For market equilibrium, P_d = P_s
\({ P }_{ d }=\frac { 16 }{ x+4 } =\frac { x }{ 2 } \Rightarrow 32=x(x+4)\)
⇒ 32-x²+4x
⇒ x²+4x-32 = 0
⇒ (x+8)(x-4) = 0
⇒ x = -8, x = 4
Since x = - 8 is not possible, X_o = 4
\(\therefore { p }_{ 0 }=\frac { 16 }{ 4+4 } =\frac { 16 }{ 8 } =2\)
\(\therefore { p }_{ 0 }{ x }_{ 0 }=2(4)=8\)
Consumer's Surplus \(CS=\int _{ 0 }^{ x0 }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(Cs=\int _{ 0 }^{ 4 }{ \frac { 16 }{ x+4 } dx-8 } \)
\(={ \left[ 16log(x+4) \right] }_{ 0 }^{ 4 }-8\)
\(=16\left[ log(4+4)-log(0+4) \right] -8\)
\(=16[log\quad 8-log4]-8\)
\(=16log\left( \frac { 8 }{ 4 } \right) -8\)
CS=(16 log 2-8)units.
Producer's Surplus \(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x0 }{ g(x)dx } \)
\(=8-\int _{ 0 }^{ 4 }{ \frac { x }{ 2 } dx } =8-{ \left[ \frac { { x }^{ 2 } }{ 4 } \right] }_{ 0 }^{ 4 }\)
\(=8-\left[ \frac { { 4 }^{ 2 } }{ 4 } \right] =8-\frac { 16 }{ 4 } =8-4\)
PS = 4units

For market equilibrium, P_d = p_s
⇒ 16-x² = 2x²+4
⇒ 16-4 = 2x²+x²
⇒ 3x² = 12
⇒ x² = 4
⇒ x = 土2
Since x = 2 is not possible x₀ = 2
p₀ = 16 - 2² = 16 - 4 = 12
∴ p₀x₀ = 2 x 12 = 24
Consumer's Surplus CS \(\int _{ 0 }^{ x0 }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ 2 }{ (16-{ x }^{ 2 }) } dx-24\)
\(={ \left[ 16x-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 2 }-24\)
\(=16(2)-\frac { { 2 }^{ 3 } }{ 3 } -24\)
\(=32-\frac { 8 }{ 3 } -24\)
\(=8-\frac { 8 }{ 3 } -24\)
\(=\frac { 16 }{ 3 } \) units
Producer's Surplus \(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x0 }{ g(x)dx } \)
\(=24-\int _{ 0 }^{ 2 }{ (2{ x }^{ 2 }+4)dx } \)
\(=24-{ \left[ \frac { { 2x }^{ 3 } }{ 3 } +4x \right] }_{ 0 }^{ 2 }\)
\(=24-\left[ \frac { 16 }{ 3 } +8 \right] \)
\(=24-\frac { 16 }{ 3 } -8=16-\frac { 16 }{ 3 } \)
\(=\frac { 48-16 }{ 3 } =\frac { 32 }{ 3 } \)units

Given

Marks	0-19	20-39	40-59	60-79	80-99
No.of.candidates	41	62	65	50	17

 This can be rewritten as

Marks:	Below 19	Below 39	Below 59	Below 79	Below 99
No. of.  candidates:	41	(41 + 62) =103	(41 + 62 + 65) = 168	(41 + 62 + 65 + 50) =218	(41 + 62 + 65 + 50 + 17) = 235

Since we have to find below 70, use Newton's backward interpolation formula
∴ x_n + nh = 70 ⇒ 99 + n(20) = 70
⇒ 20n = 70 - 99 = -29
⇒ n = \(\frac{-29}{20}\) = -1.45
∴ y₇₀ = y_n + \(\frac { n }{ 1! } { \triangledown y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }\)
\(=235-1.45(17)+\frac { (-1.45)(-1.45+1) }{ 2 } (-33)+\frac { (-1.45)(-1.45+2)(-18) }{ 6 } \)
\(=235-24.65+\frac { (-1.45)(-.45) }{ 2 } (-33)+\) (-1.45)(-0.45)(0.55)(-3)
= 235 - 24.65 - 12.375 - 1.125
= 196
Hence, the number of students who have scored below 70 are 196 (app).

