Since the number of years is even(eight), we can equally divide the given data it two equal parts and obtain the averages of first four years and last four years.

Year	Production	Average
1990	15	\(\frac{15+11+20+10}{4}=14\)
1991	11
1992	20
1993	10
1994	15	\(\frac{15+25+35+30}{4}=26.25\)
1995	25
1996	35
1997	30

\(\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \sin x } dx={ \left[ -\cos x \right] }_{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }\)
\(=-\left( \cos\frac { \pi }{ 3 } -\cos\frac { \pi }{ 6 } \right) \)
\(=\frac { 1 }{ 2 } (\sqrt { 3 } -1)\)

\( \int \sqrt{2 x+1} \ d x=\int(2 x+1)^{\frac{1}{2}} d x\)
\(=\frac { { \left( 2x+1 \right) }^{ \frac { 3 }{ 2 } } }{ 3 } +c\)

Let \(A=\left( \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\)  [Since minimum of (2, 2) is 2]
Consider the second order minor
\(\left| \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right| =40-42\)
= \(-2\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)

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

Since only four values of f(x) are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.
(ie) Δ⁴y₀ = 0, ∴ (E−1)⁴y = 0
(E⁴ - 4E³ + 6E² - 4E +1)y₀ = 0
E⁴y₀ - 4E³y₀+ 6E²y₀ - 4Ey₀ +y₀  = 0
y₄ − 4y₃ + 6y₂ − 4y₁ + y₀ = 0
67.4 − 4y₃+ 6(54.1) − 4(4.2) + 45 = 0
240.2 = 4y₃ ∴ y₃ = 60.05

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

Here only 3 tasks can be assigned to 3 men.
Step 1: Its not necessary, since each row contains zero entry. Go to Step 2.
Step 2:

Step 3 (Assignment) :

Since each row and each columncontains exactly one assignment,all the three men have been assigned a task. But task S is not assigned to any Man. The optimal assignment schedule and total cost is

Task	Men	cost
P	1	9
Q	3	6
R	2	20
s	d	0
Total cost		35

The optimal assignment (minimum) cost = Rs. 35

yx² dx = -e^-x dy
⇒ \(\frac { { x }^{ 2 } }{ { e }^{ -x } } dx=-\frac { dy }{ y } \)
⇒ x² e^x dx =\(-\frac { dy }{ y } \)
Integrating both sides,
\(\int { { x }^{ 2 }{ e }^{ x } } dx=-\int { \frac { dy }{ y } } \)
put u = x²
u^'= 2x
u^'' = 2
dv = e^x dx
v = e^x
v₁ = e^x
v₂ = e^x
Using Bernoulli's formula,
\(\int { u } dv\) = uv-u^'v₁ + u^''v₂
∴ \(\int { { x }^{ 2 }{ e }^{ x } } \)dx = x²e^x-2xe^x+2e^x
∴ From (1), x²e^x-2xe^x + 2e^x = -log y+C
⇒ e^x(x²-2x+2)+log y = C

Separating the variables, we get,
\(\frac { dy }{ x } \)= sin 2x dx
Integrating both sides we get,
\(\int { \frac { dy }{ y } } =\int { \sin 2x } \)
⇒ log y = \(\frac { -\cos 2x }{ 1 }\)+c

\(\int { \sqrt { { x }^{ 2 }+x+1 } } dx\)
\(=\int { \sqrt { { x }^{ 2 }+x+\frac { 1 }{ 4 } -\frac { 1 }{ 4 } +1 } } dx\)
Adding and subtracting
\(\frac { 1 }{ 2 } { \left[ \text{co-effective of x} \right] }^{ 2 }\)
\(={ \left[ \frac { 1 }{ 2 } (-1) \right] }^{ 2 }=\frac { 1 }{ 4 } \)
\(=\int { \sqrt { { \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }+\frac { 3 }{ 4 } } } dx\)
\(\int { \sqrt { { \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 } } } dx\)
\(\left[ \because \int { \sqrt { { x }^{ 2 }+{ a }^{ 2 } } dx=\frac { x }{ 2 } \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } +\frac { { a }^{ 2 } }{ 2 } \log { \left| x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right| +c } \right] \)
\(=\frac { 1 }{ 2 } \left( x+\frac { 1 }{ 2 } \right) \sqrt { { x }^{ 2 }+x+1 } +\frac { 3 }{ 4(2) } \log { \left| \left( x+\frac { 1 }{ 2 } \right) +\sqrt { { x }^{ 2 }+x+1 } \right| } +c\)
\(=\frac { x+\frac { 1 }{ 2 } }{ 2 } \sqrt { { x }^{ 2 }+x+1 } +\frac { 3 }{ 8 } \log { \left| \left( x+\frac { 1 }{ 2 } \right) +\sqrt { { x }^{ 2 }+x+1 } \right| } +c\)

