Equation of a circle when end points of the diameter are given is
(x - x₁) (x - x₂) + (y - y₁) (y - y₂) = 0
Here (x₁, y₁) = (4,7) and (x₂, y₂)  = (-2,5) 
(x - 4) (x + 2) + (y - 7) (y - 5 ) = 0
x² + 2x - 4x -8 + y² - 5y - 7y + 35 = 0
x² + y² - 2x - 12y + 27 = 0

Four dice are rolled, total number of possible outcomes = 6⁴
Number of possible outcomes in which 2 does not appear on any dice = 5⁴
∴ Required number of possible outcomes = Total number of possible outcomes - number of possible outcomes in which 2 does not appear on any dice.
= 6⁴ - 5⁴ = 1296 - 625
= 671

\(\frac{9 !}{6 ! 3 !}=\frac{9 \times 8 \times 7 \times 6 !}{6 ! \times 3 !}=\frac{9 \times 8 \times 7}{3 \times 2 \times 1}=84\)

Let A = \(\begin{vmatrix}5&20\\ 0&-1 \end{vmatrix}\)
Minor of 5 = M₁₁ = -1
Minor of 20 = M₁₂ = 0
Minor of 0 = M₂₁ = 20
Minor of -1 = M₂₂ = 5
Co-factor of 5 = A₁₁ = -1
Co-factor of 20 = A₁₂ = 0
Co-factor of 0 = A₂₁ = -20
Co-factor of -1 = A₂₂ = 5

let P (x₁,y₁)  be any point on the locus and A (3, -2) the given point
PA = 4
\(\Rightarrow\) PA² = 16
\(\Rightarrow\)  (x₁ - 3)² + (y₁ +2)²  = 16
\(\Rightarrow\)   \({ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }\) - 6x₁ + 4y₁ + 13 -16 = 0
\(\Rightarrow\)   \({ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }\) - 6x₁ + 4y₁ - 3 = 0
Locus of (x₁,y₁) is x² + y² - 6x + 4y - 3 = 0

\(a_{ 11 }=15,\quad { a }_{ 12 }=10,\quad { x }_{ 1 }=35\)
\(a_{ 21 }=20,\quad { a }_{ 22 }=30,\quad { x }_{ 2 }=65\)
\(b_{ 11 }=\frac { { a }_{ 11 } }{ { x }_{ 1 } } =\frac { 15 }{ 35 } =\frac { 3 }{ 7 } ;\quad b_{ 12 }=\frac { { a }_{ 12 } }{ { x }_{ 2 } } =\frac { 10 }{ 65 } =\frac { 2 }{ 13 } \)
\(b_{ 21 }=\frac { { a }_{ 21 } }{ { x }_{ 1 } } =\frac { 20 }{ 35 } =\frac { 4 }{ 7 } ;\quad b_{ 22 }=\frac { { a }_{ 12 } }{ { x }_{ 2 } } =\frac { 30 }{ 65 } =\frac { 6 }{ 13 } \)
B = \(\begin{bmatrix} \frac { 3 }{ 7 } & \frac { 2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 6 }{ 13 } \end{bmatrix}\)
\(I-B=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} \frac { 3 }{ 7 } & \frac { 2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 6 }{ 13 } \end{bmatrix}=\begin{bmatrix} \frac { 4 }{ 7 } & \frac { -2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 7 }{ 13 } \end{bmatrix}\)
\(|I-B|=\frac{4}{7} \times \frac{7}{13}-\left(\frac{2}{13} \times \frac{4}{7}\right)=\frac{28-8}{91}=\frac{20}{91}\)
Since the diagonal elements of I - B are positive and |I - B| is positive the system is available
 \((I-B)^{-1}=\frac{1}{|I-B|} \operatorname{adj}(\mathrm{I}-\mathrm{B})=\frac{91}{20}\left(\begin{array}{cc} \frac{7}{13} & \frac{2}{13} \\ \frac{4}{7} & \frac{4}{7} \end{array}\right)\)
Now, X = (I - B)^-1 D where D \(=\left[ \begin{matrix} 12 \\ 18 \end{matrix} \right] =\frac { 91 }{ 20 } \begin{bmatrix} \frac { 7 }{ 13 } & \frac { 2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 4 }{ 7 } \end{bmatrix}\left[ \begin{matrix} 12 \\ 18 \end{matrix} \right]=\frac{91}{20}\left(\begin{array}{l} \frac{84+36}{13} \\ \frac{48+72}{7} \end{array}\right)\)
\(=\left(\begin{array}{l} 42 \\ 78 \end{array}\right)\)
The gross output for two sectors X and Y are 42 and 78 respectively

