Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

P(X<0)=P(X=-2)
= 1/4

Given

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

 Find f(2.8)
Since the required value 2.8 is at the end of the table, apply Newton's backward interpolation formula.
x_n + nh = 2.8 ⇒ 3 + n (1) = 2.8
⇒  n = 2.8 - 3 = -0.2
The difference table is

Newton's backward interpolation formula is
y(x = x_n + nh) = \(\frac { n }{ 1! } { \triangledown y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }\)
⇒ y(2.8) = 34 + (-0.2) (23) + \(\frac { (-0.2)(-0.2+1) }{ 2 } (14)+\frac { (0.2)(-0.2+1)(-0.2+2) }{ 6 } \)(16)
⇒ y(2.8) = 34 - 4.6 + (-0.2) (0.8) (7) + (-0.2) (0.8) (1.8)
⇒ y(2.8) = 34 - 4.6 - 1.12- 0.288
⇒ y(2.8) = 27.992

Let the missing entries by y₁ and y₄
Since only four values of f(x) are given, the polynomial which fits the data is of degree 3.
Hence fourth differences are zero.
∴ Δ⁴y_k = 0 ⇒ (E - 1)⁴y_k = 0
(E⁴ - 4E³ + 6E²- 4E + 1) y_k = 0
Put k = 0 in (1) we get,
(E₄ - 4E₃ + 6E₂ - 4E + 1)y₀ = 0
⇒ y₄ - 4y₃ + 5y₂ - 4y₁ + y₀ = 0
⇒ y₄ - 4(15) + 6(8) - 4(y₁) + 0 = 0
⇒ y₄ - 60 + 48 - 4y₁ = 0
⇒ y₄ - 4y₁ - 12 = 0
⇒ y₄ - 4y₁ = 12  (2)
put k = 1 in (1) we get
(E⁴- 4E³ + 6E² - 4E + 1) y₁ = 0
⇒ y₅ - 4y₄ + 6y₃ - 4y₂ + y₁ = 0
⇒ 35 - 4y₄ + 6 (15) - 4 (8) +y₁ = 0
⇒ 35 - 4y₄ + 90 - 32 +y₁ = 0
⇒ 4y₄ +y₁ = -93  (3)

⇒ y₁ = \(\frac{-45}{-15}\) = 3
⇒ y₁ = 3.
Substituting y₁ = 3 in (2) we get,
y₄ - 4(3) = 12
y₄ - 12 = 12
⇒ y₄ = 12 + 12
⇒ y₄ = 24
Hence the missing entries are 3 and 24.

Here we find the population for year 1946. (i.e) the value of y at x = 1946. Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+..\)
To find y at x = 1946
\(\therefore\) x₀ + nh = 1946, x₀ = 1941, h = 10
1941 + n(10) = 1946 \(\Rightarrow\) n = 0.5

Here we find the population for year 1946. (i.e) the value of y at x = 1946. Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\(​​{ y }_{ \left( x=1946 \right) }=20+\frac { 0.5 }{ 1! } (4)+\frac { 0.5(0.5-1) }{ 2! } (1)+\frac { 0.5(0.5-1)(0.5-2) }{ 3! } (1)+\frac { 0.5(0.5-1)(0.5-2)(0.5-3) }{ 4! } (0)+\frac { 0.5(0.5-1)(0.5-2)(0.5-3)(0.5-4) }{ 5! } (-9)\)
= 20+2-0.125+0.0625-0.24609
= 21.69 lakhs

Here total supply = 25 + 35 + 40 = 100
total requirement = 30 + 25 + 45 100
total supply = total requirement
∴ The given problem is a balanced transportation problem.
Hence, there exists a feasible solution for the given transportation problem.
North West Corner Rule:
I - allocation:

[∵ min (25, 30) = 25]
II - allocation:

[∵ min (5, 35) = 5]
III - allocation:

[∵ min (25, 30) = 25]
IV - allocation:

[∵ min (5, 45) = 5]
V - allocation:

[∵ min (40, 40) = 40]
Thus, the allocations are

∴ The transportation schedule is
S₁ → D₁, S₂ → D₁, S₂ → D₂, S₂ → D₃, S₃ → D₃
Hence, the total transportation cost
= 25(9) + 5(6) + 25(8) + 5(4) + 40(9)
= 225 + 30 + 200 + 20 + 360
= Rs. 835

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.
The cost matrix of the given assignment problem is

Column 3 contains no zero. 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. Row B contains exactly one zero. Mark that zero by \(\square\) (i.e) Person B is assigned to Job 1. Mark other zeros in its column by ×.

