\({{x-2}\over{(x+2){(x-1)}^{2}}}={{A}\over{x+2}}+{{B}\over{x-1}}+{{C}\over{{(x-1)}^{2}}}\)
\(\frac{x-2}{(x+2)(x-1)^2}=\frac{A(x-1)^2+B(x+2)(x-1)+C(x+2)}{(x+2)(x-1)^2}\)
X - 2 = A(x - 1)²+ B(x +2)(x - 1)+ C(x + 2)     ..(1)
x = -2 in (1) we get,
-2 - 2 =  A (-3)2 \(\Rightarrow\) - 4 = 9 A \(\Rightarrow\) A = \({{-4}\over{9}}\)
x = 1 in (1) we get,
1- 2 = C(1 + 2) \(\Rightarrow\) -1 = 3C \(\Rightarrow\) C = \({{-1}\over{3}}\)
Equate co-efficient of x² on both sides of (1)
0 = A + B
\(B=-A=\frac{4}{9}\)  
\(\therefore\)   \({{x-2}\over{(x+1){(x-1)}^{2}}}-{{-{{4}\over{9}}}\over{x+2}}+{{{{4}\over{9}}}\over{x-1}}+{{-{{1}\over{3}}}\over{{(x-1)}^{2}}}+{{-4}\over{9(x+2)}}+{{4}\over{9(x-1)}}-{{1}\over{3{(x+1)}^{2}}}\)

Since \(\cfrac { 3\pi }{ 2 } <\left( A,B \right) <2\pi \) ,both A and B lie in the fourth quadrant,
 \(\therefore \) sinA and sinB are negative
Given \(\cos A=\frac{4}{5} \text { and } \cos B=\frac{12}{13}\)
Therefore, \(\sin A=-\sqrt{1-\cos ^2 A}\)
\(=-\sqrt{1-\frac{16}{25}}\)
\(=-\sqrt{\frac{25-16}{25}}\)
\(=-\frac{3}{5}\)
\(\operatorname{Sin} \mathrm{B}=-\sqrt{1-\cos ^2 B}\)
\(=-\sqrt{1-\frac{144}{169}}\)
\(=-\sqrt{\frac{169-144}{169}}\)
\(=-\frac{5}{13}\)
\( \cos (A+B)= \cos A \cos B- \sin A \sin B \)
\(=\frac{4}{5} \times \frac{12}{13}-\left(\frac{-3}{5}\right) \times\left(\frac{-5}{13}\right)\)
\(=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}\)

\(\frac{x}{(x-1)(x+1)^2}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^2}\)
x = A(x + 1)² + B(x - 1)(x + 1) + C(x - 1)  ... (1)
Put x = 1 in (1)
\(1=A{ (1+1) }^{ 2 }+B(0)+C(0)\)
\(\therefore \ A=\frac { 1 }{ 4 } \)
Put x = –1 in (1)
-1 = A(0) + B(0) + C(-1-1)
\(\therefore C=\frac { 1 }{ 2 } \)
Equating the constant term on both sides of (1),we get
A - B - C = 0
\(\Rightarrow B=A-C\)
\(\therefore B=-\frac { 1 }{ 4 } \)

The given system can be written as
\(\left[ \begin{matrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 8 \\ 1 \\ 4 \end{matrix} \right] \)
i.e., AX = B
X = A^–1B
Here \(A=\left[ \begin{matrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)and \( B=\left[ \begin{matrix} 8 \\ 1 \\ 4 \end{matrix} \right] \)
\(|A|=\left[ \begin{matrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{matrix} \right] \)
= -17 ≠ 0
A^–1 exists
A₁₁ = –1 A₁₂ = –8 A₁₃ = –10
A₂₁ = –5 A₂₂ = –6 A₂₃ = 1
A₃₁ = –1 A₃₂ = 9 A₃₃ = 7
\(|A_{ij}|=\left[ \begin{matrix} -1 & -8 & -10 \\ -5 & -6 & 1 \\ -1 & 9 & 7 \end{matrix} \right] \)
adj A = [A_ij]^T
\(=\left[ \begin{matrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{matrix} \right] \)
A^–1 = \(\frac{1}{|A|}adjA\)
\(=-\frac{1}{17}\left[ \begin{matrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{matrix} \right] \)
X = A^-1B
\(=-\frac{1}{17}\left[ \begin{matrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{matrix} \right] \)\(\left[ \begin{matrix} 8 \\ 1 \\ 4 \end{matrix} \right] \)
\(=-\frac{1}{17}\left[ \begin{matrix} -17 \\ -34 \\ -51 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\3 \end{matrix} \right] \)
\(\left[ \begin{matrix}x\\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right] \)
x = 1, y = 2 and z = 3.

