\(\Delta =\left| \begin{matrix} 2 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{matrix} \right| =2\)
\(\left| \begin{matrix} 1 & 1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -2 \end{matrix} \right| \)
= 2 (-3+2) - 1 (-3 -1) -1 (-2-1)
= 2(-1) -1 (-4) -1 (-3)
= -2 + 4 + 3 = 5.
Since\(\Delta \neq 0\),
we can apply Cramer's rule and the system is consistent with unique solution.
\(x=\left| \begin{matrix} 3 & 1 & -1 \\ 1 & 1 & 1 \\ 4 & -2 & 3 \end{matrix} \right| =3\left| \begin{matrix} 1 & 1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 4 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 4 & -2 \end{matrix} \right| \)
= 3(-3 + 2) -1(-3 -4) -1(-2 -4)
= 3 (-1) -1 (-7) -1 (-6)
= -3 + 7 + 6 = 10.
\(\Delta y=\left| \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & 4 & -3 \end{matrix} \right| =2\left| \begin{matrix} 1 & 1 \\ 4 & -3 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix} \right| \)
= 2(-3 -4) -3 (-3 -1) -1 (4-1)
= 2 (-7) -3 (-4) -1(3)
= 14 + 12 - 3 = -5
\(\Delta z=\left| \begin{matrix} 2 & 1 & 3 \\ 1 & 1 & 1 \\ 1 & -2 & 4 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 1 & 1 \\ -2 & 4 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix} \right| +3\left| \begin{matrix} 1 & 1 \\ 1 & -2 \end{matrix} \right| \)
= 2(4 + 2) -1(4 -1) + 3(-2 -1)
= 2(6) -1(3) + 3(-3)
= 12 - 3 - 9
= 0

\(z=\cfrac { \Delta z }{ \Delta } =\cfrac { 0 }{ 5 } =0\)
\(\therefore\)Solution set is (2, -1, 0)
 

Given

x	-1	0	1	2
y	202	175	82	55

Since we have to find f(0.5) which is at the beginning of the table, use Newton's forward interpolation formula.
x_n + nh = 0.5 ⇒ -1 + n(1) = 0.5
⇒ n = 0.5 + 1 = 1.5
∴ y_(0.5) = \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)
The difference table is

x	y	Δy	Δ2y	Δ3y
-1	202
0	175
1	82	-93	-66
2	55	-27	66	132

∴ y(0.5) = 202+\(\frac { 1.5 }{ 1! } (-27)+\frac { (1.5)(1.5-1) }{ 2! } (-66)+\frac { (1.5)(1.5-1)(1.5-2) }{ 3! } (132)\)
= 202 - 40.5 + (1.5) (.5) (-33) + \(\frac { (1.5)(.5)(-0.5) }{ 6(132) } \) (132)
= 202 - 40.5 - 24.75 - 8.25
= 202 -73.5
= 128.5
Hence f(0.5) = 128.5

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

Here the intervals are unequal. By Lagrange’s interpolation formula we have
x₀ = 5, x₁ = 6, x₂ = 9, x₃ = 11
y₀ = 12, y₁ = 13, y₂ = 14, y₃ = 16
\(y=f(x)=\frac { \left( x-{ x }_{ 1 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 0 }-{ x }_{ 1 } \right) \left( x_{ 0 }-{ x }_{ 2 } \right) \left( { x }_{ 0 }-{ x }_{ 3 } \right) } \times { y }_{ 0 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 1 }-{ x }_{ 0 } \right) \left( x_{ 1 }-{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 3 } \right) } \times { y }_{ 1 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 2 }-{ x }_{ 0 } \right) \left( x_{ 2 }-{ x }_{ 1 } \right) \left( { x }_{ 2 }-{ x }_{ 3 } \right) } \times { y }_{ 2 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 3 }-{ x }_{ 0 } \right) \left( x_{ 3 }-{ x }_{ 1 } \right) \left( { x }_{ 3 }-{ x }_{ 2 } \right) } \times { y }_{ 3 }\)
\(=\frac { (x-6)(x-9)(x-11) }{ (5-6)(5-6)(5-11) } (12)+\frac { (x-5)(x-9)(x-11) }{ (6-5)(6-9)(6-9) } (13)+\frac { (x-5)(x-6)(x-11) }{ (9-5)(9-6)(9-11) } (14)+\frac { (x-5)(x-6)(x-9) }{ (11-5)(11-6)(11-9) } (16)\)
Put x = 10
\({ y }_{ (10) }=f\left( 10 \right) =\frac { 4(1)(-1) }{ (-1)(-4)(-6) } (12)+\frac { (5)(1)(-1) }{ (1)(-3)(-5) } (13)+\frac { 5(4)(-1) }{ 4(3)(-2) } (14)+\frac { (5)(4)(1) }{ 6(5)(2) } (16)\)}
= \(\frac { 1 }{ 6 } \left( 12 \right) -\frac { 13 }{ 3 } +\frac { 5\left( 14 \right) }{ 3\times 2 } +\frac { 4\times 16 }{ 12 } \)
= 14.6663

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.

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: (Assignment)
Examine the rows with exactly one zero. Mark the zero by ׈ and draw a vertical line.
After examining the rows, examine the columns with exactly one zero. Mark the zero by ׈ and draw a horizontal line.

Only 4 assignments are made
Number not lying on the line are

40	4	26
5	23	5
6	1	6
7	5	17

and minimum of these numbers are 1.
This number 1 should be subtracted from the above number and 1 should be added to the numbers which are on the intersecting lines (ie. 52,43,59,30,21)and the other numbers remains the same.
A new cost matrix is as follows and repeat step 3.
New cost matrix is

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

Cities From	Cities To	Cost
1	11	15
2	8	19
3	7	17
4	9	38
5	10	16
6	12	20
Total Cost		Rs. 125

\(\Delta \)\(\left[ \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } \right] \)
By Partial fraction method
\(\frac { 5x+12 }{ { x }^{ 2 }+5x+6 } =\frac { A }{ x+3 } +\frac { B }{ x+2 } \)
\(A=\frac { 5x+12 }{ x+12 } [x=-3]=\frac { -15+12 }{ -1 } =\frac { -3 }{ -1 } =-3\)
\(B=\frac { 5x+12 }{ x+3 } \)[x = -2] \(=\frac { 2 }{ 1 } =2\)
\(\frac { 5x+12 }{ { x }^{ 2 }+5x+6 } = \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(\Delta \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } =\Delta \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(=\left[ \frac { 3 }{ x+1+3 } -\frac { 3 }{ x+3 } \right] +\left\{ \frac { 2 }{ x+1+2 } -\frac { 2 }{ x+2 } \right\} \)
\(=3\left[ \frac { 1 }{ x+4 } -\frac { 1 }{ x+3 } \right] +2\left[ \frac { 1 }{ x+3 } -\frac { 1 }{ x+2 } \right] \)
\(=\left[ \frac { -3 }{ (x+4)(x+3) } -\frac { 2 }{ (x+3)(2+3) } \right] \)
\(=\frac { -5x-14 }{ (x+2)(x+3)(x+4) } \)

