\(\left[ \begin{matrix} x+y+z \\ x+z \\ y+z \end{matrix} \right] =\left[ \begin{matrix} 9 \\ 5 \\ 7 \end{matrix} \right] \)
⇒ x + y + z = 9 ...(1)
⇒ x + z = 5..(2)
⇒ y + z = 7 ...(3)

Sub y = 4 in (3)
4 + z = 7
z = 3
Sub z = 3 in (2)
x + 3 = 5
x = 2
x = 2, y = 4, x = 3

2x² - x - 1 = 0 here, a = 2, b = -1, c = -1
α + β = \(\frac {-b}{a} = \frac {-(-1)}{2} = \frac {1}{2}\), αβ = \(\frac {c}{a} = -\frac {1}{2}\)
Given roots are α²β, β²α
Sum of the roots α²β + β²α = αβ(α + β) = \(-\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } \right) =-\frac { 1 }{ 4 } \)
Product of the roots (α²β) x (β²α) = α³β³ = (αβ)³ = \({ \left( -\frac { 1 }{ 2 } \right) }^{ 3 }=-\frac { 1 }{ 8 } \)
The required equation is x² - (Sum of the roots)x + (Product of the roots) = 0
x² - \(\left( -\frac { 1 }{ 4 } \right) x-\frac { 1 }{ 8 } \) = 0 gives 8x² + 2x - 1 = 0

3y² - 20y - 23 = 0
      a       b      c
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Here \(y=\frac { -(-20)\pm \sqrt { { (-20) }^{ 2 }-4\times 3\times -23 } }{ 2\times 3 } \)
\(=\frac { 20\pm \sqrt { 400+276 } }{ 6 } \)
\(=\frac { 20\pm \sqrt { 676 } }{ 6 } =\frac { 20\pm 26 }{ 6 } \)
\(=\frac { 46 }{ 6 } or\frac { -6 }{ 6 } \)
\(y\Rightarrow \frac { 23 }{ 3 } or-1\)

\(\frac{t_{2}}{t_{1}}=\frac{60}{120}=\frac{1}{2} \)
\(\frac{t_{3}}{t_{2}}=\frac{30}{60}=\frac{1}{2} \)
\(\frac{t_{4}}{t_{3}}=\frac{18}{30}=\frac{9}{15}=\frac{3}{5} \)
The ratios between successive terms are not equal. Therefore 120, 60, 30, 18,... are not a G.P

 \({ x }^{ 2 }-\left( -\frac { 3 }{ 5 } \right) x+\left( -\frac { 1 }{ 2 } \right) =0\Rightarrow \frac { 10{ x }^{ 2 }+6x-5 }{ 10 } \) = 0
Therefore, 10x² + 6x - 5 = 0

Here t₂ - t₁ = t₃ - t₂
13 - 6 = 20 - 13 = 7
It is an Arithmetic series
Sum \(\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(a+l)\)
a = 6 : l = 97
a + (n - 1) d = 97
6 + (n - 1) (7) =  97
(n - 1) (7) = 97 - 6
(n - 1) (7) = 91
\(n-1=\frac{91}{7}=13\)
n = 13 + 1 = 14
Now \(S_{n}=\frac{14}{2}(6+97)=7 \times 103\)
6 + 13 + 20 +...+ 97 = 721

f(k) - 2k - 1
f o f(k) = 5
f(f(k)) = f(2k - 1) = 5
⇒ 2(2k-1) -1 = 5
4k - 2 -1 = 5 ⇒ 4k = 8
k = 2

98 \(\equiv \) (x + 4) (mod 5)
98 - (x + 4) = 5n , for some integer n.
94 - x  = 5n
94- x is a multiple of 5
Therefore , the least positive value of x must be 4
Since 94 - 4 = 90  is the nearest multiple of 5 less than 94.

