\(\bar{\bar{X}}\) = \(\frac{11.2 + 11.8 + 10.8 + 11.6 + 11.0+ 9.6 + 10.4 + 9.6 + 10.6 + 10.0}{10}\)
\(=\frac{106.6}{10}=10.66\)
\(\bar{R}=\frac{7+4+8+5+7+4+8+4+7+9}{10}\)
\(=\frac{63}{10}=6.3\)
Control limits for mean chart
UCL = \(\bar{\bar{X}}\) +  A_{2^\(\bar{R}\)}
= 10.66 + .577(6.3) = 14.295
CL = \(\bar{\bar{X}}\) = 10.66
Control limits for R-chart
UCL =   D_{2^\(\bar{R}\)} = 2.115 \(\times\) 6.3
= 13.324
CL = _^\(\bar{R}\) = 6.3
LCL = D₃\(\bar{R}\) = 0

Cost of living index (C.L.I) = \(\frac{\Sigma PV}{\Sigma V}\)
= \(\frac{15800}{100}\) = 158
Hence, there is 58% increase in cost of living in 2003 compared to 2000.

Grand average = \(\frac{67 + 61 + 63.33 + 63.37}{4}\)
= \(\frac{255}{4}=63.75\)
Seasonal index (S.I) = \(\frac{Quarterly average}{Grand average}\times100\)
Hence, S.I for I quarter = \(\frac{67}{63.75}\times100\) = 105.01
S.I for II quarter = \(\frac{61}{63.75}\times100\) = 95.68
S.I for III quarter = \(\frac{63.33}{63.75}\times100\) = 99.35
S.I for IV quarter = \(\frac{63.67}{63.75}\times100\) = 99.87

Let the straight line of best fit be y = ax + b.
The normal equations are
Σy = a Σx + nb
Σxy = a Σx² + bΣx
⇒ 10a + 7b    = 16.9 ....(1)
30a + 10b  = 47.4.... (2)

Substituting b = 0.3 in (2) we get
30a + 3 = 47.4 ⇒ 30a = 44.4
\(a=\frac{44.4}{30}=1.48\)
∴ The straight line trend is y = 1.48x + 0.3

Given sample size n = 64
Sample mean \(\bar { x } \) = 5950
Population mean μ = 6000
Population variance σ² = 32400
Population Standard deviation σ =\(\sqrt { 32400 } \) =180
Null hypothesis: H₀: population mean μ = 6000 Alternative hypothese : H₁: μ ≠ 6000
The test statistic, Z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
=\(\frac { 5950-6000 }{ \frac { 180 }{ \sqrt { 64 } } } =\frac { -50 }{ \frac { 180 }{ 8 } } \)
= -50\(\left( \frac { 8 }{ 180 } \right) \) = -2.2
|z| = 2.2
As the level of significance is α =0.001, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58 Here |z| < z_α
Inference: Null hypotheses H₀ is accepted.
Hence, we can conclude that the sample of 64 persons with a mean income of Rs.5950 belong to the population.

Given population size N = 1500
Sample size n = 150
Sample proportion p =\(\frac { 10 }{ 150 } \)=0.07
∴ q = 1 - P = 1 - 0.07 = 0.93
Standard error of sample proportion =\(\sqrt { \frac { pq }{ n } } \)
=\(\sqrt { \frac { (0.07)(0.93) }{ 150 } } \)
S.E_(p) = 0.02 

Sample size n = 1000
Sample proportion of T.V. viewers
p=\(\frac { 320 }{ 1000 } \)=0.32
∴ q = 1 - P = 1 - 0.32 =0.68
Standard error = \(\sqrt { \frac { pq }{ n } } =\sqrt { \frac { (.32)(.68) }{ 1000 } } \)
S.E = 0.0147

Let p be the probability of happening of an event.
Given p = 0.002 = \(\frac { 2 }{ 1000 } \)
Also n = 1000
∴ Mean = np = \(1000\times \frac { 2 }{ 1000 } \) = 2
Hence, X follows Poisson distribution with
P(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
∴ P( event happens exactly twice)
= P(X = 2)
= \(\frac { e^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { e^{ -2 }.{ (2 }^{ 2 }) }{ 2 } \)
= e^-2(2) = 2(0.1353) = 0.2706
∴ P(X = 2) = 0.2706

