
Also R'(x) = 18
⇒ ∫R'(x) = ∫18dx
⇒ R(x) = 18x+k₂
When x = 0, R = 0 ⇒ k₂= 0
∴ R(x) = 18x   ...(2)
Profit function ⇒ P(x) = R(x) - C(x)
\(=18x-5x-\frac { 0.13{ x }^{ 2 } }{ 2 } -120\)
[from (1) & (2)]
P(x) = 13x-0.065x²-120

Given p.d.f is
\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 16 } { (3+x) }^{ 2 }, & -3\le x\le -1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 16 } (6-2{ x }^{ 2 }), & -1\le x\le 1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 16 } { (3-x) }^{ 2 }, & 1\le x\le 3 \end{matrix} \end{cases}\)
Area under the given curve
\(\int _{ -3 }^{ 3 }{ f(c)dx=\int _{ -3 }^{ -1 }{ \frac { 1 }{ 16 } { (3+x) }^{ 2 }+\int _{ -1 }^{ 1 }{ \frac { 1 }{ 16 } (6-2{ x }^{ 2 })dx } } +\int _{ 1 }^{ 3 }{ \frac { 1 }{ 16 } { (3-x) }^{ 2 }dx } } \)
\(=\frac { 1 }{ 16 } { \left[ \frac { (3+x{ ) }^{ 3 } }{ 3 } \right] }_{ -3 }^{ -1 }+\frac { 1 }{ 16 } { \left( 6x-\frac { { 2x }^{ 3 } }{ 3 } \right) }_{ -1 }^{ 1 }+\frac { 1 }{ 16 } { \left( \frac { { (3-x) }^{ 3 } }{ -3 } \right) }_{ 1 }^{ 3 }\)
\(=\frac { 1 }{ 16 } \left[ \left( \frac { { 2 }^{ 3 } }{ 3 } -0 \right) +\left( 6-\frac { 2 }{ 3 } \right) -\left( -6+\frac { 2 }{ 3 } \right) -\frac { 1 }{ 3 } \left( 0-({ 2 }^{ 3 }) \right) \right] \)
\(=\frac { 1 }{ 6 } \left[ \frac { 8 }{ 3 } +\frac { 16 }{ 3 } -\left( \frac { -16 }{ 3 } \right) +\frac { 8 }{ 3 } \right] \)
\(\\ =\frac { 1 }{ 16 } \left[ \frac { 8 }{ 3 } +\frac { 16 }{ 3 } +\frac { 16 }{ 3 } +\frac { 8 }{ 3 } \right] \)
\(=\frac { 1 }{ 16 } \left[ \frac { 8+16+16+8 }{ 3 } \right] \)
\(=\frac { 1 }{ 16 } \times \frac { 48 }{ 3 } =\frac { 1 }{ 16 } \times 16=1\)
Hence, area under the given curve is unity.

\(\frac { { E }_{ y } }{ { E }_{ x } } =\frac { x }{ x-2 } \)
\(\frac { { xd }_{ y } }{ { yd }_{ x } } =\frac { x }{ x-2 } \)
\(\frac { dy }{ y } =\frac { x }{ x-2 } .\frac { dx }{ x } \)
\(\int { \frac { dy }{ y } } =\int { \frac { dx }{ x-2 } } \)
log y = log(x − 2) + log k
y = k(x–2)
when x = 6, y = 16 ⇒ 16 = k(6–2)
k = 4
y = 4 (x–2)

Given C(x) = \(\int { C' } (x)dx+\)k₁
= \(\int { \left( 50+\frac { x }{ 50 } \right) } \)dx +k₁
C(x) = 50x + \(\frac{x^2}{100}\)+k₁
When quantity produced is zero, then the fixed cost is 200.
i.e. When x = 0, c = 200
⇒ k₁ = 200
Cost function is C(x) = 50x + \(\frac{x^2}{100}\) + 200        (1)
The Revenue R'(x) = 60
R(x) = \(\int { R' } (x)dx\) + k₂
\(\int { 60 } dx\)+ k₂
= 60x + k₂
When no product is sold, revenue = 0
i.e. When x = 0, R = 0
Revenue R(x) = 60x          (2)
Profit P = Total Revenue – Total cost
= 60x - 50x - \(\frac { { x }^{ 2 } }{ 100 } \) - 200
= 10x - \(\frac { { x }^{ 2 } }{ 100 } \) - 200
\(\frac { dp }{ dx } =10-\frac { x }{ 50 } \)
To get profit maximum, \(\frac { dp }{ dx } \) = 0 ⇒ x = 500
\(\frac { { d }^{ 2 }P }{ { dx }^{ 2 } } =\frac { -1 }{ 50 } <0\)
ஃ Profit is maximum when x = 500 and
Maximum Profit is P = 10(500) - \(\frac { { (500) }^{ 2 } }{ 100 } \) - 200
= 5000 – 2500 – 200
= 2300
Profit = Rs.  2,300.