In a throw of a pair of dice the doublets are (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)
Probability of getting a doublet p = 6/36 = 1/6
⇒ q = 1 – p = 5/6 and also n = 4 is given
The probability of successes
 \(=\left( \begin{matrix} 4 \\ x \end{matrix} \right) { (\frac { 1 }{ 6 } ) }^{ x }\left( \frac { 5 }{ 6 } \right) ^{ 4-x }\)
Therefore the probability of 2 successes are
\(P(X=2)\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) { (\frac { 1 }{ 6 } ) }^{ 2 }\left( \frac { 5 }{ 6 } \right) ^{ 4-2 }\)
\(=6\times \frac { 1 }{ 36 } \times \frac { 25 }{ 36 } \)
\(=\frac { 25 }{ 216 } \)

Given Curve is y² = 27x³.
\({ x }^{ 3 }=\frac { { y }^{ 2 } }{ 27 } \)
\(x={ \left( \frac { { y }^{ 2 } }{ 27 } \right) }^{ \frac { 1 }{ 3 } }=\frac { { y }^{ \frac { 2 }{ 3 } } }{ 3 } \)
∴ Area \(=\frac { 1 }{ 3 } \int _{ 1 }^{ 2 }{ { y }^{ \frac { 2 }{ 3 } } } dy=\frac { 1 }{ 3 } { \left( \frac { { y }^{ \frac { 2 }{ 3 } +1 } }{ \frac { 2 }{ 3 } +1 } \right) }_{ 1 }^{ 2 }\)
\(={ \left( \frac { 1 }{ 3 } .\frac { { y }^{ \frac { 5 }{ 3 } } }{ \frac { 5 }{ 3 } } \right) }_{ 1 }^{ 2 }\)
\(=\frac { 1 }{ 3 } \times \frac { 3 }{ 5 } { \left( { y }^{ \frac { 5 }{ 3 } } \right) }_{ 1 }^{ 2 }=\frac { 1 }{ 5 } \left( { 2 }^{ \frac { 5 }{ 3 } }-{ 1 }^{ \frac { 5 }{ 3 } } \right) \)
\(=\frac { 1 }{ 5 } \left( { 2 }^{ \frac { 5 }{ 3 } }-1 \right) \)sq.units

Given y = 36 − x² and y₀ = 11
11 = 36 – x²
x² = 25
x = 5
CS = \(\int _{ 0 }^{ x }{ \text{(demand }\ \text{ function)dx–(Price×quantity demanded)}}\)
= \(\int _{ 0 }^{ 5 }{ (36-{ x }^{ 2 })dx-5\times 11 } \)
= \({ \left[ 36x-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 5 }-55\)
= \(\left[ 36(5)-\frac { { 5 }^{ 3 } }{ 3 } \right] -55\)
= \(180-\frac { 125 }{ 3 } -55=\frac { 250 }{ 3 } \)
Hence the consumer’s surplus is = \(\frac { 250 }{ 3 } \)

Given marginal cost function = \(\frac { a }{ \sqrt { ax+b } } \)
\(\Rightarrow MC=\frac { a }{ \sqrt { ax+b } } \Rightarrow \frac { dC }{ dx } =\frac { a }{ \sqrt { ax+b } } \)
\(\Rightarrow dC=\frac { a }{ \sqrt { ax+b } } dx\)
\(\Rightarrow \int { dC } =a\int { \frac { dx }{ \sqrt { ax+b } } } \)
\(\Rightarrow C=a\int { { (ax+b) }^{ \frac { -1 }{ 2 } }dx } \)
\(\Rightarrow C=\not a \frac{(a x+b)^{\frac{-1}{2}+1}}{\left(-\frac{1}{2}+1\right) \not a}+k\)
\(\Rightarrow C=\frac { { (ax+b) }^{ \frac { 1 }{ 2 } } }{ 1 } +k\)
\(\Rightarrow C=2\sqrt { ax+b } +k\) ...(1)
Since the cost of output is zero.
C = 0, when x = 0
\(\therefore (1)\rightarrow 0=2\sqrt { 0+b } +k\)
\(\Rightarrow 0=2\sqrt { b } +k\)
\(\Rightarrow k=-2\sqrt { b } \)
∴(1)becomes
\(C=2\sqrt { ax+b } -2\sqrt { b } \)

