Commodity	Price		Quantity
	Base year p0	Current year p1	Base year q0	Current year q1
A	2	4	8	6
B	5	6	10	5
C	4	5	14	10
D	2	2	19	13

 

p0q0	p1q0	p0q1	p1q1
16	32	12	24
50	60	25	30
56	70	40	50
38	38	26	26
160	200	103	130

(i) Laspeyre's index number
\(P_{01}^{L} = \frac{\Sigma p_1q_0}{\Sigma p_0q_0}\times100\)
\(=\frac{200}{160}\times100=125\)
(ii) Paasche's Price index number
\(P_{01}^{P} = \frac{\Sigma p_1q_1}{\Sigma p_0q_1}\times100\)
\(=\frac{130}{103}\times100=126.21\)
(iii) Fisher's price index number
\(P_{01}^{F} =\sqrt {{P_{01}^{L}}\times{P_{01}^{P}}}=125.6\)

Let I = ഽx. log (1 + x) dx

u = log (1 + x); dv = x dx
\(du=\frac { 1 }{ 1+x } dx;\quad v=\frac { { x }^{ 2 } }{ 2 } \)
Using integration by parts we get,
I = ഽu dv = uv - ഽv du
= ഽ x log (1 + x) dx
= \(\frac { { x }^{ 2 } }{ 2 } log(1+x)-\int { \frac { { x }^{ 2 } }{ 2 } .\frac { 1 }{ 1+x } } dx\)
= \(\frac { { x }^{ 2 } }{ 2 } log(1+x)-\frac { 1 }{ 2 } \int { \frac { { x }^{ 2 } }{ 1+x } } dx\)
= \(\frac { { x }^{ 2 } }{ 2 } log(1+x)-\frac { 1 }{ 2 } \int { \frac { { x }^{ 2 }-1+1 }{ 1+x } } dx\)
[Adding & subtracting 1 in the numerator]
= \(\frac { { x }^{ 2 } }{ 2 } log(1+x)-\frac { 1 }{ 2 } \)
 
= \(\frac { { x }^{ 2 } }{ 2 } log(1+x)-\frac { 1 }{ 2 } \left[ \int { \left( x-1 \right) dx+log\left| 1+x \right| } \right] +c\)
= \(\frac { { x }^{ 2 } }{ 2 } log(x+1)-\frac { 1 }{ 2 } \left[ \frac { { x }^{ 2 } }{ 2 } -x+log\left| 1+x \right| \right] +c\)

Given \(\frac { dp }{ dx } =\frac { 2{ p }^{ 3 }-{ x }^{ 3 } }{ 3x{ p }^{ 2 } } \)
The numerator and denominator are homogeneous functions of 3,
∴ Put p=vx and \(\frac { dp }{ dx } =v+x\frac { dv }{ dx } \)

= \(\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } \)
⇒ \(\frac { dv }{ dx } =\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } -v=\frac { 2{ v }^{ 3 }-1-3{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
= \(\frac { -1-{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
⇒ \(\left( \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } \right) dv=-\frac { dx }{ x } \)
Integrating, \(\int { \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } } dv=-\int { \frac { dx }{ x } } \)
⇒ log(1+v³) = -log x + log c
⇒ 1+v³ = \(\frac { c }{ x } \)
Replacing v by \(\frac { p }{ x } \) we get
\(1+\frac { { p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \Rightarrow \frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \)
⇒ \(\frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 2 } } \)= c ⇒ x³+p³ = cx²...(1)
When x = 10, p = 20
⇒ 10³ + 20³ = c(10)² ⇒ 1000 + 8000 = 100 c
⇒ 9000 = 100 c
⇒ c = 90
∴ (1) becomes,
x³+p³ = 90x²
⇒ p³ = 90x²-x³
⇒ p³ = x²(90-x) which is the required relationship.

