y = 3cosx + 4sinx
y₁ = –3 sin x + 4 cos x
y₂ = –3 cos x – 4 sin x
y₂ = – (3 cos x + 4 sin x)
y₂ = –y (or) y₂ + y = 0

Given tanA = m tanB
\(\cfrac { sinA }{ cosA } =m\cfrac { sinB }{ cosB } \) 
\(\cfrac { sinAcosB }{ cosAsinB } =m\) 
Applying componendo and dividendo rule, we get
\(\cfrac { sinAcosB+cosAsinB }{ sinAcosB-cosAsinB } =\cfrac { m+1 }{ m-1 } \) 
\(\cfrac { sin(A+B) }{ sin(A-B) } =\cfrac { m+1 }{ m-1 } \)
which completes the proof.

\(\left( { x }^{ 2 }+\cfrac { 1 }{ { x }^{ 2 } } \right) ^{ 4 }=\left( { x }^{ 2 } \right) ^{ 4 }+4{ C }_{ 1 }\left( { x }^{ 2 } \right) ^{ 3 }\cfrac { 1 }{ { x }^{ 2 } } +{ 4C }_{ 2 }\left( { x }^{ 2 } \right) ^{ 2 }\left( \cfrac { 1 }{ { x }^{ 2 } } \right) ^{ 2 }+4{ C }_{ 3 }\left( { x }^{ 2 } \right) \left( \cfrac { 1 }{ { x }^{ 2 } } \right) ^{ 3 }+4{ C }_{ 4 }\left( \cfrac { 1 }{ { x }^{ 2 } } \right) ^{ 4 }\)
\(={ x }^{ 8 }+4{ x }^{ 4 }+6+\cfrac { 4 }{ { x }^{ 2 } } +\cfrac { 4 }{ { x }^{ 4 } } +\cfrac { 1 }{ { x }^{ 8 } } \)

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
\({ x }_{ 1 }+2{ x }_{ 2 }= 8\)

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

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

x	1/x
6	0.1667
14	0.0714
21	0.0476
30	0.0333
	\(\sum { 1/x=0.3190 } \)

Harmonic Mean = \(\frac { n }{ \sum { \frac { 1 }{ x } } } \)
\(=\frac { 4 }{ 0.3190 } =12.54\)
\(\therefore\) HM = 12.54

RX	RY	d=RX-RY	d2
1	12	-11	121
2	9	-7	49
3	6	-3	9
4	10	-6	36
5	3	2	4
6	5	1	1
7	4	3	9
8	7	1	1
9	8	1	1
10	2	8	64
11	11	0	0
12	1	11	121
			\(\sum\)d2=416

n=12
Rank correlation co-efficient
\(\rho =1-\frac { 6\sum { { d }^{ 2 } } }{ N({ N }^{ 2 }-1) } \)
=1-\(\frac { 6\times 416 }{ 12({ 12 }^{ 2 }-1) } =1-\frac { 16 }{ 11 } =\frac { -5 }{ 11 } \)
\(\rho\)=-0.45

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

(i) S = {1, 2, 3, 4, 5, 6}
n(S) = 6
(i) Let A be the event of getting a prime number
A = {2, 3, 5}
n(A) = 3
\(P(A)=\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let B be the event that the number is greater than or equal to 3.
B = {3, 4, 5, 6}    
n(B) = 4
\(P(B) =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)

% Rise	x	log x
5	105	2.0212
8	108	2.0334
77	177	2.2480
		\(\sum logx\) = 6.3026

GM = Antilog\(\left( \frac { \sum { logx } }{ n } \right) =Anitlog\left( \frac { 6.3026 }{ 3 } \right) =Antilog(2.1009)\)
GM = 126.2
\(\therefore\) Average increase of the commodity from 2004 to 2007
= 126.1 - 100 = 26.1%

a = Rs.1500; i = 12/100 = 0.12
P = \(\frac { a }{ i } =\frac { 1500 }{ 0.12 } \) = Rs.12,500
The person has to deposit Rs.12,500 to meet this expense.

Let us consider X for price P and Y for stock S. Then the mean and SD for P is considered as \(\bar { X } \) = 100 and σ_x = 8 respectively and the mean and SD of S is considered as \(\bar { Y } \) = 103 and σ_y = 4. The correlation coefficient between the series is r(X, Y) = 0.4
Let the regression line X on Y be
\(X-\bar { X } =r\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } (Y-\bar { Y } )\)
X-100 = (0.4)\(\frac{8}{4}\)(Y-103)
X–100 = 0.8(Y–103 )
X–0.8Y–17.6 = 0 (or)  X = 0.8Y+17.6
The regression line Y on X be \(Y-\bar { Y } =r\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } (X-\bar { X } )\)
Y-103 = (0.4)\(\frac{4}{8}\)(X-100)
Y–103 = 0.2 (X–100 )
Y–103 = 0.2 X–20
Y = 0.2 X + 83 (or) 0.2 X–Y + 83 = 0

