Given P= 5(L)^0.7(K)^0.3
Marginal Productivity of Labour (L) is
\({∂P\over ∂L}=5(0.7)(L)^{0.7-1} (K)^{0.3 }= 3.5(L)^{-o·3} (K)^{0.3} = 3.5\left(K\over L\right)^{0.3}\)
When L = 10 and K = 3,
\(\frac { \partial P }{ \partial L } =3.5\left( \frac { 3 }{ 10 } \right) ^{ 0.3 }=3.5(0.3)^{ 0.3 }=3.5\times 0.6968\)
= 2.438 = 2.44
\({∂P\over ∂L}=5(L)^{0.7} (0.3)(K)^{0.3-1 }= 1.5(L)^{0.7}(K)^-{0.7} = 1.5\left(L\over K\right)^{0.7}\)
When L= 10 and K=3,
\({∂P\over ∂L}=1.5\left(10\over3\right)^{0.7}=1.5(3.33)^{0.7} = 1.5(2.2322) = 3.481 = 3.48\)

X	Y	RX	RY	d=RX-RY	d2
28	87	10	3	7	49
36	54	8	8	0	0
99	94	1	1	0	0
30	63	9	6	3	9
78	71	4	4	0	0
85	65	3	5	-2	4
95	89	2	2	0	0
65	61	6	7	-1	1
68	38	5	10	-5	25
38	52	7	9	-2	4
					\(\sum\)d2=92

Rank correlation \(\rho =1-\frac { 6\sum { { d }^{ 2 } } }{ N({ N }^{ 2 }-1) } =1-\frac { 6\times 92 }{ 10\times 99 } =1-\frac { 92 }{ 165 } \)=1-0.56
\(\rho \) = 0.44

cov(x, y)=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ { n }^{ 2 } } =\frac { 10(-115)-50(-30) }{ { 10 }^{ 2 } } \)
=\(\frac { -1150+1500 }{ 100 } =\frac { 350 }{ 100 } \)=3.5
Again, co-efficient correlation
r(x,y)=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ \sqrt { n{ \sum { x } }^{ 2 }-{ (\sum { x } ) }^{ 2 } } \sqrt { n{ \sum { y } }^{ 2 }-{ (\sum { y } ) }^{ 2 } } } \)
=\(\frac { 10(-115)-50(-30) }{ \sqrt { 10(290)-{ (50) }^{ 2 } } .\sqrt { 10(300)-{ (-30) }^{ 2 } } } \)
=\(\frac { -1150+1500 }{ \sqrt { 2900-2500 } .\sqrt { 3000-900 } } =\frac { 350 }{ 20\times 10\sqrt { 21 } } \)=0.3819
∴ r(x,y)=0.3819

Class	Mid value x	f	c.f.	D=|x-28|	f|D|
0-10	5	5	5	23	115
10-20	15	8	13	13	104
20-30	25	15	28	3	45
30-40	35	16	44	7	112
40-50	45	6	50	17	102
			N = 50		\(\sum { f|D|=478 } \)

\(\frac { N }{ 2 } =\frac { 50 }{ 2 } =25\)
\(\therefore\) Median lies in the interval (20, -30) and its corresponding values are
L = 20, f = 15, pcf = 13 and c = 10
\(\therefore Median=L+\left( \frac { \frac { N }{ 2 } -pcf }{ f } \right) \times c=20+\frac { 25-13 }{ 15 } \times 10\)
\(=20+\frac { 120 }{ 15 } =20+8=28\)
Now, mean deviation from median
\(=\frac { \sum { f|D| } }{ N } =\frac { 478 }{ 50 } =9.56\)

Size x	No. of items f	Mid Value x	f/x
50-60	12	55	0.2182
60-70	15	65	0.2308
70-80	22	75	0.2933
80-90	18	85	0.2118
90-100	10	95	0.1053
	N=77		\(\sum { f/x=1.0594 } \)

\(\therefore\) Harmonic mean = \(\frac { N }{ \sum { \left( \frac { f }{ x } \right) } } =\frac { 77 }{ 1.0594 } =72.683\)
\(\therefore\) HM = 72.683

Let E₁ = missing card is a heart card
E₂ = missing card is a spade card
E₃ = missing card is a club card
E₄ = missing card is a diamond card
and A = drawing 2 heart cards from the remaining cards.
\(\therefore\) P(E₁) = P(E₂) = P(E₃) = P(E₄)
\(=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \)
P(A/E₁) = P(two heart cards given that one heart card is missing)
\(=\frac { 12{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { \frac { 12\times 11 }{ 2\times 1 } }{ \frac { 51\times 50 }{ 2\times 1 } } =\frac { 66 }{ 1275 } \)
P(A/E₂) = P(2 heart cards given that one spade card is missing)
\(=\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { \frac { 13\times 12 }{ 2 } }{ \frac { 51\times 50 }{ 2\times 1 } } =\frac { 78 }{ 1275 } \)
Similarly, P(A/E₃) = \(\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { 78 }{ 1275 } \)
and P(A/E₄) = \(\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { 78 }{ 1275 } \)
\(\therefore\) By Baye's theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 }).P\left( A/{ E }_{ 1 } \right) }{ P({ E }_{ 1 }).P\left( A/{ E }_{ 1 } \right) +P({ E }_{ 2 }).P\left( A/{ E }_{ 2 } \right) +P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } }{ \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } } \)
\(=\frac { \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } }{ \frac { 1 }{ 4 } \left[ \frac { 66 }{ 1275 } +\frac { 78 }{ 1275 } +\frac { 78 }{ 1275 } +\frac { 78 }{ 1275 } \right] } \)
\(=\frac { 66 }{ 1275 } \times \frac { 1275 }{ 66+78+78+78 } \)
\(=\frac { 66 }{ 300 } =\frac { 11 }{ 50 } \)

Let y = x³+8x²+5x-2
Differentiating w.r.t. 'x' we get
\({dy\over dx}=3x^2+16x+5\)
\({dy\over dx}=0⇒3x^2 + 16x + 5=0\)
⇒ (x+5)(3x+1)=0
⇒ x = - 5, -1/3

The possible intervals are (-∞, - 5), (-5, -1/3) and (-1/3, ∞)

Intervals	Sign of \(dy\over dx\)	Nature of Function
(-∞, - 5) say x = - 6	3(-6)2 + 16(-6) + 5 = 17 (Positive)	Increasing Function
(-5, -1/3) say x = -1	3(-1)2 + 16(-1) + 5 = - 8 (Negative)	Decreasing Function
(-1/3, ∞) say x = 0	3(0)2 + 16(0) + 5 = 5 (Positive)	Increasing Function

Hence the given function is increasing in the intervals (-∞, - 5), (-1/3, ∞) and decreasing in (-5, -1/3).