Given' R'(x) = 7 + e^-0.05x
Total revenue from sale of 100 units is
\(R=\int _{ 0 }^{ 100 }{ (7+{ e }^{ -0.05x })dx } \)
\(={ \left[ 7x+\frac { { e }^{ -0.05x } }{ -0.05 } \right] }_{ 0 }^{ 100 }\)
\(=700-\frac { 100 }{ 5 } ({ e }^{ -5 }-{ e }^{ 0 })\left[ 0.05=\frac { 5 }{ 100 } \right] \)
= 700-20(0.0067-1)[∵e⁰=1]
= 700-0.134
= 719.866
Since the revenue is given in thousands,
Total revenue = 719.866 x 1000
= Rs. 7,19,866

The given non-homogeneous equations can be written as
\(\left( \begin{matrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \)

Augmented matrix [A, B]	Elementary Transformation
\(\left( \begin{matrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{matrix}\begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 1 & k \\ 1 & k & 1 \\ k & 1 & 1 \end{matrix}\begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 1 & k \\ 0 & k-1 & 1-k \\ 0 & 1-k & 1-{ k }^{ 2 } \end{matrix}\begin{matrix} 1 \\ 0 \\ 1-k \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-R_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-k{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 1 & k \\ 0 & k-1 & 1-k \\ 0 & 0 & 2-k-{ k }^{ 2 } \end{matrix}\begin{matrix} 1 \\ 0 \\ 1-k \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }{ +R_{ 2 } }\)

Case (i):
When  \(k\neq 1\) and \(k\neq 2\)
\(\rho (A)=\rho (A,B)=3=\) Number of unknowns
\(\therefore \) The system has unique solution 
Case (ii):
When k = 1
\(\left[ A,B \right] \sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right) \)
\(\rho (A)=\rho\) (A, B) = 1
\(\therefore \) The system is consistent and has infinitely many solutions.
Case (iii):
When k = - 2
\(\left[ A,B \right] \sim \left( \begin{matrix} 1 & 1 & -2 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 0 \\ -3 \end{matrix} \right) \)
\(\rho (A)=2\rho (A,B)\)= 3
\(\Rightarrow \rho (A)\neq 2\rho (A,B)\)
\(\therefore\) The system is inconsistent and has no solution.

\(\Delta =\left| \begin{matrix} 2 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{matrix} \right| =2\)
\(\left| \begin{matrix} 1 & 1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -2 \end{matrix} \right| \)
= 2 (-3+2) - 1 (-3 -1) -1 (-2-1)
= 2(-1) -1 (-4) -1 (-3)
= -2 + 4 + 3 = 5.
Since\(\Delta \neq 0\),
we can apply Cramer's rule and the system is consistent with unique solution.
\(x=\left| \begin{matrix} 3 & 1 & -1 \\ 1 & 1 & 1 \\ 4 & -2 & 3 \end{matrix} \right| =3\left| \begin{matrix} 1 & 1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 4 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 4 & -2 \end{matrix} \right| \)
= 3(-3 + 2) -1(-3 -4) -1(-2 -4)
= 3 (-1) -1 (-7) -1 (-6)
= -3 + 7 + 6 = 10.
\(\Delta y=\left| \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & 4 & -3 \end{matrix} \right| =2\left| \begin{matrix} 1 & 1 \\ 4 & -3 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix} \right| \)
= 2(-3 -4) -3 (-3 -1) -1 (4-1)
= 2 (-7) -3 (-4) -1(3)
= 14 + 12 - 3 = -5
\(\Delta z=\left| \begin{matrix} 2 & 1 & 3 \\ 1 & 1 & 1 \\ 1 & -2 & 4 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 1 & 1 \\ -2 & 4 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix} \right| +3\left| \begin{matrix} 1 & 1 \\ 1 & -2 \end{matrix} \right| \)
= 2(4 + 2) -1(4 -1) + 3(-2 -1)
= 2(6) -1(3) + 3(-3)
= 12 - 3 - 9
= 0