u = x cos y + y cos x
\(\frac { { \partial }u }{ \partial x } =\cos \ y - y \sin x\)
\(\frac { { \partial }u }{ \partial y } =-x \sin y +\ cos \ x\)
\(L H S=\frac{\partial^2 u}{\partial x \partial y}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right) =\frac{\partial}{\partial x}(-x \sin y+\cos x) =-\sin y-\sin x \)
\(R H S=\frac{\partial^2 u}{\partial y \partial x}=\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\right) =\frac{\partial}{\partial y}(\cos y-y \sin x) =-\sin y-\sin x \)
\(\therefore \frac{\partial^2 u}{\partial x \partial y}=\frac{\partial^2 u}{\partial y \partial x} \)

u = log(x²+y²)
\(\frac { \partial u }{ \partial x } =\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } (2x)=\frac { 2x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\(\frac { { \partial }^{ 2 }u }{ \partial { x }^{ 2 } } =\frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) . 2-2x.2x }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } =\frac { 2\left( { y }^{ 2 }-{ x }^{ 2 } \right) }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } \)
\(\frac { \partial u }{ \partial y } =\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } (2y)=\frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\(\frac { { \partial }^{ 2 }u }{ \partial y^{ 2 } } =\frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) . 2-2y.2y }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } =\frac { 2\left( { y }^{ 2 }-{ x }^{ 2 } \right) }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } \)
\(\therefore \frac { { \partial }^{ 2 }u }{ { 2x }^{ 2 } } +\frac { { \partial }^{ 2 }u }{ \partial { y }^{ 2 } } =0\)

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

\(y={2x+1\over 3x+2}\)
\({dy\over dx}={(3x+2)(2)-(2x+1)(2)\over (3x+2)^2}\)\(={1\over (3x+2)^2}\)
Elasticity: \(η={x\over y}.{dx\over dx}\)\(={x\over \left(2x+1\over 3x+2\right)}.{1\over (3x+2)}\)
\(={x\over (2x+1)(3x+2)}\)
When x = 1, \(η={1\over15}\)

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.

Let y = sin x. sin 2x. sin 3x
Taking logarithms on both sides we get,
log y = log (sin x. sin 2x. sin 3x)
= log (sin x) + log (sin 2x) + log (sin 3x)    [\(\therefore\) log ab = log a + log b]
Differentiating with respect to 'x' we get,
\(\frac { 1 }{ y } \frac { dy }{ dx } =\frac { 1 }{ sin\quad x } .\frac { d }{ dx } (sin\quad x)+\frac { 1 }{ sin\quad 2x } .\frac { d }{ dx } (sin\quad 2x)+\frac { 1 }{ sin\quad 3x } .\frac { d }{ dx } (sin\quad 3x)\)
\(\frac { 1 }{ y } \frac { dy }{ dx } =\frac { cos\quad x }{ sin\quad x } +\frac { 2cos2x }{ sin\quad 2x } =3.\frac { cos\quad 3x }{ sin\quad 3x } \)
= cot x + 2 cot 2x + 3 cot 3x
\(\Rightarrow \frac { dy }{ dx } =y[cotx+2cot2x+3cot3x]\)
\(\Rightarrow \frac { dy }{ dx } \)= sin x sin 2x sin 3x [cot x + 2 cot 2x + 3 cot 3x]

Given lines are (b - c) x + (c - a) y + (a - b) = 0
(c - a) x + (a - b) y + (b - c) = 0
(a - b) x + (b - c) y + (c - a) = 0.
The condition for the given lines to be concurrent is
\(\left| \begin{matrix} b-c & c-a & a-b \\ c-a & a-b & b-c \\ a-b & b-c & c-a \end{matrix} \right| =0\)
Applying the elementary gains formation C₁⟶C₁+C₂+C₃
We get \(\left| \begin{matrix} 0 & c-a & a-b \\ 0 & a-b & b-c \\ 0 & b-c & c-a \end{matrix} \right| =0\)
Expanding along C₁ we get
\(\left| \begin{matrix} 0 & c-a & a-b \\ 0 & a-b & b-c \\ 0 & b-c & c-a \end{matrix} \right| =0\)
Hence the given lines are Concurrent

\({{(2x+1)!}\over{(x+2)!}}.{{(x-1)!}\over{(2x-1)!}}={{3}\over{5}}\)
\(\Rightarrow\)  \({{(2x+1)(2x)(2x-1)!}\over{(x+2)(x+1)(x-1)!}}.{{(x-1)!}\over{(2x-1)!}}={{3}\over{5}}\)
\(\Rightarrow\)  \({{(2x+1)(2)}\over{(x+2)(x+1)}}={{3}\over{5}}\)
\(\Rightarrow\)   10 (2x+1) = 3 (x+2) (x+1)
\(\Rightarrow\)   20x+10=3 (x²+3x+2)
\(\Rightarrow\)   20x+10+3x²-9x-6=0
\(\Rightarrow\)   -3x²+11x+4 = 0
\(\Rightarrow\)   -3x²-11x-4=0
\(\Rightarrow\)   (x-4)(3x+1)=0
\(\Rightarrow\)   x = 4 or x \(={{-1}\over{3}}\)
Since \(x={{-1}\over{3}}\) is not possible, x = 4.