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

Given Q_d = 30 - 5p + 2\(\frac { dp }{ dt } +\frac { { d }^{ 2 }p }{ dt^{ 1 } } \) and
Q_s = -6 + 3p
At equilibrium Q_d = Q_s
⇒ 30 - 5p+2\(\frac { dp }{ dt } +\frac { { d }^{ 2 }p }{ dt^{ 1 } } \) = -6+3p
⇒ \(\frac { { d }^{ 2 }p }{ dt^{ 2 } } +2\frac { dp }{ dt } \)- 5p + 30 - 6 - p = 0
⇒ \(\frac { { d }^{ 2 }p }{ dt^{ 2 } } +2\frac { dp }{ dt } \)- 8p = 24
The auxiliary equation is m² + 2m - 8 = 0
⇒ (m + 4) (m - 2) = 0
⇒ m = - 4, 2
∴  C.F. is Ae^-4t + Be^2t
Particular Integral PI = \(\frac { -24 }{ (D+4)(D-x) } \)e0x
= \(\frac { -24 }{ (0+4)(0-2) } =\frac { -24 }{ -8 } \)=3
∴ y = CF + PI
∴ The general solution is
P = Ae^-4t + Be^2t + 3.

Here the number of rows and columns are equal.
\(\therefore\)The given assignment problem is balanced.
Now let us find the solution.
Step 1: Select a smallest element in each row and subtract this from all the elements in its row.

Look for atleast one zero in each row and each column.Otherwise go to step 2.
Step 2 : Select the smallest 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 (Assignment):
Examine the rows with exactly one zero. First three rows contain more than one zero. Go to row D.
There is exactly one zero. Mark that zero by \(\square\) (i.e) job D is assigned to machine
I . Mark other zeros in its column by\(\text { X }\).

Step 4: Now examine the columns with exactly one zero. Already there is an assignment in column I. Go to the column II. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by\(\text { X }\).

Column III contains more than one zero. Therefore proceed to Column IV, there is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by \(\text { X }\).

Step 5: Again examine the rows. Row B contains exactly one zero. Mark that zero by \(\square\).

Thus all the four assignments have been made. The optimal assignment schedule and total cost is
\(\begin{array}{|c|c|c|} \hline \text { Job } & \text { Machine } & \text { cost } \\ \hline \text { A } & \text { II } & 12 \\ \hline \text { B } & \text { III } & 7 \\ \hline \text { C } & \text { IV } & 11 \\ \hline \text { D } & \text { I } & 8 \\ \hline {\text { Total cost }} \ && 38 \\ \hline \end{array}\)
The optimal assignment (minimum) cost
= Rs. 38

Sample No.	Observation					Total	x	R = Xmax - Xmin
	1	2	3	4	5
1	50	55	52	49	54	260	52	55-49 = 6
2	51	50	53	50	46	250	50	53-46 = 7
3	50	53	48	52	47	250	50	53-47 = 6
4	48	53	50	51	53	255	51	53-48 = 5
5	46	50	44	48	47	235	47	50-44 = 6
6	55	51	56	47	51	260	52	56-47=9
7	45	48	53	48	51	245	49	53-45=8
8	50	56	54	53	57	270	54	57-50 = 7
9	47	53	49	52	54	255	51	54-47 = 7
10	56	53	55	54	52	270	54	56-52 = 4
							510	65

\(\overline {\overline{X}}\) = \(\frac {510}{10}\) = 51
 \(\overline {R}\)  = \(\frac {65}{10}\) = 6.5
The control limits of mean chart are
UCL =  \(\overline {\overline{X}} + A_{2} \overline{R}\) 
= 51 + 0.577 (6.5)
= 51 + 3.75 = 57.75
CL =  \(\overline {\overline{X}}\) = 51
LCL    =  \(\overline {\overline{X}} - A_{2} \overline{R}\)
= 51 - 3.75 = 47.25
The control limits of range chart are
UCL = D₄\(\overline {R}\) = 2.114 (6.3) = 13.32
CL =  \(\overline {R}\) = 6.5 
[For n=5, A₂  = 0.577, D₃  = 0, D₄ = 2.114]
LCL = D₃\(\overline {R}\)  = 0

The given differential equation is of this form
\(\frac { dy }{ dx } \)+Py=Q where
P = tan x; Q = cos³x
∴ \(\int { p } dx=\int { \tan x } dx\) = log secx
∴ Integrating factor (I.F) = \(e^{ \int { p } dx }\) = e^{logx secx}
= sec x
Hence, the solution is
\(ye^{ \int { p } dx }=\int { Q } .e^{ \int { p } dx }dx+c\)
⇒ y sec x =\(\\ \int { cos^{ 3 }x } \)dx+c
[∴ cos 2x = 2cos²x-1]
⇒ 2 cos²x = 1 + cos2x
⇒ cos²x = \(\frac { 1+cos2x }{ 2 } \)]
⇒ y secx = \(\int { \cos^{ 3 }x.\frac { 1 }{ \cos x } } \)dx+c
⇒y sec x =\(\int { cos^{ 2 }x } dx+c\)
⇒ y sec x =\(\int { \left( \frac { 1+cos2x }{ 2 } \right) } dx+c\)
⇒ y sec x = \(\frac { 1 }{ 2 } \left[ x+\frac { \sin 2x }{ 2 } \right] \)+c

Time	Weight in ml					Total	\(\overline {X}\)	R = x max - xmin
	1	2	3	4	5
8.00 AM	43	41	42	43	41	210	\(\frac {210}{5} = 42\)	43-41 = 2
9:00 AM	40	39	40	39	44	202	\(\frac {202}{5} = 40.4\)	43-41 = 2
10.00 AM	42	42	43	38	40	205	\(\frac {205}{5} = 41\)	43-38 = 5
11:00 AM	39	43	40	39	42	203	\(\frac {203}{5} = 40.6\)	43-39 = 4
							164	16