Since five values are given, the polynomial which fits the data is of degree four.
Hence Δ⁵y_k = 0 (i.e) (E−1)⁵y_k = 0
i.e., (E⁵ - 5E⁴ + 10E³ - 10E² + 5E - 1)y_k = 0
E⁵y_k - 5E⁴y_k+ 10E³y_k- 10E²y_k+ 5Ey_k - y_k = 0 (1)
Put k = 0 in (1)
E⁵y₀ - 5E⁴y₀+ 10E³y₀- 10E²y₀+ 5Ey₀ - y₀ = 0
y₅ − 5y₄ + 10y₃ − 10y₂ +5 y₁ - y₀ = 0
y₅ − 5(350) +10y₃−10(260)+5(220)− 200 = 0
y₅ + 10y₃ = 3450 (2)

Put k = 1 in (1)
E⁵y₁ - 5E⁴y₁+ 10E³y₁- 10E²y₁+ 5Ey₁- y₀ = 0
y₆ − 5y₅ + 10y₄ −1 0y₃ − y₁ = 0
430  −5y₅ +10(350) −10y₃ + 5(260)− 220 = 0
5y₅+10y₃ = 5010 (3)
(3) – (2) ⇒ 4y₅ = 1560
y₅ = 390
From (1)  390 +10y₃ = 3450
10y₃ = 3450 – 390
y₃ ≅ 306

Let age = x and premium = y
To find y at x = 63
So apply Newton’s backward interpolation formula
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+..\)

x	y	\(\Delta y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 4 }y\)
45	114.84
		-18.68
50	96.16		5.84
		-12.84		-1.84
55	83.32		4		0.68
		-8.84		-1.16
60	74.48		2.84
		-6
65	68.48

\({ y }_{ (x=6.5) }=68.48+\frac { \frac { -2 }{ 5 } }{ 1! } \left( -6 \right) +\frac { \frac { -2 }{ 5 } \left( \frac { -2 }{ 5 } +1 \right) }{ 2! } 2.84+\frac { \frac { -2 }{ 5 } \left( \frac { -2 }{ 5 } +1 \right) \left( \frac { -2 }{ 5 } +2 \right) }{ 3! } \left( -1.16 \right) +\frac { \frac { -2 }{ 5 } \left( \frac { -2 }{ 5 } +1 \right) \left( \frac { -2 }{ 5 } +2 \right) \left( \frac { -2 }{ 5 } +3 \right) }{ 3! } (0.68)\)
= 68.48 + 2.4 – 0.3408 + 0.07424 – 0 – 0.028288
y(63) = 70.437

x	1941	1951	1961	1971	1981	1991
y	20	24	29	36	46	51

Here we find the population for year 1946. (i.e) the value of y at x = 1946. Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+..\)
To find y at x = 1946
\(\therefore\) x₀ + nh = 1946, x₀ = 1941, h = 10
1941 + n(10) = 1946 \(\Rightarrow\) n = 0.5

x	y	\(\Delta y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 3 }y\)	\(\Delta ^{ 4 }y\)	\(\Delta ^{ 5 }y\)
1941	20
		4
1951	24		1
		5		1
1961	29		2		0
		7		1		-9
1971	36		3		-9
		10		-8
1981	46		-5
		5
1991	51

Here we find the population for year 1946. (i.e) the value of y at x = 1946. Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\(​​{ y }_{ \left( x=1946 \right) }=20+\frac { 0.5 }{ 1! } (4)+\frac { 0.5(0.5-1) }{ 2! } (1)+\frac { 0.5(0.5-1)(0.5-2) }{ 3! } (1)+\frac { 0.5(0.5-1)(0.5-2)(0.5-3) }{ 4! } (0)+\frac { 0.5(0.5-1)(0.5-2)(0.5-3)(0.5-4) }{ 5! } (-9)\)
= 20+2-0.125+0.0625-0.24609
= 21.69 lakhs