(2x² - 3xy)² = (x(2x - 3y))²
= x(2x - 3y)²
(4x - 6y)³ = (2(2x - 3y))³
2³(2x - 3y)³
8x³ - 27y³ = (2x)³ - (3y)³
= (2x - 3y)(4x² + 6xy + 9y²)
= (2x - 3y)(4x² + 6xy + 9y²)
L.C.M = 2³x²(2x - 3y)³(4x² + 6xy + 9y²)

Given r = 5 and S₆ = 46872
Sum upto n terms of a G.P \(\mathrm{S}_{\mathrm{n}}=\frac{a\left(r^{n}-1\right)}{r-1}\)
\(46872 =\frac{a\left(5^{6}-1\right)}{5-1} \)
\(46872 =a \frac{(15625-1)}{4}=a \times \frac{15624}{4} \)
46872 = a x 3906
\(\frac{46872}{3906}=a\)
a = 12
First term of the G.P., a = 12

It is a geometric progression with
\(a=5, r=\frac{-3}{5} \neq 1\)
Sum upto n terms S_n = \(\frac{a\left(r^{n}-1\right)}{r-1}\)
\(=\frac{5\left[\left(\frac{-3}{5}\right)^{n}-1\right]}{\frac{-3}{5}-1}\)
\(=\frac{5\left[\left(\frac{-3}{5}\right)^{n}-1\right]}{\frac{-3-5}{5}}\)
\(=\frac{5\left[\left(\frac{-3}{5}\right)^{n}-1\right]}{\frac{-8}{5}}\)
\(=\frac{-25}{8}\left[\left(\frac{-3}{5}\right)^{n}-1\right] \)
\(=\frac{25}{8}\left[1-\left(\frac{-3}{5}\right)^{n}\right] \)

Let x² = a. Then, (x²)² - 13x² + 42 = a² - 13a + 42 = (a - 7)(a - 6)
Given. (a - 7)(a - 6) = 0 we get, a = 7 or 6.
Since a = x², x² = 7 then, x = 土\(\sqrt {7}\) or x² = 6 we get, x = 土\(\sqrt {6}\)
Therefore the roots are x = 土\(\sqrt {7}\), 土\(\sqrt {6}\)

\(\angle\)OPQ = 90^o - 50^o = 40^o (angle between the radius and tangent is 90^o)
OP = OQ (Radii of a circle are equal)
\(\angle\)OPQ = \(\angle\)OQP = 40^o (\(\triangle\)OPQ is isosceles)
\(\angle POQ={ 180 }^{ 0 }-\angle OPQ-\angle OQP\)
\(\angle\)POQ  = 180^o - 40^o- 40^o = 100^o.

f(x) = x² - 1
(a) f o f = f(f(x ) = f(x² - 1)
= (x² - 1)² -1; = x⁴ - 2x² + 1 - 1 = x⁴ - 2x
(b) f o f o f = f [f (f(x))] = f [f(x² - 1)] = f [x⁴ - 2x²]
= (x⁴ - 2x²)² - 1 = x⁸ - 4x⁶ + 4x⁴ - 1

Here the first term a = 8, common difference d = \(7\frac { 1 }{ 4 } \) -8 = -\(\frac { 3 }{ 4 } \).,
Sum of first n terms of an A.P S_n = \(\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
S₁₅ = \(\frac { 15 }{ 2 } \left[ 2\times 8+\left( 15-1 \right) \left( -\frac { 3 }{ 4 } \right) \right] \)
S₁₅ = \(\frac { 15 }{ 2 } \left[ 16-\frac { 21 }{ 2 } \right] =\frac { 165 }{ 4 } \)

We know that n^th term of an A.P. is
t_n = a + (n - 1)d
Given 9 times 9^th term = 15 times 15^th term
9 x t₉ = 15 x t₁₅
9[a + (9 - 1)d] = 15 [a + (15 - 1)d]
9(a + 8d) = 15 (a + 14d)
9a + 72d = 15a + 210 d
15a + 210 d - 9a - 72 d = 0
6a + 138 d = 0
6(a + 23 d) = 0
6[a + (24 - 1)d) = 0
6 x t₂₄ = 0
6 times 24^th term = 0

Let the distance be 'd'
Speed = 500 km/hr
Time = 't' hours
Distance = Time x Speed
d(t) = 500 t

Consider 2ⁿ \(\times\) 5^m
Since the product has 2 as a factor
2ⁿ \(\times\) 5^m is even
But if a number ends with the digit 5, then the number is an odd number
It is impossible.
For no value of m, 2ⁿ \(\times\) 5^m ends in 5.

