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


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

Since only four values are given,
(E -1)4 u₀ = 0
⇒ (E⁴ - 4E³ + 6E² - 4E + 1) u₀
= u₄ - 4 u₃ + 6 u₂ - 4 u₁ + u₀
⇒ 385 - 4(u₃) + 6(520) - 4 (556) + 560 =0
⇒ 385 - 4u₃ + 3120 - 2224 + 560 = 0
⇒ 1841 - 4u₃ = 0
⇒ 1841 = 4u₃
⇒ u₃ = \(\frac{1841}{4}\) = 460.25
∴ u₃ = 460.25

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

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

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

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

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

Person	Job	cost
A	5	1
B	1	0
C	4	2
D	3	1
E	2	5
Total cost		9

The optimal assignment (minimum) cost = Rs. 9

The given equation can be written as \(\frac { dy }{ dx } +\frac { 1 }{ { cos }^{ 2 }x } y=\frac { tanx }{ { cos }^{ 2 }x } \)
\(\frac { dy }{ dx } \) + y sec²x = tan x sec²x
It is of the form \(\frac{dy}{dx}\) + Py + Q
Here P = sec²x,Q = tanx sec²x
ഽPdx = ഽsec² x dx = tanx
I.F = eഽ^pdx = e^{tan x}
The required solution is y(I.F) = ഽQ(I.F)dx + c
ye^{tan x} = ഽtan x sec²xe^tan xdx + c
Put tan x = t
Then sec² xdx = dt
∴ ye^{tan x} = ഽte^t dt + c
= ഽtd(e^t) + c
= te^t − e^t + c
= tanx e^{tan x}−e^{tan x}+ c
ye^{tan x} = e^{tan x}(tan x − 1) + c

Given that \(MR=\frac { dy }{ dq } =\frac { { q }^{ 2 }+{ 3 }y^{ 2 } }{ 2qy } \)  (1)
Put y = vq and \(\frac { dy }{ dq } =v+q\frac { dv }{ dq } \) in (1)
Now (1) becomes
\(v+q\frac { dy }{ dq } =\frac { { q }^{ 2 }+{ 3 }v^{ 2 }{ q }^{ 2 } }{ 2yvq } \)
\(=\frac { 1+3{ v }^{ 2 } }{ 2v } \)
\(q\frac { dv }{ dq } =\frac { 1+3{ v }^{ 2 } }{ 2v } -v\)
\(=\frac { 1+3{ v }^{ 2 }-2{ v }^{ 2 } }{ 2v } \)
\(=\frac { { 1+v }^{ 2 } }{ 2v } \)
\(\frac { 2v }{ { 1+v }^{ 2 } } dv=\frac { dq }{ q } \)
On Integration
\(ഽ\frac { 2v }{ { 1+v }^{ 2 } } dv=ഽ\frac { dq }{ q } \)
log (1+ v²) = log q + log c
1+ v² = cq
Replace \(v=\frac { y }{ q } \)
\(1+\frac { { y }^{ 2 } }{ { q }^{ 2 } } \) = cq
q² + y² = c q³ (2)
Given output is 1 unit and revenue is Rs. 5
∴ (2) ⇒ 1 + 25 = c ⇒ c = 26
∴ The total revenue function is q² + y² = 26q³

Year	Quarterly production
	I	II	III	IV
2002	3.5	3.8	3.7	3.5
2003	3.6	4.2	3.4	4.1
2004	3.4	3.9	3.7	4.2
2005	4.2	4.5	3.8	4.4
2006	3.9	4.4	18.8	20.8
Quarterly Total	18.6	20.8	18.8	20.8
Average	3.72	4.16	3.76	20.8

Grand average = (3.72 + 4.16 + 3.76 + 4.16)/4 = 3.95
Seasonal Index (S.I) for I quarter = (Average of I quarter)/(Grand Average) \(\times\) 100
S.I. for I quarter = 3.72/3.95 \(\times\) 100 = 94.1772 
S.I. for II quarter = 4.16/3.95 \(\times\) 100 = 105.3165 
S.I. for III quarter = 3.76/3.95 \(\times\) 100 = 95.1899 
S.I. for IV quarter = 4.16/3.95 \(\times\) 100 = 105.3165 
Thus we obtain the average seasonal movement.