x = a cos \(\theta\), y = a sin \(\theta\)
\(\frac {dx }{ d\theta } =-a\sin { \theta } \) ; \(\frac {dy }{ d\theta } =a\cos { \theta } \)    
\(= { \frac { dy }{ dx} } =\frac { a\cos { \theta } }{ -a\sin { \theta } } =-\cot { \theta } \)
\(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(-\cot \theta)\)
\(=\frac{d}{d \theta}(-\cot \theta) \frac{d \theta}{d x}\)
\(=\operatorname{cosec}^2 \theta\left(\frac{-1}{a \sin \theta}\right)=\frac{-1}{a} \operatorname{cosec}^3 \theta\)

x² + y² - 22x - 4y + 25 = 0
2g  = - 22   \(\Rightarrow \)  g = -11
2f  = -4   \(\Rightarrow\) f = -2
c = 25 
\(\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=0\)    
Center of the circle is (-g, -f) = ( 11, 2)
Radius of the circle is  \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } =\sqrt { \left( -11 \right) ^{ 2 }+({ -2 })^{ 2 }-25 } \)
 \(r= \sqrt { 121+4-25 } =\sqrt { 100 } \) = 10 units

\( \sin ^{-1}\left(\frac{-3}{5}\right)-\sin ^{-1}\left(\frac{-8}{17}\right) =-\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{8}{17} \quad\left(\sin ^{-1}(-x)=-\sin ^{-1} x\right) \)
\(=\cos ^{-1} \frac{15}{17}-\cos ^{-1} \frac{4}{5} \ \sin ^{-1} \frac{3}{5}=\cos ^{-1} \frac{4}{5}\)
\(\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{15}{17}\)
\(=\cos ^{-1}\left[\frac{15}{17} \times \frac{4}{5}+\sqrt{1-\frac{225}{289}} \sqrt{1-\frac{16}{25}}\right]\)
\(=\cos ^{-1}\left[\frac{60}{85}+\frac{8}{17} \times \frac{3}{5}\right]=\cos ^{-1}\left[\frac{60+24}{85}\right]\)
\(=\cos ^{-1} \frac{84}{85}\)

Let A = \(\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}\)
\(\therefore |A|=\begin{bmatrix} -1 &5 \\-3 & 2 \end{bmatrix}=-2+15=13\)
Now, A₁₁ = 2, A₁₂ = (-3) = 3, A₂₁ = -5, A₂₂ = -1
\(\therefore\) adj A = \({\begin{bmatrix}2 & 3 \\-5 &-1 \end{bmatrix}}^{T}=\begin{bmatrix}2 & -5 \\3 & -1 \end{bmatrix}\)
Now \({A}^{-1}={{1}\over{|A|}}\) adj A = \({{1}\over{13}}\begin{bmatrix} 2& -5\\3 & -1 \end{bmatrix}\)

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 \(\theta\)

\(f(x)={ x }^{ 3 }-\frac { 1 }{ { x }^{ 3 } } \)
\(f\left( \frac { 1 }{ x } \right) \)= \(=\frac { 1 }{ { x }^{ 3 } } -{ x }^{ 3 }\)
\(f(x)+f{ \left( \frac { 1 }{ x } \right) }={ x }^{ 3 }-\frac { 1 }{ x^{ 3 } } +\frac { 1 }{ x^{ 3 } } -{ x }^{ 3 }=0\).

\(AB=\begin{bmatrix} 3&-1\\2&1 \end{bmatrix}\begin{bmatrix} 3&0\\1&-2 \end{bmatrix}=\begin{bmatrix} 9-1&0+2\\6+1&0-2 \end{bmatrix}=\begin{bmatrix} 8&2\\7&-2 \end{bmatrix}\)
= -16 - 14 = -30
\(\therefore\) |AB| = -30

Let B = \(\begin{vmatrix} 1 &-3 &2 \\4 &-1&2\\3&5&2 \end{vmatrix}\)
Minor of 1 = M₁₁ = \(\begin{vmatrix} -1 & 2 \\ 5 & 2 \end{vmatrix}=-2-10=-12\)
Minor of -3 = M₁₂ = \(\begin{vmatrix}4 &2 \\ 3 & 2 \end{vmatrix}=8-6=2\)
Minor of 2 = M₁₃ = \(\begin{vmatrix} 4 & -1 \\ 3 & 5\end{vmatrix}=20+3=23\)
Minor of 4 = M₂₁ = \(\begin{vmatrix} -3 & 2 \\5 & 2 \end{vmatrix}=-6+10=-16\)
Minor of -1 = M₂₂ = \(\begin{vmatrix}1 & 2 \\ 3 & 2 \end{vmatrix}=2-6=-4\)
Minor of 2 = M₂₃ = \(\begin{vmatrix} 1& -3 \\3 &5 \end{vmatrix}=5+9=14\)
Minor of 3 = M₃₁ = \(\begin{vmatrix} -3 &2 \\ -1 & 2 \end{vmatrix}=-6+2=-4\)
Minor of 3 = M₃₂ = \(\left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right|=2-8=-6\)
Minor of 2 = M₃₃ = \(\begin{vmatrix} 1 & -3 \\4 & -1 \end{vmatrix}=-1+12=11\)
Co-factor of 1 = A₁₁ = (-1)¹⁺¹ M₁₁ = -12
Co-factor of -3 = A₁₂= (-1)¹⁺² M₁₂ = -2
Co-factor of 2 = A₁₃= (-1)¹⁺³ M₁₃ = 23
Co-factor of 4 = A₂₁ = (-1)²⁺¹ M₂₁ = 16
Co-factor of -1 = A₂₂ = (-1)²⁺² M₂₂ = -4
Co-factor of 2 = A₂₃ = (-1)2+3 M₂₃ = -14
Co-factor of 3 = A₃₁ = (-1)³⁺¹ M₃₁ = -4
Co- factor of 5 = A₃₂ = (-1)³⁺² M₃₂ = 6
Co-factor of 2 = A₃₃ = (-1)³⁺³ M₃₃ = 11