\(z=\cfrac { \Delta z }{ \Delta } =\cfrac { 0 }{ 5 } =0\)
\(\therefore\)Solution set is (2, -1, 0)
 

Commodities	Price		Quantity
	Base year (p0)	Current year (p1)	q0	q1
Wheat	6	10	50	56
Ghee	2	2	60	60
Firewood	4	6	60	60
Sugar	10	12	30	24
Cloth	8	12	40	36

p0q0	p1q1	p0q1	p1q0
300	560	336	500
200	240	240	200
240	360	240	360
300	288	240	360
320	432	288	480
1360	1880	1344	1900

Fisher's price index number
\(P^{F}_{01}\) = \(\sqrt{ \frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}} \times ​\frac{​ {\sum p_{1}q_{1}} }{{\sum p_{0}q_{1}}}{}}\times 100\)
= \(\sqrt\frac {1900 \times 1880}{1360 \times 1344} \times 100\)
= \(\sqrt\frac {3572000}{1827840} \times 100\) 
= \(\sqrt {1.954} \times 100\)
\(P^{F}_{01}\)  = 139.8
Time reversal test:
\(P _{01} \times P_{10}\) = \(\sqrt{ \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}\times \sum p_{0}q_{1}\times \sum p_{0}q_{0}}{\sum p_{0}q_{0}\times \sum p_{0}q_{1}\times \sum p_{1}q_{1} \times \sum p_{1}q_{0}} ​}\)
=  
=  \(\sqrt {1}\)  = 1
∴    \(P _{01} \times P_{10}\) = 1
Factor reversal test:
\(P _{01} \times Q_{01}\) = \(\sqrt{ \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}\times \sum q_{1}p_{0}\times \sum q_{1}p_{1}}{\sum p_{0}q_{0}\times \sum p_{0}q_{1}\times \sum q_{0}p_{0} \times \sum q_{0}p_{1}} ​}\)

= \(\frac {1880}{1360} = \frac {\sum p_{1}q_{1}}{\sum p_{0}q_{0}}\)
Hence, it satisfies time reversal test and factor reversal test.

Given differential equation is of the form
\(\frac { dy }{ dx } \)+Py = Q where
P = 2 tanx; Q = sinx
\(\int { p } dx=\int { 2 } \tan xdx\) = 2(log secx) = log sec²x
∴ Integrating factor (I. F) =\(e^{ \int { p } dx }=e^{ log\quad sec^{ 2x } }\)
= sec²x
Hence the solution is
\(ye^{ \int { p } dx }=\int { Q } .e^{ \int { p } dx }dx+c\)
⇒ y sec²x =\(\int { \sin x }\ { \sec }^{ 2 }x\ dx\)+c
⇒ y sec²x =\(\int { \sin x } .\frac { 1 }{ \cos^{ 2 }x } \) dx + c
⇒ y sec²x =\(\int { \frac { \sin x }{ \cos x } .\frac { 1 }{ \cos x } } \) dx + c
⇒ y sec²x =\(\int { \tan x } \) sec x dx + c
⇒ y sec²x = sec x + c ...(1)
Also its given when x = \(\frac { \pi }{ 3 } \), y = 0
∴ 0\(({ sec }^{ 2 }\frac { \pi }{ 3 } )=sec\frac { \pi }{ 3 } +c\)
⇒ 0 = 2+c ⇒ c = -2
∴ (1) becomes
y sec 2x = sec x-2

