\(A=\left[ \begin{matrix} 500 & 1000 & 500 \\ 2500 & 1500 & 500 \end{matrix} \right] \)
\(May=2\times A=\left[ \begin{matrix} 1000 & 2000 & 3000 \\ 5000 & 3000 & 1000 \end{matrix} \right] =M\)
(i) Average \(=\frac { A+M }{ 2 } =\left[ \begin{matrix} 750 & 1500 & 2250 \\ 3750 & 4250 & 750 \end{matrix} \right] \)
(ii) May  = 2A
June = 2 x May = 4 A
July = 2 x June = 8A
August = 2 x July = 16A
⇒ August \(=\left[ \begin{matrix} 8000 & 1600 & 24000 \\ 40000 & 24000 & 8000 \end{matrix} \right] \)

f(x) = x²+px-4 = 0
If f(-4) = (-4)²+p(-4) = 16-4p-4 = 0
12-4p = 0
-4p = -12
p = 3.
x² + 3x + q = 0 has equal roots, 
∆ = b² - 4ac = 0
3²-4x1xq = 0
9-4q = 0

\(\\ \\ \\ \\ q=\frac { 9 }{ 4 } \)
p = 3, \(\\ \\ \\ \\ q=\frac { 9 }{ 4 } \)

\(f(x)=\frac { 1{ x }^{ 2 } }{ a } -\frac { 2x }{ b } +\frac { 3 }{ c } \)
Sum of the roots \((\alpha +\beta )=\frac { -b }{ a } =-\frac { (-2) }{ 1 } \)
Product of the roots \((\alpha \beta )=\frac { c }{ a } =\frac { 3 }{ 1 } =3\)
∝+2, β+2 are the roots (given)
Sum of the roots = ∝ + 2 + β + 2
= ∝ + β + 4
= 2 + 4 = 6
Product of the roots = (∝ + 2) (β + 2)
= ∝β + 2∝ + 2β + 4
= ∝β + 2(∝ + β) + 4
= 3 + 2 x 2 + 4
= 3 + 4 + 4 = 11
∴ The required equation x² - 6x + 11 = 0.

Let the no of seats in each row be x

⇒ 2x²-10x = x²+375
⇒ x²-10x-375 = 0
⇒x²-25x +15x-375 = 0
⇒ x(x-25) +15(x-25) = 0
⇒(x-25)(x+15) = 0
⇒ x = 25, x = -15, x > 0
∴ 25 rows are in the hall

\(60-t=\frac { { t }^{ 2 } }{ 4 } -3\)
⇒ t²-12 = 240-4t
⇒ t²+4t-252 = 0
⇒ t²+18t-14t-252 = 0
⇒ t(t +18)-14(t +18) = 0
⇒ (t +18)(t-14) = 0
∴ t = 14 or t = -18 is not possible

Yes, it is possible to design a rectangular park of
perimeter 320 m and area 4800 m².
If the length be 'l', breadth be 'b'
Area = lb = 4800 = 40 x 120m²
Perimeter = 2(l+b) = 2(40 + 120)
= 320m
∴ Length = 120 m
breadth = 40 m

Let the speed of boat in still water be 'v'
\(\because speed=\frac { distance }{ time } \Rightarrow time=\frac { distance }{ speed } \)
\(\therefore \frac { 36 }{ v-4 } -\frac { 36 }{ v+4 } =\frac { 96 }{ 60 } =\frac { 8 }{ 5 } (\because 1.6hrs=\frac { 96 }{ 60 } )\)
\(\Rightarrow 36(v+4)-36(v-4)=\frac { 8 }{ 5 } (v-4)(v+4)\)
\(\Rightarrow 36v+144-36v+144=\frac { 8 }{ 5 } ({ v }^{ 2 }+4v+4v-16)\)
\(\Rightarrow 288=\frac { 8 }{ 5 } { v }^{ 2 }-\frac { 128 }{ 5 } \Rightarrow 8{ v }^{ 2 }-128=1440\)
\(\Rightarrow 8{ v }^{ 2 }=1568\Rightarrow { v }^{ 2 }=196{ v }^{ 2 }=\pm 14\)
∴ Speed of the boat = 144m/hr.
(∵ speed cannot be -ve)