Given f(x) = x³-3x²-9x+5
Differentiating w.r.t. 'x' we get,
f'(x) = 3x²-6x-9
At stationary points,f'(x) = 0
∴ 3x² - 6x - 9 = 0
⇒ x²-2x-3 = 0 (Divided by 3)
⇒ (x + 1)(x - 3) = 0
The stationary points are obtained when x = -1, x = 3
when x = -1, f(-1) = (-1)³ - 3(-1)² - 9(-1) + 5 = 10
when x = 3, f(3) = (3)³ - 3(3)² - 9(3) + 5 = - 22
∴ The stationary values are 10 and - 22 and the stationary points are (-1, 10) and (3, - 22).

Total number of balls = 6 + 5 = 11
n(S) = 11C₂ = \(\frac { 11\times 10 }{ 2\times 1 } =55\)
Let A be the event of getting a black ball and B be the event of getting a red ball.
\(n(A)={6 C }_{ 2 }=\frac { 6\times 5 }{ 2\times 1 } =15\)
\(P(A)=\frac { n(A) }{ n(S) } =\frac { 15 }{ 55 } \)
\(n(B)=5{ C }_{ 2 }=\frac { 5\times 4 }{ 2\times 1 } =10\)
\(P(B)=\frac { n(B) }{ n(S) } =\frac { 10 }{ 55 } \)
Since the two balls must be of same colour \(A\cap B=\phi \)
\(\therefore P(A\cup B)=P(A)+P(B)\)
\(=\frac { 15 }{ 55 } +\frac { 10 }{ 55 } =\frac { 25 }{ 55 } \)
\(=\frac { 5 }{ 11 } \)

x	f	log x	f log x
10	2	1.0000	2.0000
12	3	1.0792	3.2376
15	10	1.1761	11.7610
20	8	1.3010	10.4080
50	2	1.6990	3.3980
	N = 25		\(\sum { f } \) log x = 30.8046

Geometric Mean = Antilog\(\left( \frac { \sum { f\ log\ x } }{ N } \right) \)
= Antilog\(\left( \frac { 30.8046 }{ 25 } \right) \)
= Antilog (1.2322)
GM = 17.07

Given data in ascending order are
21, 27, 28, 30, 31, 32, 34, 36, 37, 39, 45, 46, 57 and n = 13
Q₂ = size of \({ 2\left( \frac { N+1 }{ 4 } \right) }^{ th }\) value = Size of \(2{ \left( \frac { 13+1 }{ 4 } \right) }^{ th }\) value
= Size of 7^th value
Q₂ = 34 

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

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

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

\({ -x }_{ 1 }+{ x }_{ 2 }=0\)

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

 

The feasible region is not common. Thus, there is no maximum value of Z.

Since the decision variables are non-negative, the solution lies in the I-quadrant of the plane. Consider the equations
\({ x }_{ 1 }+2{ x }_{ 2 }= 8\)

\({ x }_{ 1 }\)	0	8
\({ x }_{ 2 }\)	4	0

\({ 3x }_{ 1 }+{ 2x }_{ 2 }=12\)

\({ x }_{ 1 }\)	0	4
\({ x }_{ 2 }\)	6	0

The feasible region is OABC and its co-ordinates are 0(0, 0) A( 4, 0) qo, 4) and B is the point of intersection of the lines
\({ x }_{ 1 }+2{ x }_{ 2 }=8\)   ... (1) \(and\quad { 3x }_{ 1 }+{ 2x }_{ 2 }=12\)   ...(2)
Verification of B:
\((1)\Rightarrow { x }_{ 1 }+2{ x }_{ 2 }=8\\ \quad \quad (-)\quad (-)\quad \quad (-)\\ (2)\Rightarrow 3{ x }_{ 1 }+2{ x }_{ 2 }=12\\ -----------\\ -2{ x }_{ 1 }=-4 \Rightarrow { x }_{ 1 }=2\)
\(From(1), 2+2{ x }_{ 2 }=8\Rightarrow 2{ x }_{ 2 }=6\Rightarrow { x }_{ 2 }=3\)

∴ B is (2,3) 

Corner Points	\(Z=-3{ x }_{ 1 }+4{ x }_{ 2 }\)
O(0,0)	0
A(4, 0)	-12
B (2, 3)	6
C(0,4)	16

Minimum of Z occurs at A(4, 0).
Hence, the solution is x₁= 4, x₂ = 0 and Z_min = - 12.

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 \({ x }_{ 1 }=0,{ x }_{ 2 }=\frac { 10 }{ 3 } \quad and\quad { Z }_{ max }=\frac { 20 }{ 3 } \)

Since the decision variables x₁,x₂ are non-negative, the solution lies in the I quadrant.
Consider the equations
\(3{ x }_{ 1 }+5{ x }_{ 2 }=15\)                                             

\({ x }_{ 1 }\)	0	5
\({ x }_{ 2 }\)	3	0

\(5{ x }_{ 1 }+2{ x }_{ 2 }=10\)

\({ x }_{ 1 }\)	0	2
\({ x }_{ 2 }\)	5	0

The feasible region is OABC and its co-ordinates are O(0, 0) A(2, 0) C(0, 3) and B is the point of intersection of the lines
\(3{ x }_{ 1 }+5{ x }_{ 2 }=15\) ----(1)
and \(5{ x }_{ 1 }+2{ x }_{ 2 }=10\)   ... (2)
Verification of B:
\((1)\times 5\Rightarrow 15{ x }_{ 1 }+25{ x }_{ 2 }=75\\ \quad \quad (-)\quad (-)\quad \quad (-) \\ (2)\times 3\Rightarrow 15{ x }_{ 1 }+6{ x }_{ 2 }=30\\ --------------\\ 19{ x }_{ 2 }=45 \Rightarrow { x }_{ 2 }=\frac { 45 }{ 19 } \)
\(From(1), 3{ x }_{ 1 }+5\left( \frac { 45 }{ 19 } \right) =15\)
\(\Rightarrow 3{ x }_{ 1 }=15-\frac { 225 }{ 19 } \)
\(\Rightarrow { x }_{ 1 }=\frac { 20 }{ 19 } \)
\(\therefore \ B \ is\ \left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)

Corner Points	\(Z=6{ x }_{ 1 }+5{ x }_{ 2 }\)
O(0,0)	0
A(2,0)	12
B\(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)	\(6\times \frac { 20 }{ 19 } +5\times \frac { 45 }{ 19 } =\frac { 345 }{ 19 } \)
C(0,3)	15

Maximum of Z occurs at B \(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)
Hence, the solution is \({ x }_{ 1 }=\frac { 20 }{ 19 } ,{ x }_{ 2 }=\frac { 45 }{ 19 } \ and\ { Z }_{ max }=\frac { 345 }{ 19 } \) 

Let us assume that the investment in each stock be Rs.(80 x 96)
Case (i) Income from 7% stock
\(=\frac { \text {Investment }}{\text { MV } }\times {\text { Divicend Rate}}\)

case(ii)
Income from 9% stock at 96
\(=\frac { \text {Investment }}{\text { MV } }\times {\text { Divicend Rate}}\)
=
since income from 90% stock at 96 is more than the income from 7% stock at 80,9% stock at 96 is the better investment.