Given \(u=tan^{-1}, \left(x^2+y^2\over x+y\right)\)
\(⇒\ tan\ u={x^2+y^2\over x+y}\)
Let f= tan u = \(x^2+y^2\over x+y\)
\(f(tx, ty)={(tx)^2+(ty)^2\over tx+ty}={t^2(x^2+y^2)\over t(x+y)}=t\left(x^2+y^2\over x+y\right)=t^1f(x,y)\)
∴ f is a homogeneous function of degree 1.
By Euler's theorem
\(x{∂f\over ∂x}+y{∂f\over ∂y}=nf\)
\(⇒\ x{∂f\over ∂x}+y{∂f\over ∂y}=1f\)
\(⇒x.{∂\over ∂ x}(tan\ u)+y.{∂\over ∂y}(tan\ u)=tan\ u\)
\(⇒\ x.sec^2u.{∂u\over ∂x}+y.sec^2u.{∂u\over ∂y}=tan\ u\)
Dividing throughout by sec²u we get,
\(x.{∂u\over ∂x}+y.{∂u\over ∂y}={tan\ u\over sec^2u}={sinu\over cosu}/{1\over cos^2u}={sinu\over cosu}\times cos^2u=sinu\ cosu\)
\(={2sinu\ cosu\over 2}={1\over 2}sin2u\) [∵ sin2x=2sinx cosx]
\(x.{∂u\over ∂ x}+y.{∂u\over ∂y}={1\over 2}sin 2u\)
Hence Proved.

Given C= 2000 + 1800x - 75x² + x³
\(MC={dC\over dx}={d\over dx}( C = 2000 + 1800x - 75x^2 + x^3)=1800-150x+3x^2\)
Let y = 3x² - 150x+ 1800
Differentiating w.r.t 'x' we get,
\({dy\over dx}=6x-150\)
\({dy\over dx}=0⇒6x-150 = 0\)
⇒ x=25

Intervals	Sign of \({dy\over dx}\)	Nature of Function
(0,25) say x =1	6(1) - 150 = -144 (Negative)	Decreasing function
(25, ∞) say x = 30	6(30) - 150 = 30 (Positive)	Increasing function

Hence, marginal cost function is decreasing in (0, 25) and increasing in (25, ∞).

Given \(P={4000x\over 500+x}-x\)
Differentiating w.r.t. 'x' we get
\(\frac { dp }{ dx } =\frac { { (500+x)(4000)-(4000x)(1) } }{ (50+x)^{ 2 } } -1\)
\(={4000(500+x-x)\over (500+x)^2}-1={2000000\over(500+x)^2}-1\)
Profit is maximum when \({dP\over dx}=0\) and \({d^2P\over dx}<0\)
\({dP\over dx}=0⇒{2000000\over (500+x)^2}-1=0\)
\(⇒{2000000\over (500+x)^2}=1⇒(500+x)^2=2000000\)
\(⇒500+x=1000\times\sqrt2=1000\times1.414\)
⇒ x = 1414 - 500 = 914
Also \({d^2P\over dx^2} = -{4000000\over (500+x)^3}\)
When x = 914 \({d^2P\over dx^2}<0\)
∴ Profit is maximum, when x = 914

