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

X=x	-2	0	10
P(X=x)	\(\frac{1}{4}\)	\(\frac{1}{4}\)	\(\frac{1}{2}\)

P(|X|≤2)=P(-2 =P(X=-2)+(P(X=0)
\(\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

Given

x	0	1	3	4
y	-12	0	6	12

Here the intervals are unequal
∴ By Lagranges interpolation formula, we have
x₀ = 0, x₁ = 1, x₂ = 3, x₃ = 4
y₀ = -12, y₁ = 0, y₂ = 6, y₃ = 12 and x = x.
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (x-1)(x-3)(x-4) }{ (0-1)(0-3)(0-4) } (-12)+\frac { (x-0)(x-3)(x-4) }{ (1-0)(1-3)(1-4) } (0)+\frac { (x-0)(x-1)(x-4) }{ (3-0)(3-1)(3-4) } (6)+\frac { (x-1)(x-3)(x-4) }{ (4-0)(4-1)(4-3) } (12)\)
= \(\frac { (x-1)(x-3)(x-4) }{ (-1)(-3)(-4) } (-12)+0+\frac { x(x-1)(x-4) }{ (3)(2)(-1) } (6)+\frac { x(x-1)(x-3) }{ (4)(3)(1) } (12)\)
= +[(x - 1)(x - 3)(x - 4)] - x (x - 1)(x -  4) + x(x - 1)(x - 3)
= +[(x³-4x+3)(x-4)] -x(x²-5x+4) + x(x²-4x + 3)
= - (x³ - 8x²+ 19x - 12) - 4x² + 3x
= (x - 4)(x² - 4x + 3) - x (x² - 5x + 4) + x(x² - 4x + 3)

= x³ - 7x² + 19x - 12.

Given

Year	1974	1978	1982	1990
Output in 1000 tones	25	60	80	170

 Here the intervals are unequal.
∴ By Lagranges interpolation formula, we have
x₀ = 1974, x₁ = 1978, x₂ = 1982, x₃ = 1990
y₀ = 25, y₁ = 60, y₂ = 80, y₃ = 170 and x = 1986.
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
\(\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times25+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times60+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times80+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times 70+\frac { (1986-1974)(1986-1982)(1986-1990) }{ (1990-1974)(1990-1978)(1990-1982) } \times 170\)

= 6.25 - 60 + 120 + 42.5
y = 108.75

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

\(\frac { dy }{ dx } +\frac { 2 }{ y } y\) = x³
The given differential equation is of this form [Divided by x]
\(\frac { dy }{ dx } \)+Py = Q where
P =\(\int { \frac { 2 }{ x } } \) and Q = x³
∴ \(\int { P } dx=\int { \frac { 2 }{ x } dx } \) = 2 log x = log x²
∴ Integrating factor (I. F) =\(e^{ \int { p } dx }=e^{ logx^{ 2 } }\)= x²
Hence the solution is
\(ye^{ \int { p } dx }=\int { Q } .e^{ \int { p.dx } }dx+c\)
⇒ y.x² = \(\int { { x }^{ 3 }.{ x }^{ 2 } } dx+c\)
⇒ x²y =\(\\ \int { { x }^{ 5 }dx } +c\)
⇒ x²y = \(\frac { { x }^{ 6 } }{ 6 } \) + c

Given marginal cost function is (x² + xy)dy + (3xy + y²)dx = 0 
\(\frac { dy }{ dx } =\frac { -(3xy+{ y }^{ 2 }) }{ { x }^{ 2 }+xy } \)    (1)
Put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \) in (1)
\(v+x\frac { dv }{ dx } =\frac { -(3xvx+{ v }^{ 2 }{ x }^{ 2 }) }{ { x }^{ 2 }+xvx } \)
\(=\frac { -(3v+{ v }^{ 2 }) }{ 1+v } \)
Now, \(x\frac { dv }{ dx } =\frac { -3v-{ v }^{ 2 } }{ 1+v } -v\)
\(=\frac { -3v-{ v }^{ 2 }-v-{ v }^{ 2 } }{ 1+v } \)
\(x\frac { dv }{ dx } =\frac { -4v-{ 2v }^{ 2 } }{ 1+v } \)
\(\frac { 1+v }{ 4v+2{ v }^{ 2 } } dv=\frac { -dx }{ x } \)
On Integration
\(\int { \frac { 1+v }{ 4v+2{ v }^{ 2 } } } =-\int { \frac { dx }{ x } } \)
Now, multiply 4 on both sides
\(\int { \frac { 1+v }{ 4v+2{ v }^{ 2 } } } =-4\int { \frac { dx }{ x } } \)
log (4v+2v²) = −4 logx+logc
4v + 2v² = \(\frac { c }{ { x }^{ 4 } } \)
x⁴(4v + 2v²) = c
Replace \(v=\frac { y }{ x } \)
\({ x }^{ 4 }\left( 4\frac { y }{ x } +2\frac { { y }^{ 2 } }{ { x }^{ 2 } } \right) =c\)
\({ x }^{ 4 }\left[ \frac { 4xy+2{ y }^{ 2 } }{ { x }^{ 2 } } \right] \) = c
c = 2x²(2xy + y²)   (2)
Cost of producing a pair of shoes = Rs. 12
(i.e) y = 12 when x = 2
c = 8[48 + 144]= 1536
∴ The cost function is x²(2xy + y²) = 768