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

Given \(\frac { dy }{ dx } \) − (3 cot x) y = sin 2x
It is of the form \(\frac { dy }{ dx } \) + Py = Q
Here P= − 3 cot x,Q = sin 2x
ഽPdx = ഽ-3 cot xdx = -3 log sin x = - log sin³x = log \(\frac { 1 }{ { sin }^{ 3 }x } \)
I.F. = \({ e }^{ log\frac { 1 }{ { sin }^{ 3 }x } }=\frac { 1 }{ { sin }^{ 3 }x } \)
The required solution is y (I.F) = ഽQ(I.F)dx+c
\(y\frac { 1 }{ { sin }^{ 3 }x } =\int { sin2x } \frac { 1 }{ { sin }^{ 3 }x } dx+c\)
\(\int { \frac { 1 }{ { sin }^{ 3 }x } } =\int { 2sinxcosx\times \frac { 1 }{ { sin }^{ 3 }x } } dx+c\)
= \(2\int { \frac { 1 }{ sinx } \times \frac { cosx }{ sinx } } dx+c\)
= ഽcos ecx cot xdx + c
\(y\frac { 1 }{ { sin }^{ 3 }x } =-2cosecx+c\)
Now y = 2 when x = \(\frac { \pi }{ 2 } \)
(1) ⇒ 2\(\frac {1 }{ 1 } \) = −2×1+c ⇒ c= 4
∴ (1) ⇒ \(y\frac { 1 }{ { sin }^{ 3 }x } \) = −2cosecx + 4

Given P_d = 18− 2x − x² ; P_s = 2x − 3
We know that at equilibrium prices p_d = p_s
18− 2x − x² = 2x – 3
x² + 4x −21 = 0
(x − 3) (x + 7) = 0
x = –7 or 3
The value of x cannot be negative, x = 3
When x₀ = 3
ஃ p₀ = 18 − 2(3)  − (3)² = 3
CS = \(\int _{ 0 }^{ { x }_{ o } }{ f(x) } \) dx - x₀p₀
= \(\int _{ 0 }^{ 3 }{ (18-2x-{ x }^{ 2 }) } \)dx - 3 x 3
= \({ \left[ 18x-{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }\)- 9
= 18(3) - (3)² - \(\left( \frac { { 3 }^{ 3 } }{ 3 } \right) \) - 9
CS = 27 units
PS = x₀P₀ - \(\int _{ 0 }^{ { x }_{ o } }{ g(x) } \)
= (3 \(\times\) 3) - \(\int _{ 0 }^{ 3 }{ (2x-3) } \)
= 9 - \(({ { { x }^{ 2 }-3x) } }_{ 0 }^{ 3 }\)
= 9 units
Hence at equilibrium price,
(i) the consumer’s surplus is 27 units
(ii) the producer’s surplus is 9 units.

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

Given MC = 8 + 6x
\(C(x)=\int { (8+6x)dx } \) + k₁
= 8x + 3x² + k_{1 ...}(1)
But given when x = 0, C = 0 ⇒ k₁ = 0
ஃ C(x) = 8x + 3x²  ....(2)
Given that MR = 24
R(x) = \(\int { MR } \) dx + k₂
= \(\int { 24 } \) dx + k₂
= \(\int { 24 } \) + k₂
Revenue = 0, when x = 0 ⇒ k₂ = 0
R(x) = 24x ...(3)
Total Profit functions P(x) = R(x) – C(x)
P(x) = 24x − 8x − 3x²
= 16x − 3x²

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 \(P=\frac { R }{ x } \)
\(P=\frac { a }{ b(x+b) } -c\)
\(P+c=\frac { a }{ b(x+b) } \)
\(b(x+b)=\frac { a }{ P+c } \)
\(x=\frac { a }{ b(P+c) } -b\)