x²dy + y(x + y) dx = 0
x²dy = −y( x+y) dx
\(\frac { dy }{ dx } =\frac { -(xy+{ y }^{ 2 }) }{ { x }^{ 2 } } \)
Put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \) in (1)
\(v+x\frac { dv }{ dx } =\frac { (xvx+{ v }^{ 2 }{ x }^{ 2 }) }{ { x }^{ 2 } } \)
= −(v + v²)
\(x\frac { dv }{ dx } \) = −v²−v− v
= −(v² + 2v)
On separating the variables
\(\frac { dv }{ { v }^{ 2 }+2v } =-\frac { dx }{ x } \)
\(\frac { dv }{ v(v+2) } =\frac { -dx }{ x } \)
\(\frac { 1 }{ 2 } \left[ \frac { (v+2)-v }{ v(v+2) } \right] dv=\frac { -dx }{ x } \)
\(\frac { 1 }{ 2 } ഽ\left[ \frac { 1 }{ v } -\frac { 1 }{ v+2 } \right] dv=-ഽ\frac { dx }{ x } \)
\(\frac12\) [log v - log (v + 2)] = - log x + log c
\(\frac { 1 }{ 2 } log\frac { v }{ v+2 } =log\frac { c }{ x } \)
We have
\(\frac { v }{ v+2 } =\frac { { c }^{ 2 } }{ { x }^{ 2 } } \)
Replace v = \(\frac{y}{x}\), we get
\(\frac { y }{ x\left( \frac { y }{ x } +2 \right) } =\frac { k }{ { x }^{ 2 } } \)  where c² =k
\(\frac { y{ x }^{ 2 } }{ y+2x } =k\)
When x = 1, y = 1
∴ (2) ⇒ k = \(\frac { 1 }{ 1+2 } \)
k = \(\frac { 1 }{ 3 } \)
∴ The solution is 3x² y = 2x + y

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

The present value of payment for t year = \(\int _{ 0 }^{ t }{ { 72000e }^{ -0.09t } } dt\)
Present value of 12 years = \(\int _{ 0 }^{ t }{ { 72000e }^{ -0.09t } } dt\)
= 72000\({ \left[ \frac { { e }^{ -0.09t } }{ { -0.09 } } \right] }_{ 0 }^{ \\ 12 }\)
= \(\frac { 72000 }{ -0.09 } \left[ { e }^{ -0.09(12) }-{ e }^{ 0 } \right] \)
= \(-8,00,000[{ e }^{ -1.08 }-{ e }^{ 0 }]\)
= −8,00,000 [0.3396 −1]
= 5,28,320
Cost of the machine = 5,00,000 − 30,000
= 4,70,000
Hence it not advisable to rent the machine
It is better to buy the machine.

\(\int { { \left( \log x \right) }^{ 2 } } dx= \int { udv } \)
= uv − \(\int { } \)vdu
= x (log x)² − 2\(\int { } \) logxdx...(*)
\(=x(\log x)^{ 2 }-2\int { udv } \)
\(=x(\log x)^{ 2 }-2[uv-\int { udv } ]\)
\(=x(\log x)^{ 2 }-2[x \log x-\int { dx] } \)
\(=x(\log x)^{ 2 }-2x \log x+x+c\)
\(=x[(\log{ ) }^{ 2 }-\log{ x }^{ 2 }+2]+c\)

For \(\int { } \)log x dx in (*)
Take u = (log x) Differentiate \(du=\frac { 1 }{ x } dx\)	and dv = dx Integrate v = x

Transition probability matrix
(A B) T = (A B)