Now, Row C contains exactly one zero. Mark that zero by \(\square\). Mark other zeros in its column by × .

Now, Row D contains exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its column by × .

Row E contains more than one zero, now proceed column wise. In column 1, there is an assignment. Go to column 2. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by × .

There is an assignment in Column 3 and column 4. Go to Column 5. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by × .

Thus all the five assignments have been made. The Optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = Rs. 9

\(\frac { dy }{ dx } \)+y cosx = 2 cosx
This is of the form \(\frac { dy }{ dx } \)+Py = Q where
P = cos x and Q = 2 cos x
\(\int { P } dx=\int { cosx } dx\) = sinx
Integrating factor (I.F.) = \(e^{ \int { P } dm }=e^{ sinx }\)
∴ The Solution is
\(ye^{ \int { P } dm }=\int { Q } e^{ \int { P } dm }dx\)
⇒ y e^sinx =\(\int { (2cosx){ e }^{ sinx } } dx\)+C
⇒ y e^sinx = 2\(\int { cosx } e^{ sinx }dx\)+C
⇒ y e^sinx = 2I₁+C  ..(1)
I₁=\(\int { cosx } e^{ sinx }dx\)
put sin x = t ⇒ cos dx = dt
∴ I_t =\(\int { { e }^{ t } } dt\) = et = e^sinx
∴ (1) becomes, y e^sinx = 2.e^sinx + 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

Fisher's price index number
\(P^{F}_{01}\) = \(\sqrt{ \frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}} \times ​\frac{​ {\sum p_{1}q_{1}} }{{\sum p_{0}q_{1}}}{}}\times 100\)
= \(\sqrt\frac {1900 \times 1880}{1360 \times 1344} \times 100\)
= \(\sqrt\frac {3572000}{1827840} \times 100\) 
= \(\sqrt {1.954} \times 100\)
\(P^{F}_{01}\)  = 139.8
Time reversal test:
\(P _{01} \times P_{10}\) = \(\sqrt{ \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}\times \sum p_{0}q_{1}\times \sum p_{0}q_{0}}{\sum p_{0}q_{0}\times \sum p_{0}q_{1}\times \sum p_{1}q_{1} \times \sum p_{1}q_{0}} ​}\)
=  
=  \(\sqrt {1}\)  = 1
∴    \(P _{01} \times P_{10}\) = 1
Factor reversal test:
\(P _{01} \times Q_{01}\) = \(\sqrt{ \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}\times \sum q_{1}p_{0}\times \sum q_{1}p_{1}}{\sum p_{0}q_{0}\times \sum p_{0}q_{1}\times \sum q_{0}p_{0} \times \sum q_{0}p_{1}} ​}\)

= \(\frac {1880}{1360} = \frac {\sum p_{1}q_{1}}{\sum p_{0}q_{0}}\)
Hence, it satisfies time reversal test and factor reversal test.

The given differential equation is of the form
\(\frac { dy }{ dx } \)+ Py = Q where
P =\(\frac { 1 }{ x } \); Q = xe^x
∴ \(\int { p } dx=\int { \frac { 1 }{ x } } \) = logx
∴ Integratin'cg tactor (I. F) = \(e^{ \int { p } dx }=e^{ logx }=x\)
Hence the solution is
\(ye^{ \int { p } dx }=\int { Q } .e^{ \int { p } dx }dx+c\)
⇒ yx =\(\int { x{ e }^{ x }dx+c } \)
⇒ xy = \(\int { x{ e }^{ x }dx+c } \) ....(1)
Let u = x²; dv = e^x
u' = 2x; v = e^x
u" = 2; v₁ = e^x
v₂ = e^x
Using Bernoulli's formula,
\(\int { u } dv\) = uv - u'v₁+ u"v₂
∴ (1) becomes
xy = x²e^x - 2xe^x + 2e^x+c
⇒ xy = e^x(x²-2x+2) + c