The probability function for the normal distribution is
f(x) = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } e^{ -\frac { 1 }{ 2 } \left( \frac { x-\mu }{ \sigma } \right) ^{ 2 } }\), -∞ Given f(x) = \(C{ e }^{ -\left( \frac { x-100 }{ 50 } \right) ^{ 2 } }\)
= \(C{ e }^{ -\frac { 1 }{ 2 } \left( \frac { x-100 }{ 25 } \right) ^{ 2 } }\)
= \(C{ e }^{ -\frac { 1 }{ 2 } \left( \frac { x-100 }{ 5 } \right) ^{ 2 } }\) ...(2)
Comparing (1) and (2), μ =100, σ = 5 and
C = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } =\frac { 1 }{ 5\sqrt { 2\pi } } \)
∴ C = \(\frac { 1 }{ 5\sqrt { 2\pi } } \).

Given n = 10
Probability of selecting a defective pen = p
= \(\frac { 10 }{ 50 } =\frac { 1 }{ 5 } \)
q = 1-p = \(1-\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
∴ P(X = x) = ⁿC_x p^xq^n-x
P (at least one pen is defective)
= P(X≥1) = 1-P(X<1)
= 1-P(X = 0)
= 1-10C₀ \(\left( \frac { 1 }{ 5 } \right) ^{ 0 }\left( \frac { 4 }{ 5 } \right) ^{ 10 }\)
= 1-\(\frac { { 4 }^{ 10 } }{ { 5 }^{ 10 } } \)

Given n = 10, P(H) = \(\frac { 1 }{ 2 } \) ⇒ p =\(\frac { 1 }{ 2 } \)
q=1-p = \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
P(X = x) = nC_x p^xq^n-x
∴ P(X = 5) = 10C₅ p⁵q⁵
= \(\frac { 10\times 9\times 8\times 7\times 6 }{ 5\times 4\times 3\times 2\times 1 } \left( \frac { 1 }{ 2 } \right) ^{ 5 }\left( \frac { 1 }{ 2 } \right) ^{ 10 }\)
= \(\frac { 6\times 7\times 6 }{ 2^{ 10 } } \)
= \(\frac { 2\times 3\times 7\times 2\times 3 }{ 2^{ 10 } } =\frac { 63 }{ { 2 }^{ 8 } } \)
= \(\frac { 63 }{ 256 } \).

Since X is a binomial variate X ~ B (5, p)
n = 5 and P(X = x) = nC_x p^x q^n-x
Given P[X = 2] = P[X = 3]
⇒ 5C₂ p²q³ = 5C₃ p³q²

⇒ q=p
we know p+q = 1 ⇒ p+p =1
⇒ 2p = 1⇒ p =\(\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \).

Given mean 20 ⇒ np = 20
S.D = 4 ⇒ \(\sqrt { npq } \) = 4
∴ \(\frac { npq }{ np } =\frac { 16 }{ 20 } \Rightarrow q=\frac { 4 }{ 5 } \)
P = 1-q =\(1-\frac { 4 }{ 5 } =\frac { 1 }{ 5 } \)
Substitutingp and q in npq = 16, we get
\(n\times \frac { 1 }{ 5 } \times \frac { 4 }{ 5 } \) =16
n = \(\frac { 16\times 5\times 5 }{ 4 } \)  = 100
∴ Number of events = 100

Given n = 9 and mean = 6 ⇒ np = 6
9p = 6 ⇒ \(\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)
∴ q=1-p = \(1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
Variance = npq = \(9\times \frac { 2 }{ 3 } \times \frac { 1 }{ 3 } \) = 2