Where A represents the percent of people those who bought soap A and B represents the percent of people those who bought soap B.
By the given data
A = 15% = ·15
and B = 85% = ·85
Percentage after one year is
\(\left( \cdot 15\quad \cdot 85 \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) \)
= ((.15)(·65) + (·85)(-45) ·15(-35)+ ·85(-55))
= (-0975 + ·3825 ·0525 + -4675)
= (-48 ·52)
Hence, market share after one year is 48% and 52% At equilibrium,
\(\left( A\quad B \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) =(A\quad B)\)
(-65A + A5B ·35A +·55B) = (A B)
Equating the corresponding entries on both sides we get
\(\Rightarrow \cdot 65A+\cdot 45B=A\)
\(\Rightarrow \cdot 65A+\cdot 45(1-A)=A\)
[Since A+B = 1   B = 1-A]
\(\Rightarrow \cdot 65A+\cdot 45-\cdot 45A=A\)
\(\Rightarrow \cdot 45=A-\cdot 65A+\cdot 45A\)
\(\Rightarrow \cdot 45=A\left( \cdot 35+45 \right) \)
\(\Rightarrow \cdot 45=A(\cdot 35+45)\)
\(\Rightarrow \cdot 45=A(-8)\)
\(\Rightarrow A=\cfrac { \cdot 45 }{ \cdot 8 } =\cdot 5625=56.25\)
\(\therefore B=1-A=1-\cdot 5625=\cdot 4375\)
= 43.75%
\(\therefore\) Equilibrium is reached when A = 56.25% and B = 43.75%

Let the weight assigned to the three varieties be Rs. x, Rs. y and Rs. z respectively By the given data,
x + 2y + 3z = 11
2x + 4y + 5z = 21
3x + 5y + 6z = 27
\(\Delta =\left| \begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{matrix} \right| =1\left| \begin{matrix} 4 & 5 \\ 5 & 6 \end{matrix} \right| -2\left| \begin{matrix} 2 & 5 \\ 3 & 6 \end{matrix} \right| +3\left| \begin{matrix} 2 & 4 \\ 3 & 5 \end{matrix} \right| \)
= 1(24 - 25) -2(12 - 15) + 3(10 - 12)
= 1(-1) -2 (-3) + 3(-2)
= -1+6 - 6 = -1\(\neq \) 0.
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied.
\(\Delta x=\left| \begin{matrix} 11 & 2 & 3 \\ 21 & 4 & 5 \\ 27 & 5 & 6 \end{matrix} \right| \)
\(=11\left| \begin{matrix} 4 & 5 \\ 5 & 6 \end{matrix} \right| -2\left| \begin{matrix} 21 & 5 \\ 27 & 6 \end{matrix} \right| +3\left| \begin{matrix} 21 & 4 \\ 27 & 5 \end{matrix} \right| \)
= 11(24 - 25) - 2(126 - 135) + 3(105 - 108)
= 11(-1) - 2(-9) + 3 (-3)
= 11+18-9
= -2
\(\Delta y=\left| \begin{matrix} 1 & 11 & 3 \\ 2 & 21 & 5 \\ 3 & 27 & 6 \end{matrix} \right| \)
= \(\left| \begin{matrix} 21 & 5 \\ 27 & 6 \end{matrix} \right| -11\left| \begin{matrix} 2 & 5 \\ 3 & 6 \end{matrix} \right| +3\left| \begin{matrix} 2 & 21 \\ 3 & 27 \end{matrix} \right| \)
= 1(126 - 135) - 11(12 -15) + 3(54 - 63)
= - 9 - 11(-3) + 3(-9)
= - 9 + 33 - 27
= 3
\(\Delta z=\left| \begin{matrix} 1 & 2 & 11 \\ 2 & 4 & 21 \\ 3 & 5 & 27 \end{matrix} \right| =1\left| \begin{matrix} 4 & 21 \\ 5 & 27 \end{matrix} \right| -2\left| \begin{matrix} 2 & 21 \\ 3 & 27 \end{matrix} \right| +11\left| \begin{matrix} 2 & 4 \\ 3 & 5 \end{matrix} \right| \)
= 1(108 - 105) - 2(54 - 63) + 11(10 - 12)
= 1(3) - 2(-9) + 11(-2)
= 3 + 18 - 22
= -1
\(x=\cfrac { \Delta x }{ \Delta } =\cfrac { -2 }{ -1 } =2\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -3 }{ 1 } =3\)
and \(z=\cfrac { \Delta z }{ \Delta } =\cfrac { -1 }{ -1 } =1\)
Hence, the weights assigned to the three varieties are 2, 3 and 1 respectively