\(\int { \frac { { 3x }^{ 2 }+6x+1 }{ \left( x+3 \right) \left( { x }^{ 2 }+1 \right) } } dx =\int { \left[ \frac { 1 }{ \left( x+3 \right) } \frac { 2x }{ \left( { x }^{ 2 }+1 \right) } \right] } dx\)
\(\int { \frac { dx }{ \left( x+3 \right) } +\int { \frac { 2x }{ \left( { x }^{ 2 }+1 \right) } } } dx\)
\(=\log\left| x+3 \right| +\log\left| { x }^{ 2 }+1 \right| +c\)
\(=\log\left| \left( x+3 \right) \left( { x }^{ 2 }+1 \right) \right| +c\)
\(=\log\left| { x }^{ 3 }+{ 3x }^{ 2 }+x+3 \right| +c\)
[ By partial fractions,
\(\frac { { 3x }^{ 2 }+6x+1 }{ \left( x+3 \right) \left( { x }^{ 2 }+1 \right) } =\frac { A }{ \left( x+3 \right) } +\frac { Bx+C }{ \left( { x }^{ 2 }+1 \right) } \Rightarrow \frac { { 3x }^{ 2 }+6x+1 }{ \left( x+3 \right) \left( { x }^{ 2 }+1 \right) } =\frac { 1 }{ \left( x+3 \right) } +\frac { 2x }{ \left( { x }^{ 2 }+1 \right) } \)

\(\int { \frac { 3x+2 }{ { \left( x-2 \right) }^{ 2 }\left( x-3 \right) } dx } =\int { \left[ \frac { 11 }{ \left( x-2 \right) } -\frac { 8 }{ { \left( x-2 \right) }^{ 2 } } +\frac { 11 }{ \left( x-3 \right) } \right] } dx\)
\(=11\int { \frac { dx }{ \left( x-2 \right) } -8 } \int { \frac { dx }{ { \left( x-2 \right) }^{ 2 } } +11 } \int { \frac { dx }{ \left( x-3 \right) } } \)
\(=11\log\left| x-2 \right| +\frac { 8 }{ x-2 } +11\log\left| x-3 \right| +c\)
\(=11\log\left| \frac { x-3 }{ x-2 } \right| +\frac { 8 }{ x-2 } +c\)
[ By partial fractions,
\(\frac { 3x+2 }{ { (x-2) }^{ 2 }(x-3) } =\frac { A }{ (x-2) } +\frac { B }{ (x-2)^{ 2 } } +\frac { C }{ (x-3) } \Rightarrow \frac { 3x+2 }{ { (x-2) }^{ 2 }(x-3) } =- \frac { 11 }{ (x-2) } - \frac { 8 }{ (x-2)^{ 2 } } +\frac { 11 }{ (x-3) } \)]

Let the rate of commission on the items A, B and C be x, y and z respectively.
By the given data, the non-homogeneous equations are
90x + 100y + 20z = 800
\(\Rightarrow\)9x + 10y + 2z = 80
130x + 50y + 40z = 900
\(\Rightarrow\) 13x + 5y + 4z = 90
60x + 100y + 30z = 850
\(\Rightarrow\) 6x + 10y + 3z = 85

\(\Delta =\left| \begin{matrix} 9 & 10 & 2 \\ 13 & 5 & 4 \\ 6 & 10 & 3 \end{matrix} \right| \)
\(9\left| \begin{matrix} 5 & 4 \\ 10 & 3 \end{matrix} \right| -10\left| \begin{matrix} 13 & 4 \\ 6 & 3 \end{matrix} \right| +2\left| \begin{matrix} 13 & 5 \\ 6 & 10 \end{matrix} \right| \)
= 9 (15 - 40) - 10 (39 - 24) + 2(130 - 30)
= 9 (- 25) - 10(15) + 2(100)
= - 225 - 150 + 200
= -175
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system is consistent with unique solution.
\(\Delta x=\left| \begin{matrix} 80 & 10 & 2 \\ 90 & 5 & 4 \\ 85 & 10 & 3 \end{matrix} \right| \)
= \(80\left| \begin{matrix} 5 & 4 \\ 10 & 3 \end{matrix} \right| -10\left| \begin{matrix} 90 & 4 \\ 85 & 3 \end{matrix} \right| +2\left| \begin{matrix} 90 & 5 \\ 85 & 10 \end{matrix} \right| \)
= 80(15 - 40) - 10(270 - 340) + 2(900 - 425)
= 80 (- 25) - 10 (- 70) + 2 (475)
= - 2000 + 700 + 950
= -350
\(\Delta y=\left| \begin{matrix} 9 & 80 & 2 \\ 13 & 90 & 4 \\ 6 & 85 & 3 \end{matrix} \right| \)
= \(9\left| \begin{matrix} 90 & 4 \\ 85 & 3 \end{matrix} \right| -80\left| \begin{matrix} 13 & 4 \\ 6 & 3 \end{matrix} \right| +2\left| \begin{matrix} 13 & 90 \\ 6 & 85 \end{matrix} \right| \)
= 9(270 - 340) - 80(39 - 24) +2(1105 - 540)
= 9(- 70) - 80(15) + 2(565)
= - 630 - 1200 + 1130
= -700
\(\Delta z=\left| \begin{matrix} 9 & 10 & 80 \\ 13 & 5 & 90 \\ 6 & 10 & 85 \end{matrix} \right| \)
= \(9\left| \begin{matrix} 5 & 90 \\ 10 & 85 \end{matrix} \right| -10\left| \begin{matrix} 13 & 90 \\ 6 & 85 \end{matrix} \right| +80\left| \begin{matrix} 13 & 5 \\ 6 & 10 \end{matrix} \right| \)
= 9(425 - 900) - 10(1105 - 540) + 80(130 - 30)
= 9(- 475) - 10(565) + 80 (100)
= - 4275 - 5650 + 8000
= - 1925

