Grand average = \(\frac{42.4+36.2+37.8+40.2}{4}\)
= 39.15
Seasonal Index (S.1) = \(\frac{Quarterlyaverage}{Grand average}\times 100\)
Hence,
S.1 or I quarter = \(\frac{42.4}{39.15}\times100=108.30\)
S.1 or II quarter = \(\frac{36.2}{39.15}\times100=92.54\)
S.1 or III quarter = \(\frac{37.8}{39.15}\times100=96.55\)
S.1 or IV quarter = \(\frac{40.2}{39.15}\times100=102.68\)

Let the required equation of the straight line trend is
y = a + bX
Since Σx = 0. \(a = \frac{\Sigma y}{x} = \frac{16.9}{5}\)
\(b = \frac{\Sigma xy}{\Sigma x^2} = \frac{13.3}{10} = 1.33\)
Hence, the straight line trend is
y = 3.38 + 1.33 (x - 2002)
∴ The trend for 2007 is
y_t = 3.38 + 1.33 (2007 - 2002)
⇒  y_t = 3.38 + 1.33 (5)
⇒  y_t = 3.38 + 6.65
⇒  y_t = 10.03

Given p = \(\frac { 55 }{ 100 } \)
∴ q = \(\frac { 45 }{ 100 } \) and n = 100
Standard error = \(\sqrt { \frac { pq }{ n } } =\sqrt { \frac { \frac { 55 }{ 100 } \times \frac { 45 }{ 100 } }{ 100 } } \)
= 0.0497
(a) As the level of significance α = 0.05 \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
∴ 95% confidence limits for proportion is given by \(p-Z_{ \frac { \alpha }{ 2 } }(S.E)\le p\ge p+Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.55 - (1.96) (0.0497) ≤ p ≤ 0.55 + (1.96) (0.0497)
⇒ 0.453 ≤ p ≤ 0.647
∴ 95% confidence interval for proportion is (0.45, 0.65)
(b) As the level of significance is α = 0.01, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58
∴ 99% confidence limits for proportion is given by  \(p-Z_{ \frac { \alpha }{ 2 } }(S.E)\le p\ge p+Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.55 - (2.58) (0.0497) ≤ p ≤ 0.55 + (2.58) (0.0497)
⇒ 0.422 ≤ p ≤ 0.678
Hence, 99% confidence interval for proportion is (0.42, 0.68).

Given sample size n = 200
Sample mean \(\bar { x } \) = 0.824
Sample S.D. s = 0.042
Standard error = \(\frac { s }{ \sqrt { n } } =\frac { 0.042 }{ \sqrt { 200 } } \)
= \(\frac { 0.042 }{ 14.14 } \) = 0.00270
(a) As the level of significance is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) =  1.96
∴ 95% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (1.96) (0.00270) ≤ μ ≤ 0.824 + (1.96) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.00582
⇒ 0.818 ≤ μ ≤ 0.832
Hence, the 95% confidence limits for μ is (0.818,0.832)
(b) As the level of significance is α =0.001, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58
∴ 99% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (2.58) (0.00270) ≤ μ ≤ 0.824 + (2.58) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.005825
⇒ 0.816 ≤ μ ≤ 0.832
Hence, the 99% confidence limits for μ is (0.816, 0.832)

Let X denote the height of the student
Given P (X < 40) = 10% = \(\frac { 10 }{ 100 } \) =0.1
P(X> 90) = 10% = \(\frac { 10 }{ 100 } \) =0.1

∴ P(40 < X < 90) = P(-∞ < X < ∞) - [P(X < 40) + P(X < 90)]
= 1 - (0.1 + 0.1)
= 1 - 0.2 = 0.8
∴ out of 800 students, number of students scored between 40 and 90 = 800 x 0.8
= 640 students.