Given \(\frac { dy }{ dx } \) − (3 cot x) y = sin 2x
It is of the form \(\frac { dy }{ dx } \) + Py = Q
Here P= − 3 cot x,Q = sin 2x
ഽPdx = ഽ-3 cot xdx = -3 log sin x = - log sin³x = log \(\frac { 1 }{ { sin }^{ 3 }x } \)
I.F. = \({ e }^{ log\frac { 1 }{ { sin }^{ 3 }x } }=\frac { 1 }{ { sin }^{ 3 }x } \)
The required solution is y (I.F) = ഽQ(I.F)dx+c
\(y\frac { 1 }{ { sin }^{ 3 }x } =\int { sin2x } \frac { 1 }{ { sin }^{ 3 }x } dx+c\)
\(\int { \frac { 1 }{ { sin }^{ 3 }x } } =\int { 2sinxcosx\times \frac { 1 }{ { sin }^{ 3 }x } } dx+c\)
= \(2\int { \frac { 1 }{ sinx } \times \frac { cosx }{ sinx } } dx+c\)
= ഽcos ecx cot xdx + c
\(y\frac { 1 }{ { sin }^{ 3 }x } =-2cosecx+c\)
Now y = 2 when x = \(\frac { \pi }{ 2 } \)
(1) ⇒ 2\(\frac {1 }{ 1 } \) = −2×1+c ⇒ c= 4
∴ (1) ⇒ \(y\frac { 1 }{ { sin }^{ 3 }x } \) = −2cosecx + 4

\(\overline {\overline{X}} = \frac {125.1}{10}\) = 12.51
\(\overline{R} = \frac {3.7}{10}\) = 0.37
Control limits for \(\overline {X}\) - chart
UCL =  \(\overline {\overline{X}} + A_{2} \overline {R}\)
= [when n = 4, A₂ = .729]
= 12.51 + .729 (0.37)
= 12.51 + 0.27 = 12.78
CL = \(\overline {\overline{X}}\) = 12.51
LCL = \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 12.51 - 0.27 = 12.24
Control limits for R-chart
UCL = D₄ \(\overline{R}\)
= 2.282(0.37) = 0.84
 [when n = 4, D₄ = 2.282]
CL = \(\overline{R}\) = 0.37
LCL = D₃ \(\overline{R}\) = 0
[When n = 4, D₃ = 0]
\(\overline {X}\) - 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.

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
=  \(\sqrt \frac {97\times117}{116\times140} \times 100\)
= \(\sqrt \frac {11349}{16240} \times 100\)
= \(\sqrt {0.6988}\times100\)
\(P^{F}_{01}\)   =  83.59

\(\frac { dy }{ dx } =\frac { x+3y }{ x-y } \)
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 = \(\frac { 1 }{ c } \).

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

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

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

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.

Fisher’s price index number
\({ P }_{ 01 }^{ F }=\left( \sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 10 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \right) \times 100=\left( \sqrt { \frac { 537\times 742 }{ 442\times 613 } } \right) \times 100=121.2684\)
Time Reversal Test:P₀₁\(\times\) P₁₀ = 1
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { 537\times 742\times 613\times 442 }{ 442\times 613\times 742\times 537 } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=1\)
Factor Reversal Test
\({ P }_{ 01 }\times { Q }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { 537\times 742\times 613\times 742 }{ 442\times 613\times 442\times 537 } \right) } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { 742\times 742 }{ 442\times 442 } \right) } =\frac { 742 }{ 442 } \Rightarrow { P }_{ 01 }\times { P }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)

Slope of the normal at any point P(x, y) = -\(\frac { dx }{ dy } \)
Let Q be (1, 0)
Slope of the normal PQ is \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
i.e, \(\frac { y-0 }{ x-1 } =\frac { y }{ x-1 } \)
∴ \(\frac { dx }{ dy } =\frac { y }{ x-1 } \) ⇒ \(\frac { dx }{ dy } =\frac { y }{ 1-x } \), which is the differential equation
i.e., (1− x)dx = ydy
ഽ(1−x)dx = ഽydy + c
\(x-\frac { { x }^{ 2 } }{ 2 } =\frac { { y }^{ 2 } }{ 2 } +c\)   ....(1)
Since it passes through (1,2)
1 - \(\frac { 1 }{ 2 } =\frac { 4 }{ 2 } +c\)
\(c=\frac { 1 }{ 2 } -2=\frac { 4 }{ 2 } +c\)
Put \(c = \frac { -3 }{ 2 } \) in (1)
\(x-\frac { { x }^{ 2 } }{ 2 } =\frac { { y }^{ 2 } }{ 2 } -\frac { 3 }{ 2 } \)
2x − x² = y² − 3
⇒ y² = 2x−x² + 3, which is the equation of the curve