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

Let I = \(\int _{ 0 }^{ 1 }{ \frac { x }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } } dx\)
\(=-\int _{ 0 }^{ 1 }{ \frac { -x }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } } dx\)
[Multiply and divide by -1]
\(=-\int _{ 0 }^{ 1 }{ \frac { 1-x-1 }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } } dx\)
[Adding and subtracting 1 in the numberator]
\(=\int _{ 1 }^{ 0 }{ \frac { 1-x-1 }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } } dx\)
\(\left[ \because \int _{ a }^{ b }{ f\left( x \right) } dx=-\int _{ b }^{ a }{ f\left( x \right) dx } \right] \)
\(=\int _{ 1 }^{ 0 }{ \left( \frac { 1-x }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } -\frac { 1 }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } \right) } dx\)
\(=\int _{ 1 }^{ 0 }{ \left( { \left( 1-x \right) }^{ 1-\frac { 3 }{ 4 } }-{ \left( 1-x \right) }^{ -\frac { 3 }{ 4 } } \right) } dx\)
\(=\int _{ 1 }^{ 0 }{ { \left( 1-x \right) }^{ \frac { 1 }{ 4 } }dx-\int _{ 1 }^{ 0 }{ { \left( 1-x \right) }^{ -\frac { 3 }{ 4 } } } } dx\)
\(={ \left[ \frac { { \left( 1-x \right) }^{ \frac { 1 }{ 4 } +1 } }{ -1\left( \frac { 1 }{ 4 } +1 \right) } -\frac { { \left( 1-x \right) }^{ -\frac { 3 }{ 4 } +1 } }{ -1\left( -\frac { 3 }{ 4 } +1 \right) } \right] }_{ 1 }^{ 0 }\)
\(={ \left[ -\frac { { { \left( 1-x \right) }^{ \frac { 5 }{ 4 } } } }{ \frac { 5 }{ 4 } } +\frac { { \left( 1-x \right) }^{ \frac { 1 }{ 4 } } }{ \frac { 1 }{ 4 } } \right] }_{ 1 }^{ 0 }\)
\(={ \left[ -\frac { 4 }{ 5 } { \left( 1-x \right) }^{ \frac { 5 }{ 4 } }+4{ \left( 1-x \right) }^{ \frac { 1 }{ 4 } } \right] }_{ 1 }^{ 0 }\)
\(=-\frac { 4 }{ 5 } \left( { 1 }^{ \frac { 5 }{ 4 } } \right) +4\left( { 1 }^{ \frac { 1 }{ 4 } } \right) -0\)
\(=-\frac { 4 }{ 5 } (1)+4(1)=-\frac { 4 }{ 5 } +4\)
\(=\frac { -4+20 }{ 5 } =\frac { 16 }{ 5 } \)

The mean of the binomial distribution
Ex = \(\overset { n }{ \underset { x=0 }{ \Sigma } } x.p(x)=\overset { n }{ \underset { x=0 }{ \Sigma } } \left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
= \(p.\overset { n }{ \underset { x=0 }{ \Sigma } } x.\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x-1 }q^{ n-x }\)
[Take p common]

