Here N=5

X	Y	X2	Y2	XY
6	9	36	81	54
2	11	4	121	22
10	5	100	25	50
4	8	16	64	32
8	7	64	49	56
30	40	220	340	214

\(\overline { X } =\frac { \sum { X } }{ N } =\frac { 30 }{ 5 } \)=6
\(\overline { Y } =\frac { \sum { Y } }{ N } =\frac { 40 }{ 5 } \)=8
\({ b }_{ xy }=\frac { N\sum { XY-(\sum { X } )(Y) } }{ N\sum { { X }^{ 2 }-{ (\sum { Y } ) }^{ 2 } } } =\frac { 5(214)-(30)(40) }{ 5(340)-{ (40) }^{ 2 } } \)
=\(\frac { 1070-1200 }{ 1700-1600 } =-\frac { 130 }{ 100 } \)=-1.3
b_yx=\({ b }_{ xy }=\frac { N\sum { XY-(\sum { X } )(Y) } }{ N\sum { { X }^{ 2 }-{ (\sum { X } ) }^{ 2 } } } =\frac { 5(214)-(30)(40) }{ 5(2200)-{ (30) }^{ 2 } } \)
\(=\frac { 1070-1200 }{ 1700-900 } =\frac { -130 }{ 200 } \)=-0.65
Regression equation of X on Y is
\(\Rightarrow\)X-\(\bar{X}\)=b_xy(Y-\(\bar{Y}\))
X-6=-1.3(Y-8)
X=-1.3Y+10.4+6
X=-1.3Y+16.40
Regression equation of Y on X is
Y-\(\bar{Y}\) =b_yx(X-\(\bar{X}\))
Y-8=-0.65(X-6)
Y=-0.065X+3.9+8
Y=-0.65X+11.90

Total Number of bolts produced = 1000 + 2000 + 3000 = 6000
\(P({ A }_{ 1 })=\frac { 1000 }{ 6000 } =\frac { 1 }{ 6 } \)
\(P({ A }_{ 2 })=\frac { 2000 }{ 6000 } =\frac { 1 }{ 3 } \)
\(P({ A }_{ 3 })=\frac { 3000 }{ 6000 } =\frac { 1 }{ 2 } \)
Let B be the event of selecting defective bolts.
\(\therefore \) P(B/A₁) = 1% = \(\frac { 1 }{ 100 } \) = 0.01
P(B/A₂) = 1.5% = 0.015
and (P(B/A₃) = 2% = 0.02
\(\therefore P(A_{ 1 }/B)=\frac { P({ A }_{ 1 }).P\left( B/{ A }_{ 1 } \right) }{ P({ A }_{ 1 }).P(B/{ A }_{ 1 })+P({ A }_{ 2 }).P\left( B/{ A }_{ 2 } \right) +P({ A }_{ 3 }).P\left( B/{ A }_{ 3 } \right) } \)
\(=\frac { \frac { 1 }{ 6 } \times 0.01 }{ \frac { 1 }{ 6 } \times 0.01+\frac { 1 }{ 3 } \times 0.015+\frac { 1 }{ 2 } \times 0.02 } =\frac { 1 }{ 600 } \times 60=\frac { 1 }{ 10 } \)
\(\therefore P(A_{ 1 }/B)=0.1\)

Let Reshma mix x₁ kg of food P and x₂ kg of food Q to make the mixture.
Let Z be the total cost of mixture

	Food P	Food Q	Minimum requirement
Vitamin A	3	4	8
Vitamin B	5	2	11
Cost	Rs.60	Rs.80

Thus, the mathematical formation of the given LPP is minimize Z = 60x₁+ 80x₂
Subject to the constraints
\(3{ x }_{ 1 }+4{ x }_{ 2 }\ge 8\quad 5{ x }_{ 1 }+2{ x }_{ 2 }\ge 11\quad and\quad { x }_{ 1 },{ x }_{ 2 }\ge 0\)
Consider the equations
\(3{ x }_{ 1 }+4{ x }_{ 2 }=8\)