Squaring both sides(\(\sqrt { y+1 } +\sqrt { 2y-5 } \) )² = 3²
y+1+2y-5+2(\(\sqrt { y+1 } +\sqrt { 2y-5 } \)) = 9
3y-4-9 = -2\(\sqrt { y+1 } +\sqrt { 2y-5 } \)
Again squaring both sides
(3y-13)²(-2\(\sqrt { y+1 } +\sqrt { 2y-5 } \))²
9y²-78y+169 = 4(y+1)(2y-5)
9y²-78y+169 = 4(2y²+2y-5y-5)
9y²-78y+169 = 8y²+8y-20y-20
9y²-78y+169-8y²+12y+20 = 0
y²-66y+189 = 0
y²-63-3y+189 = 0
y(y-63)-3(y-63) = 0
(y-63)(y-3) = 0
y = 63, 3

|17x² - 18x + 19|

Let Arul's speed of working be x
Let Mohan's speed of working be y
Let Ram's speed of working be z
given that they are working together.
Let 'w' be the quantum of work.
Also given that Mohan takes twice the time as Arul for finishing the work.
\(\therefore \frac { w }{ y } =2\times \frac { w }{ x } \therefore x=2y\)
\(\therefore y=\frac { x }{ 2 } \)   (2)
Also Ram takes 3 times the time as Arul for finishing the work.
\(\therefore \frac { w }{ z } =3\times \frac { w }{ x } \)
\(\therefore x=3z\quad \therefore z=\frac { x }{ 3 } \)
Substitute (2) and (3) in (1),
\(x+\frac { x }{ 2 } +\frac { x }{ 3 } =\frac { w }{ 6 } \)
∴ 6x + 3x + 2x = w
11x = w
\(x=\frac { w }{ 11 } ,y=\frac { w }{ 22 } ,z=\frac { w }{ 33 } \)
Working alone time taken as
\(Arul:\frac { w }{ x } =\frac { w }{ w/11 } =11hrs.\)
\(Mohan:\frac { w }{ y } =\frac { w }{ w/22 } =22hrs.\)
\(Ram:\frac { w }{ z } =\frac { w }{ w/33 } =33hrs\)

\(=\frac { (q+r)+p }{ (q+r)-p } \times \frac { (q+r)+p }{ (q+r)+p } \times \frac { 2qr+{ q }^{ 2 }+{ r }^{ 2 }-{ p }^{ 2 } }{ 2qr } \)

\(=\frac { { (q+r+p) }^{ 2 } }{ 2qr } =\frac { 1 }{ 2qr } \)

\(\frac { { x }^{ 3a }-8 }{ { x }^{ 2a }+2{ x }^{ a }+4 } \)
\(=\frac { { ({ x }^{ a }) }^{ 3 }-8 }{ { ({ x }^{ a }) }^{ 2 }+{ 2x }^{ a }+4 } =\frac { { ({ x }^{ a }) }^{ 3 }-{ 2 }^{ 3 } }{ { x }^{ 2a }+{ 2x }^{ a }+4 } \)
\(=\frac { ({ x }^{ a }-2)\left( { x }^{ 2a }+2{ x }^{ a }+4 \right) }{ { x }^{ 2a }+{ 2x }^{ a }+4 } \)
\(=\frac { ({ x }^{ a }-2)\left( { x }^{ 2a }+2{ x }^{ a }+4 \right) }{ { x }^{ 2a }+{ 2x }^{ a }+4 } ({ x }^{ a }-2)\)

Let f(x) = 2x⁴ + 13x³ + 27x² + 23x + 7,
g(x) = x³ + 3x² + 3x + 1,
h(x) = x² + 2x + 1
which is the least degree polynomial.
Now Dividing f (x) by h(x)

Since the Remainder is zero, h (x) is the GCD of f (x) and h (x)
Now, dividing g (x) by h (x)

Remainder is zero, h (x) is the GCD of g (x) and h (x)
h (x) divides both f(x) and g (x) completely
GCD = x² + 2x + 1

Let the number of students in sections A, B and C be 'x', 'y' and 'z' respectively
Given x + y + z = 150           ........(1)
x - 6 = z + 6
x - z = 12            .....(2)
4z = x + y
x + y - 4z = 0         ..........(3)
consider (1) and (3)

\(z=\frac{150}{5}=30\)
substituting in (2)
x - 30 = 12
x = 30 + 12
x = 42
substituting x = 42, z = 30 in (1)
42 + y + 30 = 150
y = 150 - 72
y = 78
The number of students in sections A, B and C are 42, 78, 30 respectively.

\(\frac{1}{3}(x+y-5)=y-z\)
x + y - 5 = 3y - 3z
x - 2y + 3z = 5          ..............(1)
y - z = 2x - 11
2x - y + z = 11     ........(2)
2x - 11 = 9 - (x + 2z)
2x + x + 2z = 9 + 11
3x + 2z  = 20             ......(3)
Consider (1) and (2)

z = 3/3 = 1
Substituting z = 1 in (3)
3x + 7 (1) = 20
3x = 20 - 2 = 18
\(x=\frac{18}{3}=6\)
Substituting x = 6,z = 1 is                   ...........(1)
6 - 2y + 3(1) = 5
6 + 3 - 5 = 2y
4 = 2y
y = 2y
\(y=\frac{4}{2}=2\)
Solution: x = 6,y = 2, z = 1.