\(\therefore\) The rate of commission on the items A, Band Care 2%, 4% and 11%

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%

(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.

Given that the price of commodities X, Y and Z are x, y and z respectively 
By the given data

Transaction	x	y	z	Earning
Mr. Anand	+2	+3	-6	Rs.5000
Mr. Amar	+3	-1	+2	Rs.2000
Mr. Amit	-1	+3	+1	Rs.5500

Here, purchasing is taken as negative symbol and selling is taken as positive symbol
Thus, the non-homogeneous equations are
2x + 3y - 6z = 5000
3x - y + 2z = 2000
-x + 3y + z = 550
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 2 & 3 & -6 \\ 3 & -1 & 2 \\ -1 & 3 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 5000 \\ 2000 \\ 5500 \end{matrix} \right) \)

Augmented matrix	Elementary Transformation
\(\left( \begin{matrix} 2 & 3 & -6 \\ 3 & -1 & 2 \\ -1 & 3 & 1 \end{matrix}\begin{matrix} 5000 \\ 2000 \\ 5500 \end{matrix} \right) \)
\(\left( \begin{matrix} -1 & 3 & 1 \\ 3 & -1 & 2 \\ 2 & 3 & -6 \end{matrix}\begin{matrix} 5500 \\ 2000 \\ 5000 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & -3 & -1 \\ 3 & -1 & 2000 \\ 2 & 3 & -6 \end{matrix}\begin{matrix} -5000 \\ 2000 \\ 5500 \end{matrix} \right) \)	\({ R }_{ 1 }\rightarrow { R }_{ 1 }\left( -1 \right) \)
\(\left( \begin{matrix} 1 & -3 & -1 \\ 0 & 8 & 5 \\ 0 & 9 & -4 \end{matrix}\begin{matrix} -5500 \\ 18500 \\ 16000 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 }\)
\(\left( \begin{matrix} 1 & -3 & -1 \\ 0 & 1 & \frac { 65 }{ 8 } \\ 0 & 1 & \frac { -4 }{ 9 } \end{matrix}\begin{matrix} -5500 \\ \frac { 18500 }{ 8 } \\ \frac { 16000 }{ 9 } \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }\div 8\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }\div 9\)
\(\left( \begin{matrix} 1 & -3 & -1 \\ 0 & 1 & \frac { 5 }{ 8 } \\ 0 & 0 & \frac { -4 }{ 9 } -\frac { 5 }{ 8 } \end{matrix}\begin{matrix} -5500 \\ \frac { 18500 }{ 8 } \\ \frac { 16000 }{ 9 } -\cfrac { 18500 }{ 8 } \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow R_{ 3 }-{ R }_{ 2 }\)
\(\left( \begin{matrix} 1 & -3 & -1 \\ 0 & 1 & \frac { 5 }{ 8 } \\ 0 & 0 & \frac { -77 }{ 72 } \end{matrix}\begin{matrix} -5500 \\ \frac { 18500 }{ 8 } \\ \frac { -38500 }{ 72 } \end{matrix} \right) \)