\(\overline {\overline{X}}\)  =  \(\frac {164}{4}\) = 41
\(\overline {R}\)  = \(\frac {16}{4}\) = 4
Control limits for \(\overline {X}\) - chart
UCL = \(\overline {\overline{X}} + A_{2} \overline {R}\)
= 41 + 0.577 (4)
= 41 + 2.308 = 43.31
CL =  \(\overline {\overline{X}} \) = 41
[when n = 5, A₂= 0.577, D₃ = 0, D₄ = 2.114]
LCL = \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 41 - 2.308 = 38.69
The control limits of R-chart are
UCL =  D₄ \(\overline{R}\) = 2.114 (4) = 8.44
CL =  \(\overline{R}\) = 4
LCL = D3 \(\overline{R}=0\)

\(\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text { Sample number } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \text { Total } \\ \hline \bar{X} & 49 & 45 & 48 & 53 & 39 & 47 & 46 & 39 & 51 & 45 & 462 \\ \hline \mathbf{R} & 7 & 5 & 7 & 9 & 5 & 8 & 8 & 6 & 7 & 6 & 68 \\ \hline \end{array}\)
\(\overline{\bar{X}}=\frac{\sum \bar{X}}{10}=\frac{462}{10}=46.2 \)
\(\bar{R}=\frac{\sum R}{10}=\frac{68}{10}=6.8 \)
\(\text { The control limits for } \overline{\mathrm{X}} \text { chart is }\)
\(\mathrm{UCL}=\overline{\overline{\mathrm{X}}}+\mathrm{A}_{2} \overline{\mathrm{R}}=46.2+(0.58)(6.8)=50.14\)
\(\mathrm{CL}=46.2\)
\(\mathrm{LCL}=\overline{\mathrm{X}}-\mathrm{A}_{2} \overline{\mathrm{R}}=46.2-(0.58)(6.8)=42.26 \) 

The control limits for range chart is
\(\mathrm{UCL}=\mathrm{D}_{4} \overline{\mathrm{R}}=(2.115)(6.8)=14.38 \)
\(\mathrm{CL}=\overline{\mathrm{R}}=6.8 \)
\(\mathrm{LCL}=\mathrm{D}_{3} \overline{\mathrm{R}}=0(6.8)=0\)

From the \(\overline X\) chart, we see that 4 points are outside the control limit lines. So we say that the process is out of control.
Conclusion: The above diagram shows all the three control lines with the data points plotted, since 2 points fall out of the control limits, we can say that the process is out of control.

Commodity	2016		2017
	p0	q0	p1	q1
Food	40	12	65	14
Fuel	72	14	78	20
Clothing	36	10	36	15
Wheat	20	6	42	4
Others	46	8	52	6

p0q0	p1q1	p1q0	p0q1
480	910	780	560
1008	1560	1092	1440
120	168	252	80
368	312	416	276
2336	3490	2900	2896

Fisher's price index number 
\(P^{F}_{01}\) = \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}}{{\sum p_{0}q_{0}\times \sum p_{0}q_{1}}}\) × 100
= \(\sqrt \frac {2900\times3490}{2336\times2896} \times 100\)
= \(\sqrt \frac{10121000}{6765056} \times{100}\) 
= \(\sqrt{1.496}\times{100}\)
\(P^{F}_{01}\) = 122.31
Time Reversal Test
P₀₁ × P₁₀  =  \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}\times{\sum p_{0}q_{1}\times \sum p_{0}q_{0}}}{{\sum p_{0}q_{0}\times \sum p_{0}q_{1}\times \sum p_{1}q_{1}\times \sum p_{1}q_{0}}}\)
=  \(\sqrt \frac {2900\times3490\times{2896\times2336}}{2336\times2896\times3490\times2900}\)
= \(\sqrt {1}\) = 1
Hence, it satisfies time reversal test.

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

\(\frac { dy }{ dx } =\frac { x+3y }{ x-y } \)
Since the numerator and denominator are homogeneous functions of degree 1,
put y = vx and \(\frac { dy }{ dx } \) = v + x\(\frac { dv}{ dx } \)

⇒ x\(\frac { dv }{ dx } =\frac { 1+3v }{ 1-v } -v=\frac { 1+3v-v(1-v) }{ 1-v } \)
= \(\frac { 1+3v-v+v^{ 2 } }{ 1-v } \)
⇒ \(x\frac { dv }{ dx } =\frac { 1+2v+v^{ 2 } }{ 1-v } \)
separating the variables we get,
\(\frac { 1-v }{ 1+2v+{ v }^{ 2 } } dv=\frac { dx }{ x } \)
⇒ \(\int { \frac { (1-v)dv }{ { v }^{ 2 }+2v+1 } } =\int { \frac { dx }{ x } } \)
⇒ \(\int { \frac { (1-v)dv }{ (v+1)^{ 2 } } } \) = log x + log c
\(\frac { 1-v }{ (v+1)^{ 2 } } =\frac { A }{ v+1 } +\frac { B }{ (v+1)^{ 2 } } \)
1-v = A(v+1)+B
put v = -1,
2 = B
1 = A + B ⇒ 1 = A+2
⇒ A = -1
∴ \(\frac { 1-v }{ (v+1)^{ 2 } } =\frac { -1 }{ v+1 } +\frac { 2 }{ (v+1) } \)
⇒ \(\int { \frac { -1 }{ v+1 } } dv+\int { \frac { 2 }{ (v+1)^{ 2 } } } dv\) = log xc
⇒ -log (v+1) - \(\frac { 2 }{ v+1 } \) = log xc
⇒ \(\frac { -2 }{ v+1 } \) = log xc + log(v+1)
⇒ \(\frac { -2 }{ v+1 } \) = log xc(v+1)
Replace v by \(\frac{y}{x}\) we get,
\(\frac { -2 }{ \frac { y }{ x } +1 } =logxc\left( \frac { y }{ x } +1 \right) \)

⇒ \(\frac { -2x }{ x+y } \) = logc (x+y)
⇒ e^-2x/x+y  = c(x+y)
⇒ x+y = \(\frac { 1 }{ c } \) e^-2x/x+y
⇒ x+y = ke^-2x+x+y where k = \(\frac { 1 }{ c } \).