Given a=Rs.250,r=\(\cfrac { 6 }{ 100 } \times \cfrac { 1 }{ 12 } \)=0.005,n =5 x12 = 60
P = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
=\(\cfrac { 250 }{ 0.005 } \)[1-(1.005)^-60]
= 50,000[1-0.7414]
= 50,000[0.2586]
= Rs.12930
(1.005)⁶⁰=-60 log (1.005)
= -60(0.00216)
= -0.1299-1+1
= (1-0.1299)-1
=\(\bar { 1 } .8701\)
Antilog of \(\bar { 1 } .8701\) is 0.7414

Given a = Rs.1000,i =\(\cfrac { 12 }{ 12 } \)% = 1% = 0.01,n =12
A = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
= \(\cfrac { 1000 }{ 0.01 } \left[ \left( 1.01 \right) ^{ 12 }-1 \right] \)
= 100000[1.127-1]
= Rs.12,700
(1.01)¹²=12 log (1.01)
=12(0.0043)
= 0.0516
Antilog of 0.0516 is 1.127

FV = 100
Market Price = Rs.80
Dividend Rate = 20
Yield = \(\frac { FV }{ \text {Market Price } }\) x 20
=\(\frac { 100 }{ 80 } \) x 20 = 25%

Given A = Rs.3783,i = 0.05,n = 3
A = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
3783 = \(\cfrac { a }{ i } \) [(1.05)³-1]
3783 x 0.05 = a[1.1576-1]
a = \(\cfrac { 189.15 }{ 0.1576 } \) = 1200.19
\(\therefore\) a =Rs.1200
(1.05)³ = 3 log (1.05)
= 3(0.212)
= 0.636
Antilog of 0.636 is 1.1576

Given a = Rs.200, I = 0.04 , n = 2
P = \(\cfrac { a }{ i } \left( 1+i \right) \left[ \left( 1+i \right) ^{ n }-1 \right] \)
=\(\cfrac { 200 }{ 0.04 } \) (1.04)[1-(1.04)^-2]
= 5200[1-0.9247]
= 5200(0.09247)
= Rs.391.56
(1.04)^-2  =  -2log(1.04)
= -2(0.0170)
= -0.340+1-1
Antilog of \(\bar { 1 } .9660\) is 0.9247

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

Given P = Rs.5000,r =\(\cfrac { 8 }{ 2 } \) x = 4% = 0.04,n =10
P=\(\cfrac { a }{ i } \left[ 1-\left( 1+i \right) ^{ -n } \right] \)
5000 =\(\cfrac { a }{ 0.04 } \left[ 1-\left( 1.04 \right) ^{ -10 } \right] \)
5000 x 0.04 =a[1-0.6761]
200 = a(0.3239)
a=\(\cfrac { 200 }{ 0.3239 }\)=Rs.617.50
(1.04)^-10=-10log (1.04)
= -10(0.0170)
= -0.1700
= 0.1700+(1-1)
= \(\bar { 1 } .\left( 1-0.1700 \right) \)
=\(\bar { 1 } .8300\)
Antilog of 0.8300 is 0.6761

x	log x
3	0.4771
6	0.7782
24	1.3802
48	1.6812

\(\sum { log } \) = 4.3167
Geometric Mean = Antilog\(\left( \frac { \sum { log\quad x } }{ N } \right) =Antilog\left( \frac { 4.3167 }{ 4 } \right) \)
= Antilog (1.0791)
GM = 11.99

Given b_xy =2 and b_yx=0.45
We know r= \(\sqrt { { b }_{ xy }.{ b }_{ yx } } =\sqrt { 2(0.45) } =\sqrt { 0.9 } \)=0.949
\(\therefore\)r=0.949

We haveP(not A or not B) = 1/4
\(P(\overline { A } U\overline { B } )=\frac { 1 }{ 4 } \) \(\Rightarrow P(\overline { A\cap B } )=\frac { 1 }{ 4 } \)
\(\Rightarrow 1-P(A\cap B)=\frac { 1 }{ 4 } \)   \(\Rightarrow P(A\cap B)=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
\(\therefore P(A\cap B)=\frac { 3 }{ 4 } \)
Now, P(A). P(B) = \(\frac { 1 }{ 2 } \times \frac { 7 }{ 12 } =\frac { 7 }{ 24 } \)
\(\therefore P(A\cap B)\neq P(A).P(B)\)
So, A and B are not independent events.

Let 2 and 4 be the assumed means for the X and Y series respectively.

X	Y	dx=X-2	dy=Y-4	dx2	dy2	dxdy
-3	9	-5	5	25	25	-25
-2	4	-4	0	16	0	0
-1	1	-3	-3	9	9	9
0	0	-2	-4	4	16	8
1	1	-1	-3	1	9	3
2	4	0	0	0	0	0
3	9	1	5	1	25	5
		-14	0	56	84	0

Correlation Co-efficient
r(X,Y) =\(​​​​\frac { N\sum { dXdY-(\sum { dX)(\sum { Y) } d } } }{ \sqrt { n.\sum { d{ x }^{ 2 }-{ (\sum { dx } })^{ 2 } } } \sqrt { n\sum { d{ y }^{ 2 }-(\sum { d{ y) }^{ 2 } } } } } \)
=\(\frac { 7(0)-(-14)(0) }{ \sqrt { 56(7)-({ -14) }^{ 2 } } \sqrt { 84(7)-0 } } \)=0
 ∴ r(X, Y)=0

Given n(A) = 3, n(B) = 2 and n(C) = 4
A\(\cap \)B = {3}
A\(\cap \)C = {2, 5}
\(\therefore\) n(A\(\cap \)B) = 1
and n(A\(\cap \)C) = 2
(i) P(A/B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { n(A\cap B) }{ n(B) } =\frac { 1 }{ 2 } \)
\(P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { n(A\cap B) }{ n(A) } =\frac { 1 }{ 3 } \)
(ii) \(P(A/C)=\frac { P(A\cap C) }{ P(C) } =\frac { n(A\cap C) }{ n(C) } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)
\(P(C/A)=\frac { P(A\cap C) }{ P(A) } =\frac { n(A\cap C) }{ n(A) } =\frac { 2 }{ 3 } \)

X	Y	X2	Y2	XY
4	2	16	4	8
2	3	4	9	6
3	2	9	4	6
4	4	16	16	16
2	4	4	16	8
15	15	49	49	44

b_yx=\(\frac { n\sum { XY-(\sum { X)(\sum { Y) } } } }{ n{ \sum { X } }^{ 2 }-\left( \sum { { X }^{ 2 } } \right) } =\frac { 5(44)-(15)(15) }{ 5(49)-({ 15) }^{ 2 } } =\frac { -1 }{ 5 } \)
b_xy=\(\frac { n\sum { XY-(\sum { X)(\sum { Y) } } } }{ n{ \sum { Y } }^{ 2 }-\left( \sum { { Y) }^{ 2 } } \right) } =\frac { 5(44)-(15)(15) }{ 5(49)-({ 15) }^{ 2 } } =\frac { -1 }{ 5 } \)
The co-efficient of correlation is
r= \(\sqrt { { b }_{ xy }.{ b }_{ yx } } =\sqrt { \left( -\frac { 1 }{ 5 } \right) \left( -\frac { 1 }{ 5 } \right) } \)=0.2
As b_xy and b_yx are both negative, r=-0.2

The regression co-efficient are given by
b_xy= \(r.\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } \)and b_yx=\(r.\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } \)
where r is the co-efficient of correlation.