Computation of trend values by the method of least squares.
In case of EVEN number of years, let us consider
\(X=\frac{\text{(x-Arithimetic mean of two middle years)}}{0.5}\)

\(a=\frac { \sum { Y } }{ n } =\frac { 47.8 }{ 8 } =5.975;\quad b=\frac { \sum { XY } }{ { \sum { X } }^{ 2 } } =\frac { 8.6 }{ 168 } =0.05119\)
Therefore, the required equation of the straight line trend is given by
Y = a + bX; Y = 5.975 + 0.05119 X.
When X = 1995, Y_t = 5.975 + 0.05119\(\left( \frac { 1995-1998.5 }{ 0.5 } \right) =5.6166\)
When X = 1996, Y_t = 5.975 + 0.05119\(\left( \frac { 1996-1998.5 }{ 0.5 } \right) =5.7190\)
similarly other values can be obtained.

Sample size n = 60
Population standard deviation σ = 3
The standard error for sample standard deviation
\(=\sqrt { \frac { { \sigma }^{ 2 } }{ 2n } } \)
\(=\sqrt { \frac { { 3 }^{ 2 } }{ 2(60) } } =\sqrt { \frac { 9 }{ 120 } } =\sqrt { 0.075 } \)
S.E.of sample standard deviation = 0.2739

\(\int _{ a }^{ b }{ f\left( x \right) } dx=\lim _{ n\rightarrow \infty \\ h\rightarrow 0 }{ \sum _{ r=1 }^{ n }{ h.f(a+rh) } } \)
Here a = 1, b = 3
\(h=\frac { b-a }{ n } =\frac { 3-1 }{ n } =\frac { 2 }{ n } \)
and f (x) = x
Now \(f(a+rh)=f\left( 1+r\cfrac { 2 }{ n } \right) =f\left( 1+\cfrac { 2r }{ n } \right) \)
\(\therefore \int _{ 1 }^{ 3 }{ xdx } =\underset { n\rightarrow \infty }{ lim } .\cfrac { 2 }{ n } \left( 1+\cfrac { 2r }{ n } \right) \)
= \(\underset { n\rightarrow \infty }{ lim } \sum _{ r=1 }^{ n }{ \left( \cfrac { 2 }{ n } +\cfrac { 4r }{ { n }^{ 2 } } \right) } \)
= \(\underset { n\rightarrow \infty }{ lim } \left( \cfrac { 2 }{ n } .\sum _{ r=1 }^{ n }{ .1+\cfrac { 4 }{ { n }^{ 2 } } .\sum _{ r=1 }^{ n }{ .r } } \right) \)
\(\left( \because \sum _{ r=1 }^{ n }{ 1=n } \right) \)

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

= \(2+\underset { n\rightarrow \infty }{ lim } \left( 1+\cfrac { 1 }{ n } \right) ^{ 2 }\)
= \(2+n\underset { n\rightarrow \infty }{ lim } \left( 2+\cfrac { 2 }{ n } \right) \)
= \(2+2\underset { n\rightarrow \infty }{ +lim } \left( 2+\cfrac { 2 }{ n } \right) \)

= 2 + 2 + 0
= 4

Here φ is the mean length of the components in the population.
The formula for the confidence interval is
\(\bar{x}-Z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+Z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}\)
\({ Here } \ \sigma=1.6, Z_{\alpha / 2}=1.96, \bar{x}=90 \text { and } \mathrm{n}=64\)
Then \(S.E=\frac { \sigma }{ \sqrt { n } } =\frac { 1.6 }{ \sqrt { 64 } } =0.2\)
Therefore, 90 - (1.96 x 0.2)\(\le φ \le\) 90 + (1.96 x 0.2)
\(\text { i.e. } \ (89.61 \leq \varphi \leq 90.39)\)
This implies that the probability that the true value of the population mean length of the components will fall in this interval (89.61,90.39) at 95% . Hence we concluded that 95% confidence interval ensures acceptance of the component by the consumer.