Year X	Production in tonnes (Y)	X = x-1998	X2	XY
1995	155	-3	9	-465
1996	162	-2	4	-324
1997	171	-1	1	-171
1998	182	0	0	0
1999	158	1	1	158
2000	180	2	4	360
20001	178	3	9	534
	1186	0	28	92

 Since \(\sum\)X = 0, a = \(\frac {\sum Y}{n}\) = \(\frac {1186}{7}\) = 169.428.
   b = \(\frac {\sum XY}{\sum X^2}\) \(\frac {92}{28}\) = 3.285
∴ The required equation of the straight line trend is given by Y = a + bX
 \(\Rightarrow \) Y= 69.428 + 3.285 (x - 1998)

\(\frac { dy }{ dx } =\frac { x-y }{ x+y } \)  ..... (1)
This is a homogeneous differential equation.
Now put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ (1) ⇒ \(v+x\frac { dv }{ dx } =\frac { x-vx }{ x+vx } \)
\(=\frac { 1-v }{ 1+v } \)
\(x\frac { dv }{ dx } =\frac { 1-v }{ 1+v } -v\)
\(=\frac { 1-2v-{ v }^{ 2 } }{ 1+v } \)
\(\frac { 1+v }{ { v }^{ 2 }+2v-1 } dv=\frac { -dx }{ x } \)
Multiply 2 on both sides
\(\frac { 2+v }{ { v }^{ 2 }+2v-1 } dv=-2\frac { -dx }{ x } \)
On Integration
ഽ\(\frac { 2+v }{ { v }^{ 2 }+2v-1 } \)dv = -2ഽ\(\frac { -dx }{ x } \)
log(v²+2v − 1) = −2log x + log c
v²+2v − 1 = \(\frac { c }{ { x }^{ 2 } } \)
x²(v²+2v−1) = c
Now, Replace \(v=\frac { y }{ x } \)
\({ x }^{ 2 }\left[ \frac { { { y }^{ 2 } } }{ { x }^{ 2 } } +\frac { 2y }{ x } -1 \right] =c\)
y² + 2xy − x² = c is the solution

Here the base year quantities are given, therefore we can apply Aggregate Expenditure Method.

Commodities	Price		Quantity (q0)	p0q0	p1q0
	Base year (P0)	Current year (P1)
Rice	32	48	25	800	1200
Sugar	25	42	10	250	420
Oil	54	85	6	324	510
Coffe	250	460	1	250	460
Tea	175	275	2	350	550
			Total	1974	3140

Cost of Living Index Number=\(\frac { \sum { { p }_{ 1 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \times 100=\frac { 3140 }{ 1974 } \times 100 = 159.0679\)
Hence, the Cost of Living Index Number for a particular class of people for the year 2016 is increased by 59.0679 % as compared to the year 2011.

Commodities	Base Year		Current Year		p0q0	p0q1	p1q0	p1q1
	Price (p0)	Quantity (q1)	Price ((p0))	Quantity (q1)
Rice	10	5	11	6	50	60	55	66
Wheat	12	6	13	4	72	48	78	52
Rent	14	8	15	7	112	98	120	105
Fuel	16	9	17	8	144	128	153	136
Transport	18	7	19	5	126	90	133	95
Miscellaneous	20	4	21	3	80	60	84	63
				Total	584	484	623	517

Fisher’s price index number
\({ P }_{ 01 }^{ F }=\left( \sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 10 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \right) \times 100=\left( \sqrt { \frac { 623\times 517 }{ 584\times 484 } } \right) \times 100=106.74\)
Time Reversal Test: P₀₁ × P₁₀ = 1
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { 623\times 517\times 484\times 584 }{ 584\times 487\times 517\times 623 } \right) } \)
P₀₁ x P₁₀ = 1
Factor Reversal Test
\({ P }_{ 01 }\times { Q }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { 623\times 517\times 484\times 517 }{ 584\times 484\times 584\times 623 } \right) } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { 517\times 517 }{ 585\times 584 } \right) } =\frac { 517 }{ 584 } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)

Given 3e^x tan y dx + (1 + e^x)sec²y dy = 0
3e^x tan y dx = −(1 + e^x)sec² y dy
\(\frac { { 3e }^{ x } }{ 1+{ e }^{ x } } dx=-\frac { { sec }^{ 2 }y }{ tany } dy\)
Integrating, we get 3ഽ\(\frac { { 3e }^{ x } }{ 1+{ e }^{ x } } dx\) = -ഽ\(\frac { { sec }^{ 2 }y }{ tany } dy\) + c
3log(1+ e^x ) = −log tan y + log c   \(\left[ \therefore \int { \frac { f'(x) }{ f(x) } dx=logf(x) } \right] \)
log(1+e^x)³ + log tan y = log c
log [(1 + e^x)³ tan y ] = log c
(1+e^x)³ tan y = c (1)
Given y(0) = \(\frac { \pi }{ 4 } \) (i.e) y = \(\frac { \pi }{ 4 } \) at x =0
(1) ⇒ (1 +e⁰ )³ tan \(\frac { \pi }{ 4 } \) = c
2³ (1) = c
⇒ c = 8
Hence the required solution is (1  +e^x)³ tan y = 8

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 = 50,
Sample mean \(\bar { X } \) = 1850
Population mean μ = 1800
Population standard deviation σ = 100
Null Hypotheses H₀:
μ = 1800 (i.e., To claim that the breaking strength of the cables have increased)
Alternative Hypotheses H₁:
μ≠1800(To claim that the breaking strength of the cables have not increased)
The level of significance ∝ = 1% =.001
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 1850-1800 }{ \frac { 100 }{ \sqrt { 50 } } } =\frac { 50 }{ \frac { 100 }{ 7.07 } } =\frac { 50 }{ 14.144 } =3.535\)
\(\Rightarrow \therefore Z=3.535\)
The Significant value \({ Z }_{ \frac { \alpha }{ 2 } }=2.58\)
Here \(Z<{ Z }_{ \frac { \alpha }{ 2 } }i.e.,3.535<2.58\)
Inference: Since \(Z<{ Z }_{ \frac { \alpha }{ 2 } }\) at 1% level of significance, the null hypothesis Ho is rejected.
Hence, we conclude that μ≠ 1800 and we cannot support the claim that the breaking strength of the cables have increased.

\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty \\ h\rightarrow 0 }{ lim } \sum _{ r=1 }^{ n }{ h.f\left( a+rh \right) } \)
Here a = 0, b = 1
\(h=\cfrac { b-a }{ n } =\cfrac { 1-0 }{ n } =\cfrac { 1 }{ n } \)
and f(x) = x²
Now^, \(f(a+rh)=f\left( 0+r.\cfrac { 1 }{ n } \right) =f\left( \cfrac { r }{ n } \right) \)
⁼ \(\left( \cfrac { r }{ n } \right) ^{ 2 }=\cfrac { { r }^{ 2 } }{ { n }^{ 2 } } \)
\(\therefore \int _{ 0 }^{ 1 }{ { x }^{ 2 }dx } =\underset { n\rightarrow \infty }{ lim } \sum _{ r=1 }^{ n }{ \frac { 1 }{ n } } \left( \frac { { r }^{ 2 } }{ { n }^{ 2 } } \right) \)
= \(\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 3 } } .\frac { n(n+1)(2n+1) }{ 6 } \)
\(\left[ \therefore \sum _{ r=1 }^{ n }{ { r }^{ 2 }=\cfrac { n(n+1)(2n+1) }{ 6 } } \right] \)