Mean = E(X)=\(\int _{ -\infty }^{ \infty }{ x.f(x)dx=\int _{ -12 }^{ 12 }{ x.\left( \frac { 1 }{ 24 } \right) } dx } \)
\(=\frac { 1 }{ 24 } \int _{ -12 }^{ 12 }{ x.dx } \)
\(=0[\because \int _{ -a }^{ a }{ f(x)dx=0 } when\ f(x)\ is\ an\ odd\ function]\)
\(\therefore E(X)=0\)

Let X denote the amount
∴ X is a random variable. taking the values 10 and -5 when 3 coins are tossed, sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH,TTT}
∴ P(X = 10) = P (getting 3 heads or 3 tails)
\(=\frac { 2 }{ 8 } =\frac { 1 }{ 4 } \)
P(X = -5) = p(getting 1head or 2heads)
\(=\frac { 6 }{ 8 } =\frac { 3 }{ 4 } \)
∴ Probability distribution function is

∴ Expected gain E(X) = Σx_ip_i = 10(\(\frac{1}{4}\))-5(\(\frac{3}{4}\))
\(=\frac { 10 }{ 4 } -\frac { 15 }{ 4 } =-\frac { 5 }{ 4 } =-1.25\)
∴E(X) = -1.25 [A loss of Rs. 1.25]

Let X be a random variable. That denote the mark obtained by a student for answering a question.
∴ X can take values 1 and -\(\frac{1}{2}\)
∴ P(X = 1) = P (answering a question correctly) 
= \(\frac{1}{4}\)
P(X = -\(\frac{1}{2}\)) = P(answering a question wrongly)
=\(1-\frac{1}{4}=\frac{3}{4}\)
∴ Probability distribution function is

\(\therefore E(x)=\sum { xp(x)=1(\frac { 1 }{ 4 } )-\frac { 1 }{ 2 } \left( \frac { 3 }{ 4 } \right) =\frac { 1 }{ 4 } -\frac { 3 }{ 8 } } \)
\(=\frac { 2-3 }{ 8 } =-\frac { 1 }{ 8 } \)
∴ Expectation of mark for answering a single question is -\(\frac{1}{8}\)
∴ Expectation of mark for answering 120 questions = 120(-\(\frac{1}{8}\)) = -15.

Since f(x) is a probability density function,
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 0 }^{ \infty }{ { ce }^{ -ax }dx=1 } \Rightarrow C\int _{ 0 }^{ \infty }{ { e }^{ -ax }dx=1 } \)
\(\Rightarrow c{ \left[ \frac { { e }^{ -ax } }{ -a } \right] }_{ 0 }^{ \infty }=1\Rightarrow \frac { -c }{ a } [{ e }^{ -\infty }-{ e }^{ 0 }]\)
\(\Rightarrow \frac { -c }{ a } [0-1]=1\quad [\because { e }^{ -\infty }=0,{ e }^{ 0 }=1]\)
\(\Rightarrow \frac { c }{ a } =1\Rightarrow C=a\quad \therefore C=a\)

Clearly f(x) ≥0 for all real values of x
\(\therefore \int _{ -\infty }^{ \infty }{ f(x)dx } =\int _{ 0 }^{ 3 }{ \frac { 2x }{ 9 } dx=\frac { 2 }{ 9 } \int _{ 0 }^{ 3 }{ xdx } } \)
\(=\frac { 2 }{ 9 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 3 }=\frac { 2 }{ 9 } \left[ \frac { 9 }{ 2 } -0 \right] =\frac { 2 }{ 9 } \times \frac { 9 }{ 2 } =1\)
∴ f(x) is a probability density function.

A bag contains 2 bad eggs and 8 good eggs
∴ Total number of eggs = 10
We are going to select 3 eggs, out of that 2 must be bad eggs.
∴ Required probability \(=\frac { { 2C }_{ 2 }\times { 8C }_{ 1 } }{ 10{ C }_{ 3 } } =\frac { 1\times 8 }{ \frac { 10\times 9\times 8 }{ 3\times 2\times 1 } } \)
\(=\frac { 1\times 8\times 3\times 2\times 1 }{ 10\times 9\times 8 } =\frac { 1 }{ 15 } \)
∴ Probability of getting two bad eggs = \(\frac{1}{15}.\)

When a die is rolled, sample space 
S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6
∴P(X = 0) = Probability of getting an odd number = \(\frac{3}{6}\)[∵ Their are 3 favourable events]
= \(\frac{1}{2}\)
Thus, the probability distribution of the random variable X is given by

P(X = 0) + P(X = 1) + P(X = 2)
= 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence the given distribution of probabilities is not a probability distribution.