(2a - 3b)4
\( (x+a)^n=n C_0 x^n+n C_1 x^{n-1} a+n C_2 x^{n-2} a^2 +\ldots n C_{n-1} x a^{n-1}+n C_n a^n \)
\((2 a-3 b)^4= 4 C_0(2 a)^4-4 C_1(2 a)^3(3 b) +4 C_2(2 a)^2(3 b)^2 -4 C_3(2 a)(3 b)^3+4 C_4(3 b)^4\)
= 16a⁴ - 4 (8a³) (3b) + 6 (4a²) (9b²) - 4 (2a)(27b³) + 81b⁴
= 16a⁴ - 96 a³ b + 216 a² b² - 216 ab³ + 81b⁴

\(\underset { x\rightarrow 2 }{ lim } \frac { { x }^{ 2 }-4x+6 }{ x+2 } =\cfrac { \underset { x-2 }{ lim } \left( { x }^{ 2 }+4x+6 \right) }{ \underset { x\rightarrow 2 }{ lim } \left( x+2 \right) } \)
\( =\cfrac { \left( 2 \right) ^{ 2 }-4\left( 2 \right) +6 }{ 2+2 } =\cfrac { 1 }{ 2 } \)

Let \(\left(\cos ^{-1} \frac{8}{17}\right)=\theta\) 
\(\cos \theta=\frac{8}{17}\) 
\(\sin\theta =\sqrt { 1-{ \cos }^{ 2 }\theta } \)
= \(\sqrt { 1-\cfrac { 64 }{ 289 } } \) 
= \(\cfrac { 15 }{ 17 } \)   ...(2)
From (1) and (2), we get
Now \(\tan\left( \cos\cfrac { 8 }{ 17 } \right) =\tan\theta =\cfrac { \sin\theta }{ \cos\theta } \)
\(=\cfrac { \frac { 15 }{ 17 } }{ \frac { 8 }{ 17 } } =\cfrac { 15 }{ 8 } \)

\(\cfrac { sin2\theta }{ 1+cos2\theta } =\cfrac { 2sin\theta cos\theta }{ { 2cos }^{ 2 }\theta } =\cfrac { sin\theta }{ cos\theta } =tan\theta \)

Equation of circle is
\(\left( x-h \right) ^{ 2 }+\left( y-k \right) ^{ 2 }={ r }^{ 2 }\) 
Here \(\left( h,k \right) =(3,-1)\) and r = 4
Equation of circle is
\(\left( x-3 \right) ^{ 2 }+\left( y+1 \right) ^{ 2 }=16\)
\({ x }^{ 2 }-6x+9+{ y }^{ 2 }+2y+1=16\)
\({ x }^{ 2 }+{ y }^{ 2 }-6x+2y-6=0\)

The number of combination = \({ 32C }_{ 4 }\) 
\( =\cfrac { 32! }{ 4!(32-4)! } \)
\(=\cfrac { 32! }{ 4!\left( 28 \right) ! } \)

Let \(\frac { 1 }{ \left( { x }^{ 2 }-1 \right) } =\frac { A }{ x-1 } +\frac { B }{ x+1 } \)
Multiplying both sides by (x - 1)(x + 1), we get
1 = A(x +1) + ( Bx -1) ... (1)
Put x =1 in (1) we get, 1 = A(2)
 \(\therefore\) A = \(\frac{1}{2}\)
Put x = -1 in (1) we get, 1 = A(0) + B(-2)
\(\therefore\)  B = - \(\frac{1}{2}\)

Let A = \(\left[ \begin{matrix} 8 & 2 \\ 4 & 3 \end{matrix} \right] \)
|A| = \(\left| \begin{matrix} 8 & 2 \\ 4 & 3 \end{matrix} \right| \)
= 24 – 8 = 16 ≠ 0
\(\therefore\) A is a non-singular matrix

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

(i) Variables:
Let x₁ and x₂ represent the number of fans and sewing machines.
(ii) Objective functions:
Let Z be the profit of the dealer.
∴ Maximize Z = 22x₁ + 18x₂ is the objective function.
(iii) Constraints:
\({ x }_{ 1 }+{ x }_{ 2 }\le 20\)
\({ 360x }_{ 1 }+{ 240x }_{ 2 }\le 5760\)
(iv) Non-negative restrictions:
Since the number of fans and sewing machine cannot be negative, we have x₁, x₂ ≥ 0.
Hence, mathematical formulation of the LPP is
Maximize \(Z=22{ x }_{ 1 }+18{ x }_{ 2 }\)
Subject to the constraints
\({ x }_{ 1 }+{ x }_{ 2 }\le 20\)
\({ 360x }_{ 1 }+{ 240x }_{ 2 }\le 5760\)
and x₁, x₂ ≥ 0.