Given n = 10, p = \(\frac { 20 }{ 100 } =\frac { 1 }{ 5 } \)
∴ q = 1-p = \(1-\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
Let
X denote the number of defective bolts chosen
∴ X = 2
(i) Using binomial distribution
P(X = 2) = \(10{ C }_{ 2 }\left( \frac { 1 }{ 5 } \right) ^{ 2 }\left( \frac { 4 }{ 5 } \right) ^{ 8 }\)
= \(\frac { 10\times 9 }{ 2\times 1 } \left( \frac { { 4 }^{ 8 } }{ { 5 }^{ 10 } } \right) =45\left( \frac { 4^{ 8 } }{ { 5 }^{ 10 } } \right) \)
(ii) Using Poisson distribution
λ = np = 10 \(\times\) \(\frac { 1 }{ 5 } \) = 2
P(X = x) \(\times\)\(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
x = 0,1,2,......n
∴ P(X = 2) = \(\frac { e^{ -2 }(2^{ 2 }) }{ 2 } =e^{ -2 }\left( \frac { 4 }{ 2 } \right) \)
= 2e^-2
= 2(0.1353) = 0.2706
∴ P(X = 2) = 0.2706

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

Here only 3 tasks can be assigned to 3 men.
Step 1 :
Select the smallest element in each row and subtract it will all the elements in its row.

Here each row and column has atleast one zero.
Step 2:
Examine the row with only one zero, mark that zero by \(\Box\) and draw a vertical line.
After examining all the rows, examine the column with single zero, mark that zero by \(\Box\) and draw a horizontal line.
Step 3:
Only two assignment have been made.
The elements not lying on the line are
\(\begin{matrix} 6 & 10 & 14 \\ 5 & 9 & 11 \\ 5 & 5 & 12 \end{matrix}\)
and minimum is 5.
Subtract 5 from all these numbers. Other numbers remains the same.
A new cost matrix will be found .and repeat step 2.
∴ The new cost matrix is

Thus, 3 assignments have been made.
The optical assignment schedule and total cost is

Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
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

Here column IV has 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:
Examine the row with exactly one zero. Mark the zero by \(\Box \) and draw a vertical line. After examining all the rows examine the column with one zero. mark the zero by \(\Box\) and draw a horizontal line.

Here only 4 assignments have been made.
The numbers not lying on the line are

and min. of these numbers is 1.
Now subtract 1 from all these numbers and add 1 to the numbers on the intersecting line (ie. 6, 7, 2). Other numbers remains the same.
∴ The new cost matrix is

Now, repeat Step 3.
Thus, all the 5 assignments have been made.
The optimal assignment schedule and total cost is

E(X) = Σx_i²p_i
\(\Rightarrow E(X)=1(\frac { 1 }{ 2 } )+2\left( \frac { 1 }{ 5 } \right) +4\left( \frac { 3 }{ 25 } \right) +2A\left( \frac { 1 }{ 10 } \right) +3A\left( \frac { 1 }{ 25 } \right) +5A\left( \frac { 1 }{ 25 } \right) \)
\(\Rightarrow \frac { 1 }{ 2 } +\frac { 2 }{ 5 } +\frac { 12 }{ 25 } +\frac { A }{ 5 } +\frac { 3A }{ 25 } +\frac { A }{ 5 } \)
\(=\frac { 69 }{ 50 } +\frac { 13A }{ 25 } \)
Since E(X) = 2.94
\(\frac { 69 }{ 50 } +\frac { 13A }{ 25 } =2.94\Rightarrow \frac { 13A }{ 25 } =2.94-\frac { 69 }{ 50 } \)
= 2.94 - 1.38 = 1.56
\(\Rightarrow A=\frac { 1.56\times 25 }{ 13 } =\frac { 39 }{ 13 } =3\)
\(E({ X }^{ 2 })=1\left( \frac { 1 }{ 2 } \right) +4\left( \frac { 1 }{ 5 } \right) +16\left( \frac { 3 }{ 25 } \right) +{ (2A) }^{ 2 }\left( \frac { 1 }{ 10 } \right) { +(3A) }^{ 2 }\left( \frac { 1 }{ 25 } \right) +{ (5A) }^{ 2 }\left( \frac { 1 }{ 25 } \right) [\because A=3]\)
\(=\frac { 1 }{ 2 } +\frac { 4 }{ 5 } +\frac { 48 }{ 25 } +36\left( \frac { 1 }{ 10 } \right) \)
\(E({ X }^{ 2 })=\frac { 25+40+96+180+162+450 }{ 50 } \)
\(=\frac { 953 }{ 50 } =19.06\)
\(V(X)=E({ X }^{ 2 })-{ [E(X)] }^{ 2 }\)
\(=19.06-{ (2.94) }^{ 2 }[\because E(X)=2.94]\)
= 19.06 - 8.6436
V(X) = 10.4164