Let p be the probability getting his piston rejected
Given p = 12% = \(\frac { 12 }{ 100 } \) = 0.12
∴ q = 1-p = 1-0.12 = 0.88
n = 10
(a) P (not more than 2 rejects)
= P(X≤2) = P(X = 0) + P(X = 1) + P(X = 2)
= 10C₀(0.12)⁰ (0.88)¹⁰ + 10C₁ (0.12)¹ (0.88)⁹ + 10C₂ (0.12)² (0.88)⁸
[∵ P(x) = nC_x p^xq^n-x]

= (0.88)⁸ [(0.88)² + 10(0.12) (0.88) + 45 (0.12)²]
= (0.88)⁸ [0.7744 + 1.056 + 0.648]
= (0.3596) (2.4784) = 0.8913
∴ Probability of not more than 2 rejects = 0.8913
b) P(at least 2 rejects)
= P(X ≥ 2) = 1 - P (X < 2)
= 1 - [P(X = 0) + P (X = 1)]
= 1-[10C₀(0.12)⁰ (0.88)¹⁰ + 10C₁ (0.12)¹ (0.88)⁹ ]
= 1 - [(0.88)¹⁰ + 10 (0.12) (0.88)⁹]
= 1 - (0.88)⁹ [0.88 + 1.2] = 1 - (0.31647) (2.08)
= 1 - 0.6583 = 0.34173
P (atleast 2 rejects) = 0.34173

Given average number of customers, appear in a counter
λ = 2
∴ X follows a poisson distribution with
p(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
(i) P (no customer appears)
= P(X = 0) = \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } \) = e^-λ = e^-2 = 0.1353   [∵ e^-2 = 0.1353]
(ii) P(3 or more customer appear)
= P(X≥3)
= 1-P(X<3)
= 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^-2 (1+2+\(\frac { 4 }{ 2 } \))
= 1-e^-2 (5) = 1 - 0.1353(5)
= 1 - 0.6765 = 0.3235
Hence, probability of three or more customers appear in a counter of a certain bank is 0.3235.

Given that probability of getting defective item
p = 5% = \(\frac { 5 }{ 100 } \) ⇒ q = 1-p = \(1-\frac { 5 }{ 100 } =\frac { 95 }{ 100 } \)
n = 20
p(x) = \({ n }_{ C_{ x } }{ p }^{ x }{ q }^{ n-x }\), x = 0,1,2....n
(i) P(Exactly 3 defectives)
= \({ 20 }_{ { C }_{ 3 } }\left( \frac { 5 }{ 100 } \right) ^{ 3 }\left( \frac { 95 }{ 100 } \right) ^{ 20-3 }\)
= \(\\ { 20 }_{ { C }_{ 3 } }\)(0.05)3(0.95)17
= \(\\ \frac { 20\times 19\times 18 }{ 3\times 2\times 1 } \) (0.05)³(0.95)⁵(0.95)⁵(0.95)⁵(0.95)²
= (60 x 19) (0.000125)(0.7738)(0.7738)(0.7738)(0.9025)
= 0.059
(ii) P(atleast 2 defectives)
= P(X≥2) =1-P(X<2)
= 1-[P(X=0) + P(X=1)]
= 1-[\(\\ { 20 }_{ { C }_{ 0 } }\)(0.05)⁰(0.95)²⁰ + \(\\ { 20 }_{ { C }_{ 1 } }\)(0.05)¹(0.95)¹⁹]
= 1 - [(0.95)²⁰ + 20 (0.05) (0.95)¹⁹]
= 1 - [0.3585 + (0.3774)]
= 1 - [0.7359] = 0.2641.
(iii) P (exactly 4 defectives)
P(X = 4) = \(\\ { 20 }_{ { C }_{ 4 } }\) (0.05)4(0.95)16

= (15 x 17 x 19) (0.00000625) (0.4402)
= 0.0133
(iv) Find the mean and variance
Mean = np =\(20\times \frac { 5 }{ 100 } =\frac { 100 }{ 100 } \) = 1
Variance =npq = \(20\times \frac { 5 }{ 100 } =\frac { 95 }{ 100 } \) = 0.95