=  \(np.\overset { n }{ \underset { x=1 }{ \Sigma } } \left( \begin{matrix} n-1 \\ x-1 \end{matrix} \right) { p }^{ x-1 }q^{ n-x }\)
= np (q +p)^n-1 [using binomial theorem]
(x+a)ⁿ = \({ x }^{ n }+{ n }_{ { C }_{ 1 } }{ x }^{ n-1 }{ a }^{ 1 }+...+{ a }^{ n }\)
= np(1)^n-1 [∵ p+q=1]
= np
∴ Mean = E(x) = np..(1)
Now, E(X²) =\(\overset { n }{ \underset { x=0 }{ \Sigma } } { x }^{ 2 }.\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
=\(\overset { n }{ \underset { x=0 }{ \Sigma } } \{ x(x-1)+x\} \left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
=\(\overset { n }{ \underset { x=0 }{ \Sigma } } x(x-1).\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }+{ \Sigma }_{ x }.\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
=\(\overset { n }{ \underset { x=2 }{ \Sigma } } x(x-1)\frac { n(n-1) }{ x(x-1) } \left( \begin{matrix} n-2 \\ n-2 \end{matrix} \right) { p }^{ x-2 }{ q }^{ n-x }\)+\(\Sigma x\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
= \(n(n-1){ p }^{ 2 }\left\{ \Sigma \left( \begin{matrix} n-2 \\ n-2 \end{matrix} \right) { p }^{ x-2 }{ q }^{ n-x } \right\} \) +np (using (1))
= n(n-1)p² (q+p)^n-2 + np
[using binomial theory]
= n(n-1)p² (1) + np  ∴ p+ q = 1
= n(n-1)p²+ np .... (2)
Variance = E(X²) - [E(X)]²
= n(n-1)p²+np-(np)²
[From (1) & (2)]

= np (1-p) = npq       [ ∵ p+q = 1⇒ q = 1-p]
∴ Mean = np and variance = npq

Let X denote the length of life of dry battery cells follows normal distribution with mean 12 and SD 3 hours

(i) more than 13 hours
P(X > 13)
When X = 13
\(Z=\frac { X-\mu }{ \sigma } =\frac { 13-12 }{ 3 } =0.333\)
P(X > 13) = P(Z > 0.333) = 0.5 – 0.1293 = 0.3707
The expected battery cells life to have more than 13 hours is 125 × 0.3707 = 46.34%

(ii) less than 5 hours
P(X < 5)
When X = 5
\(Z=\frac { X-\mu }{ \sigma } =\frac { 5-12 }{ 3 } =-2.333\)
P(X < 5) = P(Z < –2.333) = P(Z > 2.333)
= 0.5 – 0.4901 = 0.0099
The expected battery cells life to have more than 13 hours is 125 × 0.0099 = 1.23%

(iii) between 9 and 14 hours
When X = 9
\(Z=\frac { X-\mu }{ \sigma } =\frac { 9-12 }{ 3 } =-1\)
When X = 14
\(Z=\frac { X-\mu }{ \sigma } =\frac { 14-12 }{ 3 } =0.667\)
P(9 < X < 14) = P(–1 < Z < 0.667)
= P(0 < Z < 1) + P(0 < Z < 0.667)
= 0.3413 + 0.2486
= 0.5899
The expected battery cells life to have more than 13 hours is 125 x 0.5899 = 73.73%

Let X be the waiting time of a customer in the queue and it is normally distributed with mean 5 and SD 0.7.

(i) for less than 6 minutes
\(Z=\frac { X-\mu }{ \sigma } =\frac { 6-5 }{ 0.7 } =1.4285\)
P(X < 6) = P(Z < 1.43)
= 0.5 + 0.4236
= 0.9236
(ii) between 3.5 and 6.5 minutes
When X = 3.5
\(Z=\frac { X-\mu }{ \sigma } =\frac { 3.5-5 }{ 0.7 } =2.1429\)
When X = 6.5
\(Z=\frac { X-\mu }{ \sigma } =\frac { 6.5-5 }{ 0.7 } =2.1429\)
P(3.5 < X < 6.5)
= P (–2.1429 < Z < 2.1429)
= P(0 < Z < 2.1429) + P(0 < Z < 2.1429)
= 2 P(0 < Z < 2.1429)
= 2 x .4838
= 0.9676

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

We know that, the expected random variable 
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x) } dx\)
\(=\int _{ 100 }^{ \infty }{ { xe }^{ -\frac { x }{ 100 } }dx } \)
\(=\left\{ { \left[ x\left( \frac { { e }^{ \frac { x }{ 100 } } }{ -\frac { 1 }{ 100 } } \right) \right] }_{ 100 }^{ \infty }-\int _{ 100 }^{ \infty }{ \left( \frac { { e }^{ \frac { x }{ 100 } } }{ -\frac { 1 }{ 100 } } \right) dx } \right\} (\because \int { udv=uv-\int { vdu } } )\)
\(=\left[ \left( 10000 \right) \left( { e }^{ -1 } \right) +\left( 10000 \right) \left( { e }^{ -1 } \right) \right] \)
\(=\left[ \left( 10000 \right) \left( 0.3679 \right) +\left( 10000 \right) (0.3679) \right] \)
= 7358 hours 
Therefore, the mean life of a radio tube is 7,358 hours.

