B = \(\left[ \begin{matrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{matrix} \right] \)
I - B=\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)-\(\left[ \begin{matrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 0.2 & -0.2 \\ - 0.9 & 0.3 \end{matrix} \right] \)
|I - B|=  \(\left[ \begin{matrix} 0.2 & -0.2 \\ - 0.9 & 0.3 \end{matrix} \right] \)
= (0.2)(0.3) - (-0.2)(-0.9)
= 0.06 - 0.18
= 0.12 < 0
Since |I - B| is negative, Hawkins – Simon conditions are not satisfied.
Therefore, the given system is not viable.

Mean \(\bar { x } =\frac { 2+3+5+6+9+10+12+17+20+26 }{ 10 } \)
\(=\frac { 110 }{ 10 } =11\)

x	|D|=|x-\(\bar { x } \)| = |x-11|
2	9
3	8
5	6
6	5
9	2
10	1
12	1
17	6
20	9
26	15
	\(\sum { |D|=62 } \)

\(\therefore \) Mean deviation from mean = \(\frac { \sum { |D| } }{ n } =\frac { 62 }{ 10 } =6.2\)

Since the decision variables x₁ ,x₂ are non-negative, the solution lies in the I quadrant of the plane.
Consider the equations
\(-{ x }_{ 1 }+3{ x }_{ 2 }=10\)

\({ x }_{ 1 }\)	0	2
\({ x }_{ 2 }\)	10/3	4

\({ x }_{ 1 }+{ x }_{ 2 }=6\)

\({ x }_{ 1 }\)	0	6
\({ x }_{ 2 }\)	6	6

\({ x }_{ 1 }{ -x }_{ 2 }=2\)

\({ x }_{ 1 }\)	4	2
\({ x }_{ 2 }\)	2	0

The feasible region is OABCD and its co-ordinates are O(0, 0)A(2, 0) B(4, 2) C(2, 4) and D(0, 10/3)

Corner Points	\(Z=-{ x }_{ 1 }+2{ x }_{ 2 }\)
0(0,0)	0
A(2, 0)	-2
B (4, 2)	0
C(2,4)	6
D\(\left( 0,\frac { 10 }{ 3 } \right) \)	\(\frac { 20 }{ 3 } \)

Maximum of Z occurs at\(D\left( 0,\frac { 10 }{ 3 } \right) \). Hence, the solution is \({ x }_{ 1 }=0,{ x }_{ 2 }=\frac { 10 }{ 3 } \quad and\quad { Z }_{ max }=\frac { 20 }{ 3 } \)

Given a = Rs.500,i = 5% = 0.05,n =10
A =\(\cfrac { a }{ i } \left( 1+i \right) \left[ \left( 1+i \right) ^{ n }-1 \right] \)
\(=\frac { 500 }{ 0.05 } (1.05)\left[ (1.05)^{ 10 }-1 \right] \)
\(=10,500[1.629-1]\)
\(=10,500(0.629)\)
\(\therefore\) = Rs. 6604.50
 At the end of 10 years, I will receive  Rs. 6604.50.
\((1.05)^{ 10 }=10log(1.05)\)
\(=0.2120\)
Antilog of 0.2120 is 1.629

Let E₁ = the first compartment of the purse is chosen
E₂ = the second compartment of the purse is chosen
A = a rupee coin is drawn from the purse.
Since one of the two compartments is chosen randomly
\(P({ E }_{ 1 })=\frac { 1 }{ 2 } =P({ E }_{ 2 })\)
Also, P(A/E₁) = P(Drawing a rupee coin from the I compartment) = \(\frac { 2 }{ 5 } \)
P(A/E₂) = P(Drawing a rupee coin from the II compartment) = \(\frac { 3 }{ 5 } \)
\(\therefore\) By the law of total probability,
P(drawing a one-rupee coin) = P(A) = P(E₁).P(A/E₁)+P(E₂).P(A/E₂)
\(=\frac { 1 }{ 2 } \times \frac { 2 }{ 5 } +\frac { 1 }{ 2 } \times \frac { 3 }{ 5 } =\frac { 2 }{ 10 } +\frac { 3 }{ 10 } =\frac { 5 }{ 10 } =\frac { 1 }{ 2 } \)

Correlation co-efficient
r(X, Y) =\(\frac { N\sum { XY-(\sum { X)(\sum { Y) } } } }{ \sqrt { n\sum { { X }^{ 2 }-{ (\sum { X) } }^{ 2 } } } \sqrt { n\sum { { Y }^{ 2 }-({ \sum { Y) } }^{ 2 } } } } \)
=\(\frac { 25(50)-(125)(100) }{ \sqrt { (1650)(25)-{ (125) }^{ 2 } } \sqrt { (1460)(25)-{ (100) }^{ 2 } } } \)
=\(\frac { -450 }{ \sqrt { 1125\sqrt { 1060 } } } \)
r =-0.41

x	f	c.f
0	150	150
10	142	292
20	130	422
30	120	542
40	72	614
50	30	644
60	12	656
70	4	660
	N = 660