(i) Variables:
Let x₁, x₂ represent the number of units of X and Y. 
(ii) Constraints:

	Labour	Capital
X	2	3
Y	3	1

 

∴ 2x₁ + 3x₂ ≤ 30 and 3x₁ + x₂ ≤ 17
(iii) Non-negative restrictions:
Since the number of units of X and Y cannot be negative,x₁, x₂ ≥ 0.
Hence, the mathematical formulation of the LPP is maximize \(Z=100{ x }_{ 1 }+120{ x }_{ 2 }\)
Subject to the constraints
\( { 2x }_{ 1 }+3{ x }_{ 2 }\le 30\)
\({ 3x }_{ 1 }+{ x }_{ 2 }\le 17\)
and x₁, x₂ ≥ 0.

a = 50, i = \(\frac{5}{100}\) = .05
A = \(\cfrac { a }{ i } =\cfrac { 50 }{ 0.05 } \) Rs.1000

Given Σxy = 120, Σx² = 90, Σy² = 640
Then r = \(\frac { \Sigma xy }{ \sqrt { \Sigma { x }^{ 2 }\Sigma { y }^{ 2 } } } =\frac { 120 }{ \sqrt { 90(640) } } =\frac { 120 }{ \sqrt { 57600 } } =\frac { 120 }{ 240 } \) = 0.5

\(p(x)=\frac { { x }^{ 3 } }{ 3 } +{ x }^{ 2 }+x\)
p(x) = x² + 2x + 1 = (x +1)²
It is clear that P'(x)>0 for all x.
\(\therefore\) The firm is running a profitable business.

\(\cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}\)
\(\cos 20^{\circ}+\left(\cos 100^{\circ}+\cos 140^{\circ}\right) =\cos 20^{\circ}+2 \cos \left(\frac{100^{\circ}+140^{\circ}}{2}\right) \cos \left(\frac{100^{\circ}-140^{\circ}}{2}\right) \)
\(=\cos 20^{\circ}+2 \cos 120^{\circ} \cos \left(-20^{\circ}\right)\)
\(=\cos 20^{\circ}+2 \cos \left(180^{\circ}-60\right) \cos 20^{\circ}\)
\(=\cos 20^{\circ}-2 \cos 60^{\circ} \cos 20^{\circ}\)
\(=\cos 20^{\circ}-2\left(\frac{1}{2}\right) \cos 20^{\circ}\)
\(=\cos 20^{\circ}-\cos 20^{\circ}=0\)

\(\mathrm{LHS}=\frac{\cos 2 A-\cos 3 A}{\sin 2 A+\sin 3 A}\)
\(=\frac{-2 \sin \frac{2 A-3 A}{2} \sin \frac{2 A+3 A}{2}}{2 \sin \frac{2 A+3 A}{2} \cos \frac{2 A-3 A}{2}}\)
\(=\frac{\sin A / 2}{\cos A / 2}=\tan A / 2\)
= RHS
Hence proved.

\(lety=\frac { { x }^{ 2 }cos\frac { \pi }{ 4 } }{ sinx } ={ x }^{ 2 }\times \frac { 1 }{ \sqrt { 2 } sinx } \)
\(\therefore y=\frac { 1 }{ \sqrt { 2 } } \frac { { x }^{ 2 } }{ sinx } \)
Differentiating with respect to 'x' we get
\(\frac { dy }{ dx } =\frac { 1 }{ \sqrt { 2 } } \left[ \frac { sinx.(2x)-{ x }^{ 2 }cosx }{ sin^{ 2 }x } \right] \)[by quotient rule]
= \(\frac { x }{ \sqrt { 2 } } \left[ \frac { 2sinx-xcosx }{ { sin }^{ 2 }x } \right] \)
= \(\frac { x }{ \sqrt { 2 } } \left[ \frac { 2sinx }{ { sin }^{ 2 }x } -\frac { xcosx }{ sin^{ 2 }x } \right] =\frac { x }{ \sqrt { 2 } } \)[cosec x-x cot x cosec x]
\(\frac { x }{ \sqrt { 2 } } cosecx[2-cotx]\)

LHS=\(\frac{\tan 69^o+\tan 66^o}{1-\tan 69^o\tan 66^o}\left[\because\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan(A+B)\right]\)
= tan (69° + 66°) = tan (135°)
= tan (180 - 45°) = - tan 45° = -1 = RHS.
Hence proved.