y² = 4 ax is the right open
⇒ y = \(\sqrt { 4ax } \) parabola.
The limits are from x = 0 to x = a
∴ Area \(=2\int _{ 0 }^{ a }{ ydx } \)
\(=2\int _{ 0 }^{ a }{ \sqrt { 4ax } } dx\)
\(=2\int _{ 0 }^{ a }{ 2\sqrt { a } \sqrt { x } dx } \)
\(=4\sqrt { a } \int _{ 0 }^{ a }{ \sqrt { x } dx } =4\sqrt { a } \int _{ 0 }^{ a }{ { x }^{ \frac { 1 }{ 2 } }dx } \)
\(=4\sqrt { 3 } { \left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ a }=4\sqrt { a } \times \frac { 2 }{ 3 } { \left( { x }^{ \frac { 3 }{ 2 } } \right) }_{ 0 }^{ a }\)
\(=8\frac { \sqrt { a } }{ 3 } \left( { a }^{ \frac { 2 }{ 3 } }-0 \right) \)
\(=\frac { 8\sqrt { a } }{ 3 } (a\sqrt { a } )\)
\(=\frac { 8 }{ 3 } { a }^{ 2 }\) sq.units

\(\int { \frac { { 4x }^{ 2 }+2x+6 }{ { \left( x+1 \right) }^{ 2 }-(x-3) } } dx\)
\(=\int { \left( \frac { A }{ x+1 } +\frac { B }{ { \left( x+1 \right) }^{ 2 } } +\frac { C }{ x-3 } \right) } dx\)
\(=\left( \frac { 1 }{ x+1 } +\frac { -2 }{ { \left( x+1 \right) }^{ 2 } } +\frac { 3 }{ x-3 } \right) dx\)
4x²+2x+6 = A (x+1) (x-3) + B(x-3) + (x+1)²
⇒ Putting x = -1, 4-2+6 = B(-4)
⇒ 8 = B(-4) ⇒ B = -2
Putting x = 3
36 + 6 + 6 = C (16)
⇒ 48 = 16C
⇒ C = 3
Putting x = 0,
6 = -3A -3B + C
⇒ 6 = -3A + 6 + 3
⇒ 3A = 3
⇒ A = 1
\(=\log { \left| x+1 \right| } -2\int { { \left( x+1 \right) }^{ -2 } } dx+3\log { \left| x-3 \right| } +c\)
\(=\log { \left| x+1 \right| } -2\frac { { \left( x+1 \right) }^{ -2+1 } }{ -2+1 } +3\log { \left| x-3 \right| } +c\)
\(=\log { \left| x+1 \right| } +2{ \left( x+1 \right) }^{ -1 }+3\log { \left| x-3 \right| } +c\)
\(=\log { \left| x+1 \right| } +\frac { 2 }{ x+1 } +3\log { \left| x-3 \right| } +c\)