\(r=\sqrt { { b }_{ xy }.{ b }_{ yx } } \)

r(x, y) =\(\frac { n\sum { xy-(\sum { x)(\sum { y) } } } }{ \sqrt { n\sum { { x }^{ 2 }-{ (\sum { x) } }^{ 2 } } } .\sqrt { n\sum { { y }^{ 2 }-{ (\sum { y } }^{ 2 }) } } } \)
=\(\frac { 20(75)-10(70) }{ \sqrt { 20(250)-{ (10) }^{ 2 }.\sqrt { 20(300)-{ (70) }^{ 2 } } } } \)
=\(\frac { 1500-700 }{ \sqrt { 5000-100.\sqrt { 6000-4900 } } } \)
=\(\frac { 800 }{ (70)(33.166) } =\frac { 800 }{ 2321.62 } \)=0.345

X	Y	XY
1	5	5
2	4	8
3	3	9
4	2	8
5	1	5
15	15	35

cov(X, Y) =\(\frac { 1 }{ n } \left[ \sum { XY-\frac { 1 }{ n } (\sum { X)(\sum { Y) } } } \right] \)
=\(\frac { 1 }{ 5 } \left[ 35-\frac { 1 }{ 5 } (15)(15) \right] \)
= \(\frac { 1 }{ 5 } [35-45]=\frac { -10 }{ 5 } \)=-2

N=7

X	Y	x2	y2	xy
12	14	144	196	168
9	8	81	64	72
8	6	64	36	48
10	9	100	81	90
11	11	121	121	121
13	12	169	144	156
7	3	49	9	21
70	63	728	651	676

Karl Pearson correlation co-efficient
r(x, y) =\(\frac { N\sum { XY-(\sum { X)(\sum { Y) } } } }{ \sqrt { N.\sum { { X }^{ 2 }-({ \sum { X) } }^{ 2 } } } .\sqrt { N.{ \sum { Y } }^{ 2 }-({ \sum { Y) } }^{ 2 } } } \)
=\(\frac { 7(676)-(70)(63) }{ \sqrt { 7(728)-{ (70) }^{ 2 } } \sqrt { 7(651)-{ (63) }^{ 2 } } } \)
=\(\frac { 4732-4410 }{ (14)(24.2487) } =\frac { 322 }{ 229.4818 } \)=0.9485

X	Y	X2	Y2	XY
1	9	1	81	9
2	8	4	64	16
3	10	9	100	30
4	12	16	144	48
5	11	25	121	55
6	13	36	169	78
7	14	49	196	98
8	16	64	256	128
9	15	81	225	135
45	108	285	1356	597

r(x,y) =\(\frac { N\sum { XY-(\sum { X)(\sum { Y) } } } }{ \sqrt { N.{ \sum { X } }^{ 2 }-({ \sum { X) } }^{ 2 } } .\sqrt { N.{ \sum { Y } }^{ 2 }-{ (\sum { Y) } }^{ 2 } } } \)
=\(\frac { 9(597)-45(108) }{ \sqrt { 9(285)-{ (45) }^{ 2 } } .\sqrt { 9(1356)-({ 108) }^{ 2 } } } \)
=0.95
\(\therefore\)X and Y are highly positively correlated.

\(\Rightarrow\)   \(\alpha+2\beta={{\pi}\over{4}}\)

LHS\(=\cos^{-1}\left(\frac{3}{5}\cos x+\frac{4}{5}\sin x\right)\)
put \(\frac{3}{5}=\cos\theta\)
\(\therefore\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-\left(\frac35\right)^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac45\)
Now, \(\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac45}{\frac35}=\frac43\Rightarrow\theta=\tan^{-1}\left(\frac43\right)\)
\(\therefore\cos^{-1}\left(\frac{3}{5}\cos x+\frac45\sin x\right)\)
= cos^-1 (cos \(\theta\) cos x + sin \(\theta\) sin x)
= cos^-1 [cos (x - \(\theta\) )]    [\(\therefore\) cos (A- B) = cos A cos B + sin A sin B]
= x - \(\theta\)
= \(x-{\tan}^{-1}\left({{4}\over{3}} \right)\) = RHS
Hence proved.

LHS\(=\frac{\sin5x-2\sin3x+\sin x}{\cos 5x-\cos x}\)
\(=\frac{(\sin5x+\sin x)-2\sin3x}{\cos5x-\cos x}\)
\(=\frac{2\sin\left(\frac{5x+x}{2}\right)\cos\left(\frac{5x-x}{2}\right)-2\sin3x}{-2\sin\left(\frac{5x+x}{2}\right)\sin\left(\frac{5x-x}{2}\right)}\)
\(=\frac{2\sin3x\cos2x-2\sin3x}{-2\sin3x.\sin2x}\)
\(\frac{-2\sin3x(1-\cos2x)}{-2\sin3x.\sin2x}=\frac{1-\cos2x}{\sin2x}\)
\(=\frac{2\sin^2x}{2\sin x\cos x}=\frac{\sin x}{\cos x}\tan x=RHS\)
Hence proved.

LHS=\(\frac{\sin(x+y)}{\sin(x-y)}=\frac{\sin x\cos y+\cos x+sin y}{\sin x\cos y-\cos x\sin y}\)
Dividing the numerator and denominator by cos x cos y,
We get LHS,
\(\frac{\frac{\sin x\cos y}{\cos x\cos y}+\frac{\cos x\sin y}{\cos x\cos y}}{\frac{\sin x\cos y}{\cos x\cos y}-\frac{\cos x\sin y}{\cos x\cos y}}=\frac{\tan x+\tan y}{\tan x-\tan y}=RHS\)
Hence proved.