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

Commodities	Base Year		Current Year		p0q0	p0q1	p1q0	p1q1
	Price (p0)	Quantity (q1)	Price ((p0))	Quantity (q1)
Rice	15	5	16	8	75	120	80	128
Wheat	10	6	18	9	60	90	108	162
Rent	8	7	15	8	56	64	105	120
Fuel	9	5	12	6	45	54	60	72
Transport	11	4	11	7	44	77	44	77
Miscellaneous	16	6	15	10	96	160	90	150
				Total	376	565	487	709

Laspeyre’s price index number
\({ P }_{ 01 }^{ L }=\frac { \sum { { p }_{ 1 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \times 100=\frac { 487}{ 376} \times 100=129.5212\)
Paasche’s price index number
\({ P }_{ 01 }^{ P }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 1 } } } \times 100=\frac { 709 }{ 565 } \times 100=125.4867\)
Fisher’s price index number
\({ P }_{ 01 }^{ F }=\sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \times 100=\sqrt { \frac { 487\times 709}{ 376\times 565} } \times 100=127.4879\)
On an average, there is an increase of 29.52%, 25.48% and 27.48% in the price of the commodities by Laspeyre’s, Paasche’s, Fisher’s price index number respectively, when the base year compared with the current year.

\(\int _{ a }^{ b }{ f(x) } =\lim _{ n\rightarrow \infty \\ h\rightarrow 0 }{ \sum _{ r=1 }^{ n }{ h } f(a+rh) } \)
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

\(\int _{ a }^{ b }{ f(x) } dx=\lim _{ n\rightarrow \infty \\ h\rightarrow 0 }{ \sum _{ r=1 }^{ n }{ h } f(a+rh) } \)
Here a = 0, b = 1, \(h=\frac { b-a }{ n } =\frac { 1-0 }{ n } =\frac { 1 }{ n } \) and f(x) = x
Now f(a+rh) = \(f (0+ \frac{r}{h}) = f (\frac{r}{n}) = r/n\)
On substituting in (1) we have
\(\int _{ 0 }^{ 1 }{ x } dx=\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { 1 }{ n } . } \frac { r }{ n } } \)
\(=\lim _{ n\rightarrow \infty }{ \frac { 1 }{ { n }^{ 2 } } \sum _{ r=1 }^{ n }{ r } } \)
\(=\lim _{ n\rightarrow \infty }{ \frac { 1 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } } \)
\(=\lim _{ n\rightarrow \infty }{ \frac { 1 }{ { n }^{ 2 } } } .\frac { { n }^{ 2 }\left( 1+\frac { 1 }{ n } \right) }{ 2 } \)
\(=\frac { 1+0 }{ 2 } =\frac { 1 }{ 2 } \)
∴ \(\int _{ 0 }^{ 1 }{ x } dx=\frac { 1 }{ 2 } \)

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.

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

Given probability function is

Value of X = x	0	1	2	3	4	5	6	7
P(x)	0	k	2k	2k	3k	k2	2k2	7k2+k

i) Since the given function is a probability function, each ρ_i>0 and Σρ_i=1
⇒ 0+k+2k+2k+3k+k²+2k²+7k²+k = 1
⇒ 10k²+9k = 1
⇒10k²+9k-1 = 0
on factoring we get
(k+1)(10k-1) = 0
⇒k = -1 or k = \(\frac{1}{10}\)
k = -1 is not possible [each ρ_i>0]
we get k = \(\frac{1}{10}\)
ii) P(X<6) = P(X = 0)+P(X = 1)+P(X=2+P(X=3)+P(X=4)+P(X=5)
P(X<6) = 0 + k + 2k + 2k + 3k + k²
= 8k+k²
= \(8\left( \frac { 1 }{ 10 } \right) { \left( \frac { 1 }{ 10 } \right) }^{ 2 }\)
\(=\frac { 8 }{ 10 } +\frac { 1 }{ 100 } =\frac { 80+1 }{ 100 } =\frac { 81 }{ 100 } \)
\(\therefore P(X<6)=\frac { 81 }{ 100 } \)
Now P(X≥6) = P(X=6)+P(X=7)
= 2k²+7k²+k
= 9k²+k
\(=9{ \left( \frac { 1 }{ 10 } \right) }^{ 2 }+\frac { 1 }{ 10 } \)
\(=\frac { 9 }{ 100 } +\frac { 1 }{ 10 } =\frac { 9+10 }{ 100 } =\frac { 19 }{ 100 } \)
\(\therefore P(X\ge 6)=\frac { 19 }{ 100 } \)
And P(0X<5) = P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)
= k+2k+2k+3k
\(8k=8\left( \frac { 1 }{ 10 } \right) =\frac { 8 }{ 10 } \)
\( \therefore P(0\))
iii) Given P(X ≤ x) ≥ \(\frac{1}{2}\)
⇒P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)
= 0+k+2k+2k+3k
\(=8k=8\left( \frac { 1 }{ 10 } \right) =\frac { 8 }{ 10 } =\frac { 4 }{ 5 } >\frac { 1 }{ 2 } \)
\(\therefore P(X\le 4)\frac { 1 }{ 2 } \le x=4\)
∴ The minimum value of x is 4.