\(\Delta =\left| \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 2 \\ 1 & 1 & -2 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -1 & 2 \\ 1 & -2 \end{matrix} \right| -2\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| \)
= 1(2 -2) - 2(-4 -2) + 1(2 + 1)
= 1(0)-2(-6)+1(3)
= 12 + 3 = 15\(\neq \)0.
Since \(\Delta \neq 0\) Cramer's rule can be applied and thesystem is consistent with unique solution.
\({ \Delta }x=\left| \begin{matrix} 7 & 2 & 1 \\ 4 & -1 & 2 \\ -1 & 1 & -2 \end{matrix} \right| \)
= \(7\left| \begin{matrix} -1 & 2 \\ 1 & -2 \end{matrix} \right| -2\left| \begin{matrix} 4 & 2 \\ -1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 4 & -1 \\ -1 & 1 \end{matrix} \right| \)
= 7 (2 -2) -2 (-8 + 2) + 1 (4 - 1)
= 7 (0) - 2(-6) + 1(3)
= 12 + 3 = 15
\(\Delta y=\left| \begin{matrix} 1 & 7 & 1 \\ 2 & 4 & 2 \\ 1 & -1 & -2 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 4 & 2 \\ -1 & -2 \end{matrix} \right| -7\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 2 & 4 \\ 1 & -1 \end{matrix} \right| \)
= 1 (- 8 + 2) -7(-4 -2) + 1(-2 -4)
= 1 (-6) -7 (-6) + 1 (-6)
= - 6 + 42 - 6 = 30
\(\Delta z=\left| \begin{matrix} 1 & 2 & 7 \\ 2 & -1 & 4 \\ 1 & 1 & -1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -1 & 4 \\ 1 & -1 \end{matrix} \right| -2\left| \begin{matrix} 2 & 4 \\ 1 & -1 \end{matrix} \right| +7\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| \)
= 1 (1 - 4) - 2(- 2 - 4) + 7(2 + 1)
= 1(-3)-2(-6)+7(3)
= - 3 + 12 + 21 = 30

\(\therefore\) Solution set is {1, 2, 2}

Transition probability matrix
(A B) T = (A B)

Where A represents the percent of people those who bought soap A and B represents the percent of people those who bought soap B.
By the given data
A = 15% = ·15
and B = 85% = ·85
Percentage after one year is
\(\left( \cdot 15\quad \cdot 85 \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) \)
= ((.15)(·65) + (·85)(-45) ·15(-35)+ ·85(-55))
= (-0975 + ·3825 ·0525 + -4675)
= (-48 ·52)
Hence, market share after one year is 48% and 52% At equilibrium,
\(\left( A\quad B \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) =(A\quad B)\)
(-65A + A5B ·35A +·55B) = (A B)
Equating the corresponding entries on both sides we get
\(\Rightarrow \cdot 65A+\cdot 45B=A\)
\(\Rightarrow \cdot 65A+\cdot 45(1-A)=A\)
[Since A+B = 1   B = 1-A]
\(\Rightarrow \cdot 65A+\cdot 45-\cdot 45A=A\)
\(\Rightarrow \cdot 45=A-\cdot 65A+\cdot 45A\)
\(\Rightarrow \cdot 45=A\left( \cdot 35+45 \right) \)
\(\Rightarrow \cdot 45=A(\cdot 35+45)\)
\(\Rightarrow \cdot 45=A(-8)\)
\(\Rightarrow A=\cfrac { \cdot 45 }{ \cdot 8 } =\cdot 5625=56.25\)
\(\therefore B=1-A=1-\cdot 5625=\cdot 4375\)
= 43.75%
\(\therefore\) Equilibrium is reached when A = 56.25% and B = 43.75%

Let ‘x’ be the number of cars of type C₁
Let ‘y’ be the number of cars of type C₂
Let ‘z’ be the number of cars of type C₃
3x + 2y + 4z = 28
x + y + 2z =13
2x + 2y + z =14
Here \({ \triangle }=\left| \begin{matrix} 3 & 2 & 4 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right| =-3\neq 0\)
\({ \triangle }_{ x }=\left| \begin{matrix} 28 & 2 & 4 \\ 13 & 1 & 2 \\ 14 & 2 & 1 \end{matrix} \right| =-6\)
 \({ \triangle }_{ y }=\left| \begin{matrix} 3 & 28 & 4 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right| =-9\)
\({ \triangle }_{ z }=\left| \begin{matrix} 3 & 2 & 28 \\ 1 & 1 & 13 \\ 2 & 2 & 14 \end{matrix} \right| =-12\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -6 }{ -3 } =2\) 
\(y=\frac { { \triangle }y }{ { \triangle } } =\frac { -9 }{ -3 } =3\)
\(z=\frac { { \triangle }z }{ { \triangle } } =\frac { -12 }{ -3 } =4\)
\(\therefore \) The number of cars of each type which can be produced are 2, 3 and 4.