.Clearly the last equivalent matrix is in echelon form and it has three non-zero rows 
\(\therefore \rho (A)=\rho \left( \left[ A,B \right] \right) =3\) Number of unknowns.
\(\therefore\) The given system is consistent and has unique solution. To find the solution, let us rewrite the above : echelon form into the matrix form.
\(\left( \begin{matrix} 1 & -3 & -1 \\ 0 & 1 & \frac { 5 }{ 8 } \\ 0 & 0 & \frac { -77 }{ 72 } \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) \left( \begin{matrix} -5000 \\ \frac { 18500 }{ 8 } \\ \frac { -38500 }{ 72 } \end{matrix} \right) \)
\(\Rightarrow x-3y-z=-5500\)
\(y+\cfrac { 5 }{ 8 } z=\cfrac { 18500 }{ 8 } \)
\(\cfrac { -77 }{ 72 } z=\cfrac { 38500 }{ 72 } \)

\(\Rightarrow z=\cfrac { -38500 }{ -77 } \)
\(\Rightarrow z=500\)
\((2)\Rightarrow y+\cfrac { 5 }{ 8 } \left( 500 \right) =\cfrac { 18500 }{ 8 } \)
\(y=\cfrac { 18500 }{ 8 } -\cfrac { 2500 }{ 8 } \)

\(\Rightarrow y=2000\)
\((1)\Rightarrow x-3\left( 2000 \right) -500=-5500\)
\(\Rightarrow x-6000-500=-5500\)
\(\Rightarrow x-6000-500=-5500\)
\(\Rightarrow x=-5500+6500\)
\(\Rightarrow x=1000\)
Hence, the prices per unit of three commodities are Rs.1000, Rs. 2000 and Rs. 500 respectively

We have, P = a + bl + cm
Putting above values we have
6,950 = a + 40b + 10c
6,725 = a + 35b + 9c
7,100 = a + 40b + 12c
The Matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 40 & 10 \\ 1 & 35 & 9 \\ 1 & 40 & 12 \end{matrix} \right) \left( \begin{matrix} a \\ b \\ c \end{matrix} \right) =\left( \begin{matrix} 6950 \\ 6725 \\ 7100 \end{matrix} \right) \)

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 1 & 40 & 10 \\ 1 & 35 & 9 \\ 1 & 40 & 12 \end{matrix}\begin{matrix} 6950 \\ 6725 \\ 7100 \end{matrix} \right) \) \(\left( \begin{matrix} 1 & 40 & 10 \\ 0 & -5 & -1 \\ 0 & 0 & 2 \end{matrix}\begin{matrix} 6950 \\ -225 \\ 150 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }\)
\(\rho (A)=3,\rho ([A,B])=3\)

\(\therefore \) The given system is equivalent to the matrix equation
\(\left( \begin{matrix} 1 & 40 & 10 \\ 0 & -5 & -1 \\ 0 & 0 & 2 \end{matrix} \right) \left( \begin{matrix} a \\ b \\ c \end{matrix} \right) =\left( \begin{matrix} 6950 \\ -225 \\ 150 \end{matrix} \right) \)
a + 40b + 10c = 6950    (1)
-5b - c = -225    (2)
2c = 150  (3)
c- = 75
Now, (2) \(\Rightarrow \) -5b - 75 = -225
b = 30
and (1) \(\Rightarrow \) a + 1200 + 750 = 6950
a = 5000
a = 5000, b = 30, c = 75
\(\therefore \) The production equation is P = 5000 + 301 + 75m
\(\therefore \) P_{at l = 50, m = 15} = 5000 + 30(50) + 75(15)
= 7625 units.
\(\therefore \) The production = 7,625 units.

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
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 6 \\ 14 \\ 30 \end{matrix} \right) \)
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.