Given f'(x) = x + b, f(1) = 5 and f(2) = 13
f'(x) = x + b
\(\Rightarrow \int { f'(x)dx=\int { \left( x+b \right) dx } } \)
[ ∴ Integration is the reverse process of differentiation]
\(\Rightarrow f(0)=\frac { { x }^{ 2 } }{ 2 } +bx+c\)...(1)
Given f(1) = 5
\(\Rightarrow 5=\frac { { 1 }^{ 2 } }{ 2 } +b(1)+c\)
\(\Rightarrow 5=\frac { 1 }{ 2 } +b(1)+c\Rightarrow 5-\frac { 1 }{ 2 } =b+c\)
\(\Rightarrow \frac { 10-2 }{ 2 } =b+c\Rightarrow b+c=\frac { 9 }{ 2 } \)
\(\Rightarrow 2b+2c=9\)..(2)
\(Also\quad f(2)=13\Rightarrow 13=\frac { { 2 }^{ 2 } }{ 2 } +b(2)+c\)
⇒ 13 = 2 + 2b + c
⇒ 13-2 = 2b + c
⇒ 2b + c = 11    ---(3)
(2) - (3) ⟶ 2b + 2c = 9
-2b + -c = -11
c = -2
Substituting c = -2 in (3) we get
2b - 2 = 11⇒ 2b = 11+2 ⇒ 2b = 13
\(\Rightarrow b=\frac { 13 }{ 2 } \)
Substituting \(b=\frac { 13 }{ 2 } \), c = -2 in(1) we get,
\(f(x)=\frac { { x }^{ 2 } }{ 2 } +\frac { 13 }{ 2 } x-2\)

The equations are
2x + 3y = 7
3x + 5y = 9
Here \(\triangle =\left| \begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix} \right| =1\)
\(\neq 0\)
\(\therefore \) we can apply Cramer’s Rule
Now \({ \triangle }_{ x }=\left| \begin{matrix} 7 & 3 \\ 9 & 5 \end{matrix} \right| =8\)  \({ \triangle }_{ y }=\left| \begin{matrix} 2 & 7 \\ 3 & 9 \end{matrix} \right| =-3\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }_{ X } }{ \triangle } =\frac { 8 }{ 1 } =8\)   \(y=\frac { { \triangle }_{ y } }{ \triangle } =\frac { -3 }{ 1 } =-3\)
\(\therefore \) Solution is x = 8, y = −3

Let x be the marks and y be the number of students
By converting the given series into cumulative frequency distribution, the difference table is as follows.

x	y	\(\Delta y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 4 }y\)
Less than 40	31
		42
50	73		9
		51		–25
60	124		-16
		35		12
70	159		-4
		31
80	190

\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
To find y at x = 45
\(\therefore\) x₀+nh = 45 , x0 = 40, h = 10 \(\Rightarrow n=\frac { 1 }{ 2 } \)
y_{(x = 45)} = \(31+\frac { 1 }{ 2 } \times 42+\frac { \frac { 1 }{ 2 } \left( \frac { -1 }{ 2 } \right) }{ 2 } (9)+\cfrac { \frac { 1 }{ 2 } \left( \frac { -1 }{ 2 } \right) \left( \frac { -3 }{ 2 } \right) }{ 6 } \times \left( -25 \right) +\frac { \frac { 1 }{ 2 } \left( \frac { -1 }{ 2 } \right) \left( \frac { -3 }{ 2 } \right) \left( \frac { -5 }{ 2 } \right) }{ 24 } \times \left( -37 \right) \)
= \(31+21-\frac { 9 }{ 8 } -\frac { 25 }{ 16 } -\frac { 37\times 15 }{ 384 } \)
= 47.867 ≅ 48

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

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

Sample size n = 400 stores
Sample mean \(\bar x\) = 153.7 bars
Sample SD s = 17.2 bars
Population mean m = 146.3 bars
Since population SD is unknown we can consider the sample SD s = \(\sigma\)
Null Hypothesis :
The advertising campaign is not successful i.e, H₀: \(\mu\) = 146.3 
(There is no significant difference between the mean weekly sales of soap bars in department stores before and after advertising campaign)
Alternative Hypothesis H₁:
\(\mu\) >143.3 (Right tail test). The advertising campaign was successful
Level of significance \(\sigma\) = 0.05
Test statistic :
\(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
\(Z=\frac { 153.7-146.3 }{ \frac { 17.2 }{ \sqrt { 400 } } } \)
\(=\frac { 7.4 }{ 0.86 } =8.605\)
\(\therefore\) Z = 8.605
Comparing the calculated value Z=8.605 and the significant value or table value \({ Z }_{ \alpha }=1.645.\) We get 8.605 > 1.645 
Inference:
Since, the calculated value is much greater than table value i.e., Z > \({ Z }_{ \alpha }\), it is highly significant at 5% level of significance.
Hence we reject the null hypothesis H₀ and conclude that the advertising campaign was definitely successful in promoting sales.

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