Since X is the random variable taking values
1, 2,....7
P(X=1)+P(X=2)+...P(X=7) = 1
⇒ c+2c+2c+3c+c²+2c²+7c²+c = 1
⇒ 10c²+ 9c +1 = 0
⇒ (c+1)(10c-1)= 0
⇒ c = \(\frac{1}{10}\)[∵ c-1 is not possible]
Now, E(X) = Σxipi
= 1(c) + 2(2c) + 3(2c) + 4(3c) + 5c² + 12c² + 7(7c² + c)
= 66c²+ 30c = 66(\(\frac{1}{10}\))² + 30(\(\frac{1}{10}\))
= \(\frac{66}{100}+3=\frac{366}{100}=3.66\)
∴ E(X) = 3.66

Let I = \(\int _{ 2 }^{ 4 }{ (2x-1) } dx\)
Here a = 2, b = 4 ⇒ \(h=\frac { b-a }{ n } =\frac { 4-2 }{ n } =\frac { 2 }{ n } \)
Given f(x) = 2x - 1
\(f(a+rh)=f\left( 2+r.\frac { 2 }{ n } \right) \)
= \(2\left( 2+\frac { 2r }{ n } \right) -1\)
= \(4+\frac { 4r }{ n } -1=3+\frac { 4r }{ n } \)
\(\therefore f(a+rh)=3+\frac { 4r }{ n } \)
\(\therefore \int _{ 2 }^{ 4 }{ (2x-1) } =\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { 2 }{ n } } } \left( 3+\frac { 4r }{ n } \right) \)
= \(\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { 2 }{ n } } } \left( 3+\frac { 4r }{ n } \right) \)
= \(\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \left( \frac { 6 }{ n } +\frac { 8r }{ { n }^{ 2 } } \right) } } \)
= \(\lim _{ n\rightarrow \infty }{ \left( \frac { 6 }{ n } .\sum _{ r=1 }^{ n }{ .1 } +\frac { 8 }{ { n }^{ 2 } } .\sum _{ r=1 }^{ n }{ r } \right) } \)
= \(\lim _{ n\rightarrow \infty }{ \left( \frac { 6 }{ n } .n+\frac { 8 }{ { n }^{ 2 } } .\frac { n(n+1) }{ 2 } \right) } \)

= \(6+4\lim _{ n\rightarrow \infty }{ \left( 1+\frac { 1 }{ n } \right) } \)
= \(\therefore \int _{ 2 }^{ 4 }{ (2x-1) } dx=10\)
= \(\left[ \because \sum _{ r=1 }^{ n }{ r } =n\& \sum _{ r=1 }^{ n }{ r } n=\frac { n(n+1) }{ 2 } \right] \)

Let I = \(\int { \frac { 1 }{ { 3x }^{ 2 }+13x-10 } } dx\)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { x }^{ 2 }+\frac { 13 }{ 3 } x-\frac { 10 }{ 3 } } } \)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { x }^{ 2 }+\frac { 13 }{ 3 } x+\frac { 169 }{ 36 } -\frac { 169 }{ 36 } -\frac { 10 }{ 3 } } } \)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { \left( x+\frac { 13 }{ 6 } \right) }^{ 2 }-\frac { 289 }{ 36 } } } \)

= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { \left( x+\frac { 13 }{ 6 } \right) }^{ 2 }-{ \left( \frac { 17 }{ 16 } \right) }^{ 2 } } } \)
= \(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| } +c \right] \)
= \(\frac { 1 }{ 3 } \times \frac { 1 }{ 2\times \frac { 17 }{ 6 } } log\left| \frac { x+\frac { 13 }{ 6 } -\frac { 17 }{ 6 } }{ x+\frac { 13 }{ 6 } +\frac { 17 }{ 6 } } \right| +c\)
= \(\frac { 1 }{ 17 } log\left| \frac { x-\frac { 2 }{ 5 } }{ x+5 } \right| +c\)
= \(\frac { 1 }{ 17 } log\left| \frac { 3x-2 }{ 3(x+5) } \right| +c\)