Q₁ = Size of \({ \left( \frac { N+1 }{ 4 } \right) }^{ th }\) value
= Size of (165.25)^th value = 10
Q₃ = Size of 3\({ \left( \frac { N+1 }{ 4 } \right) }^{ th }\)value
= size of (495.75)^th value = 30
Q.D = \(\frac { 1 }{ 2 } ({ Q }_{ 3 }-{ Q }_{ 1 })=\frac {30-10}{2}=\frac { 20 }{ 2 } =10\)
Co-efficient of Q.D = \(\frac { { Q }_{ 3 }-{ Q }_{ 1 } }{ { Q }_{ 3 }+{ Q }_{ 1 } } =\frac { 20 }{ 40 } =\frac { 1 }{ 2 } \) = 0.5

Let us define the following events.
E₁: X speaks truth
E₂: X tells a lie
E: X reports a six
From the data given in the problem, we have
P(E₁) = \(\frac{4}{5}\); P(E₂) = \(\frac{1}{5}\);
P(E/E₁) = \(\frac{1}{6}\); P(E/E₂) = \(\frac{5}{6}\)
The required probability that actually there was six (by Bayes theorem) is 
\(P(E_1/E)=\frac{P(E_1)P(E/E_2)}{P(E_1)P(E/E_1)+P(E_2)P(E/E_2)}\) = \(\frac{\frac{4}{5}\times \frac{1}{6}}{(\frac{4}{5}\times \frac{1}{6})+(\frac{1}{5}\times \frac{5}{6})}=\frac{4}{9}\)

A = 30,200 i = \(\frac{8}{400}\) = 0.02, n = 10 x 4 = 40
A = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
30,200 ​\( =\cfrac { a }{ 0.02 } \left[ \left( 1.02 \right) ^{ 40 }-1 \right] \)
30,200 x 0.02 = a[2.2080-1]   
a =​​​​​​ ​\(\cfrac { 604 }{ 1.2080 } \) = Rs. 500

Calculation of Harmonic Mean

Marks X	No. of Students f	\(\frac{f}{x}\)
10	20	2.000
20	30	1.500
25	50	2.000
40	15	0.375
50	5	0.100
	N = 120	\(\sum { \left( \frac { 1 }{ X } \right) } \)= 5.975

\(HM=\frac { n }{ \sum { \left( \frac { f }{ X } \right) } } =\frac {120 }{ 5.975 } =20.08\)

Marks are arranged in ascending order
12 15 20 28 30 40 50
n = number of observations = 7
Q₁ = Size of \(\left( \frac { n+1 }{ 4 } \right) ^{th}\)  value
= Size of \(\left( \frac { 7+1 }{ 4 } \right) ^{th}\) value
= Size of 2^nd value = 15
Q₃ = Size of \(\left( \frac { 3(n+1) }{ 4 } \right) ^{th}\) value
= Size of \(\\ \left( \frac { 3\times 8 }{ 4 } \right) ^{th}\) value
= Size of 6^th value = 40
D₆ = Size of \(\left( \frac { 6(n+1) }{ 10 } \right) \)^th value
= Size of \(\\ \left( \frac { 6\times 8 }{ 10 } \right) ^{th}\) value
= Size of 4.8^th value
= Size of 5^th value = 30
P₅₀ = Size of \(\left( \frac {50(n+1) }{ 100 } \right) ^{th}\) value
= Size of 4^th value = 28 
Hence Q₁ = 15, Q₃ = 40, D₆ = 30 and P₅₀ = 28

Using the precedence relationships and following the rules of network construction, the required network diagram is shown in following figure.

(i) Variables: Let x₁, x₂ and x₃ be the number of units of products P₁, P₂ and P₃ to be produced.
(ii) Objective function: Profit on x₁ units of the product P₁ = 20 x₁
Profit on x₂ units of the product P₂ = 25 x₂
Profit on x₃ units of the product P₃ = 15 x₃
Total profit = 20 x₁ + 25 x₂ + 15 x₃
Since the total profit is to be maximized, we have to maximize Z = 20 x₁ + 25 x₂ + 15 x₃
Constraints: 6x₁ + 3x₂ + 12x₃ ≤ 200
2x₁ + 5x₂ + 4x₃ ≤ 350
x₁ + 2x₂ + x₃ ≤ 100
Non-negative restrictions: Since the number of units of the products A, B and C cannot be negative, we have x₁, x₂, x₃ ≥ 0
Thus, we have the following linear programming model.
Maximize Z = 20 x₁ + 25 x₂ + 15 x₃
Subject to 6 x₁ + 3 x₂ + 12 x₃ ≤ 200
2x₁ + 5x₂ + 4x₃ ≤ 350
x₁ + 2x₂ + x₃ ≤ 100
x₁, x₂, x₃ ≥ 0