Consider a 2,000 individuals getting injection of a given serum , n = 2000
Let X be the number of individuals suffering a bad reaction.
Let p be the probability that an individual suffers a bad reaction = 0.001
and q = 1– p = 1– 0.001 = 0.999
Since n is large and p is small, Binomial Distribtuion approximated to poisson distribution
So, λ = np = 2000 × 0.001 = 2
(i) Probability out of 2000, exactly 3 will suffer a bad reaction is
\(P(X=3)=\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } =\frac { { e }^{ -2 }{ 2 }^{ 3 } }{ 3! } =0.1804\)
(ii) Probability out of 2000, more than 2 individuals will suffer a bad reaction
= P(X > 2)
1-[P(X\(\le\)2)]
= 1 – [P(x = 0) + P(x = 1) + P(x = 2)]
\(=1-\left[ \frac { { e }^{ -2 }{ 2 }^{ 0 } }{ 0! } +\frac { { e }^{ -2 }{ 2 }^{ 1 } }{ 1! } +\frac { { e }^{ -2 }{ 2 }^{ 2 } }{ 2! } \right] \)
\(=1-{ e }^{ 2 }\left( \frac { { 2 }^{ 0 } }{ 0! } +\frac { { 2 }^{ 1 } }{ 1! } +\frac { { 2 }^{ 2 } }{ 2! } \right) \)
= 0.323

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

From the given data, you first calculate the probability distribution of the random variable. Then using it you calculate mean and variance.
X p(x)
1  F(1) = \(\frac{1}{6}\)
2  F(2)-F(1) = \(\frac{2}{6}\)-\(\frac{1}{6}\) = \(\frac{1}{6}\)
3  F(3)-F(2) = \(\frac{3}{6}\)-\(\frac{2}{6}\) = \(\frac{1}{6}\)
4  F(4)-F(3) = \(\frac{4}{6}\)-\(\frac{3}{6}\) = \(\frac{1}{6}\)
5  F(5)-F(4) = \(\frac{5}{6}\)-\(\frac{4}{6}\) = \(\frac{1}{6}\)
6  F(6)-F(5) = 1-\(\frac{5}{6}\) = \(\frac{1}{6}\)
The probability mass function is

Mean of the random variable X = E(X)\(\sum _{ x }^{ }{ x } { P }_{ X }(x)\)
\(=\left( 1\times \frac { 1 }{ 6 } \right) +\left( 2\times \frac { 1 }{ 6 } \right) +\left( 3\times \frac { 1 }{ 6 } \right) +\left( 4\times \frac { 1 }{ 6 } \right) +\left( 5\times \frac { 1 }{ 6 } \right) +\left( 6\times \frac { 1 }{ 6 } \right) \)
\(=\frac { 1 }{ 6 } (1+2+3+4+5+6)\)
= 7/2
\(E({ X }^{ 2 })=\sum _{ x }^{ }{ { x }^{ 2 } } { P }_{ X }(x)\)
\(=\left( { 1 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 2 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 3 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 4 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 5 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 6 }^{ 2 }\times \frac { 1 }{ 6 } \right) \)
\(=\frac { 1 }{ 6 } ({ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 }+{ 6 }^{ 2 })\)
\(=\frac { 91 }{ 6 } \)
Variance of the Random Variable \(X=V(X)=E\left( { X }^{ 2 } \right) -{ [E(X)] }^{ 2 }\)
\(=\frac { 91 }{ 6 } -{ \left( \frac { 7 }{ 2 } \right) }^{ 2 }\)
\(=\frac { 35 }{ 12 } \)

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

We have F(x) = f(x) ≥ 0, where F(x) is the distribution function and f(x) is the probability density function.
Here F(x) = 0 for x ≤ 1  f(x) = 0 for x ≤ 1 
Again F(x) = 1 for x > 3 
f(x) = d/dx (1) = 0 for x > 3 
In 1 < x ≤ 3, F(x) = k(x – 1)4 
f(x) = d/dx (k(x – 1)4) = 4k(x – 1)3 
\(\therefore f(x)=4k{ (x-1) }^{ 3 }for\quad 1\le x\le 3\)
i) Since f(x) is a probability density function,
\(\int _{ 1 }^{ 3 }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 1 }^{ 3 }{ { 4k(x-1) }^{ 3 }dx=1 } \)