\({ x }^{ 2 }\frac { dc }{ dx } -10xc=-10\)
\(\div { x }^{ 2 },\frac { dc }{ dx } -\frac { 10c }{ x } =\frac { 10 }{ { x }^{ 2 } } \frac { dc }{ dx } +Pc=Q\)
This is a first order linear differential equation of the form \(\frac{dc}{dx}\) + Pc = Q where
\(P=\frac { 10 }{ x }\) and \(Q=-\frac { 10 }{ { x }^{ 2 } } \)
\(\int { pdx } =-\int { \frac { 10 }{ x } =10logx=log\left( \frac { 1 }{ { x }^{ 10 } } \right) } \)
∴ I.F. = \({ e }^{ \int { pdf } }{ = }^{ { e }^{ log{ 1/x }^{ 10 } } }=\frac { 1 }{ { x }^{ 10 } } \)
∴ General solution is
\({ Ce }^{ \int { px } }=\int { Q.{ e }^{ \int { pdf } }dx+k } \)
\(\Rightarrow c.\left( \frac { 1 }{ { x }^{ 10 } } \right) =\int { -\frac { 10 }{ { x }^{ 2 } } . } \frac { 1 }{ { x }^{ 10 } } dx+k\)
\(=-10\int { \frac { 1 }{ { x }^{ 12 } } } dx+k\)
\(\Rightarrow \frac { c }{ { x }^{ 10 } } =-10\int { { x }^{ -12 } } dx+k\)
\(\Rightarrow \frac { c }{ { x }^{ 10 } } =\frac { 10 }{ 11 } \left( \frac { 1 }{ { x }^{ 11 } } \right) +k\)
\(\Rightarrow \frac { { C }_{ 0 } }{ { x }_{ 0 } } =\frac { 10 }{ 11 } \left( \frac { 1 }{ { x }_{ 0 }^{ 11 } } \right) +k\)
When c = c₀, x = x₀
\(\Rightarrow k=\frac { c }{ { x }_{ 0 }^{ 10 } } -\frac { 10 }{ 11.{ x }_{ 0 }^{ 11 } } \)
∴ The solution is
\(\Rightarrow \frac { c }{ x^{ 10 } } =\frac { 10 }{ 11 } \left( \frac { 1 }{ { x }^{ 11 } } \right) +k\left( \frac { c }{ { x }_{ 0 }^{ 10 } } -\frac { 10 }{ 11{ x }_{ 0 }^{ 11 } } \right) \)
\(\Rightarrow \frac { c }{ x^{ 10 } } -\frac { c }{ { x }_{ 0 }^{ 10 } } =\frac { 10 }{ 11 } \left( \frac { 1 }{ { x }^{ 11 } } -\frac { 10 }{ { x }_{ 0 }^{ 11 } } \right) \).

Given x²\(\frac { dy }{ dx } \)= y²+2xy
⇒ \(\frac { dy }{ dx } =\frac { { y }^{ 2 }+2xy }{ x^{ 2 } } \)
The numerator and denominator are homogeneous function of degree
∴ put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)

= v²+2v
⇒ v + x\(\frac { dv }{ dx } \) = v²+ 2v
⇒ x\(\frac { dv }{ dx } \) = v²+ 2v-v = v²+ v
Separating the variables,
\(\frac { dv }{ { v }^{ 2 }+v } =\frac { dx }{ x } \Rightarrow \frac { dv }{ v(v+1) } =\frac { dx }{ x } \)
[\(\frac { 1 }{ v(v+1) } =\frac { A }{ v } +\frac { B }{ v+1 } \)
1 = A(v+1)+Bv
put v = -1
1 = -B
put v = 0
1 = A
∴ \(\left( \frac { 1 }{ v } +\frac { 1 }{ v+v } \right) dv=\frac { dx }{ x } \)
Integrating
\(\int { \frac { dv }{ v } } -\int { \frac { dv }{ v+1 } } =\int { \frac { dx }{ x } } \)
⇒ log v - log (v + 1) - log x + log c
⇒ log\(\left( \frac { v }{ v+1 } \right) \)= log(xc)
⇒ \(\frac { v }{ v+1 } \)
Replacing v by \(\frac { y }{ x } \) we get,
\(\frac { \frac { y }{ x } }{ \frac { y }{ x } +1 } =xc\Rightarrow \frac { \frac { y }{ x } }{ \frac { x+y }{ x } } \) = xc
⇒ \(\frac { y }{ x+y } \) = xc
⇒ y = cx(x+y)  ....(1)
Given, when x = -1, y = 1
∴ 1 = c(1) (1+1) ⇒ = 2c ⇒ c = \(\frac { 1 }{ 2 } \)
∴ (1) becomes, y = \(\frac { x }{ 2 } \)(x+y)
⇒ 2y = x(x+y)