We will find the correct values of \(\sum\)x, \(\sum\)x², \(\sum\)y², and \(\sum\)xy by delecting the old values and adding new ones.
\(\therefore\)\(\sum\)x=125-(6+8)+(6+8)=125
\(\sum\)y=100-(14+6)+(12+8)=100
x²=650-(6²+8²)+(8²+6²)=650
y²=460-(14²+6²)+(12²+8²)=436
and xy =508-(14x6+8x6)+(12x8+6x8)=520
\(\therefore\) Correlation Co-efficient
r(x,y)=\(\frac { N\sum { xy } -(\sum { x } )(\sum { y } ) }{ \sqrt { N{ \sum { x } }^{ 2 }-{ (\sum { x } ) }^{ 2 } } \sqrt { N{ \sum { y } }^{ 2 }-{ (\sum { y } ) }^{ 2 } } } \)
\(\Rightarrow\)\(\frac { 25(520)-125(100) }{ \sqrt { 25(650)-{ (125) }^{ 2 } } \sqrt { 25(436)-{ (100) }^{ 2 } } } \)
\(\Rightarrow\) r(x,y)=0.66

E1 =0	L6 =22
E2 = 0+7=7	L5 =(22 - 3) = 19
E3 =(7 + 3) or (0 + 6)  Whichever is maximum = 0	L4 =22 - 6 = 16 .
E4 =(7 + 9) or (10 + 2)  Whichever is maximum=16	L3 =(22 - 9) or (16 - 8) Whichever is minimum=13
E5 =(0+ 11)or(10+4)  Whichever is maximum=14	L2 =16 - 9 = 7
E6 =(16 + 6) or (10 + 9) or (14 + 3)  Whichever is maximum = 22	L1 =7-7=0

Activity	Duration	EST	EFT=EST+tij	LST	LFT
1-2	7	0	7	7-7=0	7
1-3	6	0	6	13-6=7	13
1-5	11	0	11	19-11=8	19
2-3	3	7	10	13-3=10	13
2-4	9	7	16	16-9=7	16
3-4	2	13	15	19-4=15	16
3-5	4	13	17	19-4=15	19
3-6	9	10	19	22-9=13	22
4-6	6	16	22	22-6=16	22
5-6	3	14	17	22-3=19	22

EFT and LFT are same in the activity, 1 - 2, 2 - 4 and 4 - 6.
Hence the critical path is 1 - 2 - 4 - 6 and the project completion time is 22 Weeks.

Let x₁ g of wheat and x₂ g of rice be mixed in the daily diet. Let Z be the minimum cost of diet.

	Proteins	Carbohydrates	Cost
1 g of Wheat	0.1 g	0.25 g	Rs.4/kg
1 g of Rice	0.05 g	0.5 g	Rs.6/kg
Minimum Requirement	50 g	200 g

Thus, the mathematical formulation of the LPP
Minimize  \(Z=\frac { 4{ x }_{ 1 } }{ 1000 } +\frac { 6{ x }_{ 2 } }{ 1000 } \quad \Rightarrow \quad Z=\frac { { x }_{ 1 } }{ 250 } +\frac { 3{ x }_{ 2 } }{ 500 } \)
Subject to the constraints
\(0.1{ x }_{ 1 }+0.05{ x }_{ 2}\ge 50 \Rightarrow 2{ x }_{ 1 }+{ x }_{ 2 }\ge 1000\)
\( 0.25{ x }_{ 1 }+0.5{ x }_{ 2\quad }\ge 200 \Rightarrow { x }_{ 1 }+2{ x }_{ 2 }\ge 800\)
\(and\ { x }_{ 1 },{ x }_{ 2 }\ge 0\)
Consider the equation
\(2{ x }_{ 1 }+{ x }_{ 2 }= 1000\)

\({ x }_{ 1 }\)	0	500
\({ x }_{ 2 }\)	1000	0

\({ x }_{ 1 }+2{ x }_{ 2 }= 800\)

\({ x }_{ 1 }\)	0	500
\({ x }_{ 2 }\)	1000	0

 

 

The feasible region is ABC an its co-ordinates are A(800, 0), C(0, 1000) and B is the point of intersection of the lines 2x₁+ x₂ = 1000 ... (1) x₁+ 2x₂ = 800 ... (2)
Verification of B:
\(\Rightarrow (1)\times 2 4{ x }_{ 1 }+2{ x }_{ 2 }=2000\)
\(\quad (-)\quad (-)\quad \quad (-)\)
\(\Rightarrow (2) 15{ x }_{ 1 }+6{ x }_{ 2 }=30 \Rightarrow { x }_{ 1 }=400\)
\(--------------\)
Substracting,     \(3{ x }_{ 1 }=2000\)
From (2),    \(\quad 400+2{ x }_{ 2 }=800\)
\(2{ x }_{ 2 }=400\quad \Rightarrow \quad { x }_{ 2 }=200\)  