\(\Rightarrow k[{ (3-1) }^{ 4 }-{ (0) }^{ 4 }]=1\)
\(\Rightarrow k({ 2 }^{ 4 })=1\Rightarrow k(16)=1\)
\(\Rightarrow k=\frac { 1 }{ 16 } \)
ii) \(\therefore\)p.d.f
\(f(x)=\frac { 4\times 1 }{ 16 } { (x-1) }^{ 3 }for\quad 1\le x\le 3\)
\(f(x)=\frac { 1 }{ 4 } { (x-1) }^{ 3 }for\quad 1\le x\le 3\)

The present value of payment for t year = \(\int _{ 0 }^{ t }{ { 72000e }^{ -0.09t } } dt\)
Present value of 12 years = \(\int _{ 0 }^{ t }{ { 72000e }^{ -0.09t } } dt\)
= 72000\({ \left[ \frac { { e }^{ -0.09t } }{ { -0.09 } } \right] }_{ 0 }^{ \\ 12 }\)
= \(\frac { 72000 }{ -0.09 } \left[ { e }^{ -0.09(12) }-{ e }^{ 0 } \right] \)
= \(-8,00,000[{ e }^{ -1.08 }-{ e }^{ 0 }]\)
= −8,00,000 [0.3396 −1]
= 5,28,320
Cost of the machine = 5,00,000 − 30,000
= 4,70,000
Hence it not advisable to rent the machine
It is better to buy the machine.

i) Since P[X\(\le\)a₁] = P[X>a₁] 
\(P[X\le { a }_{ 1 }]=\frac { 1 }{ 2 } \)
\(i.e., \int _{ 0 }^{ { a }_{ 1 } }{ f(x)dx } =\frac { 1 }{ 2 } \)
\(i.e., \int _{ 0 }^{ { a }_{ 1 } }{ { 5x }^{ 4 } } dx=\frac { 1 }{ 2 } \)
\(5{ { \left[ \frac { { x }^{ 5 } }{ 5 } \right] } }_{ { 0 }_{ } }^{ a1 }=1/2\)
\({ a }_{ 1 }={ (0.5) }^{ \frac { 1 }{ 5 } }\)
ii) \(P[X>{ a }_{ 2 }]=0.05\)
\(\int _{ { a }_{ 2 } }^{ 1 }{ { f(x })dx=0.05 } \)
\(\int _{ { a }_{ 2 } }^{ 1 }{ { 5x }^{ 4 }dx=0.05 } \)
\(5{ { \left[ \frac { { x }^{ 5 } }{ 5 } \right] } }_{ { a }_{ 2 } }^{ 1 }=0.05\)
\({ a }_{ 2 }={ [0.95] }^{ \frac { 1 }{ 5 } }\)

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

\(\int { \frac { { 3x }^{ 2 }-2x56 }{ \left( x-1 \right) -\left( { x }^{ 2 }+5 \right) } } dx\)
⇒ 3x²-2x+5 = A (x²+5) + (Bx+c) (x-1)
Putting x =1,
3 - 2+5 = A (1+ 5)
⇒ 6 = A (6)
⇒ A = 1
Putting x = 0,
5 = 5 A - C
⇒ 5 = 5 - C    [∵ A = 1]
⇒ C = 5 - 5 ⇒ C = 0
Putting x = -1,
3 + 2+ 5 = A (6) + (C-B) (-2)
⇒ 10 = 6A + 2B - 2C
⇒ 10 = 6 + 2B + 0
⇒ 10 - 6 = 2B ⇒ 4 = 2B
⇒ B = 2
\(=\int { \left( \frac { A }{ x-1 } +\frac { Bx+c }{ { x }^{ 2 }+5 } \right) } dx\)
\(=\int { \left( \frac { 1 }{ x-1 } +\frac { 2x+0 }{ { x }^{ 2 }+5 } \right) } dx\)
\(=\int { \frac { 1 }{ x-1 } } dx+\int { \frac { 2x }{ { x }^{ 2 }+5 } } dx\)
\(=\log { \left| x-1 \right| } +\log { \left| { x }^{ 2 }+5 \right| } +c\)
\(\left[ \because \int { \frac { { f }^{ 1 }(x) }{ f(x) } dx=\log { \left| f\left( x \right) \right| +c } } \right] \)
\(=\log { \left| \left( { x }^{ 2 }+5 \right) \left( x-1 \right) \right| } +c\)
[∵ log m + log n = log mn]
\(=\log { \left| { x }^{ 3 }-{ x }^{ 2 }+5x-5 \right| } +c\)
\(=\frac { { 3x }^{ 2 }-2x+5 }{ \left( x-1 \right) \left( { x }^{ 2 }+5 \right) } =\frac { A }{ (x-1) } +\frac { Bx+C }{ \left( { x }^{ 2 }+5 \right) } \)