In \(\Delta BAC\) and \(\Delta PRQ,\quad \frac { AB }{ RQ } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
\(\frac { BC }{ QP } =\frac { 6 }{ 12 } =\frac { 1 }{ 2 } ;\frac { CA }{ PR } =\frac { 3\sqrt { 3 } }{ 6\sqrt { 3 } } =\frac { 1 }{ 2 } \)
Therefore, \(\frac { AB }{ QP } =\frac { BC }{ QP } =\frac { CA }{ PR } \)
By SSS similarity, we have \(\triangle\)BAC~\(\triangle\)QRB
\(\angle\)P =\(\angle\)C (since the corresponding parts of similar triangle)
\(\angle\)P =\(\angle\)C 180^o -\((\angle A+\angle B)={ 180 }^{ 0 }-({ 90 }^{ 0 }+{ 60 }^{ 0 })\)
\(\angle\)P = 180^o - 150^o = 30^o

A x B = {(3,1), (3,7), (3,10), (4,1), (4,7), (4,10), (7,1), (7,7), (7,10), (8,1), (8,7), (8,10)}
We note that, R₁ ⊆  A x B. Thus, R₁ is a relation from A to B.

A = {1,2,3} B = {2,3,5,7} 
A x B = {(1,2),(1,3),(1,5),(1,7),(2,2),(2,3),(2,5),(2,7),(3,2),(3,3),(3,5),(3,7)}
B x A = {(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(5,1),(5,2),(5,3),(7,1),(7,2),(7,3)}

A x B = {(3,2), (3,4), (5,2), (5,4)}
We have A = {set of all first coordinates of elements of A x B}. Therefore, A = {3,5}
B = {set of all second coordinates of elements of A x B}. Therefore, B = {2,4}
Thus A = {3,5} and B = {2,4}.

\(\left(\begin{matrix} x-3 & 3x-z \\ x+y+7 & x+y+z \end{matrix} \right) =\left( \begin{matrix} 1 & 0 \\ 1 & 6 \end{matrix} \right) \)
x-3 =1 ⇒ x = 4
3x-z = 0
3(4)-z = 0
-z=-12 ⇒ z = 12
x + y + 7 = 1
x + y = -6
4 + y = -6
y = -10
x = 4, y = -10, z = 12

a_ij = |i - 2j|
a₁₁ = |1-2 x 1| = |1-2| = |-1| = 1
a₁₂ = |1-2 x 2| = |1-4| = |-3| = 3
a₁₃ = |1-2 x 3| = |1-6| = |-5| = 5
a₂₁ = |2-1 x 1| = |2-2| = 0
a₂₂ = |2-2 x 2| = |-2| = 2
a₂₃ = |2-2 x 3| = |-4| = 4
a₃₁= |3-2 x 1| = |1| = 1
a₃₂ = |3-2 x 2| = |-1| = 1
a₃₃ = |3-2 x 3| = |-3| = 3
\(\therefore \left[ \begin{matrix} 1 & 3 & 5 \\ 0 & 2 & 4 \\ 1 & 1 & 6 \end{matrix} \right] \) is the required 3x3 matrix.

Let Senthil’s house number be x.
It is given that 1 + 2 + 3 +...  + (x -1) = (x + 1) + (x + 2) + .... + 49
1 + 2 + 3 + ... + (x - 1) = [1 + 2 + 3...+49] - [1 + 2 + 3+ ...+ x] 
\(\frac { x-1 }{ 2 } \left[ 1+\left( x-1 \right) \right] =\frac { 49 }{ 2 } \left[ 1+49 \right] -\frac { x }{ 2 } \left[ 1+x \right] \)
\(\frac { x\left( x-1 \right) }{ 2 } =\frac { 49\times 50 }{ 2 } -\frac { x\left( x+1 \right) }{ 2 } \)
x² - x = 2450 - x2 - x \(\Rightarrow\) 2 x 2 = 2450
x² = 1225 gives x = 35
Therefore, Senthil’s house number is 35.