\(y=\sin ^{ -1 }{ \left( \sqrt { \cos { x } } \right) } \)
Differentiating with respect to 'x' we have
\(\frac { dy }{ dx } =\frac { d }{ dx } .\sin ^{ -1 }{ { \left( \cos { x } \right) }^{ \frac { 1 }{ 2 } } } =\frac { 1 }{ \sqrt { 1-{ \left( \sqrt { \cos { x } } \right) }^{ 2 } } } .\frac { d }{ dx } { \left( \cos { x } \right) }^{ \frac { 1 }{ 2 } }\)
\(=\frac { 1 }{ \sqrt { 1-\cos { x } } } .\frac { 1 }{ 2\sqrt { 1-\cos { x } } } .\frac { d }{ dx } \left( \cos { x } \right) \)
\(=\frac { 1 }{ 2\sqrt { \cos { x } } .\sqrt { 1-\cos { x } } } \left( -\sin { x } \right) =\frac { -\sin { x } }{ 2\sqrt { \cos { x } } .\sqrt { 1-\cos { x } } } \)

The circle touches the co-ordinate axes and the line x = a is shown in the diagram.
\(\therefore \ centre\ is\ \left( \frac { a }{ 2 } ,\frac { a }{ 2 } \right) and\ r=\frac { a }{ 2 } \)
\(\therefore\) There may be two such circles, one lying

above X-axis and other below X-axis.
Equation of the circle lying above the X-axis is
\({ \left( x-\frac { a }{ 2 } \right) }^{ 2 }+{ \left( y+\frac { a }{ 2 } \right) }^{ 2 }={ \left( \frac { a }{ 2 } \right) }^{ 2 }\)

Equation of the circle lying below the X-axis is \({ \left( x-\frac { a }{ 2 } \right) }^{ 2 }+{ \left( y+\frac { a }{ 2 } \right) }^{ 2 }={ \left( \frac { a }{ 2 } \right) }^{ 2 }\)

P(n): "n² + n is even"
Given that P(k) is true.
\(\Rightarrow \)   k² + k is even
\(\Rightarrow \)  k² + k = 2\(\lambda \) for some \(\lambda \)  ....(1)
To prove that P (k + 1) is true.
P (k + 1) : (k + 1)²+ (k + 1) is even
Consider (k + 1)² + (k + 1)
= k²+2k+ 1 +k+ 1
= k²+2k+ 1 +k+ 1
= (k²+ k) + (2 k + 2)
= 2\(\lambda \) + 2 (k + 1)   [from (1)]
= even + even
= Sum of two even numbers is always an even number .
∴  P (k + 1) is true.

Let \(\triangle=\begin{vmatrix} 6&-1&5\\3&0&4\\-2&7&-3 \end{vmatrix}\)
Now M₁₂ = \(\begin{vmatrix} 3&4\\-2&-3 \end{vmatrix}=-9-(-8)=-1\)
M₂₂ = \(\begin{vmatrix} 6&5\\-2&-3 \end{vmatrix}=-18-(-10)=-8\)
M₃₂ = \(\begin{vmatrix} 6&5\\3&4 \end{vmatrix}=24-15=9\)
Now expansion of |A| using co-factors of elements of second column we get,
|A| = a₁₂ A₁₂ + a₂₁ A₂₁ + a₃₁A₃₁
A₁₂ = (-1)^{1 + 2} M₁₂ (-1) (-1) = 1
A₂₂ = (-1)^{2 + 2} M₂₂ = 1(-8) = -8
A₃₂ = (-1)^{3 + 2} M₃₂ = -(9) = -9
\(\therefore\) |A| = -1(1) + 0(-8) + 7(-9) = -1 - 63 = -64.

Let A = \(\left| \begin{matrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1 \\ cos\theta & 1 & x \end{matrix} \right| \)
Expanding along R₁ we get
|A| = x\(\left| \begin{matrix} -x & 1 \\ 1 & x \end{matrix} \right| -sin\theta \left| \begin{matrix} -sin\theta & 1 \\ cos\theta & x \end{matrix} \right| +cos​​\theta \begin{vmatrix} -sin\theta & -x \\ cos\theta & 1 \end{vmatrix}\)
= x(-x² - 1) - sin \(\theta\) (-x sin \(\theta\) - cos \(\theta\)) + cos \(\theta\) (-sin \(\theta\) + x cos \(\theta\))
\(=-x^{ 3 }-x+xsin^{ 2 }\theta +sin\theta cos\theta +xcos^{ 2 }\theta \)
= -x³ - x + x(sin²\(\theta\) + cos²\(\theta\))
= -x³ - x + x(1)    [\(\because\) sin²\(\theta\) + cos²\(\theta\) ] = -1
= -x³ which is independent of \(\theta\)

\(=\left|\begin{array}{ccc} 5 & 5^2 & 5^3 \\ 5^2 & 5^3 & 5^4 \\ 5^4 & 5^5 & 5^6 \end{array}\right|\)
Taking 5 and 5² common from R₁ and R₂
\(5 \times 5^2\left|\begin{array}{ccc} 1 & 5 & 5^2 \\ 1 & 5 & 5^2 \\ 5^4 & 5^5 & 5^6 \end{array}\right|=0\left(\text {Since } R_1=R_2\right)\)

\(\left|\begin{array}{ccc} 3 & -2 & 4 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right|=3[0-2]+2[6-1]+4[4-0]\)
\(=-6+10+16=20\)