\(\eta _{ d }=\frac { p+2{ p }^{ 2 } }{ 100-p-{ p }^{ 2 } } \)
\(\frac { -p }{ x } \frac { dx }{ dp } =\frac { p(2p+1) }{ 100-p-{ p }^{ 2 } } \)
\(\frac { -dx }{ x } =\frac { -(2p+1) }{ { p }^{ 2 }+p-100 } dp\)
\(\int { \frac { dx }{ x } } =\int { \frac { 2p+1 }{ { p }^{ 2 }+p-100 } } dp\)
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

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

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

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

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

\(\Gamma \left( \frac { 9 }{ 2 } \right) =\frac { 7 }{ 2 } \Gamma \left( \frac { 7 }{ 2 } \right) \)
\(=\frac { 7 }{ 2 } \Gamma \left( \frac { 5 }{ 2 } \right) \)
\(=\frac { 7 }{ 2 } \times \frac { 5 }{ 2 } \times \Gamma \left( \frac { 3 }{ 2 } \right) \)
\(=\frac { 7 }{ 2 } \times \frac { 5 }{ 2 } \times \left( \frac { 3 }{ 2 } \right) \times \Gamma \left( \frac { 1 }{ 2 } \right) \)
\(=\frac { 7 }{ 2 } \times \frac { 5 }{ 2 } \times \frac { 3 }{ 2 } \times \frac { 1 }{ 2 } \sqrt { \pi } \)
\(=\frac { 105 }{ 16 } \sqrt { \pi } \)

There many ways to select 15 random samples from the given Tippet’s random number table. Since the population size is 83(two-digit number). Here the door numbers are assigned from 1 to 83. Assume that at random we first choose 2^nd column. So the first sample is 66 and other 14 samples are 74, 52, 39, 15, 34, 11, 14, 13, 27, 61, 79, 72, 35, and 60. If the numbers are above 83, choose the next number ranging from 1 to 83.

2952	6641	3992	9792	7969	5911	3170	5624
4167	9524	1545	1396	7203	5356	1300	2693
2670	7483	3408	2762	3563	1089	6913	7991
0560	5246	1112	6107	6008	8125	4233	8776
2754	9143	1405	9025	7002	6111	8816	6446

Given A =\(\left( \begin{matrix} 1 & -3 \\ 9 & 1 \end{matrix}\begin{matrix} 4 & 7 \\ 2 & 0 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & -3 \\ 0 & 28 \end{matrix}\begin{matrix} 4 & 0 \\ -34 & -63 \end{matrix} \right) { R }_{ 2 }\rightarrow { R }_{ 2 }-9{ R }_{ 1 }\)
\(-\left( \begin{matrix} 1 & -3 \\ 0 & 0 \end{matrix}\begin{matrix} 4 & 0 \\ \frac { 10 }{ 3 } & -63 \end{matrix} \right) { R }_{ 2 }\rightarrow { R }_{ 2 }+\frac { 28 }{ 3 } .{ R }_{ 1 }\)
The last equivalent matrix is in echelon form and there are 2 non-zero rows
\(\therefore \rho (A)=2\)

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

Action	States				Minimum
	(s1)	(s2)	(s3)	(s4)
A1	5	10	18	25	5
A2	8	7	8	23	7
A3	21	18	12	21	12
A4	30	22	19	15	15

Max (5,7,12,15) = 15 ஃ Action A4 is the best

Computation of four- yearly moving averages.