Corner Points	\( Z=\frac { { x }_{ 1 } }{ 250 } +\frac { 3{ x }_{ 2 } }{ 500 } \)
A(800,0)	\(\frac { 800 }{ 250 } =3.2\)
B(400, 200)	\(\frac { 400 }{ 250 } +\frac { 600 }{ 500 } =2.8\)
C(0,1000)	\(\frac { 3000 }{ 500 } =6\)

Minimum of Z occurs at B(400, 200). Hence, the solution is x₁= 400, x₂ = 200 and Z_rnin = 2.8

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 E₁, E₂ and A be defined as
E₁ = First bag is drawn
E₂ = Second bag is drawn
A = Red ball is drawn
\(\therefore\) P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A/E₁) = \(\frac{3}{7}\), P(A/E₂) = \(\frac{5}{11}\)
By Baye's theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { 1/2\times 3/7 }{ 1/2\times 3/7+1/2\times 5/11 } \)
\(=\frac { 3/7 }{ 3/7+5/11 } =\frac { 3/7 }{ \frac { 33+35 }{ 77 } } =\frac { 3}{7 }\times \frac { 77 }{ 68 } =\frac { 33 }{ 68 } \)

Let investment in 6% stock be x
\(\therefore\) Investment in 5% stock = 13,500 - x.
At 6% stock
If Investment = 140, Income = 6
If Investment = x, Income \(=\frac { 6x }{ 140 }=\frac { 3x }{ 70 } \)
At 5% stock
If Investment = 125, Income = 5
If Investment = 13,500 - x,
Income = \(\frac { 5(13,500-x )}{ 125 } \)  
\(=\frac { 13,500 -x}{ 25 } =540-\frac { x }{ 25 } \)
Total income = 560
\( \frac { 3x }{ 70 } +540-\frac { x }{ 25 } =560\)
\(\frac { 3x }{ 70 } -\frac { x }{ 25 } =20\)
\(\frac { 15x-14x }{ 350 } =20\) 
x = 350 x 20 = Rs. 7000
\(\therefore\) Amount invested in 6% stock = Rs. 7000
Amount invested in 5% stock = 13500 - 7000
= Rs. 6500

X	Y	X2	Y2	XY
1	4	1	16	4
2	8	4	64	16
3	2	9	4	6
4	12	16	144	48
5	10	25	100	50
6	14	36	196	84
7	16	49	256	112
8	6	64	36	48
9	18	81	324	162
​\(\Sigma X\)​ = 45	​\(\Sigma Y\)​ = 90	​\(\Sigma X^2\)​ = 285	​\(\Sigma Y^2\)​ = 1140	​\(\Sigma XY\)​ = 530

\(\bar{X} =\frac{45}{9}=5 \quad \bar{Y}=\frac{90}{9}=10 \)
\(b_{y x} =\frac{N \Sigma X Y-\Sigma X \Sigma Y}{N \Sigma X^2-(\Sigma X)^2} \)
\(=\frac{9(530)-4050}{9(285)-2025}=\frac{720}{540}=1.33\)
Regression equation of Y on X
\(Y-\overset{-}{Y}=b_{yx}(X-\overset{-}{X})\)
Y - 10 = 1.33(X - 5)
Y = 1.33X - 6.65 + 10
Y = 1.33X + 3.35

Calculation of Regression equation

X	x = (X-13)	x2	Y	y = (Y-41)	y2	xy
10	-3	9	40	-1	1	3
12	-1	1	38	-3	9	3
13	0	0	43	2	4	0
12	-1	1	45	4	16	-4
16	3	9	37	-4	16	-12
15	2	4	43	2	4	4
ΣX = 78	Σx = 0	Σx2 = 24	ΣY = 246	Σy = 0	Σy2 = 50	Σxy = -6