Using \(\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)\), we get
\(LHS=\sin^2\left(\frac{\pi}{8}+\frac{x}{2}\right)-\sin^2\left(\frac{\pi}{8}-\frac{x}{2}\right)=\sin\left(\frac{\pi}{8}+\frac x2+\frac{\pi}{8}-\frac x2\right).\sin\left(\frac{\pi}{8}+\frac x2-\frac{\pi}{8}+\frac x2\right)\)
\(=\sin\left(\frac{2\pi}{8}\right).\sin\left(\frac{2x}{2}\right)=\sin\left(\frac{\pi}{4}\right).\sin x=\frac{1}{\sqrt2}.\sin x=RHS\)
Hence proved.

Given pair of lines is
3x²+10xy+8y²+14x+22y+15=0
2h=10
Here a=3, h=5, b=8,
Let \(\theta\) be the angle between the pair of lines
Then \(tan\quad \theta =\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } =\frac { \pm 2\sqrt { 25-3(8) } }{ 3+8 } =\frac { \pm 2\sqrt { 1 } }{ 11 } =\frac { \pm 2 }{ 11 } \)
\(\therefore \quad tan\quad \theta =\frac { 2 }{ 11 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 2 }{ 11 } \right) \)

The length of the tangent to the circle x² + y² +2gx + 2fy + C = 0 from a point
(x₁, y₁)  is \(\sqrt { { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+2g{ x }_{ 1 }+2{ fy }_{ 1 }+C } \)
Length of tangent = \(\sqrt { { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+2{ x }_{ 1 }+4{ y }_{ 1 }+9 } \)
= \(\sqrt { 1+4-2+8+9 } \)
= \(\sqrt{20}\) units
[Here (x₁, y₁) = (1, 2)]

LHS 10P₃ = 10 x 9 x 8 = 720
RHS 9P₃ + 3. 9P₂ = 9 x 8 x 7 + 3 x 9 x 8
= 9 x 8 (7 + 3) = 72 (10) = 720
LHS= RHS  Hence proved.

LHS\(=\tan^{-1}\left(\frac mn\right)-\tan^{-1}\left(\frac{m-n}{m+n}\right)=\tan^{-1}\left(\frac{\frac{m}{n}-\frac{m-n}{m+n}}{1+\left(\frac mn\right)\left(\frac{m-n}{m+n}\right)}\right)\)
\(=\tan^{-1}\left(\frac{\frac{m^2+mn-mn+n^2}{n(m+n)}}{\frac{n(m+n)+m(m-n)}{n(m+n)}}\right)\)
\(=\tan ^{-1}\left(\frac{m^2+n^2}{m^2+n^2}\right)=\tan ^{-1}(1)=\frac{\pi}{4}=\mathrm{RHS}\)
Hence proved.

\(\frac { 7! }{ 6! } =\frac { 7\times 6! }{ 6! } =7\)

\(\sin { { 300 }^{ o } } \)
\(=\sin { 300 }^{ o } =\sin { \left( { 360 }^{ o }-{ 60 }^{ o } \right) } \) (IV quad)
\(= \sin 60^o=-\frac { \sqrt { 3 } }{ 2 } \)

Let |A| = \(\begin{vmatrix} 6 &5 &12 \\ 2 & 4 &4 \\2 & 1 & 4 \end{vmatrix}\)
Taking 2 common from C₁ and 4 common from C₃, we get,
\(|A|=2\times4\begin{vmatrix} 3 & 5&3 \\ 1 & 4 & 1\\1 &1 &1\end{vmatrix}=8\times 0\ [\because C_1\equiv C_3]=0\)

AB = \(\begin{bmatrix} 1 \\ -4 \\3 \end{bmatrix}\begin{bmatrix} -1 &2 & 1 \end{bmatrix}=\begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\-3 & 6 & 3 \end{bmatrix}\)
\(\therefore\)  \({(AB)}^{T}=\begin{bmatrix} -1 &4&-3 \\ 2 & -8&6\\1&-4&3 \end{bmatrix}\)  ....(1)
\({B}^{T}=\begin{bmatrix} -1 & 2 & 1 \end{bmatrix}^{T}=\begin{bmatrix} -1\\2\\1\end{bmatrix}\)and \({A}^{T}={\begin{bmatrix} 1\\-4\\3\end{bmatrix}}^{T}=\begin{bmatrix} 1&-4&3 \end{bmatrix}\)
\(\therefore\)  \({B}^{T}{A}^{T}=\begin{bmatrix} -1\\2\\1 \end{bmatrix}\begin{bmatrix} 1&-4&3 \end{bmatrix}=\begin{bmatrix} -4 & 4&-3 \\ 2&-8 &6\\1&-4&3 \end{bmatrix}\)  ....(2)
From (1) and (2), (AB)^T = B^T . A^T

\(\mathrm{RHS}=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
Let \(x=\tan \theta \Rightarrow \theta=\tan ^{-1} x\)
\(=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=\sin ^{-1}(\sin 2 \theta)\)
\(=2 \theta=2 \tan ^{-1} x=\text { LHS }\)
Hence proved.