Let ‘x’ be the cost of a Business Mathematics book
Let ‘y’ be the cost of a Accountancy book.
Let ‘z’ be the cost of a Commerce book.
\(\therefore \) 3x + 2y + z = 840
2x + y + z = 570
x + y + 2z = 630
Here \({ \triangle }=\left| \begin{matrix} 3 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{matrix} \right| =-2\neq 0\)
\({ \triangle }_{ x }=\left| \begin{matrix} 840 & 2 & 1 \\ 570 & 1 & 1 \\ 630 & 1 & 2 \end{matrix}\begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right| =-240 \)
\({ \triangle }_{ y }=\left| \begin{matrix} 3 & 840 & 1 \\ 2 & 570 & 1 \\ 1 & 630 & 2 \end{matrix} \right| =-300 \)
\({ \triangle }_{ z }=\left| \begin{matrix} 3 & 2 & 840 \\ 2 & 1 & 570 \\ 1 & 1 & 630 \end{matrix} \right| =-360\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -240 }{ -2 } =120 \)
\(y=\frac { { \triangle }y }{ { \triangle } } =\frac { 300 }{ -2 } =150 \)
\(z=\frac { { \triangle }z }{ { \triangle } } =\frac { 360 }{ -2 } =180\) 
\(\therefore \) The cost of a Business Mathematics book is Rs. 120,
the cost of a Accountancy book is Rs. 150 and
the cost of a Commerce book is Rs. 180.

Given non-homogeneous equations are
5x+ 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 4 \\ 9 \\ 5 \end{matrix} \right) \)

Augmented matrix [A, B]	Elementary Transformation
\(\left( \begin{matrix} 5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10 \end{matrix}\begin{matrix} 4 \\ 9 \\ 5 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 3 & 26 & 2 \\ 5 & 3 & 7 \\ 7 & 2 & 10 \end{matrix}\begin{matrix} 9 \\ 4 \\ 5 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 2 }\)
\(\left( \begin{matrix} 1 & \frac { 26 }{ 3 } & \frac { 2 }{ 3 } \\ 5 & 3 & 7 \\ 7 & 2 & 10 \end{matrix}\begin{matrix} 3 \\ 4 \\ 5 \end{matrix} \right) \)	\({ R }_{ 1 }\rightarrow { R }_{ 1 }\div 3\)
\(\left( \begin{matrix} 1 & \frac { 26 }{ 3 } & \frac { 2 }{ 3 } \\ 0 & \frac { -121 }{ 3 } & \frac { 11 }{ 3 } \\ 7 & 2 & 5 \end{matrix}\begin{matrix} 3 \\ -11 \\ 5 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-5{ R }_{ 1 }\)
\(\left( \begin{matrix} 1 & \frac { 26 }{ 3 } & \frac { 2 }{ 3 } \\ 0 & \frac { -121 }{ 3 } & \frac { 11 }{ 3 } \\ 0 & \frac { -176 }{ 3 } & \frac { 16 }{ 3 } \end{matrix}\begin{matrix} 3 \\ -11 \\ -16 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-7{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & \frac { 26 }{ 3 } & \frac { 2 }{ 3 } \\ 0 & \frac { -11 }{ 3 } & \frac { 1 }{ 3 } \\ 0 & \frac { -11 }{ 3 } & \frac { 1 }{ 3 } \end{matrix}\begin{matrix} 3 \\ -1 \\ -1 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }\div 11\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }\div 16\)
\(\left( \begin{matrix} 1 & \frac { 26 }{ 3 } & \frac { 2 }{ 3 } \\ 0 & \frac { -11 }{ 3 } & \frac { 1 }{ 3 } \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 3 \\ -1 \\ 0 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