(i) Regression equation of X on Y
\(X-\bar { X } =r\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } (Y-\bar { Y } )\)
\(\bar { x } =\frac { 78 }{ 6 } \)=13, \(\bar { Y } =\frac { 246 }{ 6 } \) = 41
b_xy  = \(r\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } =\frac { \Sigma xy }{ \Sigma { y }^{ 2 } } =\frac { -6 }{ 50 } \) = -0.12
X–13 = –0.12 (Y–41 )
X–13 = –0.12Y + 4.92
X = –0.12Y + 17.92
(ii) Regression Equation of Y on X
\(Y-\bar { Y } =r\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } (X-\bar { X } )\)
b_yx = \(r\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } =\frac { \Sigma xy }{ \Sigma x^{ 2 } } =-\frac { 6 }{ 24 } \) = -0.25
Y–41 = –0.25 (X–13 )
Y–41 = –0.25 X + 3.25
Y = –0.25 X + 44.25
When X is 20, Y will be
Y = –0.25 (20)+44.25
= –5 + 44.25
= 39.25 (when the price is Rs. 20, the likely demand is 39.25)

u = log\(\frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \)
e^u = \(\frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \) = f(x, y) ... (1)
Consider f(x, y)  = \(\frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \)
f(tx, ty) = \(\frac { { t }^{ 4 }{ x }^{ 4 }+{ t }^{ 4 }{ y }^{ 4 } }{ tx+ty } ={ t }^{ 3 }\left( \frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \right) ={ t }^{ 3 }f(x,y)\)
\(\therefore\)  f is a homogeneous function of degree 3.
Using Euler’s theorem we get
\(x.\frac { \partial u }{ \partial u } +y.\frac { \partial u }{ \partial y } =3f\)
Consider f(x, y) = e^u
\(x.\frac { \partial u }{ \partial u } +y.\frac { \partial u }{ \partial y } =3e\)^u
\(∴ { e }^{ u }x.\frac { \partial u }{ \partial u } +{ e }^{ u }y.\frac { \partial u }{ \partial y } =3{ e }^{ u }\)
\(x.\frac { \partial u }{ \partial u } +y.\frac { \partial u }{ \partial y } =3\)

E1 = 0	L5 = 32
E2 = 22 + 0 = 22	L4 = 52 - 12 = 40
E3 = max of {0 + 27, 22  + 12} = 34	L3 = (40 - 6) = 34
E4 = max of {22 + 14, 34 + 6} = 40	L2 = min of {34 - 12, 40 14} = 22
E5 = 40 + 12 = 52	L1 = 0

Activity	Duration tij	EST	EFT = EST + tij	LST = LFT - tij	LFT
1 -2	22	0	22	22 - 22 = 0	22
1 - 3	27	0	27	34 -27 = 7	34
2 - 3	12	22	34	34 - 12 = 22	34
2 - 4	14	22	36	40 - 14 = 26	40
3 - 4	6	34	40	40 - 6 = 34	40
4 -5	12	40	52	52 - 12 = 40	52

\(\because \) EFT and LFT are same in 1 - 2, 2 - 3, 3 - 4 and 4 - 5. the Critical path 1-2-3-4-5 and duration time taken is 52 days.

Given f(x) = 2x³ + 3x² – 12x … (1)
f'(x) = 6x² + 6x – 12
f"(x) = 12x + 6
f'(x) = 0 \(\Rightarrow\)6x² + 6x – 12 = 0
\(\Rightarrow\) 6(x² + x – 2) = 0
\(\Rightarrow\) 6(x + 2)(x – 1) = 0
\(\Rightarrow\) x = –2 ; x = 1
When x = –2, f''(–2) = 12(–2) + 6 = –18 < 0
\(\therefore\) f(x) attains local maximum at x = – 2 and local maximum value is obtained from (1) by substituting the value x = – 2
f(–2) = 2 (–2)³ + 3(–2)² – 12(–2)
= –16 + 12 + 24 = 20.
When x = 1, f"(1) = 12(1) + 6 = 18.
f(x) attains local minimum at x = 1 and the local minimum value is obtained by substituting x = 1 in (1).
f(1) = 2(1) + 3(1) – 12 (1) = –7
Extremum values are – 7 and 20.