B \(=\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}\)
I - B = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}=\begin{bmatrix} 0.50 & -0.30 \\ -0.41 & 0.67 \end{bmatrix}\)
= (0.50) (0.67) - (0.30) (0.41)
\(|I-B|\) = 0.335 - 0.123 = 0.212 >  0
Since the main diagonal elements of I - B are positive and |I-B| is positive. Hawkins Simon conditions are satisfied. Therefore given system is viable

Let \(y=\sqrt { \cfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3x }^{ 2 }+4x+5 } } \) 
= \(\left| \cfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3x }^{ 2 }+4x+5 } \right| ^{ \frac { 1 }{ 2 } }\) 
Taking logarithm on both sides,
\(logy=\cfrac { 1 }{ 2 } \left[ log\left( x-3 \right) +log\left( { x }^{ 2 }+4 \right) -log\left( { 3x }^{ 2 }+4x+5 \right) \right] \)\({[\because \log a b} =\log a+\log b \text { and } \log \frac{a}{b} =\log a-\log b]\) 
Differentiating with respect to x
\(\cfrac { 1 }{ y } .\cfrac { dy }{ dx } =\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ x-3 } +\cfrac { 2x }{ { x }^{ 2 }+4 } -\cfrac { 6x+4 }{ { 3x }^{ 2 }+4x+5 } \right] \)
\(\cfrac { dy }{ dx } =\cfrac { 1 }{ 2 } \sqrt { \cfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3x }^{ 2 }+4x+5 } } \)\( \left[ \cfrac { 1 }{ x-3 } +\cfrac { 2x }{ { x }^{ 2 }+4 } -\cfrac { 6x+4 }{ { 3x }^{ 2 }+4x+5 } \right] \)

Let y= 75-12x+6x²-x³
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=0-12+12x-3x^2=-3(x^2-4x+4)=-3(x-2)^2\)
\(\Rightarrow{dy\over dx}\le 0\) for all \(x \in (-\infty , \infty)\)
\(\therefore \) y decreases as x increases.

Let the number of shares be x.
Market value of 1 share = 100 - 10 = 90
Market value of x shares = 90x
Income from 10% discount shares
If Investment = 33, Income = 15
If Investment = 90x, Income  \(=\cfrac { 15 \times90x }{ 33 } \)   
Income from 10% premium shares
= \(\cfrac { 110x }{ 33 } \times 15\) 
\(\cfrac { 110x }{ 33 } \times 15-\cfrac { 90x }{ 33 } \times 15=450\)
\(\cfrac { 15x }{ 33 } {( 110 - 90) } =450\)
x = \(\cfrac { 450\times 33 }{ 15\times 20 } \) = 49.5~50 shares

C = 15 + 9x - 6x² + x³
\({dC\over dx}=9-12x+3x^2\)
\({dC\over dx}=-12x+6x\)
For minimum cost \({dC\over dx}=0\)
\(\Rightarrow 3x^2-12x+9=0\)
\(\Rightarrow x^2-4x+3=0\) (Divided by 3)
\(\Rightarrow (x-1)(x-3)=0\)
\(\Rightarrow x=1,3\)
At x = 1, \({d^2C\over dx^2}=-12+6 = -6 < 0\)
At x = 3, \({d^2C\over dx^2}=-12+6 = -6 > 0\)
At x = 3 cost is minimum.

Since both the decision variables x₁, x₂ are non-negative, the solution lies in the first quadrant of the plane.
Consider the equations x₁ – x₂ = –1 and – x₁ + x₂ = 0
x₁ – x₂ = –1 is a line passing through the points (0,1) and (–1,0)
–x₁ + x₂ = 0 is a line passing through the point (0,0)
Now we draw the graph satisfying the conditions x₁ – x₂ ≤ –1; –x₁ + x₂ ≤ 0 and x₁, x₂ ≥ 0

There is no common region(feasible region) satisfying all the given conditions. Hence the given LPP has no solution.

We know that average cost [AC] is minimum when average cost [AC] = marginal cost [MC].
Cost: \(C={1\over3}x^3-3x^2+9x\)
AC = \({1\over3}-3x^2+9\) and MC = x² - 6x + 9
Now, AC = MC ⇒ \({1\over 3}x^2-3x+9-x^2-6x+9\)
⇒ 2x² - 9x = 0 ⇒ \(x={9\over 2}\) unit (∵ x > 0)

\(X=\left(\begin{array}{ccc} 8 & -1 & -3 \\ -5 & 1 & 2 \\ 10 & -1 & -4 \end{array}\right)\)
\(|X|=8(-4+2)+1(20-20)-3(5-10)\)
\(=-16+15=-1 \neq 0\)
\(X^{-1} \text { exists }\)
\(A_{11}=\text {Co-factor of } 8=-4+2=-2 \)
\(A_{12}=\text {Co-factor of }-1=-(20-20)=0\)
\(A_{13}=\text {Co-factor of }-3=5-10=-5\)
\(\mathrm{A}_{21}=\text {Co-factor of }-5=-(4-3)=-1 \)
\(\mathrm{A}_{22}=\text {Co-factor of } 1=-32+30=-2 \)
\(\mathrm{A}_{23}=\text {Co-factor of } 2=-(-8+10)=-2\)
\(\mathrm{A}_{31}=\text {Co-factor of } 10=-2+3=1 \)
\(\mathrm{A}_{32}=\text {Co-factor of }-1=-(16-15)=-1 \)
\(\mathrm{A}_{33}=\text {Co-factor of }-4=8-5=+3\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -2 & 0 & -5 \\ -1 & -2 & -2 \\ 1 & -1 & +3 \end{array}\right)\)
\(\operatorname{Adj} X=\left(\begin{array}{ccc} -2 & -1 & 1 \\ 0 & -2 & -1 \\ -5 & -2 & +3 \end{array}\right)\)
\(X^{-1}=\frac{1}{|X|} \operatorname{adj} X\)
\(=\frac{-1}{1}\left(\begin{array}{ccc} -2 & -1 & 1 \\ 0 & -2 & -1 \\ -5 & -2 & +3 \end{array}\right)\)
\(\text {Given } Y=X^{-1}\)
\(\left(\begin{array}{ccc} 2 & 1 & -1 \\ 0 & 2 & 1 \\ 5 & p & q \end{array}\right)=\left(\begin{array}{ccc} 2 & 1 & -1 \\ 0 & 2 & 1 \\ 5 & 2 & -3 \end{array}\right)\)
\(\mathrm{p}=2 ; \mathrm{q}=-3\)