Using Newton's backward interpolation formula,
we can find y when x = 58.
∴ x + nh = x ⇒ 60 + n(5) = 58
⇒ 5n = 58 - 60 = -2
⇒ n = \(\frac{-2}{5}\)= -0.4
and y(58) = \({ y }_{ n }+\frac { n }{ n! } \nabla { y }_{ n }+\frac { n(n+1) }{ 2! } { \nabla }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n-2) }{ 3! } { \nabla }^{ 3 }{ (y }_{ n })+\) .....
The difference table is

\(y(58)=68.48+\frac { (-0.4) }{ 1! } (-6)+\frac { (0.4)(-0.4+1) }{ 2! } (2.84)+\frac { (-0.4)(-0.4+1)(-0.4+2) }{ 3! } (-1.16)+\frac { (-0.4)(-0.4+2)(-0.4+2)(-0.4+3) }{ 3! } (0.68)\)
=68.48 + (0.4)(6) + \(\frac { (-0.4)(0.6) }{ 2 } (2.84)+\frac { (-0.4)(0.6)(1.6) }{ 6 } (-1.16)+\frac { (-0.4)(0.6)(1.6)(2.6) }{ 24 } (0.68)\)
= 68.48 + 2.4 - 0.3408 + 0.07424 - 0.028288
= 70.5851052
⇒ y(58) = 70.59
∴ Hence, premium for a policy maluting at the age of 58 is 70.59.

Since e^1.75 lies at the beginning of the table, we can use Newton's forward interpolation formula
∴ x_o + nh = x ⇒ 1.7 + n(0.1) = 1.75
⇒ n(0.1) = 1.75 - 1.7 = 0.05
⇒ n = \(\frac{0.05}{0.1}\) = 0.5
\({ y }_{ x }={ y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })+.......\)
The difference table is

∴ \(y\left( { e }^{ 1.75 } \right) =5.74+\frac { 0.5 }{ 1! } (0.576)+\frac { (0.5)(0.5-1) }{ 2! } (0.06)+\frac { (0.5)(0.5-1)(0.5-2) }{ 3! } (0.007)\)
y(e^1.75) = 5.474 + 0.288 - 0.0075 + 0.0004375
= 5.7549375

Given \(C'(x)=20+\frac { x }{ 20 } \)
\(\Rightarrow \int { C'(x)dx= } \int { \left( 20+\frac { x }{ 20 } \right) dx } \)
\(C(x)=20x+\frac { { x }^{ 2 } }{ 40 } +{ k }_{ 1 }\)
Given when x = 0, C = 200
⇒ k₁ = 200
\(\therefore C(x)=20x+\frac { { x }^{ 2 } }{ 40 } +200\quad ---(1)\)
Also, R'(x) = 30
\(R(x)=\int { 30dx } +{ k }_{ 2 }=30x+{ k }_{ 2 }\)
When x = 0 R = 0 ⇒ K₂ = 0
∴ R(x) = 30x   ----(2)
Profit = Total revenue - total cost
\(=30x-20x-\frac { { x }^{ 2 } }{ 40 } -200\)
\(\therefore P=10x-\frac { { x }^{ 2 } }{ 40 } -200---(3)\)
\(\frac { dp }{ dx } =10-\frac { 2x }{ 40 } =10-\frac { x }{ 20 } \)
\(\frac { dp }{ dx } =0\)
\(\Rightarrow 10-\frac { x }{ 20 } =0\)
\(\Rightarrow 10=\frac { x }{ 20 } \Rightarrow x=200\)
\(\frac { d^{ 2 }p }{ { dx }^{ 2 } } =-\frac { 1 }{ 20 } <0\)
∴ Profit is maximum when x= 200
∴ maximum profit \(P=10(200)-\frac { { (200) }^{ 2 } }{ 40 } -200[From(3)]\)
P = 2000 -1000 - 200
2000 - 1200
P = Rs .800.
Hence, the maximum profit is Rs. 800.