First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant.
Write all the inequalities of the constraints in the form of equations.
Therefore we have the lines x₁ + 4x₂ = 24 ; 3x₁ + x₂ = 21; x₁ + x₂= 9
x₁ + 4x₂ = 24 is a line passing through the points (0, 6) and (24, 0)
[(0, 6) is obtained by taking x₁ = 0 in x₁ + 4x₂ = 24 , (24, 0) is obtained by taking x₂ = 0 in x₁+ 4x₂ = 24].
Any point lying on or below the line x₁ + 4x₂ = 24 satisfies the constraint x₁ + 4x₂ ≤ 24 .
3 x₁ + x₂ = 21 is a line passing through the points (0, 21) and (7, 0). Any point lying on or below the line 3 x₁ + x₂ = 21 satisfies the constraint 3 x₁ + x₂ ≤ 21.
x₁ + x₂ = 9 is a line passing through the points (0, 9) and (9, 0). Any point lying on or below the line x₁ + x₂ = 9 satisfies the constraint x₁ + x₂ ≤ 9.
Now we draw the graph

The feasible region satisfying all the conditions is OABCD.The co-ordinates of the points are O(0,0) A(7,0);B(6,3) [the point B is the intersection of two lines x₁ + x₂ = 9 and 3 x₁ + x₂ = 21]; C(4,5) [the point C is the intersection of two lines x₁ + x₂ = 9 and x₁ + 4x₂ = 24] and D(0,6).
Corner points Z = 2x₁ + 5x₂ O(0,0) 0 A(7,0) 14 B(6,3) 27 C(4,5) 33 D(0,6) 30 Maximum value of Z occurs at C. Therefore the solution is x₁ = 4, x₂ = 5, Z_max = 33

p = 50 – 3x
\({dp\over dx}=-3\) ⇒ \({dx\over dp}=-{1\over 3}\)
Elasticity of demand: \(η_d=-{p\over x}.{dx\over dp}\)
\(=-{50-3x\over x}\left(-{1\over3}\right)\)\(={50-3x\over 3x}\ \ ...(1)\)
Now, Revenue: R = px
= (50 - 3x)x = 50x - 3x²
Average revenue: AR = p = 50 - 3x
Marginal revenue: \(MR={dR\over dx} = 50 - 6x\)
\({AR\over AR-MR}={50-3x\over (50-3x)-(50-6x)}\)
\(={50-3x\over 3x}\ \ ...(2)\)
From (1) and (2), we get
\(η_d={AR\over AR-MR}\), Hence verified.

LHS = cos 6x
= cos 3(2x) = 4 cos³2x- 3cos2x    [∴ cos 3\(\theta\) = 4 cos³\(\theta\) - 3 cos\(\theta\)]
= 4 (2cos²x - 1)³- 3(2cos²x - 1)  [∴ cos2x = 2cos²x - 1]
= 4 (8 cos⁶x-12 cos⁴x + 6 cos²x-1) - 6 cos²x + 3
= 32 cos⁶x - 48 cos⁴x + 18 cos²x - 1
= RHS. Hence proved.

RHS = tan 4x = tan2(2x)
= \(\frac { 2\quad tan\quad 2x }{ 1-{ tan }^{ 2 }2x } \left[ \because tan2x=\frac { 2\quad tan\quad x }{ 1-{ tan }^{ 2 }x } \right] \)
= \(\frac { 2.\frac { 2\quad tan\quad x }{ 1-{ tan }^{ 2 }x } }{ 1-\left( \frac { 2\quad tan\quad x }{ 1-{ tan }^{ 2 }x } \right) ^{ 2 } } =\frac { \frac { 4\quad tan\quad x }{ 1-{ tan }^{ 2 }x } }{ \frac { (1-{ tan }^{ 2 }x)^{ 2 }-4{ tan }^{ 2 }x }{ (1-{ tan }^{ 2 }x)^{ 2 } } } \)
= \(\frac { 4\quad tan\quad x }{ 1-{ tan }^{ 2 }\quad x } \times \frac { (1-{ tan }^{ 2 }\quad { x })^{ 2 } }{ 1+{ tan }^{ 4 }x-2\quad { tan }^{ 2 }x-4{ tan }^{ 2 }x } \)
=  \(\frac { 4 tanx(1-{ tan }^{ 2 }\quad x) }{ 1+{ tan }^{ 4 }x-6\ { tan }^{ 2 }x } \)  = LHS