\(\underset { x\rightarrow 0 }{ lim } \frac { 2sinx-sin2x }{ { x }^{ 3 } } \)= \(\underset { x\rightarrow 0 }{ lim } \frac { 2sinx-2sinx\quad cosx }{ { x }^{ 3 } } \)
= \(\underset { x\rightarrow 0 }{ lim } \frac { 2sinx(1-cosx) }{ { x }^{ 3 } } =2.\underset { x\rightarrow 0 }{ lim } \frac { sinx }{ x } \underset { x\rightarrow 0 }{ lim } \frac { { 2sin }^{ 2 }\frac { x }{ 2 } }{ { x }^{ 2 } } \)  \(\left[ \therefore 1-cosx=2{ sin }^{ 2 }\frac { x }{ 2 } \right] \) 
= \(4(1).\underset { x\rightarrow 0 }{ lim } \frac { { sin }^{ 2 }\frac { x }{ 2 } }{ \left( \frac { x }{ 2 } \right) ^{ 2 } \times \ \ 4 } \left[ \therefore \underset { \phi \rightarrow 0 }{ lim } \quad \frac { sin\phi }{ \phi } =1 \right] \) [Multiplying and dividing by 4in the denominator]
= \(\frac { 4 }{ 4 } .\underset { x\rightarrow 0 }{ lim } \left( \frac { sin\frac { 2 }{ x } }{ \frac { 2 }{ x } } \right) \)
=1 x 1 = 1

LHS\(=\left\{1+\cot x-\sec\left(\frac{\pi}{2}+x\right)\right\}\left\{1+\cot x+\sec\left(\frac{\pi}{2}+x\right)\right\}\)
=(1 + cot x + cosec x) (1 + cot x - cosec x)
(1 + cot x)² - (cosec²x) [\(\therefore\)(a + b) (a - b) = a² - b²]
1 + cot²x + 2 cot x - cosec²x
= cosec²x + 2 cot x - cosec²x [\(\therefore\) 1 + cot²x = cosec²x]
= 2 cot x = RHS. Hence proved.

We know that \(\cos^2x+\sin^2x=1\)
\(\Rightarrow\cos x=\pm\sqrt{1-\sin^2x}\)
In the III quadrant, cos x is negative
\(\therefore\cos x=-\sqrt{1-\sin^2x}=-\sqrt{1-\frac{24}{25}}=-\frac15\left[\because\sin^2x=\frac{4(6)}{25}=\frac{24}{25}\right]\)
In the III quadrant, tan x is positive
\(\therefore\tan x=\frac{\sin x}{\cos x}=\frac{-2\sqrt6}{5}\times\frac{-5}{1}=2\sqrt{6}\)
\(cosec x=\frac{1}{\sin x}=\frac{-5}{2\sqrt6}\)
\(\sec x=\frac{1}{\cos x}=-5\) and
\(\cot x=\frac{1}{\tan x}=\frac{1}{2\sqrt6}\)

Given e^y(x+1)=1    ....(1)
Differentiating with respect to 'x' we get,
\({ e }^{ y }(1)+(x+1){ e }^{ y }\frac { dy }{ dx } =0\)   [product rule]
\(\Rightarrow { e }^{ y }+(1)\frac { dy }{ dx } =0\quad [using\quad (1)]\)
\(\Rightarrow \frac { dy }{ dx } =-{ e }^{ y }\)...(2)
Differentiating again with respect to 'x' we get,
\(\frac { d }{ dx } \left( \frac { dy }{ dx } \right) =\frac { d }{ dx } \left( -{ e }^{ y } \right) \)
\(\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-{ e }^{ y }.\frac { dy }{ dx } \)
\(=\left( \frac { dy }{ dx } \right) \left( \frac { dy }{ dx } \right) \) [using (2)]
\(={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
Hence proved.

Given sin² x + cos² y = 1.
Differentiating with respect to 'x' we get,
\(2\sin { x } .\frac { d }{ dx } \left( \sin { x } \right) +2\cos { y } .\frac { d }{ dx } \left( \cos { y } \right) =0\)    
\(\Rightarrow 2\sin { x } \cos { x } +2\cos { y } \left( -\sin { y } \right) \frac { dy }{ dx } =0\)
\(\Rightarrow \sin { 2x } -2\sin { y } \cos { y } \left( \frac { dy }{ dx } \right) =0\)
\(\Rightarrow \sin { 2x } -\sin { 2y } \left( \frac { dy }{ dx } \right) =0\)
\(\Rightarrow \sin { 2x } =\sin { 2y } \left( \frac { dy }{ dx } \right) \)
\(\Rightarrow \left( \frac { dy }{ dx } \right) =\frac { \sin { 2x } }{ \sin { 2y } } \)

Let (x₁, y₁) be the point of contact.
\(\therefore\) Equation of tangent at (x₁,y₁) to the circle is xx₁ + yy₁ = 9
Its slope is -\(\frac { { x }_{ 1 } }{ { y }_{ 1 } } \)
Given that the tangent is parallel to 2x + y - 3 = 0
\(\therefore\) Their slopes must be equal
\(\therefore \quad \frac { -{ x }_{ 1 } }{ { y }_{ 1 } } =\frac { -2 }{ 1 } \Rightarrow { x }_{ 1 }=2{ y }_{ 1 }\)
But \({ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }=9\)
\(\Rightarrow { (2{ y }_{ 1 }) }^{ 2 }+{ y }_{ 1 }^{ 2 }=9\Rightarrow 4{ y }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }=9\)
\(\Rightarrow 5{ y }_{ 1 }^{ 2 }=9\Rightarrow { y }_{ 1 }=\pm \frac { 3 }{ \sqrt { 5 } } \)
\(\therefore x=\frac { 6 }{ \sqrt { 5 } } \)
Hence \(\frac { { 6x }_{ 1 } }{ \sqrt { 5 } } =\frac { 3{ y }_{ 1 } }{ \sqrt { 5 } } =9\Rightarrow 6x+3y=9\sqrt { 5 } \)
\(\Rightarrow 2x+y=3\sqrt { 5 } \)

Given f(x) = x² -x² -sin x + 5
\(\underset { x\rightarrow \pi }{ lim } f(x)=\underset { x\rightarrow \pi }{ lim } ({ x }^{ 2 }-sinx+5)\)
\(={ \pi }^{ 2 }-sin\quad \pi +5={ \pi }^{ 2 }-0+5+5=5{ \pi }^{ 2 }\)  \([\therefore sin \ \pi=0]\)
Also, f(\(\pi\)) = \({ \pi }^{ 2 }-sin\quad \pi +5=\pi \quad -0+5=\pi +5\)
ஃ  \(\underset { x\rightarrow \pi }{ lim } f(x)=f(x)\)
ஃ  f(x) is continuous at x = \(\pi\)

The given line is 2x + y = 1
Any line parallel to this is 2x + y + k = 0
Since this passes through (0,0),2(0) + 0+ k = 0 \(\Rightarrow\) k = 0
\(\therefore\)  Equation of parallel line is 2x + y = 0 .....(1)
Any line perpendicular to 2x + y = 1 is x - 2y + k₁ = 0
Since this passes through (0, 0), k₁ = 0
\(\therefore\)  Perpendicular line is x - 2y = 0 ......(2)
\(\therefore\) Required combined equation is
(2x + y)(x - 2y) = 0
\(\Rightarrow\) 2x² - 4xy + xy - 2y² = 0
\(\Rightarrow\) 2x² - 3xy - 2y² = 0.