Let A represents the percent of commuters who use the transit system and B represents the percent of commuters who use their own car. Transition probability matrix

Given 50% of commuters use the transit system and 50% of the commuters use their own car this year.
(i) Percentage of commuters after one year
\(\left( \cdot 5\cdot 5 \right) \left( \begin{matrix} \cdot 9 & \cdot 1 \\ \cdot 2 & \cdot 8 \end{matrix} \right) \)
= (-5\(\times\)·9+·5\(\times\).2 ·5\(\times\)·1+·5\(\times\)·8)
= (-45 + ·10 ·05 +.40)
= (-55 - 45)
A = 55% and B = 45%
(ii) Equilibrium will be reached in the long run at equilibrium, we must have
(A B)T = (A B) wher A+B = 1
\(\Rightarrow \left( \begin{matrix} A & B \end{matrix} \right) \left( \begin{matrix} \cdot 9 & \cdot 1 \\ \cdot 2 & \cdot 8 \end{matrix} \right) =\left( \begin{matrix} A & B \end{matrix} \right) \)
\(\left( \begin{matrix} \cdot 9A+\cdot 2B & \cdot 1A+8B \end{matrix} \right) =\left( \begin{matrix} A & B \end{matrix} \right) \)
Equating the corresponding entries on both sides we get,
\(\cdot 9A+\cdot 2B=A\Rightarrow \cdot 9A+\cdot 2(1-A)=A\)
[Since A + B = 1, B = 1 -A]
\(\Rightarrow \cdot 9A+\cdot 2-\cdot 2A=A\)
\(\Rightarrow \cdot 2=A-\cdot 9A+\cdot 2A\)
\(\Rightarrow \cdot 2=A(1-\cdot 9+\cdot 2)\)
\(\Rightarrow \cdot 2=A=(\cdot 3)\)
\(\therefore\) 67% of the commuters will be using the transit system in the long run.

Given that MR \(=\frac { { e }^{ x } }{ 100 } +x+{ x }^{ 2 }\)
\(\int { MR } =\int { \left( \frac { { e }^{ x } }{ 100 } +x+{ x }^{ 2 } \right) } dx\)
\(R=\frac { { e }^{ x } }{ 100 } +\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +k\)
When x=0, R=0
\(\Rightarrow 0=\frac { { e }^{ 0 } }{ 100 } +0+0+k\)
\(k=-\frac { 1 }{ 100 } [\because { e }^{ 0 }=1]\)
∴ Revenue \(R=\frac { { e }^{ x } }{ 100 } +\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { 1 }{ 100 } \)

Let the three numbers be x, y and z respectively
Given
x + y + z = 6
x + 2z = 7
3x + y + z = 12

The last equivalent matrix is in echelon form
\(\rho (A)=3\) and \(\rho (A,B)=3\)
\(\therefore \rho (A)=\rho (A,B)=3=Numberofunknowns\)
\(\therefore\) The system is consistent and has unique solution.
To find the solutions, let us rewrite the echelon form into matrix form
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & -4 \end{matrix}\begin{matrix} x \\ y \\ z \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 6 \\ 1 \\ -8 \end{matrix} \right) \)
x + y + z = 6    ..(1)
-y + z = 1
-4z = -8
From (3),\(-4z=-8\Rightarrow z=\cfrac { -8 }{ -4 } =2\)
Substituting z = 2 in (2) we get
\(-y+2=1\Rightarrow -y=1-2\Rightarrow -y=-1\)
\(\Rightarrow y=1\)
Substitutingy = 1 andz = 2 in (1) we get
\(x+1+2=6\Rightarrow x+3=6\Rightarrow x=6-3\)
\(\Rightarrow x=3\)
Hence, the numbers are 3, 1, 2.

\(\Delta =\left| \begin{matrix} 2 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{matrix} \right| =2\)
\(\left| \begin{matrix} 1 & 1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -2 \end{matrix} \right| \)
= 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

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.

Using interpolation, find the value of f(x) when x =1
Here the intervals are unequal. By Lagrange's interpolation formula, we have
x₀=3, x₁ = 7, x₂ = 11, x₃ = 19
y₀ = 42, y₁ = 43, y₂ = 47, y₃ = 60 and x = 15
∴ Y =f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (15-7)(15-11)(15-19) }{ (3-7)(3-11)(3-19) } \times 42+\frac { (15-3)(15-7)(15-19) }{ (11-3)(11-7)(11-19) } \times 43+\frac { (15-3)(15-7)(15-19) }{ (11-3)(11-7)(11-19) } \times 47+\frac { (15-3)(15-7)(15-11) }{ (19-3)(19-7)(19-11) } \times 60\)

= \(\frac{21}{2}-43+70.5+15\)
= 10.5 - 43 + 70.5 + 15
y= 53
Hence when x = 15, f(x) = 53.