Max (9, 8, 5) = 9
∴ d₁ is the best strategy using maximin principle.

Min (45, 70, 53) = 45
∴ Choosing Bonds is the best decision under minimax principle.

Here total availability = 150 + 100 + 250 = 500
total requirement = 50 + 150 + 300 = 500
∴ Total availability = total requirement
∴ The given problem is a balanced transportation problem
Hence, there exists a feasible solution to the given problem
I - allocation:

[∵ least cost is 4 & min (50,150) = 50]
II - allocation:

[∵ least cost is 6 & min (150, 250) = 150]
III - allocation:

[∵ least cost is 8 & min (300, 100) = 100]
IV - allocation:

[∵ least cost is 9 & min (200,100) = 100]
V - allocation:

[∵ min (100, 100) = 100]
Thus, the allocations are

∴ The transportation schedule is
O₁ → D₁, O₁ → D₃, O₂ → D₃, O₃ → D₂, O₃ → D₃
Hence, the total transportation cost is
= 50(4) + 100(8) + 100(11) + 150(6) + 100(9)
= 200 + 800 + 1100 + 900 + 900
= Rs. 3900

Here, total supply = 10 + 5 + 3 = 18
total demand = 5 + 4 + 6 + 3 = 18
∴ Total supply = total demand
∴ The given problem is a balanced transportation problem.
∴ We can find an initial basic feasible solution to the given problem.
I - allocation:

[∵ min (5, 10) = 5]
II - allocation:

[∵ min (4, 5) = 4]
III - allocation:

[∵ min (1, 6) = 1]
IV - allocation:

[∵ min (5, 5) = 5]
V - allocation:

[∵ min (3, 3) = 3]
Thus, the allocations are

∴ The transportation schedule is
1 → A, 1 → B, 1 → C, 2 → C, 3 → D
Hence, the total transportation cost
= 5(3) + 4(1) + 1(7) + 5(5) + 3(2)
= 15 + 4 + 7 + 25 + 6 = Rs. 57

Given \(\int _{ 0 }^{ a }{ { 3x }^{ 2 } } dx=8\)

⇒ \({ \left[ { x }^{ 3 } \right] }_{ 0 }^{ a }=8\)
⇒ a³ - 0 = 8
⇒ a³ = 8
⇒ a³ = 2³
⇒ a = 2
∴ a = 2

 \(\int { \frac { 2+3cosx }{ { sin }^{ 2 }x } } dx\) = \(\int { \frac { 2 }{ { sin }^{ 2 }x } } dx\) + \(\int { \frac { 3cosx }{ { sin }^{ 2 }x } } dx\)
= 2 \(\int { \frac { 1 }{ { sin }^{ 2 }x } } dx\) + 3 \(\int { \frac { cosx }{ { sin }x } .\frac { 1 }{ sinx } } dx\)
= 2 ∫ cosec² x dx + 3 ∫ cot x cosec x dx
= -2 cot x - 3 cosec x + c

∫ tan²x dx = ∫ (sec² x-1) dx
[∵ 1 + tan² x = sec² x]
= ∫ sec²x dx - ∫ 1 dx
= tan x - x +c.