Slope of the line 3x-y+5=0 is
\(\Rightarrow \quad { m }_{ 2 }=-\frac { co-efficient\quad of\quad x }{ co-efficient\quad of\quad y } =\frac { -3 }{ -1 } =3\)
Let m₁ = m and \(\theta =45\)
\(\therefore \quad tan\theta =\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \)
\(\Rightarrow \quad tan\quad 45=\left| \frac { m-3 }{ 1+3m } \right| \)
\(\Rightarrow \quad 1=\left| \frac { m-3 }{ 1+3m } \right| \)
\(\Rightarrow \quad 1|1+3m|=|m-3|\)
\(\Rightarrow \quad 1+3m=\pm (m-3)\)
\(\Rightarrow \quad 1+3m=m-3\quad or\quad 1+3m=-m+3\)
\(\Rightarrow\)   2m = - 4 or  4m = 2
\(\Rightarrow \)  m = - 2  or  \(m=\frac { 2 }{ 4 } =\frac { 1 }{ 2 }​​​​​​​\)
\(\therefore \)   m =  - 2 or \(\frac { 1 }{ 2 } \)

The given lines are
y = m₁x + C₁ \(\Rightarrow\) m₁ x - y + C₁ = 0
y = m₂x + C₂ \(\Rightarrow\) m₂x - y + C₂ = 0
y = m₃x + C₃ \(\Rightarrow\) m₃x - y + C₃ = 0.
The condition for these lines to be concurrent is \(\left| \begin{matrix} { m }_{ 1 } & -1 & { C }_{ 1 } \\ { m }_{ 2 } & -1 & { C }_{ 2 } \\ { m }_{ 3 } & -1 & { C }_{ 3 } \end{matrix} \right| =0\)
Interchanging C₂ and C₃ we get,
\(\left| \begin{matrix} { m }_{ 1 } & { C }_{ 1 } & 1 \\ { m }_{ 2 } & { C }_{ 2 } & 1 \\ { m }_{ 3 } & { C }_{ 3 } & 1 \end{matrix} \right| =0\)    [Taking (-1) common from C₃]
Expanding along C₁ we get
\({ m }_{ 1 }\left| \begin{matrix} { C }_{ 2 } & 1 \\ { C }_{ 3 } & 1 \end{matrix} \right| -{ m }_{ 2 }\begin{vmatrix} { C }_{ 1 } & 1 \\ { C }_{ 3 } & 1 \end{vmatrix}+{ m }_{ 3 }\begin{vmatrix} { C }_{ 1 } & 1 \\ { C }_{ 2 } & 1 \end{vmatrix}=0\)
\(\Rightarrow\) m₁(C₂ - C₃) - m₂(C₁ - C₃) + m₃(C₁ - C₂) = 0
\(\Rightarrow\) m₁(C₂ - C₃) + m₂(C₃ - C₁) + m₃(C₁ - C₂) = 0

Given equation is
3x² + 7xy + 2y² + 5x + 5y + 2 = 0
Factorizing 3x² + 7xy + 2y² = (x + 2y) (3x + y)
\(\therefore \) 3x² + 7xy + 2y² + 5x + 5y + 2 = (x + 2y + l)(3x + y + m)
Equating the x and y Co-ordinates both sides,

We get 5 = m + 3l  ...(1)
5 = 2m + l   ....(2)

(1) \(\times\) (2) \(\rightarrow \) 10	= 2m + 6l
(2) \(\rightarrow \) 5	= 2m + 1l
5	= 0 + l \(\Rightarrow \)  l = 1

Substituting l = 1 in (2) we get,
5 = 2m + 1 \(\Rightarrow \) 2m = 4 \(\Rightarrow \) m = 2.
Hence the separate equations are
x + 2y + 1 = 0 and 3x + y + 2 = 0.

Given \(y=\sqrt { x } +\frac { 1 }{ \sqrt { x } } .\Rightarrow y=\frac { x+1 }{ \sqrt { x } } \)  ......(1)
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } =\frac { \sqrt { x } .\frac { d }{ dx } (x+1)-(x+1).\frac { d }{ dx } (\sqrt { x } ) }{ { (\sqrt { x } })^{ 2 } } \)
\(=\frac { \sqrt { x } (1)-(x+1).\frac { 1 }{ 2\sqrt { x } } }{ x } =\frac { \frac { 2x-(x+1) }{ 2\sqrt { x } } }{ x } \)
\(\frac { dy }{ dx } =\frac { 2x-x-1 }{ 2x.\sqrt { x } } =\frac { x-1 }{ 2x\sqrt { x } } \)...(2)
LHS=\(2x\left( \frac { dy }{ dx } \right) +y\)
\(=2x\left( \frac { x-1 }{ 2x\sqrt { x } } \right) +\frac { x+1 }{ \sqrt { x } }\)
 [From (1) and (2)]
\(=\frac { x-1 }{ \sqrt { x } } +\frac { x+1 }{ \sqrt { x } } =\frac { x-1+x+1 }{ \sqrt { x } } =\frac { 2x }{ \sqrt { x } } =\frac { 2\sqrt { x } .\sqrt { x } }{ \sqrt { x } } =2\sqrt { x } \) = RHS.
Hence proved

\(\Rightarrow\) Let the co-ordinates of the centre of the required circle be C(a,0). Since it passes through P(2,3) = 25
CP = radius = 5
\(\Rightarrow \sqrt { ({ a-2) }^{ 2 }+{ (0-3) }^{ 2 } } =5\)
\(\Rightarrow\) (a- 2 )² + 9 = 25 \(\Rightarrow\) (a - 2)² = 16
\(\Rightarrow\) (a - 2)² = (\(\pm \)4)²

\(\Rightarrow\) a - 2 = \(\pm \)4
\(\Rightarrow\) a =   \(\pm \) 4 + 2
\(\Rightarrow\) a = 4 + 2 or -4 + 2
\(\Rightarrow\) a = 6 or -2
Thus the Co-ordinates of the centre are (6, 0) or (-2, 0)
Hence the equations of the required circle are
(x - 6)² + (y - 0)² = 5² \(\Rightarrow\) x² + 36 - 12x + y² = 25
\(\Rightarrow\) x² + y² - 12x + 11 = 0 (OR)
(x + 2)² + (y - 0)² = 5²
\(\Rightarrow\) x² + 4x +4 + y² = 25
\(\Rightarrow\)  x² + y² + 4x - 21 = 0.

Let p(x₁,y₁) be the point on the locus and A(-1, 0) B(0, 2) are the fixed points
Given pA = 3 pB
\(\Rightarrow\)  pA² = 9 pB²
\(\Rightarrow ({ x }_{ 1 }+1)^{ 2 }+({ y }_{ 1 }-0)^{ 2 }=9[({ x }_{ 1 }-0)^{ 2 }+({ y }_{ 1 }-2)^{ 2 }]\)
\(\Rightarrow { x }_{ 1 }^{ 2 }+2{ x }_{ 1 }+1+{ y }_{ 1 }^{ 2 }=9[{ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }-4{ y }_{ 1 }+4]\)
\(\Rightarrow { x }_{ 1 }^{ 2 }+2{ x }_{ 1 }+1+{ y }_{ 1 }^{ 2 }=9{ x }_{ 1 }^{ 2 }+9{ y }_{ 1 }^{ 2 }-36{ y }_{ 1 }+36\)
\(\Rightarrow 8{ x }_{ 1 }^{ 2 }+8{ y }_{ 1 }^{ 2 }-{ 2x }_{ 1 }-36{ y }_{ 1 }+35=0\)
\(\therefore Locus\quad of\quad (x_{ 1 },{ y }_{ 1 })\quad is\) 8x² + 8y² - 2x - 36y + 35 = 0

Here 12 things are to be divided equally among 4 persons, hence the groups are regarded as distinct.
Hence required number of ways = \(\frac { 12! }{ { (3!) }^{ 4 } } \)
\(=\frac { 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 }{ 3\times 2\times 3\times 2\times 3\times 2\times 3\times 2 } \)
=3696000

No. of ways of giving I prize = n
No. of ways of giving II prize = n
.............................................
Similarly no. of ways of giving n^th prize = n
Therefore By fundamental principle of counting, the required number of ways 

Any number between 100 and 1000 is a 3 - digit number

1. Hundred place can be filled in 6 ways using the numbers 1,2,3,4,5,6 (excluding 0)
2. Tens place can be filled in 6 ways (including 0)
3. Ones place can be filled in 5 ways.
\(\therefore\)  By fundamental principle of counting, total number of 3 digit numbers = 6 x 6 x 5 = 180.