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

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 the amount invested in the rate of 2%, 3% and 6% be Rs. x, Rs. y and Rs. z respectively
By the given data,
x+ y+z = 8500
\(\cfrac { 2x }{ 100 } +\cfrac { 3y }{ 100 } +\cfrac { 6z }{ 100 } =380\)
\(\Rightarrow \cfrac { 2x+3y+6z }{ 100 } =380\)
\(\because Interest=\cfrac { PNR }{ 100 } =\cfrac { x\times 1\times 2 }{ 100 } =\cfrac { 2x }{ 100 } \)
\(\Rightarrow 2x+3y+6z=38000\)
Also,z = x+y
x+y-z  =0
\(\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{matrix} \right| \)
\(1\left| \begin{matrix} 3 & 6 \\ 1 & -1 \end{matrix} \right| -1\left| \begin{matrix} 2 & 6 \\ 1 & -1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \)
= 1(-3 - 6) - 1(-2 -6) + 1 (2 - 3)
= 1(-9) - 1 (-8) + 1(-1)
= -9 + 8 - 1 = 2 \(\neq \) 0 
Since \(\Delta \neq 0\),Cramer's rule can be applied and the system is consistent with unique solution
\({ \Delta x }=\left| \begin{matrix} 8500 & 1 & 1 \\ 38000 & 3 & 6 \\ 0 & 1 & -1 \end{matrix} \right| \)
= \(8500\left| \begin{matrix} 3 & 6 \\ 1 & -1 \end{matrix} \right| -1\left| \begin{matrix} 38000 & 6 \\ 1 & -1 \end{matrix} \right| +1\left| \begin{matrix} 38000 & 3 \\ 0 & 1 \end{matrix} \right| \)
= 8500 (- 3 - 6) - 1(-38000 -0) + 1(38000 - 0)
= 8500(-9) - 1(-38000) + 1(38000)
= - 76500 + 38000 + 38000
= -500
\(\Delta y=\left| \begin{matrix} 1 & 8500 & 1 \\ 2 & 38000 & 6 \\ 1 & 0 & -1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 38000 & 6 \\ 0 & -1 \end{matrix} \right| -8500\left| \begin{matrix} 2 & 6 \\ 1 & -1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 38000 \\ 1 & 0 \end{matrix} \right| \)
= 1 (-38000 - 0) - 8500 (-2 -6) + 1(0 - 38000)
= - 38000 - 8500 (-8) - 38000
= - 38000 + 68000 - 38000
= - 8000
\(\Delta z=\left| \begin{matrix} 1 & 1 & 8500 \\ 2 & 3 & 38000 \\ 1 & 1 & 0 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 38000 & 6 \\ 0 & -1 \end{matrix} \right| -1\left| \begin{matrix} 2 & 38000 \\ 1 & 0 \end{matrix} \right| +8500\left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \)
= 1 (0 - 38000) - 1(0 -38000) +85000 (2 - 3)
= - 38000 + 38000 + 8500 (-1)
= - 8500


Hence, the amount invested in the three accounts are Rs. 250, Rs. 4000 and Rs. 4250 respectively.

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

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

\(\int _{ 1 }^{ e }{ \log x } dx=\int _{ 1 }^{ e }{ udv } \)
= \({ \left[ uv \right] }_{ 1 }^{ e }-\int _{ 1 }^{ e }{ vdu } \)
= \({ \left[ x\log x \right] }_{ 1 }^{ e }-\int _{ 1 }^{ e }{ x\frac { 1 }{ x } } dx\)
= ( e log e −1 log 1) - \({ [x] }_{ 1 }^{ e }\)
= (e − 0) − (e −1)
= 1