Given demand equation is \(x=\sqrt { 100-p } \) and
Supply equation is \(x=\frac { p }{ 2 } -10\)
At market equilibrium, \(\sqrt { 100-p } =\frac { p }{ 2 } -10\)  Squaring both sides,
\(100-p={ \left( \frac { p }{ 2 } -10 \right) }^{ 2 }\)
\(\frac { { p }^{ 2 } }{ 4 } -10p+p=0\)
\(\frac { { p }^{ 2 } }{ 4 } -9p=0\)
\({ p }^{ 2 }-36p=0\)
p(p-36) = 0
p = 0 or p = 36
Since p cannot be zero, p = 36
\(\therefore \ { x }_{ 0 }=\sqrt { 100-36 } =\sqrt { 64 } =8\)
∴ p₀x₀ = 36 x 8 = 288
Given demand equation is \(x=\sqrt { 100-p } \)
X² = 100-P(Squaring both sides)
p = 100-x²
Consumer's Surplus
\(CS=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(CS=\int _{ 0 }^{ 8 }{ (100-{ x }^{ 2 }) } dx-288\)
\(={ \left[ 100x-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 8 }-288\)
\(=100(8)-\frac { { 8 }^{ 3 } }{ 3 } -288\)
\(=800-\frac { 512 }{ 3 } -288\)
\(=512-\frac { 512 }{ 3 } \)
\(=\frac { 1536-512 }{ 3 } \)
\(CS=\frac { 1024 }{ 3 } \)units
Supply equation is given as
\(x=\frac { p }{ 2 } -10\)
\(x=\frac { p-20 }{ 2 } \)
2x = p - 20
p = 2x + 20
∴ Producer's surplus
\(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x }{ g(x)dx } \)
\(=288-\int _{ 0 }^{ 8 }{ (2x+20)dx } \)
\(=288-{ \left( \frac { { 2x }^{ 2 } }{ 2 } +20x \right) }_{ 0 }^{ 8 }\)
\(=288-{ \left( { x }^{ 2 }+20x \right) }_{ 0 }^{ 8 }\)
= 288 - (8² + 20(8))
= 288 - (64 + 160)
= 288 - 224
PS = 64 units

Given probability distribution function
\(F(x)\begin{cases} 0,\quad ​​​​\text{for}\quad x<0 \\ \frac { 1 }{ 2 } ,\quad ​​​​\text{for}\quad 0\le x<1 \\ 0,\quad ​​​​\text{for}\quad 1\le x<2\quad \\ \frac { 1 }{ 4 } ,\quad ​​​​\text{for}\quad 2\le x<4 \\ 0,\quad ​​​​\text{for}\quad x\ge 4 \end{cases}\)
a) The distribution function F(x) is continuous since it is a step function
We know f'(x) = f(x)
∴ Probability density function
\(F(x)\begin{cases} 0,\quad \text{for}\quad x\le 0 \\ \frac { 1 }{ 2 } ,\quad \text{for}\quad 0\le x<1 \\ 0,\quad \text{for}\quad 1\le x<2\quad \\ \frac { 1 }{ 4 } ,\quad \text{for}\quad 2\le x<4 \\ 0,\quad \text{for}\quad x\ge 4 \end{cases}...(1)\)
(b) (i) Probability that a person will have to wait more than 3 minutes is P(X > 3)
∴ P(X>3) = P(3≤X<4)+P(X≥4)
\(\frac{1}{4}+0=\frac{1}{4}\) [from (1)]
ii) Probability that a person will have to wait less than 3 minutes is P(X < 3)
∴ P(X<3) = P(X≤0)+P(0≤x≤1)+P(1≤x≤2)+P(2≤x≤3)
0 + \(\frac{1}{2}+0+\frac{1}{4}\)[from (1)]
∴ P(X<3) = \(\frac{3}{4}\)
= P(1≤X<2)+P(2≤X<3)
= 0 +\(\frac{1}{4}=\frac{1}{4}\) [from (1)]