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 5 \\ 4 \\ 1 \end{matrix} \right) \)
A X = B

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{matrix}\begin{matrix} 5 \\ 4 \\ 1 \end{matrix} \right) \)  \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{matrix}\begin{matrix} 4 \\ 5 \\ 1 \end{matrix} \right) \) \( \left( \begin{matrix} 1 & 1 & 1 \\ 0 & -1 & -1 \\ 1 & -2 & 1 \end{matrix}\begin{matrix} 4 \\ -3 \\ -3 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & -1 \\ 0 & 0 & 3 \end{matrix}\begin{matrix} 4 \\ -3 \\ 3 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 2 }\) \({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ 2R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ 2R }_{ 2 }\)
\(\rho (A)=3,\) \(\rho ([A,B])=3\)

Obviously the last equivalent matrix is in the echelon form. It has three non-zero rows.
\(\rho (A)=3,\) \(\rho ([A,B])=3\) = Number of unknowns .
The given system is consistent and has unique solution.
To find the solution, let us rewrite the above echelon form into the matrix form.  
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & -1 & -1 \\ 0 & 0 & 3 \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 4 \\ -3 \\ 3 \end{matrix} \right) \)
x + y + z = 4   (1)
y + z = 3   (2)
3z = 3   (3)
\((3)\Rightarrow z=1\)
\((2)\Rightarrow y=3-z=2 \)
\((1)\Rightarrow x=4-y-z\)
x = 1
\(\therefore \)  x = 1, y = 2, z = 1

(i) \(\int _{ -2 }^{ 1 }{ f(x) } dx=\int _{ -2 }^{ 1 }{ { x }^{ 2 }dx } ={ \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ -2 }^{ 1 }=\frac { 1 }{ 3 } -\left( \frac { -8 }{ 3 } \right) =3\)
(ii) \(\int _{ 1 }^{ 2 }{ f(x) } dx=\int _{ 1 }^{ 2 }{ xdx } ={ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 1 }^{ 2 }=\frac { 4 }{ 2 } -\frac { 1 }{ 2 } =\frac { 3 }{ 2 } \)
(iii) \(\int _{ 2 }^{ 3 }{ f(x) } dx=\int _{ 2 }^{ 3 }{ (x-4) } dx{ \left[ \frac { { x }^{ 2 } }{ 2 } -4x \right] }_{ 2 }^{ 3 }=\left( \frac { 9 }{ 2 } -12 \right) -\left( \frac { 4 }{ 2 } -8 \right) \) \(=\frac { 15 }{ 2 } +6=\frac { -3 }{ 2 } \)
(iv) \(\int_{-2}^{1.5} f(x) d x\) = \(\int_{-2}^{1} f(x) d x\) + \(\int_{1}^{1.5} f(x) d x\)
= 3 + \({ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 1 }^{ 1.5 } \) 
=  3 + \(\frac { 225 }{ 2 } -\frac { 1 }{ 2 } = 3 + \frac { 125 }{ 2 } = 3.625\)
(v) \(\int_{1}^{3} f(x) d x\) = \(\int_{1}^{2} f(x) d x\) + \(\int_{2}^{3} f(x) d x\)
= \(\frac { 3 }{ 2 } + (\frac { -3 }{ 2 }) = 0\) using (ii) and (iii)

(i) Take a suitable scale for the values of x and y, and plot the various points on the graph paper for given values of x and y.
(ii) Draw a suitable curve passing through the plotted points.
(iii) Find the point corresponding to the value x = 38 on the curve and then read the corresponding value of y on the y- axis, which will be the required interpolated value.
From the graph in Figure we find that for x = 38, the value of y is equal to 35

Given equation can be written as \(\frac { ydx-xdy }{ { y }^{ 2 } } -{ 3x }^{ 2 }{ e }^{ { x }^{ 2 } }dx=0\)
Integrating, ഽ\(\frac { ydx-xdy }{ { y }^{ 2 } } \) - ഽ3x²e^x3 dx = c
ഽ\(d\left( \frac { x }{ y } \right) \) - ഽe^tdt = c  
(where t = x³ and dt = 3x²dx )
\(\frac { x }{ y } \) - e^t = c
\(\frac { x }{ y } \) - \({ e }^{ { x }^{ 2 } }\) = c