\(m=\frac{-a}{b}=\frac{-7}{5}\)

Equality coefficient of x in the numerator
kx = 4 (2n) + 1 (x)
kx = 9x

Since if n = 1 then (1) (2) (3) (4) = 24 is divisible by = 24

nP₂ = 20
n(n - 1) = 5 x 4
n = 5

5x - 16 = 0
5x = 16
\(x=\frac{16}{5}\)

\(A=\left(\begin{array}{cc} \frac{4}{5} & \frac{-5}{12} \\ \frac{-2}{5} & \frac{1}{2} \end{array}\right)\)
\(|A|=\frac{2}{5}-\frac{1}{6}=\frac{12-5}{30}=\frac{7}{30}\)
\(A^{-1}=\frac{1}{|A|} \text { adjA }=\frac{30}{7}\left(\begin{array}{ll} \frac{1}{2} & \frac{5}{12} \\ \frac{2}{5} & \frac{4}{5} \end{array}\right)\)

Equation of parabola \(y^2=-4 a x\)
It passes through (-2, -3)
\(9=-4 a(-2) \Rightarrow 4 a=\frac{9}{2}\)
Required equation \(y^2=\frac{-9 x}{2}\)

Let P (n) denote the statement. 4 + 8 + 12 + ...... + 4n = 2n(n + 1)
Put n = 1
LHS = 4
RHS ⇒ 2(2) = 4
LHS = RHS
∴ P(1) is true.
Let us assume that P(k) is true.
p(k) : 4 + 8 + 12 + ..... + 4 = 2k(k + 1)
To prove that P(k + 1) is true
4 + 8 + .... + 4k + 4(k + 1) = 2(k + 1) = 2(k + 1)(k + 2)
P(k) + 4(k + 1)
= 2k(k + 1) + 4(k + 1)
= (k + 1)(2k + 4) [using (1)]
= 2(k + 1)(k + 2) = RHS
∴ P (k + 1) is true whenever P(k) is true.
∴ p(n) is true for all \(n\in N\).

We consider selling the units to be a positive earning and buying as a negative earning
\(3 x+5 y-4 z =6000\)
\(2 x-3 y+z =5000 \)
\(-x+4 y+6 z =13000\)
The given system can be written as
\(\left(\begin{array}{ccc} 3 & 5 & -4 \\ 2 & -3 & 1 \\ -1 & 4 & 6 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 6000 \\ 5000 \\ 13000 \end{array}\right)\)
\(A X=\mathrm{B} \Rightarrow X=A^{-1} B\)
\(\text {Where } \mathrm{A}=\left(\begin{array}{ccc} 3 & 5 & -4 \\ 2 & -3 & 1 \\ -1 & 4 & 6 \end{array}\right), X=\left(\begin{array}{l} x \\ y \\ z \end{array}\right)\)
\(B=\left(\begin{array}{c} 6000 \\ 5000 \\ 13000 \end{array}\right)\)
\(|A|=3(-18-4)-5(12+1)-4(8-3)\)
\(=-66-65-20=-151 \neq 0\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -22 & -13 & 5 \\ -46 & 14 & -17 \\ -7 & -11 & -19 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \text { adj } A=\frac{-1}{151}\left(\begin{array}{ccc} -22 & -13 & 5 \\ -46 & 14 & -17 \\ -7 & -11 & -19 \end{array}\right)^T\)
\(X=\mathrm{A}^{-1} \mathrm{~B}\)
\(=\frac{-1}{151}\left(\begin{array}{ccc} -22 & -46 & -7 \\ -13 & 14 & -11 \\ 5 & -17 & -19 \end{array}\right)\left(\begin{array}{c} 6000 \\ 5000 \\ 13000 \end{array}\right)\)
\(=\frac{-1}{151}\left(\begin{array}{ccc} -132000 & -230000 & -91000 \\ -78000 & +70000 & -143000 \\ 30000 & -85000 & -247000 \end{array}\right)\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\frac{-1}{151}\left(\begin{array}{l} -453000 \\ -151000 \\ -302000 \end{array}\right)=\left(\begin{array}{l} 3000 \\ 1000 \\ 2000 \end{array}\right)\)
\(x=Rs.3000, \mathrm{y}=Rs.1000, \mathrm{z}=Rs.2000\)