Year	Sales	4-yearly centered moving total	4-yearly moving Average	4-yearly centered moving Average
2001	124	---	--	--
2002	120	--	--	--
		519	129.75
2003	135	--		139.37
		540	135
2004	140	--		139.75
		578	144.50
2005	145	--		147.87
		605	151.25
2006	158	--		162.50
		635	158.75
2007	162	--		162.50
		665	166.25
2008	170	--	--	-
2009	175	--	--	-

Equation of the family of paraboles with latus rectum 4a and whose axes are parallel to the x-axis is (y- k)² = 4a(x- h)
[Where (h, k) is the centre of the parabola]
Differentiating w.r.t. 'x' we get,
2(y-k)\(\left( \frac { dy }{ dx } \right) \) = 4a(1)
⇒ 2(y-k)\(\left( \frac { dy }{ dx } \right) \) = 4a           (1)
Differentiating again w.r.t x we get,
2(y-k)\(\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) (2)\frac { dy }{ dx } \) = 0  (Product rule)
(y-k)\(\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) ^{ 2 }\) = 0 [Divided by 2]
y-k= \(\frac { -\left( \frac { dy }{ dx } \right) ^{ 2 } }{ \frac { { d }^{ 2 }y }{ dx^{ 2 } } } \)    (2)
Substituting (2) in (1) we get,
2\(\frac { -\left( \frac { dy }{ dx } \right) ^{ 2 } }{ \frac { { d }^{ 2 }y }{ dx^{ 2 } } } \left( \frac { dy }{ dx } \right) \)= 4a
⇒ \(-2\left( \frac { dy }{ dx } \right) ^{ 3 }=4a\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) \)
⇒ \(4a\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) ^{ 3 }\)= 0
\(2a\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 3 }=0\).

1. It can be kept small in size without losing its accuracy.
2. It is easy to administer, if the population under study is sub-divided.
3. It reduces the time and expenses in dividing the strata into geographical divisions, since the government itself had divided the geographical areas.

\(\int _{ 0 }^{ 1 }{ ({ e }^{ x }-{ 4a }^{ x }+2+\sqrt [ 3 ]{ x } } )dx\) 
= \({ \left[ { e }^{ x }-\frac { 4a }{ \log a } +2x+3\frac { { x }^{ \frac { 4 }{ 3 } } }{ 4 } -1 \right] }_{ 0 }^{ 1 }\)
= \(e-\frac { 4a }{ \log a } +2+\frac { 3 }{ 4 } -1+\frac { 4 }{ \log a } \)
= \(e+\frac { 4(1-a) }{ \log a } +\frac { 7 }{ 4 } \)

Probability of getting a degree p = 0.4
∴ q = 1– p
= 1 - 0.4
= 0.6
(i) P (one will be a graduate) = P(X = 1) = 5C₁ (0.4)(0.6)⁴
= 0.2592
(ii) P ( atleast one will be a graduate) = 1–P (none will be a graduate)
= 1-5C₀(P⁰)(Q)5-0
= 1-5C₀(0.4)⁰(0.6)5
= 1-0.0777
= 0.9222

We know that
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \),since f(x) is a density function
\(\int _{ 0 }^{ \infty }{ { ke }^{ -2x } } dx=1\)
\(k\int _{ 0 }^{ \infty }{ { ke }^{ -2x } } dx=1\)
\(k{ \left[ \frac { { e }^{ -2x } }{ -2 } \right] }_{ 0 }^{ \infty }=1\)
⇒ k = 2
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(=\int _{ 0 }^{ \infty }{ x{ e }^{ -2x } } dx\)
\(=2\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx } \)
\(=2\left\{ { \left[ \frac { { xe }^{ -2x } }{ -2 } \right] }_{ 0 }^{ \infty }\int _{ 0 }^{ \infty }{ \frac { { e }^{ -2x } }{ -2 } dx } \right\} \)
\((\because \int { udv=uv-\int { udv } } )\)
\(=\int _{ 0 }^{ \infty }{ { e }^{ -2x }dx } \) \(=\frac { 1 }{ 2 } \)

Given
\(MR=10+3x-{ x }^{ 2 }\)
\(\frac { dR }{ dx } =10+3x-{ x }^{ 2 }\)
\(\Rightarrow dR=(10+3x-{ x }^{ 2 })dx\)
\(\Rightarrow \int { dR } =\int { (10+3x-{ x }^{ 2 }) } dx\)
\(\Rightarrow R=10x+\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } +k\)
When x = 0, R = 0 ⇒ k = 0
\(\Rightarrow R=10x+\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 2 } }{ 3 } \)
Demand function\(P=\frac { R }{ x } =\frac { 10x+\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } }{ x } \)
P = 10 + \(\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \)

Here I(x) = 200 – 0.2x
C₁ = Rs. 3.5
T = 30
Total inventory carrying cost = C₁\(\int _{ 0 }^{ r }{ I(x) } dx=3.5\int _{ 0 }^{ 30 }{ (200-0.2x) } dx\)
= \(3.5{ \left( 200x-\frac { { 0.2x }^{ 2 } }{ 2 } \right) }_{ 0 }^{ 30 }\) = 20,685