[Taking n common from each bracket]
= \(\cfrac { 1 }{ 6 } .\underset { n\rightarrow \infty }{ lim } \left( 1+\cfrac { 1 }{ n } \right) \left( 2+\cfrac { 1 }{ n } \right) \)
= \(\cfrac { 1 }{ 6 } \left( 1+0 \right) \left( 2+0 \right) \)


= \(\cfrac { 1 }{ 3 } \)

\(\int _{ a }^{ b }{ f(x) } =\lim _{ n\rightarrow \infty \\ h\rightarrow 0 }{ \sum _{ r=1 }^{ n }{ h } f(a+rh) } \)
Here a = 1, b = 2, \(h=\frac { b-a }{ n } =\frac { 2-1 }{ n } =\frac { 1 }{ n } \) and f(x) = 2x + 1
f (a + rh) = f\(\left( 1+\frac { r }{ n } \right) \)
= \(2\left( 1+\frac { r }{ n } \right) +1\)
= \(2+\frac { 2r }{ n } +1\)
f (a + rh) = 3 + \(\frac {2r }{ n } \)
\(\int _{ 1 }^{ 2 }{ (x) } dx=\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { 1 }{ n } \left( 3+\frac { 2r }{ n } \right) } } \)
\(=\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { }{ } \left( \frac{3}{n}+\frac { 2r }{ n } \right) } } \)
=\(\overset { lt }{ n\rightarrow \infty } \left[ \frac { 3 }{ n } \sum _{ r=1 }^{ n }{ 1 } +\frac { 2 }{ { n }^{ 2 } } \sum _{ r=1 }^{ n }{ r } \right] \)
= \(\lim _{ n\leftarrow \infty }{ \left[ \frac { 3 }{ n } n+\frac { 2 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } \right] } \)
= \(3+\lim _{ n\rightarrow \infty }{ \left( 1+\frac { 1 }{ n } \right) } \)
\(\int _{ 1 }^{ 2 }{ f(x) } dx\) = 3 +1 = 4

Given, sample size = 100, \(\bar x\) = 7.4, since σ is unknown but s = 1.2 is known.
In this problem, we consider \(\breve { \sigma } =s\quad { Z }_{ \frac { \alpha }{ 2 } }=1.96\)
\(S.E=\frac { \breve { \sigma } }{ \sqrt { n } } =\frac { s }{ \sqrt { n } } =\frac { 1.2 }{ \sqrt { 100 } } =0.12\)
Hence 95% confidence limits for the population mean are
\(\bar { x } -{ Z }_{ \frac { \alpha }{ 2 } }\frac { \sigma }{ \sqrt { n } } <\mu <\bar { x } +{ Z }_{ \frac { \alpha }{ 2 } }\frac { \sigma }{ \sqrt { n } } \)
\(7.4-(1.96\times 0.12)\le \mu \le 7.4+(1.96\times 0.12)\)
\(7.4-0.2352\le \mu \le 7.4+0.2352\)
\(7.165\le \mu \le 7.635\)
This implies that the probability that the true value of the population mean breaking strength of the cotton threads will fall in this interval (7.165,7.635) at 95%.

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin x }{ \sin x+ \cos x } } \)dx  .......(1)
I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin\left( \frac { \pi }{ 2 } -x \right) }{ \sin\left( \frac { \pi }{ 2 } -x \right) +\cos\left( \frac { \pi }{ 2 } -x \right) } } \) dx  \(\left[∵ \int _{ 0 }^{ a }{ f(x) } dx=\int _{ 0 }^{ a }{ f(a-x)dx } \right] \)
\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \cos x }{ \cos x+\sin x } } \)dx ....(2)
(1) + (2) ⇒
\(2I=\int _{ \frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \left[ \frac { \sin x }{ \sin x+\cos x } +\frac { \cos x }{ \cos x+ \sin x } \right] } \)dx
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ dx } ={ [x] }_{ 0 }^{ \frac { \pi }{ 2 } }=\frac { \pi }{ 2 } \)
\(\therefore I=\frac { \pi }{ 4 } \)

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

Let p be the probability of success and q be the probability of failure.
Given p = 2q
p = 2(1-p)    [∵ p +  q = 1]
p = 2-2p ⇒ 3p = 2
p = \(\frac{2}{3}\)
∴ q = 1-p = 1-\(\frac{2}{3}\) = \(\frac{1}{3}\) and n = 5
(i) P (3 successes) = P(X = 3)
= ⁵C₃ \(\left( \frac { 2 }{ 3 } \right) ^{ 3 }\left( \frac { 1 }{ 3 } \right) ^{ 2 }\)      ∵ p(x) = ⁿC_x p^xq^n-x, n = 5, x = 3

=\(\frac { 10\times 8 }{ 243 } =\frac { 80 }{ 243 } \)
(ii) P (atleast 3 successes)
= P(X ≥ 3)
= 1-P(X < 3)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= 1-[⁵C₀\(\left( \frac { 2 }{ 3 } \right) ^{ 0 }\left( \frac { 1 }{ 3 } \right) ^{ 5 }\) + ⁵C₁\(\left( \frac { 2 }{ 3 } \right) ^{ 1 }\left( \frac { 1 }{ 3 } \right) ^{ 4 }\) + ⁵C₂\(\left( \frac { 2 }{ 3 } \right) ^{ 2 }\left( \frac { 1 }{ 3 } \right) ^{ 3 }\)]
= 1-\(\left[ \left( \frac { 1 }{ 3 } \right) ^{ 5 }+5\left( \frac { 2 }{ 3 } \right) \left( \frac { 1 }{ { 3 }^{ 4 } } \right) +10\left( \frac { { 2 }^{ 2 } }{ { 3 }^{ 2 } } \right) \left( \frac { 1 }{ { 3 }^{ 3 } } \right) \right] \)
= 1-\(\frac { 1 }{ { 3 }^{ 5 } } \)(1+10+40) = \(1-\frac { 1 }{ 243 } \)(51)
= \(\frac { 243-51 }{ 243 } =\frac { 192 }{ 243 } \).