We know that log_a^x = x
∴  ∫ a^3log_ax dx = ∫ a^log_ax³ dx = ∫ x³ dx
= \(\frac { { x }^{ 4 } }{ 4 } +c\)

The auxiliary equation is m² - 6m + 25 = 0
Here a = 1, b = -6, c = 25
∴ m = \(\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } =\frac { 6\pm \sqrt { 36-4(1)(25) } }{ 2 } \)
= \(\frac { 6\pm \sqrt { 36-100 } }{ 2 } =\frac { 6\pm \sqrt { -64 } }{ 2 } =\frac { 6\pm 8i }{ 2 }\)

= 3 ± 4i
∴ α = 3, β = 4
Complementary function CF is e^ax
[A cosβx + B sinβx]
⇒ CF = e^3x[A cos4x + B sn4x]
∴ The general solution is e^3x
[A cos4x + B sin4x]

The auxiliary equation is 3m² - 5m + 2 = 0
⇒ (m - 1)(3m - 2) = 0
⇒ m = 1, \(\frac { 2 }{ 3 } \)
The roots are real and different
∴ Complementary function CF is Ae^x + \({ Be }^{ \frac { 2 }{ 3 } x }\)
∴ The general solution is y = Ae^x + \({ Be }^{ \frac { 2 }{ 3 } x }\).

Given \(\frac { dC }{ dq } =a-\frac { c }{ q } \)
⇒ \(\frac { dC }{ dq } +\frac { C }{ q } \) = a
The given differential equation is of the form
\(\frac { dC }{ dq } \)+PC = Q where
P = \(\frac { 1 }{ q } \) and Q = a
\(\int { p } dq=\int { \frac { 1 }{ q } } \)
Integrating factor I.F = e^{log q} =q
∴ The solution is C y\(e^{ \int { P } dq }=\int { Q } e^{ \int { P } dq }\)+C
⇒ C(q) = \(\int { a } .qdq+C\)
⇒ C.q = a\(\left( \frac { { q }^{ 2 } }{ 2 } \right) \)+C
⇒ 2Cq = aq² + K  where K = 2C.

The given equation is of the form \(\frac { dy }{ dx } \)+ py = Q
where P = a and Q = e^x
\(\int { P } dx=\int { a } dx\) = ax
Integrating factor (L.F.) = \(e^{ \int { P } dx }\) = eax
∴ The solution is y \(e^{ \int { P } dx }=\int { Q } e^{ \int { P } dx }dx\)+C
⇒ y.e^ax =\(\int { { e }^{ x }.{ e }^{ ax } } dx+C\)
⇒ y.e^ax = \(\int { e^{ (a+1)x }dx+C } \)
⇒ y.e^ax = \(\frac { { e }^{ (a+1)x } }{ a+1 } \)+C

x dy = -y dx
Separating the variables we get
\(\frac { dy }{ y } =-\frac { dx }{ x } \)
Integrating, \(\int { \frac { dy }{ y } } =-\int { \frac { dx }{ x } } \)
⇒ log y = -log x + log C
⇒ log y = log\(\left( \frac { C }{ x } \right) \Rightarrow y=\frac { C }{ x } \) ⇒ xy = C.

Equation of family of rectangular hyperbolas whose asymptotes are the Co-ordinate axis is
xy - c²
Differentiating w.r.t. 'x' we get,
x.\(\frac { dy }{ dx } \)+y(1) = 0
⇒ x\(\left( \frac { dy }{ dx } \right) \)+y(1) = 0 which is the required differential equation.

Given y = mx + \(\frac { a }{ m } \) ...(1)
Differentiating w.r.t. 'x' we get,
\(\frac { dy }{ dx } \) = m(1)+0 ⇒ m = \(\frac { dy }{ dx } \) ...(2)
Substituting (2) in (1) we get,
y = \(\left( \frac { dy }{ dx } \right) x+\frac { a }{ \frac { dy }{ dx } } \Rightarrow y=\frac { \left( \frac { dy }{ dx } \right) ^{ 2 }x+a }{ \left( \frac { dy }{ dx } \right) } \)
⇒ y\(\left( \frac { dy }{ dx } \right) =x\left( \frac { dy }{ dx } \right) ^{ 2 }\)+ a which is the required differential equation.

We know  f(x), ∇f(x + h), ∇f(x + 2h),... are the first order differences of f(x).
= f(x) - 2f(x - h) + f(x - 2h)
Consider ∇²f(x) = ∇[∇f(x))]
= ∇[f(x) - f(x - h)]
= ∇ f(X) - ∇ f(x - h)
= [f(x) - f(x - h)] - [f(x - h) - f(x - 2h)]
= f(x) - f(x - h) - f(x - h) - f(x - 2h)
∴ ∇² f(x), ∇² f(x + h), ∇² f(x + 2h) ... are the second order differences of f(x).