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
∴ 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

\(\frac { dy }{ dx } \)+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

\(\overline {\overline{X}} = \frac {5.10+4.98+5.02+4.96+4.96+5.04+4.94+4.92+4.92+4.98}{10}\)
\(\overline {\overline{X}}\) = \(\frac {49.82}{10} = 4.982\)
\(\overline {\overline{R}} = \frac {0.3+.4+.2+.4+.1+.1+.8+.5+.3+.5}{10}\)
\(\overline {R}\) = \(\frac {3.6}{10}\) = 0.36
The control limits of mean chart are
UCL  = \(\overline {\overline{X}} + A_{2} \overline{R}\)
= 4.982 + .577(.36)
= 4.982+ .208 = 5.19
CL = 4.982
LCL =  \(\overline {\overline{X}} - A_{2} \overline{R}\)
= 4.982 - .208 = 4.774
The control limits of R-chart are
UCL = D₄\(\overline {R}\) = 2.114 (.36) = 0.761
CL = \(\overline {R}\)  = .36
LCL = D₃\(\overline {R}\)  = 0

Mean Chart

 Range chart

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

\(x\frac { dC }{ dx } =\frac { 3 }{ x } -C\)
\(x\frac { dC }{ dx } =\frac { 3 }{ x^2 } -\frac {C}{x}\)
\(\frac { dC }{ dx } +\frac { C }{ x } =\frac { 3 }{ { x }^{ 2 } } \)
i.e., \(\frac { dC }{ dx } +\frac { C }{ x } =\frac { 3 }{ { x }^{ 2 } } \)
It is of the form \(\frac { dC }{ dx } +PC=Q\)
Here, \(P=\frac { 1 }{ x } ,Q=\frac { 3 }{ { x }^{ 2 } } \)
ഽPdx = ഽ\(\frac 1 x\)dx = log x
I.F = eഽ^pdx = e^{log x} = x
The Solution is
C(I.F) = ഽQ(I.F)dx + k where k is constant
Cx = ഽ\(\frac{3}{x^2}\)xdx + k
= 3ഽ\(\frac 1 x\) dx + k
Cx = 3log x + k  (1)
Given C = 2 When x = 1
(1) ⇒ 2 × 1 ⇒ k =2
∴ The relationship between C and x is
Cx = 3 log x + 2

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
\({ P }_{ 01 }\times { Q }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { 537\times 742\times 613\times 742 }{ 442\times 613\times 442\times 537 } \right) } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { 742\times 742 }{ 442\times 442 } \right) } =\frac { 742 }{ 442 } \Rightarrow { P }_{ 01 }\times { P }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)

Sample size n = 100
Sample mean \(\bar { X } \)= 72
Population mean μ = 76
Population standard deviation σ = 8
Null Hypotheses H₀:
μ = 76(i.e., There is no Significant difference between the state scores and the national scores)
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 72-76 }{ \frac { 8 }{ \sqrt { 100 } } } =\frac { -4 }{ \frac { 8 }{ 10 } } =\frac { -4 }{ 8 } =-5\)
\(\Rightarrow |Z|=5\)
\({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
\(Z>{ Z }_{ \frac { \alpha }{ 2 } }i.e.,\ 5>1.96\)
Inference : Since \(Z>{ Z }_{ \frac { \alpha }{ 2 } }\)  at 5% level of significance, the null hypothesis H₀ is rejected. 
Hence, we conclude that there is significant difference between the state scores and the national scores.

Given Sample size n = 400
Sample mean \(\\ \bar { X } =67.47\)inches
Population mean μ = 67.39 & σ = 1.30 inches)
Null Hypotheses H_o:
μ = 67.39 inches (i.e., the sample has been drawn from the population with μ = 67.39 & σ = 1.30 inches)
Alternative Hypotheses H₁:
μ ≠ 67.39 inches (two tail test)
(i.e., the sample has not been drawn from the population with μ = 67.39 & σ = 1.30 inches) The level of significance a = 5% = 0.05
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 67.47-67.39 }{ \frac { 1.30 }{ \sqrt { 400 } } } =\frac { 0.08 }{ 0.065 } =1.2308\)
\(\therefore |Z|=1.2308\)
The significant value \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here \(Z={ Z }_{ \frac { \alpha }{ 2 } }i.e.,1.2308<1.96\)
Inference: Since \({ Z }_{ \frac { \alpha }{ 2 } }\)at 5% level of significance the null hypothesis H_o is accepted.
Hence, we conclude that the sample has been drawn from the population with mean height 67.39 inches and standard deviation 1.30 inches.