\({ x }_{ 1 }\)	0	8/3
\({ x }_{ 2 }\)	2	0

\(5{ x }_{ 1 }+2{ x }_{ 2 }=11\)

\({ x }_{ 1 }\)	0	8/3
\({ x }_{ 2 }\)	2	0

The feasible region is ABC and its co-ordinates are A\(\left( \frac { 8 }{ 3 } ,0 \right) \), C\(\left( 0,\ \frac { \pi }{ 2 } \right) \)and B is the point of intersection of the lines  3x₁ + 4x₂ = 8 ..... (1) and 5x₁ + 2x₂ = 11 .... (2)
Verification of B:
\((1) \Rightarrow 3{ x }_{ 1 }+4{ x }_{ 2 }=8\)
\( (-)\quad (-)\quad \quad (-)\)
\((2)\times 2\Rightarrow 10{ x }_{ 1 }+4{ x }_{ 2 }=22\)
\(--------------\)
\( -7x_{ 1 }=-14 \Rightarrow { x }_{ 1 }=2\)
\(From(1), 3(2)+4{ x }_{ 2 }=8\)
\(4{ x }_{ 2 }=8-6=2\Rightarrow { x }_{ 2 }=\frac { 1 }{ 2 } \)
\( \therefore \ B\ is\ \left( 2,\frac { 1 }{ 2 } \right) \) 

Corner Points	Z = 60x1+ 80x2
A(8/3,0)	\(60\times \frac { 8 }{ 3 } =160\)
B (2, 1/2)	\(120+80\times \frac { 1 }{ 2 } =160\)
C(0,11/2)	\(80\times \frac { 11 }{ 2 } =440\)

Minimum of Z occurs at \(A\left( \frac { 8 }{ 3 } ,0 \right) and\quad B\left( 2,\frac { 1 }{ 2 } \right) \)
Hence, least cost of mixture is n60 when 8/3 kg of food P and 0 kg of food Q and 2 kg of food P and 112kg of food Q are mixed

Given a = Rs.75, i=\(\cfrac { 8 }{ 12 } \)% = 4% = 0.004,n = 10 X 2 =20
A = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
=\(\cfrac { 75 }{ 0.04 } \) [(1.04)²⁰-1]
=1875 (2.1878-1)
= 1875(1.878)
= Rs.2227
(1.04)⁴⁰ = 20 log (1.04)
= 20(0.0170)
= 0.34
Antilog of 0.34 is 2.1878

Let y=x³-6x²+7
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=3x^2-12x\)
\({dy\over dx}=0\)
\(\Rightarrow3x^2-12x=0\)
\(\Rightarrow3x(x-4)=0\)
\(\Rightarrow x=0 \ or \ x=4\)
\({d^2y\over dx^2}=6x-12\)
when x=0 \({d^2y\over dx^2}=-12<0\)
\(\therefore \) y is maximum at x=0
\(\therefore \) maximum value=0³-6(0)²+7=7
when x=4, \({d^2y\over dx^2}=-6(4)-12=12>0\)
\(\therefore \) y is minimum at x=4
\(\therefore \) Minimum value =4⁴-6(4)²+7=64-96+7=-25
Hence maximum value is 7 and minimum value is -25.

Given equation of the circle is x² + y² = 169...(1)
Equation of the tangent at (x₁, y₁) to circle (1) is xx₁ + yy₁ = 169
Now, Equation of the tangent at (5,12) to circle (1) is
x(5) + y(12) = 169 ⇒ 5x + 12y - 169 = 0...(2)
and equation of the tangent at (12,-5) to circle (1) is
x(12) + y(-5) = 169 ⇒ 12x - 5y = 169 = 0...(3)
Let m₁ and m₂ be the slopes of the tangents (2) and (3)
ஃ m₁=\(\frac { -Co-efficient\quad of\quad x }{ Co-efficient\quad of\quad y } =\frac { -5 }{ 12 } \)
Similarly m₂ = \(\frac { -12 }{ -5 } =\frac { 12 }{ 5 } \)
Consider m₁m₂ = \(\left( \frac { -5 }{ 12 } \right) \left( \frac { 12 }{ 5 } \right) =-1\)
Since m₁m₂ = -1, the tangents at (5,12) and (12,-5) to the circle x² + y² = 169 are perpendicular to each other.