Let X be the normal variate of life of light bulbs with mean 125 and standard deviation 18.
(i) less than 95 days
When X = 95
\(Z=\frac{X-\mu}{\sigma}=\frac{95-125}{18}=-1.667\)
\(P(X<95)=P(Z<-1.667)\)
= 0.5 – P(0 < Z < 1.67)
= 0.5 – 0.4525
= 0.0475
No. of bulbs expected to expire in less than 95 days out of 900 bulbs 900 × .0475 = 43 bulbs

\(\int _{ 2 }^{ 3 }{ \frac { { x }^{ 4 }+1 }{ { x }^{ 2 } } } dx=\int _{ 2 }^{ 3 }{ ({ x }^{ 2 }+{ x }^{ -2 }) } dx\)
= \({ \left[ \frac { { x }^{ 3 } }{ 3 } -\frac { 1 }{ x } \right] }_{ 2 }^{ 3 }\)
= \(\left( 9-\frac { 1 }{ 3 } \right) -\left( \frac { 8 }{ 3 } -\frac { 1 }{ 2 } \right) =\frac { 13 }{ 2 } \)

(i) greater than 1.09
The total area under the curve is equal to 1 , so that the total area to the right Z = 0 is 0.5 (since the curve is symmetrical). The area between Z = 0 and 1.09 (from tables) is 0.3621
P(Z > 1.09) = 0.5000 - 0.3621 = 0.1379
The shaded area to the right of Z = 1.09 is the probability that Z will be greater than 1.09

(ii) less than –1.65
The area between -1.65 and 0 is the same as area between 0 and 1.65. In the table the area between zero and 1.65 is 0.4505 (from the table). Since the area to the left of zero is 0.5 , P(Z< 1.65) = 0.5000 – 0.4505 = 0.0495.

(iii) lying between -1.00 and 1.96
The probability that the random variable Z in between -1.00 and 1.96 is found by adding the corresponding areas :
Area between -1.00 and 1.96 = area between (-1.00 and 0) + area betwn (0 and 1.96)
P(–1.00 < Z < 1.96) = P(–1.00 < Z < 0) + P(0 < Z < 1.96)
= 0.3413 + 0.4750 (by tables)
= 0.8163

(iv) lying between 1.25 and 2.75
Area between Z = 1.25 and 2.75 = area betwn (z = 0 and z = 2.75)
– area betwn (z=0 and z = 1.25)
P(1.25 < Z < 2.75) = P(0 < Z < 2.75) – P(0 < Z < 1.25)
= 0.4970 - 0.3944 = 0.1026

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) =\(\frac{2}{25}\)

ഽ\(\left[ \frac { 1 }{ \log x } -\frac { 1 }{ { \left( \log x \right) }^{ 2 } } \right] dx\) = ഽ\(\left[ \frac { 1 }{ z } -\frac { 1 }{ { z }^{ 2 } } \right] { e }^{ z }dx\)
= ഽ e^x [ f(z) +f'(z)] dx
= e^z f(z) + c
= e^z \(\left[ \frac { 1 }{ z } \right] \) + c
= \(\frac { x }{ \log\ x } \) + c

Take z = log x ஃdz = \(\frac { 1 }{ x } \)dx ⇒ dx ex dz [∵ x = ex ] and f(z) = \(\frac { 1 }{ z } \)  ஃ f'|(z) = \(-\frac { 1 }{ { z }^{ 2 } } \)

Given demand function P_d = 25 - 3x and
Supply function P_s= 5 + 2x
At market equilibrium, P_d = P_s
⇒ 25-3x = 5+2x
⇒ 25-5 = 2x+3x
⇒ 20 = 5x
⇒ x = \(\frac{20}{5}\)
⇒ x₀ = 4
When x₀ = 4, p₀ = 25-3(4)
= 25-12 = 13
p₀ = 13
∴p₀x₀ = 13(4) = 52
∴ Consumer's surplus
\(CS=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(\\ =\int _{ 0 }^{ 4 }{ (25-3x)dx-52 } \)
\(={ \left[ 25x-\frac { { 3x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 4 }-52\)
\(=25(4)-\frac { 3\left( { 4 }^{ 2 } \right) }{ 2 } -52\)
=100-24-52
=100-76
C.S = 24 units
Producer's Surplus
\((PS)={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x }{ g(x)dx } \)
\(=52-\int _{ 0 }^{ 4 }{ (5+2x)dx } \)

= 52 - (5(4) + 4²}
= 52 - (20 + 16)
= 52 - 36
PS = 16 units

Given probability distribution function
\(F(x)\begin{cases} 0,\quad ​​​​\text{for}\quad x<0 \\ \frac { 1 }{ 2 } ,\quad ​​​​\text{for}\quad 0\le x<1 \\ 0,\quad ​​​​\text{for}\quad 1\le x<2\quad \\ \frac { 1 }{ 4 } ,\quad ​​​​\text{for}\quad 2\le x<4 \\ 0,\quad ​​​​\text{for}\quad x\ge 4 \end{cases}\)
a) The distribution function F(x) is continuous since it is a step function
We know f'(x) = f(x)
∴ Probability density function
\(F(x)\begin{cases} 0,\quad \text{for}\quad x\le 0 \\ \frac { 1 }{ 2 } ,\quad \text{for}\quad 0\le x<1 \\ 0,\quad \text{for}\quad 1\le x<2\quad \\ \frac { 1 }{ 4 } ,\quad \text{for}\quad 2\le x<4 \\ 0,\quad \text{for}\quad x\ge 4 \end{cases}...(1)\)
(b) (i) Probability that a person will have to wait more than 3 minutes is P(X > 3)
∴ P(X>3) = P(3≤X<4)+P(X≥4)
\(\frac{1}{4}+0=\frac{1}{4}\) [from (1)]
ii) Probability that a person will have to wait less than 3 minutes is P(X < 3)
∴ P(X<3) = P(X≤0)+P(0≤x≤1)+P(1≤x≤2)+P(2≤x≤3)
0 + \(\frac{1}{2}+0+\frac{1}{4}\)[from (1)]
∴ P(X<3) = \(\frac{3}{4}\)
= P(1≤X<2)+P(2≤X<3)
= 0 +\(\frac{1}{4}=\frac{1}{4}\) [from (1)]

Given marginal revenue function.
\(MR=\frac { DR }{ dx } =20{ e }^{ \frac { -x }{ 10 } }\left( 1-\frac { x }{ 10 } \right) \)
\(dR={ 20e }^{ \frac { -x }{ 10 } }\left( 1-\frac { x }{ 10 } \right) dx\)
\(R=\int { { 20e }^{ \frac { -x }{ 10 } } } \left( 1-\frac { x }{ 10 } \right) dx\)
We know that \(\int { { e }^{ ax }[af(x)+f'(x)]dx={ e }^{ ax }f(x) } +c\)
Here \(\\ a=\frac { -1 }{ 10 } ,f(x)=x,\quad f'(x)=1\)
\(R=20\int { { e }^{ \frac { -x }{ 10 } } } \left[ -\frac { 1 }{ 10 } x+1 \right] dx=20{ e }^{ \frac { -x }{ 10 } }x+k\)
\(\Rightarrow R=20{ e }^{ \frac { -x }{ 10 } }x+k\quad ...(1)\)
When x = 0, R = 0
0 = 0 + k ⇒ k = 0
(1) becomes 
\(R=20x{ e }^{ \frac { -x }{ 10 } }\)
Demand function P
\(=\frac { R }{ x } =\frac { 20x{ e }^{ \frac { -x }{ 10 } } }{ x } \)
\(\Rightarrow P=20{ e }^{ \frac { -x }{ 10 } }\)