Given that mean μ = 150 and standard deviation σ = 15
(i) when X = 125 thousand
\(Z=\frac { X-\mu }{ \sigma } =\frac { 125-150 }{ 15 } =-1.667\)
When X = 145 thousand
\(Z=\frac { X-\mu }{ \sigma } =\frac { 145-150 }{ 15 } =-0.33\)
Area between Z = 0 and Z = -1.67 is 0.4525
Area between Z = 0 and Z = -0.33 is 0.1293
P(–1.667 ≤ Z ≤ –0.33) = 0.4525 – 0.1293
= 0.3232
Therefore the number of branches having sales between 125 thousand and 145 thousand is 550 × 0.3232 = 178

(ii) When X = 140 thousand
\(Z=\frac { X-\mu }{ \sigma } =\frac { 140-150 }{ 15 } =-0.67\)
When X = 160 thousand
\(Z=\frac { X-\mu }{ \sigma } =\frac { 160-150 }{ 15 } =0.67\)
P(–0.67 < Z < 0.67) = P(–0.67 < Z < 0) + P(0 < Z < 0.67)
= P(0 < Z < 0.67) + P(0 < Z < 0.67)
= 2 P(0 < Z < < 0.67)
= 2 × 0.2486
= 0.4972
Therefore, the number of branches having sales between Rs. 140 thousand and Rs. 160 thousand = 550 × 0.4972 = 273

P = 1/5/100 = 1/500 = 0.002 n = 10  λ = np = 0.02
\(p(x)=\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } =\frac { { e }^{ -0.02 }{ (0.02) }^{ x } }{ x! } \)
(i) Number of packets containing no defective = N p(o) = 1,00,000 × e^–0.02
= 98020
(ii) Number of packets containing one defective = N p(1) = 1,00,000 × 0.9802 × 0.02
= 1960
(iii) Number of packets containing 2 defectives = N p(2) = 20

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

\(\eta _{ d }=\frac { p+2{ p }^{ 2 } }{ 100-p-{ p }^{ 2 } } \)
\(\frac { -p }{ x } \frac { dx }{ dp } =\frac { p(2p+1) }{ 100-p-{ p }^{ 2 } } \)
\(\frac { -dx }{ x } =\frac { -(2p+1) }{ { p }^{ 2 }+p-100 } dp\)
\(\int { \frac { dx }{ x } } =\int { \frac { 2p+1 }{ { p }^{ 2 }+p-100 } } dp\)
log x = log(p² + p = 100) + log k
ஃ x = k(p² + p −100)
When x = 70, p = 5,
70 = k(25 + 5 − 100)
⇒ k = –1
Hence x = 100 − p − p²
R = px
Revenue = p(100 – p – p²)

(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

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

\(\int { { \left( \log x \right) }^{ 2 } } dx= \int { udv } \)
= uv − \(\int { } \)vdu
= x (log x)² − 2\(\int { } \) logxdx...(*)
\(=x(\log x)^{ 2 }-2\int { udv } \)
\(=x(\log x)^{ 2 }-2[uv-\int { udv } ]\)
\(=x(\log x)^{ 2 }-2[x \log x-\int { dx] } \)
\(=x(\log x)^{ 2 }-2x \log x+x+c\)
\(=x[(\log{ ) }^{ 2 }-\log{ x }^{ 2 }+2]+c\)

(i) Transition probability matrix T = \(_{ E }^{ M }\left( \begin{matrix} \overset { M }{ 0.8 } & \overset { E }{ 0.2 } \\ 0.7 & 0.3 \end{matrix} \right) \) 
After one study period, \(\left( \overset { M }{ 60\quad } \overset { E }{ 40 } \right) \)  \(_{ E }^{ M }\left( \begin{matrix} \overset { M }{ 0.8 } & \overset { E }{ 0.2 } \\ 0.7 & 0.3 \end{matrix} \right) \) =\(\left( \overset { M }{ 76\quad } \overset { E }{ 24 } \right) \)
So in the very next study period, there will be 76 students do maths work and 24 students do the English work.
After two study periods,
\(\left( \overset { M }{ 76\quad } \overset { E }{ 24 } \right) \) \(_{ E }^{ M }\left( \begin{matrix} \overset { M }{ 0.8 } & \overset { E }{ 0.2 } \\ 0.7 & 0.3 \end{matrix} \right) \) 
= (60.8+16.8 15.2+7.2)
= (77.6 22.4)
After two study periods there will be 78 (approx) students do maths work and 22 (approx) students do English work.