Since three values of f(x) are given, we assume that the polynomial is of degree two.
⇒ Δ³(f(x₀)) = 0
⇒ ∆³(y_o) = 0
⇒ (E - 1)³ y_o= 0
⇒ (E³ - 3E² + 3E - 1) y_o= 0
⇒ y₃- 3y₂+ 3y₁ - y_o= 0
⇒ 157 - 3 (126) + 3y₁ - 100 = 0
⇒ y₁ = 107
∴ The missing term is 107.

Given supply function is P = x + x + 3 and X_o = 4
∴ P_o = 4² + 4 + 3
= 16+ 4 + 3 = 23
∴ p₀x₀ = 23(4) = 92
Producer's surplus
\(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x0 }{ g(x)dx } \)
\(=92-\int _{ 0 }^{ 4 }{ \left( { x }^{ 2 }+x+3 \right) dx } \)
\(=92-{ \left( \frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 2 } }{ 2 } +3x \right) }_{ 0 }^{ 4 }\)
\(=92-\left( \frac { { 4 }^{ 3 } }{ 3 } +\frac { { 4 }^{ 2 } }{ 2 } +3(4) \right) \)
\(=92-\left( \frac { 64 }{ 3 } +8+12 \right) \)
\(=92-\frac { 64 }{ 3 } -20=72-\frac { 64 }{ 3 } \)
\(=\frac { 216-64 }{ 3 } \)
\(PS=\frac { 152 }{ 3 } \) units

Given demand function is p = 25 - x - X²
and p₀ = 19
⇒ 19 = 25-x-x²
⇒ x² + x - 6 = 0
⇒ (x + 3) (x - 2) = 0
⇒ x = -3 or x = 2
Since x cannot be negative x_o = 2
p_o x_o = 19(2) = 38
\(CS=\int _{ 0 }^{ 2 }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ 2 }{ \left( 25-x-{ x }^{ 2 } \right) dx-38 } \)
\(={ \left( 25x-\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ 2 }-38\)
\(=25(2)-\frac { 4 }{ 2 } -\frac { 8 }{ 3 } -38\)
\(=50-2-\frac { 8 }{ 3 } -38\)
\(=10-\frac { 8 }{ 3 } =\frac { 30-8 }{ 3 } \)
\(CS=\frac { 22 }{ 3 } \) units

Given \(MC=\frac { 100 }{ x } \)
\(\\ \int { MC } =\int { \frac { 100 }{ x } } \)
⇒ C = 100 log x + k
Given C(16) =100 ⇒ When x = 16,
C = 100
∴ 100 = 100 log 16 + k
⇒ k = 100-100 log 16
C = 100 log x + 100 - 100 log 16
= 100 (log x - log16 + 1)
\(C=100(log\left( \frac { x }{ 16 } \right) +1)\)

Given C'(x) = 85 + \(\frac{375}{x^2}\).
We know C(x) ഽC'(x) + k
The cost of producing 10 incremental units after 15 units have been produced
\(C'(x)=85+\frac { 375 }{ { x }^{ 2 } } \)
\(C(x)=\int { C'(x) } dx\)
\(\int _{ 15 }^{ 25 }{ C'(x) } dx\)
\(\int _{ 15 }^{ 25 }{ \left( 85+\frac { 375 }{ { x }^{ 2 } } \right) } dx\)
\({ \left[ 85+\frac { 375 }{ { x } } \right] }_{ 15 }^{ 25 }\)
\(\left( 85(25)-\frac { 375 }{ 25 } \right) -\left( 85(15)-\frac { 375 }{ 15 } \right) \)
= (2125 - 15) - (1275 - 25)
= 2110 - 1250 = Rs. 860