\(\frac { EY }{ Ex } =\frac { -7x }{ (1-2x)(2+3x) } \)
\(\frac { 7 }{ (2x-1)(3x+2) } =\frac { A }{ 2x-1 } +\frac { B }{ 3x+2 } \)
7 = A(3x+2)+B(2x-1)
\(Put\ x=\frac { -2 }{ 3 } 7=B\left( \frac { -4 }{ 3 } -1 \right) 7=B\left( \frac { -7 }{ 3 } \right) \)
\(Put\ x=\frac { 1 }{ 2 } 7=A\left( \frac { 3 }{ 2 } +2 \right) \Rightarrow 7=A\left( \frac { 7 }{ 2 } \right) \)
\(\therefore \frac { 7 }{ (2x-1)(3x+2) } =\frac { 2 }{ 2x-1 } -\frac { 3 }{ 3x+2 } \)
Also, it is given that x = 2, when y \(=\frac{3}{8}\)
\(\Rightarrow \frac { x }{ y } \frac { dy }{ dx } =\frac { -7x }{ (1-2x)(2+3x) } \)

\(=\frac { -7dx }{ (1-2x)(2+3x) } \)
\(\Rightarrow \frac { dy }{ y } =\frac { 7dx }{ (2x+1)(3x+2) } \)
\(\int { \frac { dy }{ y } =\int { \frac { 7dx }{ (2x-)(3x+2) } } } \)
\(\int { \frac { dy }{ y } =\int { \left( \frac { 2 }{ 2x-1 } -\frac { 3 }{ 3x+2 } \right) dx } } \)
\(log\quad y=2\int { \frac { 1 }{ 2x-1 } dx-3\int { \frac { dx }{ 3x+2 } } } \)
\(=2\frac { log|2x-1| }{ 2 } -3\frac { log|3x+2| }{ 3 } +logc\)
\(=log\quad y-log\quad c=log\left| \frac { 2x-1 }{ 3x+2 } \right| \)
\(\Rightarrow log\left| \left( \frac { y }{ c } \right) \right| =log\left| \frac { 2x-1 }{ 3x+2 } \right| \)
\(\Rightarrow \frac { y }{ c } =\frac { 2x-1 }{ 3x+2 } y=c\left( \frac { 2x-1 }{ 3x+2 } \right) \) ....(1)
When \(x=2,y=\frac { 3 }{ 8 } \)
\(\Rightarrow \frac { 3 }{ 8 } =c\left( \frac { 4-1 }{ 8 } \right) \)
\(\frac { 3 }{ 8 } =c\left( \frac { 3 }{ 8 } \right) =c=1\)
\(y=\left( \frac { 2x-1 }{ 3x+2 } \right) \)
\(\Rightarrow y= \frac { 2x-1 }{ 3x+2 }\)

(a) We know that
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\)
\(\int _{ 0 }^{ 10 }{ Axdx } +\int _{ 10 }^{ 20 }{ A(20-x)dx=1 } \)
\(A\left\{ { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 10 }+{ \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 20 } \right\} =1\)
A[(50-0)+(400-200)-(200-50)] = 1
\(A=\frac{1}{100}\)
(b) (i) The probability that the number of pounds of bread that will be sold tomorrow is more than 10 pounds is given by
\(P(10\le X\le 20)=\int _{ 10 }^{ 20 }{ \frac { 1 }{ 100 } (20-x) } dx\)
\(=\frac { 1 }{ 100 } { \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 20 }\)
\(=\frac { 1 }{ 100 } [(400-200)-(200-50)]\)
= 0.5
(ii) The probability that the number of pounds of bread that will be sold tomorrow is less than 10 pounds, is given by
\(P(0\le X\le 20)=\int _{ 0 }^{ 10 }{ \frac { 1 }{ 100 } } xdx\)
\(=\frac { 1 }{ 100 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 10 }\)
\(=\frac { 1 }{ 100 } (50-0)\)
= 0.5
(ii) The probability that the number of pounds of bread that will be sold tomorrow is between 5 and 15 pounds is
\(P(5\le X \le15)=\int _{ 5 }^{ 10 }{ \frac { 1 }{ 100 } xdx } +\int _{ 10 }^{ 15 }{ \frac { 1 }{ 100 } (20-x)dx } \)
\(=\frac { 1 }{ 100 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 5 }^{ 10 }+\frac { 1 }{ 100 } { \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 15 }\)
= 0.75

MR = a (x + b)^-2 − c
\(R=\int { { (x+b) }^{ 2 } } \)dx - c \(\int { dx } \)
R = \(\frac { a(x+{ b })^{ -2 } }{ -1 } \) - cx + k
R = \(\frac { a }{ (x+b) } \) - cx + k
When x = 0, R = 0
\(0=\frac { a }{ b } -c(0)+k\)
\(k=\frac { a }{ b } \)
\(R=-\frac { a }{ x+b } \) - cx+\(\frac{a}{b}\)
\(=\frac { -ab+a(x+b) }{ b(x+b) } -cx\quad \)
\(R=\frac { ax }{ b(x+b) } -cx\)
Demand function \(P=\frac { R }{ x } \)
\(P=\frac { a }{ b(x+b) } -c\)
\(P+c=\frac { a }{ b(x+b) } \)
\(b(x+b)=\frac { a }{ P+c } \)
\(x=\frac { a }{ b(P+c) } -b\)

Equation of the required circle is \({ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\) (1)
put \(y=0\), \({ x }^{ 2 }={ a }^{ 2 }\)
⇒  \(x=\pm a\)
Since equation (1) is symmetrical about both the axes
The required area = 4 [Area in the first quadrant between the limit 0 and a.]
\(=4\int _{ 0 }^{ a }{ y } \ dx\)
\(=4\int _{ 0 }^{ a }\sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx=4{ \left[ \frac { x }{ 2 } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } +\frac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{ \frac { x }{ a } } \right] }_{ 0 }^{ a }\)
\(=4{ \left[ 0+\frac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{( \frac { a }{ a } )} \right] }=4{ \left[ \frac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{( 1)} \right] }=4.\frac { { a }^{ 2 } }{ 2 }\frac { {π} }{ 2 }\)
= πa² sq. units

x +y + z = 1, 3x - y - z = 4, x + 5y + 5z = k

Augmented matrix  [A,B]	Elementary Transformation
\(\left( \begin{matrix} 1 & 1 & 1 \\ 3 & -1 & -1 \\ 1 & 5 & 5 \end{matrix}\begin{matrix} 1 \\ 4 \\ k \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & -4 & -4 \\ 0 & 4 & 4 \end{matrix}\begin{matrix} 1 \\ 1 \\ k-1 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & -4 & -4 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 1 \\ k \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 }\)

Here clearly \(\rho (A)=2\)
Since the given system is inconsistent \(\rho (A)\neq \rho (A,B)\)
This can take any value other than zero.
\(\therefore\) k can take any value other than zero.