G	L	Number of ways
(6)	(4)
3	2	\(6 C_3 \times 4 C_2\)	120
2	3	\(6 C_2 \times 4 C_3\)	60
1	4	\(6 C_1 \times 4 C_4\)	6
			186

G	L	Number of ways
(6)	(4)
5	0	\(6 C_5 \times 4 C_0\)	6
4	1	\(6 C_4 \times 4 C_1\)	60
3	2	\(6 C_3 \times 4 C_2\)	120
			186

Out of 10 digits 2 digits have already been used (35) with the remaining 8 digits we have to fill 4 places (6 digit number)
No. of ways \(=8 P_4=8 \times 7 \times 6 \times 5\)
\(=1680\)

(1 + x)²ⁿ
2n is even
Middle term is \(t_{\frac{n}{2}+1}=t_{\frac{2 n}{2}+1}=t_{n+1}\)
\(r=n\)
\(t_{r+1}=n C_r x^{n-r} a^r\)
\(t_{n+1}=2 n C_n(1)^{2 n-n} x^n\)
\(=2 n C_n \cdot x^n\)
\(n C_r=\frac{n !}{r !(n-r) !}\)
\(=\frac{(2 n) !}{n !(2 n-n) !} x^n\)
\(=\frac{(2 n)(2 n-1)(2 n-2)(2 n-3) \ldots 5 \cdot 4 \cdot 3 \cdot 2.1}{n ! n !} x^n\)
\(=\frac{(2 n)(2 n-2) \ldots 4.2(2 n-1)(2 n-3) \ldots 5.3 .1}{n ! n !} x^n\)
\(=\frac{2^n(n(n-1) \ldots 2.1)(2 n-1)(2 n-3) \ldots 5.3 .1}{n ! n !} x^n\)
\(=\frac{2^n \cdot n !(2 n-1)(2 n-3) \ldots 5 \cdot 3 \cdot 1}{n ! n !} x^n\)
\(=\frac{1.3 .5 \ldots(2 n-3)(2 n-1) 2^n x^n}{n !}\)

\(A=\left(\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right)\)
\(|A|=3(12-10)+1(-30+25)+1(30-30)\)
\(=6-5=1 \neq 0\)
\(\therefore A^{-1} \text { exists }\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right)\)

\(\frac { 1 }{ (x-1)(x+2)^{ 2 } } =\frac { A }{ x-1 } +\frac { B }{ x+2 } +\frac { C }{ (x+2)^{ 2 } } \)
⇒ \(\frac { 1 }{ (x-1)(x+2)^{ 2 } } =\frac { A(x+2)^{ 2 }+B(x-1)(x+2)+C(x-1) }{ (x-1)(x+2)^{ 2 } } \)
⇒ 1 = A (x + 2)² + B (x - 1)(x + 2) + C(x - 1)  ....(1)
Putting x = 1 in (1) we get,
1 = A(3)² ⇒ 1 = 9A ⇒ A = \(\frac { 1 }{ 9 } \)
Putting x = -2 in (1) we get
1 = C(-2-1) ⇒ 1 = -3 \(\Rightarrow\) C = \(\frac { -1 }{ 3 } \)
Equate co-efficient of x² on both sides of (1)
\(0=A+B\)
\(B=-A=\frac{-1}{9}\)
\(\frac{1}{(x-1)(x+2)^2}-\frac{1}{9(x-1)}-\frac{1}{9(x+2)}-\frac{1}{3(x+2)^2}\)

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.