Given p.d.f is
\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 16 } { (3+x) }^{ 2 }, & -3\le x\le -1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 16 } (6-2{ x }^{ 2 }), & -1\le x\le 1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 16 } { (3-x) }^{ 2 }, & 1\le x\le 3 \end{matrix} \end{cases}\)
Area under the given curve
\(\int _{ -3 }^{ 3 }{ f(c)dx=\int _{ -3 }^{ -1 }{ \frac { 1 }{ 16 } { (3+x) }^{ 2 }+\int _{ -1 }^{ 1 }{ \frac { 1 }{ 16 } (6-2{ x }^{ 2 })dx } } +\int _{ 1 }^{ 3 }{ \frac { 1 }{ 16 } { (3-x) }^{ 2 }dx } } \)
\(=\frac { 1 }{ 16 } { \left[ \frac { (3+x{ ) }^{ 3 } }{ 3 } \right] }_{ -3 }^{ -1 }+\frac { 1 }{ 16 } { \left( 6x-\frac { { 2x }^{ 3 } }{ 3 } \right) }_{ -1 }^{ 1 }+\frac { 1 }{ 16 } { \left( \frac { { (3-x) }^{ 3 } }{ -3 } \right) }_{ 1 }^{ 3 }\)
\(=\frac { 1 }{ 16 } \left[ \left( \frac { { 2 }^{ 3 } }{ 3 } -0 \right) +\left( 6-\frac { 2 }{ 3 } \right) -\left( -6+\frac { 2 }{ 3 } \right) -\frac { 1 }{ 3 } \left( 0-({ 2 }^{ 3 }) \right) \right] \)
\(=\frac { 1 }{ 6 } \left[ \frac { 8 }{ 3 } +\frac { 16 }{ 3 } -\left( \frac { -16 }{ 3 } \right) +\frac { 8 }{ 3 } \right] \)
\(\\ =\frac { 1 }{ 16 } \left[ \frac { 8 }{ 3 } +\frac { 16 }{ 3 } +\frac { 16 }{ 3 } +\frac { 8 }{ 3 } \right] \)
\(=\frac { 1 }{ 16 } \left[ \frac { 8+16+16+8 }{ 3 } \right] \)
\(=\frac { 1 }{ 16 } \times \frac { 48 }{ 3 } =\frac { 1 }{ 16 } \times 16=1\)
Hence, area under the given curve is unity.

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

MR = a (x + b)^-2 − c
\(R=\int { { (x+b) }^{ 2 } } \)dx - c \(\int { dx } \)
R = \(\frac { a(x+{ b })^{ -2 } }{ -1 } \) - cx + k
R = \(\frac { a }{ (x+b) } \) - cx + k
When x = 0, R = 0
\(0=\frac { a }{ b } -c(0)+k\)
\(k=\frac { a }{ b } \)
\(R=-\frac { a }{ x+b } \) - cx+\(\frac{a}{b}\)
\(=\frac { -ab+a(x+b) }{ b(x+b) } -cx\quad \)
\(R=\frac { ax }{ b(x+b) } -cx\)
Demand function \(P=\frac { R }{ x } \)
\(P=\frac { a }{ b(x+b) } -c\)
\(P+c=\frac { a }{ b(x+b) } \)
\(b(x+b)=\frac { a }{ P+c } \)
\(x=\frac { a }{ b(P+c) } -b\)

\(y=\left| x+3 \right| =\begin{cases} x+3\quad if\quad x\ge -3\quad \\ -(x+3)\quad if\quad x<-3 \end{cases}\)
Required area = \(\int _{ b }^{ a }{ y } dx=\int _{ -6 }^{ 0 }{ y } dx\)
= \(\int _{ -6 }^{ -3 }{y}\ dx+\int _{ -3 }^{ 0 }{ y} dx\)
= \(\int _{ -6 }^{ -3 }{ -(x+3) } dx+\int _{ -3 }^{ 0 }{ (x+3) } dx\)
= \(-{ \left[ \frac { { (x+3) }^{ 2 } }{ 2 } \right] }_{ -6 }^{ -3 }{ +\left[ \frac { { (x+3) }^{ 2 } }{ 2 } \right] }_{ -3 }^{ 0 }\) 
\(=-\left[ 0-\frac { 9 }{ 2 } \right] +\left[ \frac { 9 }{ 2 } -0 \right] \)
= 9 sq. units

 