Given A = \(\begin{bmatrix}1 &3&3 \\1 &4&3\\1&3&4 \end{bmatrix} \)
\(A_{11}=\text {Cofactor of } 1=16-9=7\)
\(A_{12}=\text {Cofactor of } 3=-(4-3)=-1\)
\(A_{13}=\text {Cofactor of } 3=3-4=-1\)
\(A_{21}=\text {Cofactor of } 1=-(12-9)=-3 \)
\(A_{22}=\text {Cofactor of } 4=4-3=1 \)
\(A_{23}=\text {Cofactor of } 3=-(3-3)=0 \)
\( A_{31}=\text {Cofactor of } 1=9-12=-3 \)
\(A_{32}=\text {Cofactor of } 3=3-3=0 \)
\(A_{33}=\text {Cofactor of } 4=4-3=1\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{array}\right)\)
A^-1 = \(\begin{bmatrix} 7&-3&-3\\-1&1&0\\-1&0&1 \end{bmatrix}\)
\(|A| =1(16-9)-3(4-3)+3(3-4) \)
\(=7-3-3=1 \neq 0\)
\(\therefore \mathrm{A}^{-1} \text { exists }\)
\(\mathrm{A}(\operatorname{adj} \mathrm{A})=\left(\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right)\left(\begin{array}{ccc} 7 & -3 & -1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right)\)
\(=\left(\begin{array}{lll} 7-3-3 & -3+3+0 & -3+0+3 \\ 7-4-3 & -3+4+0 & -3+0+3 \\ 7-3-4 & -3+3+0 & -3+0+4 \end{array}\right)\)
\(=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=|A| I\)
\(=\mathrm{A}(\operatorname{adj} \mathrm{A})=|A| I\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right)\)

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

LHS = \(\begin{vmatrix}-a^2&ab&ac\\ab&-b^2&bc\\ac&bc&-c^2 \end{vmatrix}\)
Taking a, b, c common from R₁, R₂ and R₃ respectively we get.
LHS = \(abc\begin{vmatrix}-a&b&c\\a&-b&c\\a&b&-c \end{vmatrix}\)
Again taking a, b, c common from C₁ C₂ and C₃ respectively.
LHS = \(a^2b^2c^2\begin{vmatrix}-1&1&1\\1&-1&1\\1&1&-1 \end{vmatrix}\)
\(R_2 \rightarrow R_2+R_1, R_3 \rightarrow R_3+R_1\)
\(=a^2b^2c^2\begin{vmatrix}0&0&2\\1&-1&1\\1&1&-1 \end{vmatrix}\)
\(=a^2 b^2 c^2[-1(0-4)]=4 a^2 b^2 c^2\)
= RHS Hence proved.

\(\frac { { x }^{ 2 }-6x+2 }{ { x }^{ 2 }(x+2) } =\frac { A }{ x } +\frac { B }{ { x }^{ 2 } } +\frac { C }{ x+2 } \)
⇒ \(\frac { { x }^{ 2 }-6x+2 }{ { x }^{ 2 }(x+2) } =\frac { Ax(x+2)+B(x+2)+Cx^{ 2 } }{ { x }^{ 2 }(x+2) } \)
⇒ x² - 6x + 2 = Ax (x + 2) + B (x + 2)+ Cx²   ..(1)
x = -2 in (1) we get,
4 + 12+ 2 = \(C(-2)^2\) = C(4)
⇒ 18 = 4C ⇒ C = \(\frac { 18 }{ 4 } =\frac { 9 }{ 2 } \)
x = 0 in (1) we get,
2 = B (2) ⇒ B = 1
Equating co-efficient of x² on both sides of (1)
1 = A + C ⇒ = A + \(\frac { 9 }{ 2 } \) [∵ C=\(\frac { 9 }{ 2 } \)]
⇒ A = \(1-\frac { 9 }{ 2 } =\frac { 2-9 }{ 2 } =\frac { -7 }{ 2 } \)
\(\frac{x^2-6 x+2}{x^2(x+2)}=\frac{-7}{2 x}+\frac{1}{x^2}+\frac{9}{2(x+2)}\).