Given that the price of commodities X, Y and Z are x, y and z respectively 
By the given data

Transaction	x	y	z	Earning
Mr. Anand	+2	+3	-6	Rs.5000
Mr. Amar	+3	-1	+2	Rs.2000
Mr. Amit	-1	+3	+1	Rs.5500

Here, purchasing is taken as negative symbol and selling is taken as positive symbol
Thus, the non-homogeneous equations are
2x + 3y - 6z = 5000
3x - y + 2z = 2000
-x + 3y + z = 550
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 2 & 3 & -6 \\ 3 & -1 & 2 \\ -1 & 3 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 5000 \\ 2000 \\ 5500 \end{matrix} \right) \)

Augmented matrix	Elementary Transformation
\(\left( \begin{matrix} 2 & 3 & -6 \\ 3 & -1 & 2 \\ -1 & 3 & 1 \end{matrix}\begin{matrix} 5000 \\ 2000 \\ 5500 \end{matrix} \right) \)
\(\left( \begin{matrix} -1 & 3 & 1 \\ 3 & -1 & 2 \\ 2 & 3 & -6 \end{matrix}\begin{matrix} 5500 \\ 2000 \\ 5000 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & -3 & -1 \\ 3 & -1 & 2000 \\ 2 & 3 & -6 \end{matrix}\begin{matrix} -5000 \\ 2000 \\ 5500 \end{matrix} \right) \)	\({ R }_{ 1 }\rightarrow { R }_{ 1 }\left( -1 \right) \)
\(\left( \begin{matrix} 1 & -3 & -1 \\ 0 & 8 & 5 \\ 0 & 9 & -4 \end{matrix}\begin{matrix} -5500 \\ 18500 \\ 16000 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 }\)
\(\left( \begin{matrix} 1 & -3 & -1 \\ 0 & 1 & \frac { 65 }{ 8 } \\ 0 & 1 & \frac { -4 }{ 9 } \end{matrix}\begin{matrix} -5500 \\ \frac { 18500 }{ 8 } \\ \frac { 16000 }{ 9 } \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }\div 8\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }\div 9\)
\(\left( \begin{matrix} 1 & -3 & -1 \\ 0 & 1 & \frac { 5 }{ 8 } \\ 0 & 0 & \frac { -4 }{ 9 } -\frac { 5 }{ 8 } \end{matrix}\begin{matrix} -5500 \\ \frac { 18500 }{ 8 } \\ \frac { 16000 }{ 9 } -\cfrac { 18500 }{ 8 } \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow R_{ 3 }-{ R }_{ 2 }\)
\(\left( \begin{matrix} 1 & -3 & -1 \\ 0 & 1 & \frac { 5 }{ 8 } \\ 0 & 0 & \frac { -77 }{ 72 } \end{matrix}\begin{matrix} -5500 \\ \frac { 18500 }{ 8 } \\ \frac { -38500 }{ 72 } \end{matrix} \right) \)

.Clearly the last equivalent matrix is in echelon form and it has three non-zero rows 
\(\therefore \rho (A)=\rho \left( \left[ A,B \right] \right) =3\) Number of unknowns.
\(\therefore\) The given system is consistent and has unique solution. To find the solution, let us rewrite the above : echelon form into the matrix form.
\(\left( \begin{matrix} 1 & -3 & -1 \\ 0 & 1 & \frac { 5 }{ 8 } \\ 0 & 0 & \frac { -77 }{ 72 } \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) \left( \begin{matrix} -5000 \\ \frac { 18500 }{ 8 } \\ \frac { -38500 }{ 72 } \end{matrix} \right) \)
\(\Rightarrow x-3y-z=-5500\)
\(y+\cfrac { 5 }{ 8 } z=\cfrac { 18500 }{ 8 } \)
\(\cfrac { -77 }{ 72 } z=\cfrac { 38500 }{ 72 } \)

\(\Rightarrow z=\cfrac { -38500 }{ -77 } \)
\(\Rightarrow z=500\)
\((2)\Rightarrow y+\cfrac { 5 }{ 8 } \left( 500 \right) =\cfrac { 18500 }{ 8 } \)
\(y=\cfrac { 18500 }{ 8 } -\cfrac { 2500 }{ 8 } \)

\(\Rightarrow y=2000\)
\((1)\Rightarrow x-3\left( 2000 \right) -500=-5500\)
\(\Rightarrow x-6000-500=-5500\)
\(\Rightarrow x-6000-500=-5500\)
\(\Rightarrow x=-5500+6500\)
\(\Rightarrow x=1000\)
Hence, the prices per unit of three commodities are Rs.1000, Rs. 2000 and Rs. 500 respectively