5x² + 5y² +4x - 8y - 16 = 0
[Divide by 5]
x² + y² +  \(\frac { 4 }{ 5 } x-\frac { 8 }{ 5 } y-\frac { 16 }{ 5 } =0\)
Here 2g  = \(\frac { 4 }{ 5 } \) \(\Rightarrow\)  \(g=+\frac { 2 }{ 5 } \)
2f = \(-\frac { 8 }{ 5 } \)     \(\Rightarrow\) \(f=-\frac { 4 }{ 5 } \)
and c = \(-\frac { 16 }{ 5 } \)
Center of the circle is (-g, -f) \(\Rightarrow \) \(\left( -\frac { 2 }{ 5 } ,\frac { 4 }{ 5 } \right) \)
Radius of the circle is \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
\(\Rightarrow\) r = \(\sqrt { \frac { 4 }{ 25 } +\frac { 16 }{ 25 } +\frac { 16 }{ 5 } } =\sqrt { \frac { 20 }{ 25 }+ { \frac { 16 }{ 5 }} } \)
\(\Rightarrow\) \(r=\sqrt { \frac { 20 }{ 5 } } \) = \(\sqrt { 4 } \) = 2 units

2x + 3y = 5
x - 2y = -1
The given system can be written as
\(\begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix}\)\(\begin{bmatrix} X\\Y \end{bmatrix}\)=\(\begin{bmatrix} 5\\-1 \end{bmatrix}\)
AX = B \(\Rightarrow \) \(X=A^{-1} B\)
Where A =\(\begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix}\), \(X= \left[ \begin{matrix} x \\ y \end{matrix} \right] \),B=\(\begin{bmatrix} 5\\-1 \end{bmatrix}\)
\(|A|=-4-3=-7 \neq 0\)
\(\therefore\) A^-1 exists.
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{-1}{7}\left(\begin{array}{cc} -2 & -3 \\ -1 & 2 \end{array}\right)\)
\(X=A^{-1} B\)
\(=\frac{-1}{7}\left(\begin{array}{cc} -2 & -3 \\ -1 & 2 \end{array}\right)\left(\begin{array}{c} 5 \\ -1 \end{array}\right)\)
\(=\frac{-1}{7}\left(\begin{array}{cc} -10 & +3 \\ -5 & -2 \end{array}\right)=\frac{-1}{7}\left(\begin{array}{l} -7 \\ -7 \end{array}\right)\)
\(\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} 1 \\ 1 \end{array}\right)\)
\(\text {Solution is } x=1, y=1\)

(n + 2)! = 60(n - 1)!
(n + 2)(n + 1)(n)(n - 1)! = 60(n - 1)!
(n + 2)(n + 1)(n) = 60
(n + 2)(n + 1)(n) = 5 \(\times \) 4  \(\times \) 3
n = 3
(Counting 60 as product of 3 consecutive natural numbers)

The condition for concurrent line is
\(\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=0\)
\(\Rightarrow \) \(\left| \begin{matrix} 1 & 1 & -4 \\ 3 & 0 & 2 \\ 3 & -3 & 16 \end{matrix} \right| =0\)
\(\Rightarrow \)  1 (0+6) -1 (48 -6) - 4 (-9 -0)
\(\Rightarrow \) 6 - 42 + 36 = 0
\(\Rightarrow \) 42 - 42 = 0
Hence,therefore the given lines are concurrent

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

\((x-2 y)^{13}\)
\(T_{r+1}=n C_r x^{n-r} a^r\)
\(n=13, r=4\)
\(t_{r+1}=13 C_r x^{13-r}(-2 y)^r\)
\(t_5=13 C_4 x^{13-4}(-2 y)^4\)
\(=\frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} x^9(16) y^4\)
\(=11440 x^9 y^4\)