\({{x+4}\over{(x^2-4)(x+1)}}={{x+4}\over{(x+2)(x-2)(x+1)}}={{A}\over{x+2}}+{{B}\over{x-2}}+{{C}\over{x+1}}\)
\(\Rightarrow\)  \({{x+4}\over{(x^2-4)(x+1)}}=\frac { { { A(x-2)(x+1)B+(x+2)(x+1)+C(x+2)(x-2) } } }{ (x+2)(x-2)(x+1) } xx\)
\(\Rightarrow\)  x + 4 = A (x - 2) (x + 1) + B (x + 2) (x + 1) C (x +2)(x - 2)   ...(1)
Putting x = 2 in (1) we get,
6 = B (4) (3)  \(\Rightarrow\boxed{B={{1}\over{2}}}\)
Putting x = -1 in (1) we get,
3 = C (1)(-3)  \(\Rightarrow\quad \boxed{C=-1}\)
Putting x = 0 in (1) we get,
4 = - 2A + 2B - 4C
\(\Rightarrow\)   \(4=-2A+2\left( {{1}\over{2}} \right)-4(-1)\)     \(\left[ \because B={{1}\over{2}},C=-1 \right]\)
\(\Rightarrow\)   \(4=-2A+1+4\ \ \Rightarrow\ 4=-2A+5\ \Rightarrow -1\)   =-2A
\(\Rightarrow\)   \(\boxed{A={{1}\over{2}}}\)
\(\therefore\) \({{x+4}\over{(x^2-4)(x+1)}}=​​​​​​{{{{1}\over{2}}}\over{x+2}}​​+{{{{1}\over{2}}}\over{x-2}}-{{1}\over{}x+1}={{1}\over{2(x+2)}}+{{1}\over{2(2x-2)}}-{{1}\over{x+1}}\)

Given A \(=\begin{bmatrix} 3&2\\7&5 \end{bmatrix}\)
\(|A|=\begin{vmatrix} 3&2\\7&5 \end{vmatrix}=15-14=1\Rightarrow{A}^{-1}\) exists.
A₁₁ = 5, A₁₂ = -7, A₂₁ = -2, A₂₂ = 3
\(\therefore\)  adj A \({=\begin{bmatrix} 5&-7 \\ -2 & 3 \end{bmatrix}}^{T}=\begin{bmatrix} 5 & -2 \\-7 & 3 \end{bmatrix}\)
\(\therefore\ {A}^{-1}={{1}\over{|A|}}adj\ A={{1}\over{1}}\begin{bmatrix} 5&-2 \\ -7&3 \end{bmatrix}=\begin{bmatrix} 5&-2 \\ -7&3 \end{bmatrix}\)
Now B \(=\begin{bmatrix} 4&6\\ 3&2 \end{bmatrix}\)
\(|B|=\begin{vmatrix}4 &6 \\ 3 &2 \end{vmatrix}=8-18-10\Rightarrow{B}^{-1}\) exists.
B₁₁ = 2, B₁₂ = -3, B₂₁ = -6, B₂₂ = 4
\(\therefore\ adj\ B={\begin{bmatrix} 2 & -3 \\ -6 &4 \end{bmatrix}}^{T}=\begin{bmatrix} 2&-6 \\ -3 &4 \end{bmatrix}\)
\(\therefore\ {B}^{-1}={{1}\over{|B|}}adj\ B={{1}\over{10}}\begin{bmatrix} 2&-6 \\ -3&4 \end{bmatrix}={{1}\over{10}}\begin{bmatrix} -2&6 \\ 3 &-4 \end{bmatrix}\)
\(\therefore\ {B}^{-1}{A}^{-1}={{1}\over{10}}\begin{bmatrix} -2 & 6 \\3 & -4 \end{bmatrix}\begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix}={{1}\over{10}}\begin{bmatrix} -10-42& 4+18\\ 15+28 & -6-12 \end{bmatrix}={{1}\over{10}}\begin{bmatrix} -52 & 22 \\ 43 & -18 \end{bmatrix}\)........(1)
\(AB=\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\begin{bmatrix} 4 & 6 \\ 3 & 2 \end{bmatrix}=\begin{bmatrix} 12+6 & 18+4\\ 28+15 & 42+10\end{bmatrix}=\begin{bmatrix} 18 & 22\\ 43 & 52 \end{bmatrix}\)
\(|AB|=\begin{vmatrix} 18&22 \\ 43 &52 \end{vmatrix}=936-946=-10\Rightarrow{(AB)}^{-1}\) exists.
\(adj\ AB={\begin{bmatrix} 52 &-43 \\ -22& 18 \end{bmatrix}}^{T}=\begin{bmatrix} 52 & -22 \\ -43 & 18 \end{bmatrix}\)
\(\therefore \ {(AB)}^{-1}adj\ (AB)={{-1}\over{10}}\begin{bmatrix} 52&-22 \\-43 & 18 \end{bmatrix}={{1}\over{10}}\begin{bmatrix} -52 & 22 \\ 43 &-18 \end{bmatrix}\)
From (1) and (2), (AB)^-1 = B^-1 A^-1.
Hence proved.

Let A = \(\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| \)
Applying C₁➝C₁+9C₂ we get
A = \(\left| \begin{matrix} 2+63 & 7 & 65 \\ 3+72 & 7 & 65 \\ 5+81 & 9 & 86 \end{matrix} \right| =\left| \begin{matrix} 65 & 7 & 65 \\ 75 & 8 & 75 \\ 86 & 9 & 86 \end{matrix} \right| =0[ \because {C}_{1}\equiv{C}_{3}]\)

LHS = \(\begin{vmatrix}x+a &b&c \\a &x+b&c\\a&b&x+c \end{vmatrix}\)
Applying C₁ \(\rightarrow\) C₁ + C₂ + C₃ we get,
LHS = \(\begin{vmatrix}x+a+b+c &b&c \\x+a+b+c &x+b&c\\x+a+b+c&b&x+c \end{vmatrix}\)
Taking (x + a + b + c) common from C₁ we get,
= (x + a + b + c) \(\begin{vmatrix}1 &b&c \\1 &x+b&c\\1&b&x+c \end{vmatrix}\)
Applying R₂ \(\rightarrow\) R₂ - R₁ and R₃ \(\rightarrow\) R₁ we get,
= \((x+a+b+c)\begin{vmatrix}1 &b&c \\0&x&0\\0&0&x\end{vmatrix}\)
Expanding along C₁ we get,
\(=(x+a+b+c)\begin{bmatrix} 1\begin{bmatrix} x & 0 \\ 0 & x\end{bmatrix} +0+0\end{bmatrix}\)
= (x + a + b + C)(x²) = RHS
Hence proved.