\(\therefore\) The system is consistent with infinitely many solutions let us rewrite the above echelon form into matrix form
\(\left( \begin{matrix} 1 & \frac { 26 }{ 3 } & \frac { 2 }{ 3 } \\ 0 & \frac { -11 }{ 3 } & \frac { 1 }{ 3 } \\ 0 & 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 3 \\ -1 \\ 0 \end{matrix} \right) \)
\(x+\cfrac { 26 }{ 3 } y+\cfrac { 2 }{ 3 } z=3\)
\(x+\cfrac { 26 }{ 3 } y+\cfrac { 2 }{ 3 } z=3\)
let z = k where k\(\in\) R
\((2)\Rightarrow \cfrac { -11 }{ 3 } y+\cfrac { k }{ 3 } =-1\)

\(\Rightarrow \ -11y=-3-k\) 
11y = 3 + k
\(\Rightarrow \quad y=\cfrac { 1 }{ 11 } \left( 3+k \right) \)
Substituting \(y=\cfrac { 1 }{ 11 } \left( 3+k \right) \) and z = k in (1) we get,
\(x+\cfrac { 26 }{ 3 } \left( \cfrac { 3+k }{ 11 } \right) +\cfrac { 2 }{ 3 } k=3\)
\(=\cfrac { 26 }{ 3 } \left( \cfrac { 3+k }{ 11 } \right) -\cfrac { 2k }{ 3 } +3\)
\(\cfrac { 78-26k }{ 33 } -\cfrac { 2k }{ 3 } +3\)
\(\cfrac { 78-26k-22k+99 }{ 33 } \)
\(\cfrac { 78-26k-22k+99 }{ 33 } \)
\(\cfrac { 21-48k }{ 33 } =\cfrac { 3(7-16k) }{ 33 } \)
= \(\cfrac { 1 }{ 11 } (7-6k)\)
 \(\therefore\) Solution set is \(\left\{ \cfrac { 1 }{ 11 } \left( 7-16k \right), \cfrac { 1 }{ 11 } (3+k),k \right\} \)K \(\in\) R
Hence, for different values of k, we get infinitely many solutions.

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 2 & -3 \\ 3 & -1 & -2 \\ 2 & 3 & -5 \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} -2 \\ 1 \\ k \end{matrix} \right) \)
AX = B

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 1 & 2 & -3 \\ 3 & -1 & -2 \\ 2 & 3 & -5 \end{matrix}\begin{matrix} -2 \\ 1 \\ k \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 2 & -3 \\ 0 & -7 & 7 \\ 0 & -1 & 1 \end{matrix}\begin{matrix} -2 \\ 7 \\ 4+k \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 2 & -3 \\ 0 & -7 & 7 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} -2 \\ 7 \\ 21+7k \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ 3R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow 7{ R }_{ 3 }-{ R }_{ 2 }\)
\(\rho (A)=2,\rho ([A,B])=2\quad or\quad 3\)

For the equations to be consistent, \(\rho ([A,B])=\)\(\rho (A)=2\)
\(\therefore \) 21 + 7k = 0
7k = -21
k = -3

The errors that arise due to human factors which always vary from one investigator to another in selecting, estimating or using measuring instruments are called Non-Sampling errors.
It may arise in the following ways:
a) Due to negligence and carelessness of the part of either investigator or respondents.
b) Due to lack of trained and qualified investigators.
c) Due to framing of a wrong questionnaire.
d) Due to applying wrong statistical measure.
e) Due to incomplete investigation and sample survey.

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

Annual consumption at timet = 0 (In the year 2000) = p₀ = 2547 metric ton.
Total production of Gold from 2000 to 2013 = \(\int _{ 0 }^{ 13 }{ 2547e^{ 0.006t } } dt\)
= \(\frac { 2547 }{ 0.006 } \left[ e^{ 0.006t } \right] _{ 0 }^{ 13 }\)
= 424500 (e^0.078 −1)
= 34,426.95 metric tons approximately.