Given MR = 9 - 6x² + 2x
⇒ഽMR=ഽ(9 - 6x² + 2x)sx
\(\Rightarrow R=9x-\frac { { 6x }^{ 3 } }{ 3 } +\frac { { 2x }^{ 2 } }{ 2 } +k\)
⇒ R = 9x - 2x³ + x² + k
When x = 0, R = 0 ⇒ k = 0
∴ R = 9x - 2x³ + x²

Given MC = 3x² - 2x + 8
⇒ ഽMC = ഽ(3x² - 2x + 8)dx
\(\Rightarrow C=\frac { { 3x }^{ 3 } }{ 3 } -\frac { { 2x }^{ 2 } }{ 2 } +8x+k\)
⇒ C = x³ - x² + 8x + k
Since there is no fixed cost,
when x = 0, C = 0 ⇒ k = 0
∴ C = x³ - x² + 8x

Given curve is y = 4x - X²
The limits are from x = 0 to x = 3
∴ Area \(\int _{ 0 }^{ 3 }{ y } dx=\int _{ 0 }^{ 3 }{ (4x-{ x }^{ 2 }) } dx\)
\({ \left[ \frac { 4{ x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }=2(9)-\frac { 27 }{ 3 } \)
=18 - 9
Area = 9 sq.units

Area under the curve is
\(A=\int _{ c }^{ d }{ xdy } =\int _{ 2 }^{ 4 }{ \sqrt { 4y } dy } \)
\(=2\int _{ 2 }^{ 4 }{ { y }^{ \frac { 1 }{ 2 } }dy } =2{ \left( \frac { { y }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right) }_{ 2 }^{ 4 }\)
\(=\frac { 4 }{ 3 } \left( { y }^{ \frac { 3 }{ 2 } } \right) \)
\(=\frac { 4 }{ 3 } \left( { 4 }^{ \frac { 3 }{ 2 } }-{ 2 }^{ \frac { 3 }{ 2 } } \right) \)
\(=\frac { 4 }{ 3 } \left( 4\sqrt { 4 } -2\sqrt { 2 } \right) \)
\(=\frac { 4 }{ 3 } (8-2\sqrt { 2 } )\)sq.units

The matrix A is singular, if
\(\left| \begin{matrix} 1 & -2 & 3 \\ 1 & 2 & 1 \\ x & 2 & -3 \end{matrix} \right| =0\)
\(1\left| \begin{matrix} 2 & 1 \\ 2 & -3 \end{matrix} \right| +2\left| \begin{matrix} 1 & 1 \\ x & -3 \end{matrix} \right| +3\left| \begin{matrix} 1 & 2 \\ x & 2 \end{matrix} \right| =0\)
\(\Rightarrow\) (-8) -6 - 2x + 6 - 6x = 0
\(\Rightarrow\) -8-2x-6x = 0
\(\Rightarrow\) -8-8x = 0
\(\Rightarrow\) -8 = 8x
\(\Rightarrow\) \(x=\cfrac { -8 }{ 8 } =-1\)

The non-homogeneous equations are
x + 2y = 3, 2x + 4y = 6

Here \(\rho (A)=1\) and \(\rho \left( \left[ A,B \right] \right) =1\)
Since \(\rho (A)=\rho \left[ \left( A,B \right) \right] =1<\) Number of unknowns, the given system is consistent with infinitely many solutions.
To find the solution, let us rewrite the above echelon form into the matrix form, we get
\(\left( \begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 3 \\ 0 \end{matrix} \right) \)
\(\Rightarrow x+2y=3\)
let \(y=k,k\varepsilon R\)
\((1)\Rightarrow x+2k=3\Rightarrow x=3-2k\)
\(\therefore\) Solution set is \(\left\{ 3-2k,k \right\} ,k\epsilon R\)
For different values of k; we get infinite number of solutions

Let \(A=\left[ \begin{matrix} 7 & -1 \\ 2 & 1 \end{matrix} \right] \)
The order of A is 2 x 2
\(\rho (A)\le min(2,2)\)
\(\left[ \begin{matrix} 7 & -1 \\ 2 & 1 \end{matrix} \right] =7-(-2)=7+29\neq 0\)
The highest order of non-vanishing minor of A is 2
\(\therefore \rho (A)=2\)