Let ‘x’ be the cost of a Business Mathematics book
Let ‘y’ be the cost of a Accountancy book.
Let ‘z’ be the cost of a Commerce book.
\(\therefore \) 3x + 2y + z = 840
2x + y + z = 570
x + y + 2z = 630
Here \({ \triangle }=\left| \begin{matrix} 3 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{matrix} \right| =-2\neq 0\)
\({ \triangle }_{ x }=\left| \begin{matrix} 840 & 2 & 1 \\ 570 & 1 & 1 \\ 630 & 1 & 2 \end{matrix}\begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right| =-240 \)
\({ \triangle }_{ y }=\left| \begin{matrix} 3 & 840 & 1 \\ 2 & 570 & 1 \\ 1 & 630 & 2 \end{matrix} \right| =-300 \)
\({ \triangle }_{ z }=\left| \begin{matrix} 3 & 2 & 840 \\ 2 & 1 & 570 \\ 1 & 1 & 630 \end{matrix} \right| =-360\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -240 }{ -2 } =120 \)
\(y=\frac { { \triangle }y }{ { \triangle } } =\frac { 300 }{ -2 } =150 \)
\(z=\frac { { \triangle }z }{ { \triangle } } =\frac { 360 }{ -2 } =180\) 
\(\therefore \) The cost of a Business Mathematics book is Rs. 120,
the cost of a Accountancy book is Rs. 150 and
the cost of a Commerce book is Rs. 180.

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & k \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 7 \\ 18 \\ 6 \end{matrix} \right) \)
 AX = B

Augmented matrix [A,B]

Elementary Transformation

\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & k \end{matrix}\begin{matrix} 7 \\ 18 \\ 6 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & k \end{matrix}\begin{matrix} 7 \\ 11 \\ 6 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & k-2 \end{matrix}\begin{matrix} 7 \\ 11 \\ -5 \end{matrix} \right) \) ρ(A) = 2 or 3, ρ([A]) = 3

\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

For the equations to be inconsistent
\(\rho ([A,B])\neq \rho (A)\)
It is possible if k − 2 = 0.
\(\therefore \)  k = 2

\(\int _{ 1 }^{ 2 }{ \frac { 1 }{ (x+1)(x+2) } } dx=\int _{ 1 }^{ 2 }{ \left[ \frac { 1 }{ x+1 } -\frac { 1 }{ x+2 } \right] } dx\)
= \({ \left[ \log\left| x+1 \right| -\log\left| x+2 \right| \right] }_{ 1 }^{ 2 }\)
= \(\log\frac { 3 }{ 4 } -\log\frac { 2 }{ 3 } \)
= \(\log\frac { 9 }{ 8 } \)

By partial fractions,

\(\frac { 1 }{ (x+1)(x+2) } \) =  \(\frac { A }{ x+1 } +\frac { B }{ x+2 } \) ⇒\(\frac { 1 }{ (x+1)(x+2) } \) = \(\frac { 1 }{ x+1 } +\frac { 1 }{ x+2 } \)

\(\int { \frac { \left( 3x+2 \right) dx }{ \left( x-2 \right) \left( x-3 \right) } } \)
\(\frac { 3x+2 }{ \left( x-2 \right) \left( x-3 \right) } =\frac { A }{ x-2 } +\frac { B }{ x-3 } \)
⇒ 3x+2 = A (x-3)+B(x-2)
Put x = 3
9+2 = B(1) ⇒ B = 11
Put x = 2
8 = A(-1) ⇒ A = -8
\(\frac { 3x+2 }{ \left( x-2 \right) \left( x-3 \right) } =\frac { -8 }{ x-2 } +\frac { 11 }{ x-3 } \)
\(=\int { \left( \frac { -8 }{ x-2 } +\frac { 11 }{ x-3 } \right) } dx\)
\(=-8\log { \left| x-2 \right| } +11\log { \left| x-3 \right| } +c\)
\(=-11\log { \left| x-3 \right| } -8\log { \left| x-2 \right| } +c\)

Transition probability matrix

Where A represents the percentage of subscribers and B represents the percentage of non - subscribers.
A 40% = ·40
By the given data, 40% received the order of subscription = 60% are non-subscribers.
A 40% = ·40
and B 60% = ·60

((-40)(-45) + (·60)(-30) (-40)(.55) + (-60)(.70))
(-18+·18 ·22+42)
(·36 ·64)
\(\Rightarrow\) 36% of those receiving the current letter can be expected to order a subscription

Given non-homogeneous equations are
x + y + z = 3
x + 2y + 3z = 4
x + 4y + 9z = 6
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 11 & 4 & 9 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 3 \\ 4 \\ 6 \end{matrix} \right) \)
AX = B

Augmented matrix	Elementary Transformation
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{matrix}\begin{matrix} 3 \\ 4 \\ 6 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 3 & 8 \end{matrix}\begin{matrix} 3 \\ 1 \\ R \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R_{ 1 } }\)
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{matrix}\begin{matrix} 3 \\ 1 \\ 0 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

Clearly the last equivalent matrix is in echelon form and it has three non-zero rows
\(\therefore \rho (A)=3\quad \rho \left( \left[ A,B \right] \right) =3\)
\(\rho (A)=\rho \left( \left[ A,B \right] \right) =3\)
\(\therefore\) The given system is consistent and has unique solution.
To find the solution, let us rewrite the above echelon form into the matrix form
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 3 \\ 1 \\ 0 \end{matrix} \right) \)
\(\Rightarrow\) x + y + z = 3
y + 2z = 1
\((3)\Rightarrow 2x=0\Rightarrow z=\cfrac { 0 }{ 2 } =0\)
\((2)\Rightarrow y+2(0)=1\Rightarrow y+0=1\Rightarrow y=1-0=1\)
\(\left( 1 \right) \Rightarrow x+1+0=3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=3-1\)
\(\Rightarrow x=2\)
\(\therefore\) Solution set [2, 1, 0]