Given equations are 2x + 5y = 1; 3x + 2y = 7
This system of equations can be written in matrix form as \(\begin{pmatrix}2 & 5 \\3 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} 1\\7 \end{pmatrix}\Rightarrow Ax=B\)
where A = \(\begin{pmatrix} 2 & 5 \\3 & 2 \end{pmatrix},X=\begin{pmatrix} x\\y \end{pmatrix}\) and B = \(\begin{pmatrix} 1 \\ 7 \end{pmatrix}\)
\(\therefore\)  X = A^-1 B.
\(|A|=\begin{vmatrix} 2 &5 \\3 & 2 \end{vmatrix}=4-15=-11\)
A₁₁ = 2, A₁₂ = -3, A₂₁ = -5, A₂₂ = 2
\(\therefore\)   adj A = \(\begin{bmatrix} 2 & -3 \\ -5 & 2 \end{bmatrix}^{T}=\begin{bmatrix} 2 & -5 \\-3 & 2 \end{bmatrix}\)
\(\therefore\)  \({A}^{-1}={{1}\over{|A|}}\) . adj A = \({{-1}\over{11}}\begin{bmatrix} 2 & -5\\ -3& 2 \end{bmatrix}\)
X = A^{- 1} B \(={{-1}\over{11}}\begin{bmatrix} 2 & -5\\ -3&2 \end{bmatrix}\begin{bmatrix} 1\\7 \end{bmatrix}\)
\(=\frac { 1 }{ 11 } \left[ \begin{matrix} +2-35 \\ -3+14 \end{matrix} \right] ={{-1}\over{11}}\begin{bmatrix} -33\\11 \end{bmatrix}=\begin{bmatrix} 3\\-1 \end{bmatrix}\)
\(\therefore\)  x = 3.and y = -1.

Given A \(=\begin{bmatrix} 2&3\\-1&4 \end{bmatrix}\)
\(|A|=\begin{bmatrix} 2&3\\-1&4 \end{bmatrix}=8+3=11\)
Now, A₁₁ = 4, A₁₂ = - (-1) = 1, A₂₁  = 3, A₂₂ = 2
\(\therefore\ adj\ A={\begin{bmatrix} 4&1\\-3&2 \end{bmatrix}}^{T}=\begin{bmatrix} 4&-3\\1&2 \end{bmatrix} \)
\(\therefore\) A (adj A)\(=\begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} { }\begin{bmatrix} 4 & -3 \\ 1 & 2 \end{bmatrix}\)
\(=\begin{bmatrix} 8+3&-6+6\\-1+4&3+8 \end{bmatrix}=\begin{bmatrix} 11&0\\0&11 \end{bmatrix}=11\begin{bmatrix} 1&0\\0&1 \end{bmatrix}=|A|I_2\)   ...(1)
Also ( adj A ) A = \(\begin{bmatrix} 4&-3\\1&2 \end{bmatrix}\begin{bmatrix} 2&3\\-1&4 \end{bmatrix}\)
\(=\begin{bmatrix} 8+3&12-12\\2-2&3+8 \end{bmatrix}=\begin{bmatrix} 11&0\\0&11 \end{bmatrix}=11\begin{bmatrix}1&0\\0&1 \end{bmatrix}=|A|I_2\)   ....(2)
From (1) and (2), A( adj A) = (adj A) A = |A|.I₂

Given that A^T = A^-1
\(\Rightarrow\) AA^T = AA^-1 
\(\Rightarrow\) AA^{T = I}
Now, AA^T = \(\left[ \begin{matrix} cos\ \alpha & sin\ \alpha \\ -sin\ \alpha & \ cos\ \alpha \ \end{matrix} \right] \)\(\left[ \begin{matrix} cos\ \alpha & -sin\ \alpha \\ sin\ \alpha & \ cos\ \alpha \ \end{matrix} \right] \)
\(=\left[ \begin{matrix} { cos }^{ 2 }\alpha +{ sin }^{ 2 }\alpha & -sin\alpha cos\alpha +sin\alpha cos\alpha \\ -sin\alpha cos\alpha +sin\alpha cos\alpha & +{ sin }^{ 2 }\alpha +{ cos }^{ 2 }\alpha \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)[\(\because\) sin²\(\alpha\) + cos²\(\alpha\) = 1]
Thus, AA^T = I is true for all \(\alpha\).
Hence \(\alpha\)  can take any real value.

The system of equations can be written in the form AX = B where,
\(A=\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 7 \\ 11 \\ 1 \end{matrix} \right] \)
Now, |A| = \(\left| \begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{matrix} \right| =1\left| \begin{matrix} 0 & 3 \\ -3 & 0 \end{matrix} \right| -2\left| \begin{matrix} 1 & 3 \\ 2 & 0 \end{matrix} \right| +1 \left| \begin{matrix} 1 & 0 \\ 2 & -3 \end{matrix} \right| \)
= 1(0 + 9) - 2(0 - 6) + 1(-3 - 0) = 9 + 12 - 3 = 18 \(\neq \) 0 
\(\Rightarrow\) A^-1 exists.
A₁₁ = 0 + 9 = 9, A₁₂ = -(0 - 6) = 6, A₁₃ = -3 - 0 = -3
A₂₁ = -(0 + 3) = -3, A₂₂ = 0 - 2 = -2, A₂₃ = -(-3 - 4) = 7
A₃₁ = 6 - 0 = 6, A₃₂ = -(3 - 1) = -2, A₃₃ = 0 - 2 = -2
\(\therefore adj\quad A={ \left[ \begin{matrix} 9 & 6 & -3 \\ -3 & -2 & 7 \\ 6 & -2 & -2 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ |A| } adj\quad A=\frac { 1 }{ 18 } \left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \)
\(\therefore \ X={ A }^{ -1 }B=\frac { 1 }{ 18 } \left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \left[ \begin{matrix} 7 \\ 11 \\ 1 \end{matrix} \right] \)
\(=\frac { 1 }{ 18 } \left[ \begin{matrix} 63 & -33 & +6 \\ 42 & -22 & -2 \\ -21 & +77 & -2 \end{matrix} \right] =\frac { 1 }{ 18 } \left[ \begin{matrix} 36 \\ 18 \\ 54 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ 3 \end{matrix} \right] \)
\(\therefore\) x = 2, y = 1, and z = 3.

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 \(\theta\)