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

Transition probability matrix

Where A represents the percentage of people using transit system and B represents the percentage of people using metro train.
By the given data
A 60% = .60
and B 40% = ·4
= ((-6)(-7)+(-4)(-3) (-6)(-3)+(-4)(-7))
= (-42+·12 ·18+·28)
= (-54 -46)
\(\therefore\) A = 54%and B = 46%
(i) The percent of Commuters using the transit system after one year is 54% and the percent of commuters using the metro train after one year is 46%
(ii) Equilibrium will be reached in the long run. At equilibrium we must have
(A B) T = (A B)
where A+B = 1
\(\Rightarrow \left( A\quad B \right) \left( \begin{matrix} \cdot 7 & \cdot 3 \\ \cdot 3 & \cdot 7 \end{matrix} \right) =(A\quad B)\)
(-7A +·3B ·3A +.7B) = (A B)
Equaling the entries on both sides, we get
·7A+ ·3B = A
\(\Rightarrow \cdot 7A+\cdot 3(1-A)=A\)
\(\left[ \because A+B=1\Rightarrow B=1-A \right] \)
\(\Rightarrow \cdot 7A+\cdot 3(1-A)=A\)
\(\Rightarrow \cdot 3=A-\cdot 7A+\cdot 3A\)
\(\Rightarrow \cdot 3=A(\cdot 3+\cdot 3)\)
\(\Rightarrow 3=A(\cdot 6)\)
\(\Rightarrow A=\cfrac { \cdot 3 }{ \cdot 6 } =\cfrac { 1 }{ 2 } =\cdot 50\)
\(\therefore\) The percent of commuters using the transit system in the long run is 50%

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
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 6 \\ 10 \\ b \end{matrix} \right) \)
AX = B

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{matrix}\begin{matrix} 6 \\ 10 \\ b \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & a-1 \end{matrix}\begin{matrix} 6 \\ 4 \\ b-6 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & a-3 \end{matrix}\begin{matrix} 6 \\ 4 \\ b-10 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\)   \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }\)

Case (i) For no solution:
The system possesses no solution only when \(\rho (A)\neq ([A,B])\) which is possible only when a−3 = 0 and b −10 \(\neq \) 0.
Hence for a = 3, b \(\neq \) 10, the system possesses no solution.
Case (ii) For a unique solution:
The system possesses a unique solution only when \(\rho (A)= ([A,B])\)=number of unknowns.
i.e when \(\rho (A)=\rho ([A,B])\) = 3
Which is possible only when a−3 \(\neq \) 0 and b may be any real number as we can observe .
Hence for a \(\neq \) and b \(\in \) R, the system possesses a unique solution.
Case (iii) For an infinite number of solutions:
The system possesses an infinite number of solutions only when
\(\rho (A)=\rho ([A,B])\) i,e when \(\rho (A)=\rho ([A,B])\) = 2 < 3 ( number of unknowns) which is possible only when a − 3 = 0, b−10 = 0
Hence for a = 3, b = 10, the system possesses infinite number of solutions.

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & k \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 7 \\ 18 \\ 6 \end{matrix} \right) \)
 AX = B

Augmented matrix [A,B]

Elementary Transformation

\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & k \end{matrix}\begin{matrix} 7 \\ 18 \\ 6 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & k \end{matrix}\begin{matrix} 7 \\ 11 \\ 6 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & k-2 \end{matrix}\begin{matrix} 7 \\ 11 \\ -5 \end{matrix} \right) \) ρ(A) = 2 or 3, ρ([A]) = 3

\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

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

Let A represents the percent of people who subscribe the magazine and B represents the percent of people who do not subscribe the magazine.
Given 60% of people subscribe again implies 40% of people do not subscribe. And 25% of people are going to subscribe implies 75% of people are not going to subscribe.
\(\therefore\) Transition probability matrix.

Also, it is given that 40% of those received the order of subscription implies 60% are not going to receive the order.
\(\left( \begin{matrix} \cdot 4 & \cdot 6 \end{matrix} \right) \left( \begin{matrix} \cdot 6 & \cdot 4 \\ \cdot 25 & \cdot 75 \end{matrix} \right) \)
= \(\left( \left( \begin{matrix} \cdot 4 & \cdot 6 \end{matrix} \right) +\left( \cdot 6 \right) \left( \cdot 25 \right) \left( \cdot 4 \right) \left( \cdot 4 \right) +\left( \cdot 6 \right) \left( \cdot 75 \right) \right) \)
= \(\left( \cdot 24+\cdot 15\quad \cdot 16+\cdot 45 \right) =\left( \cdot 39\quad \cdot 61 \right) \)
\(\therefore\) 39% of people who received the current letter can be expected to order a subscription.