LHS = (sin 3x + sin x) sin x + (cos 3x - cos x) cos x
= sin 3x sin x + sin²x + cos 3x cos x - cos²x
= sin 3x sin x + cos 3x cos x - (cos²x - sin²x)
= cos (3x - x) - cos 2x [∴ cosA cosB + sinA sinB = cos(A - B) and cos²A - sin²A = cos²A]
= 0 = RHS. Hence proved

Given \(f(x)=\left\{ \begin{matrix} 6ax+3b & if\quad x>1 \\ ax-2b & if\quad x<1 \\ 15 & if\quad x=1 \end{matrix} \right\} \)
\(L[{ f(x)] }_{ x=1 }=\lim _{ x\rightarrow { 1 }^{ - } }{ f(x) =\lim _{ h\rightarrow 0 }{ f(1-h)=\lim _{ h\rightarrow 0 }{ a(1-h)-2b\quad [\because f(x)=ax-2b\quad if\quad x<1] } } } \)
\(=\lim _{ h\rightarrow 0 }{ a-ah-2b=a-2b } \) .....(1)
R[f(x)]_x=1=\(\lim _{ x\rightarrow { 1 }^{ + } }{ f(x)=\lim _{ h\rightarrow 0 }{ f(1+h) } } \)
\(=\lim _{ x\rightarrow { 1 }^{ + } }{ 6a(1+h)+3b\quad [\because f(x)=6ax+3b\quad if\quad x>1] } \)
= 6(91+0) + 3b = 6a + 3b   ....(2)
Also, f(1) = 15 ....(3)
Since f(X) is continuous at x =1,
L[f(x)]_x=1=R[f(x)]_x=1=f(1)
\(\Rightarrow\) a - 2b = 6a + 3b = 15 [using (1), (2) and (3)]
\(\Rightarrow\) a - 2b = 15 ....(4)
and 6a+3b=15 \(\Rightarrow\) 2a + b =5  ..(5)
(4) \(\times\) 2 \(\rightarrow \) 2a - 4b = 30
       -             -      -
(5)        \(\rightarrow \)  2a + b = 5
____________________
              -5b = 25  \(\Rightarrow\) b = -5
Substituting b = -5 in (4) we get,
a - 2(-5) = 15
\(\Rightarrow\) a + 10 = 15
\(\Rightarrow\) a = 15 - 10 = 5
\(\therefore\)    a = 5 and b = -5.   

I and II prizes Mathematics can be given in (30 x 29) ways.
I & II prizes in physics also can be given in (30 x 29) ways.
I prize in chemistry can be given in 30 ways.
I prize in English can be given in 30 ways.
Hence, the number of ways to give prizes in all the four subjects = (30 x 29) x (30 x29)x 30 x 30
= 6.8121 x 10⁸

Given \(\Delta =\left| \begin{matrix} { cosec }^{ 2 }\theta & { cot }^{ 2 }\theta & 1 \\ { cot }^{ 2 }\theta & { cosec }^{ 2 }\theta & -1 \\ 42 & 40 & 2 \end{matrix} \right| =0\)
Applying C₁\(\rightarrow\)C₁ - C₂, we get,
\(\Delta =\left| \begin{matrix} { cosec }^{ 2 }\theta -{ cot }^{ 2 }\theta & { cot }^{ 2 }\theta & 1 \\ { cot }^{ 2 }\theta -{ cosec }^{ 2 }\theta & { cosec }^{ 2 }\theta & -1 \\ 42-40 & 40 & 2 \end{matrix} \right| \)
\(=\left| \begin{matrix} 1 & { cot }^{ 2 }\theta & 1 \\ -1 & { cosec }^{ 2 }\theta & -1 \\ 2 & 40 & 2 \end{matrix} \right| \)  [\(\because\) cosec²  \(\theta\)  - cot²  \(\theta\)  =1]   
= 0
\(\Delta =0\)    [\(\because\)  C₁ \(\equiv \) C₃]