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 6 \\ 10 \\ b \end{matrix} \right) \)
AX = B

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{matrix}\begin{matrix} 6 \\ 10 \\ b \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & a-1 \end{matrix}\begin{matrix} 6 \\ 4 \\ b-6 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & a-3 \end{matrix}\begin{matrix} 6 \\ 4 \\ b-10 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\)   \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }\)

Case (i) For no solution:
The system possesses no solution only when \(\rho (A)\neq ([A,B])\) which is possible only when a−3 = 0 and b −10 \(\neq \) 0.
Hence for a = 3, b \(\neq \) 10, the system possesses no solution.
Case (ii) For a unique solution:
The system possesses a unique solution only when \(\rho (A)= ([A,B])\)=number of unknowns.
i.e when \(\rho (A)=\rho ([A,B])\) = 3
Which is possible only when a−3 \(\neq \) 0 and b may be any real number as we can observe .
Hence for a \(\neq \) and b \(\in \) R, the system possesses a unique solution.
Case (iii) For an infinite number of solutions:
The system possesses an infinite number of solutions only when
\(\rho (A)=\rho ([A,B])\) i,e when \(\rho (A)=\rho ([A,B])\) = 2 < 3 ( number of unknowns) which is possible only when a − 3 = 0, b−10 = 0
Hence for a = 3, b = 10, the system possesses infinite number of solutions.

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 6 \\ 14 \\ 30 \end{matrix} \right) \)
A X = B

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{matrix}\begin{matrix} 6 \\ 14 \\ 30 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 2 & 4 \end{matrix}\begin{matrix} 6 \\ 8 \\ 16 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 6 \\ 8 \\ 0 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ 2R }_{ 2 }\)
\(\rho (A)=2,\rho ([A,B])=2\)

Obviously the last equivalent matrix is in the echelon form. It has two non-zero rows.
\(\rho (A)=2,\rho ([A,B])=2\)
\(\rho (A)=2,\rho ([A,B])=2\)
The given system is consistent and has infinitely many solutions.
The given system is equivalent to the matrix equation,
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 6 \\ 8 \\ 0 \end{matrix} \right) \)
x + y + z = 6   (1)
y + 2z = 8   (2)
\((2)\Rightarrow \)Y = 8 - 2Z,
\((2)\Rightarrow \) X = 6 - Y - Z = 6 - (8 - 2z) - Z = z - 2
Let us take z = k,k \(\in \) R, we get x = k − 2, y = 8 − 2k, Thus by giving different values for k we get different solutions.
Hence the given system has infinitely many solutions.

Commodity	Price		P = \(\frac {p_{1}}{p_{0}}\) \(\times 100\)	V	[PV
	2012 (p0)	2015 (p0)
Rice	250	280	112	10	1120
Wheat	70	85	121.42	5	607.1
Corn	150	170	113.33	6	679.98
Oil	25	35	140	4	560
Dhal	85	90	105.88	3	317.64
				28	3284.72

Using family budget method,
  C.L.I =  \(\frac {\sum PV}{\sum V}\) = \(\frac {3284.72}{28}\) = 117.31

\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ x \sin x } dx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ udv } \)
\(={ \left( uv \right) }_{ 0 }^{ \frac { \pi }{ 2 } }-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ vdu } \)
\(={ \left[ -x \cos x \right] }_{ 0 }^{ \frac { \pi }{ 2 } }+\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos x } dx\)
\(=0+{ \left[ \sin x \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=1\)

Take u = x Differentiate du = dx

and dv = sin x dx Integrate v = − cos x

 \(\int _{ -1 }^{ 1 }{ { ({ x }^{ 3 }+{ 3x }^{ 2 }) }^{ 3 } } \) (x² +  2x)dx = \({ \left[ \frac { 1 }{ 3 } \frac { ({ { x }^{ 3 }+{ 3x }^{ 2 }) }^{ 4 } }{ 4 } \right] }_{ -1 }^{ 1 }\) 
 \(\left[ \because { \left[ f(x) \right] }^{ n }f'(x)dx=\frac { { [f(x)] }^{ n+1 } }{ n+1 } \right] \)
= \(\frac { 1 }{ 3 } \)(64 - 4)
= 20