Given y = e^a cos^-1 x    ...(1)
Differentiating with respect to 'x' we get,
\({{dy}\over{dx}}={e}^{a\ {cos}^{-1}x}.{{d}\over{dx}}\left( a\ {\cos}^{-1}x \right)={e}^{a\ {cos}^{-1}x}.\left( {{-a}\over{\sqrt{1-{x}^{2}}}} \right)\)
\(={{-ay}\over{\sqrt{1-{x}^{2}}}}\)  [ using (1) ] 
\(\Rightarrow\sqrt{1-x^2}.{{dy}\over{dx}}=-ay\)
Squaring both sides we get,
\((1-x^2).{\left( {{dy}\over{dx}} \right)}^{2}=a^2y^2\)
Differentiating again with respect to 'x' we get,
\(\left( (1-x^2).2\left( {{dy}\over{dx}} \right)\left({{d^2y}\over{dx^2}} \right) +{\left({{dy}\over{dx}} \right)}^{2}(-2x)=a^2(2y)\left( {{dy}\over{dx}} \right) \right)\)
Dividing throughout by 2 \(\left({{dy}\over{dx}} \right)\) we get,
\((1-x^2).\left({{d^2y}\over{dx^2}} \right)-x\left( {{dy}\over{dx}} \right)=a^2 y\Rightarrow(1-x^2)\left( {{d^2y}\over{dx^2}} \right)-x\left( {{dy}\over{dx}} \right)-a^2y=0.\)
Hence proved.

\(\lim _{ x\rightarrow 1 }{ \frac { { x }^{ 7 }-{ 2x }^{ 5 }+1 }{ { x }^{ 3 }-{ 3x }^{ 2 }+2 } } =\lim _{ x\rightarrow 7 }{ \frac { { x }^{ 7 }-{ x }^{ 5 }-{ x }^{ 5 }+1 }{ { x }^{ 3 }-{ x }^{ 2 }-{ 2x }^{ 2 }+2 } } \)
\(=\lim _{ x\rightarrow 7 }{ \frac { { x }^{ 5 }({ x }^{ 2 }-1)-1({ x }^{ 5 }-1) }{ { x }^{ 2 }(x-1)-2({ x }^{ 2 }-1) } } \)
Dividing the numerator and denominator by (x-1),
\(\\ =\lim _{ x\rightarrow 1 }{ \frac { \frac { { x }^{ 5 }({ x }^{ 2 }-1) }{ x-1 } -1\frac { { (x }^{ 5 }-1) }{ x-1 } }{ \frac { { x }^{ 2 }(x-1) }{ x-1) } -2\frac { ({ x }^{ 2 }-1) }{ x-1 } } } \)
\(=\frac { \lim _{ x\rightarrow 1 }{ { x }^{ 5 }(x+1)-\lim _{ x\rightarrow 1 }{ \frac { { x }^{ 5 }-{ 1 }^{ 5 } }{ x-1 } } } }{ \lim _{ x\rightarrow 1 }{ { x }^{ 2 } } -\lim _{ x\rightarrow 1 }{ 2(x+1) } } \)
\(=\frac { 1(1+1)-5(1)^{ 4 } }{ { 1 }^{ 2 }-2(1+1) } \left[ \because \lim _{ n\rightarrow a }{ \frac { { x }^{ n }-{ a }^{ n } }{ x-a } =n.{ a }^{ n-1 } } \right] \)
\(=\frac { 2-5 }{ 1-2(2) } =\frac { -3 }{ -3 } =1\)

Given \(({\sqrt{2}+1})^{5}+{(\sqrt{2}-1)}^{5}\)
=[\({ \left( \sqrt { 2 } \right) }^{ 5 }\)+ 5C_I \({ \left( \sqrt { 2 } \right) }^{ 4}\) (1)¹ + 5C₂ \({ \left( \sqrt { 2 } \right) }^{ 3 }\) .(1)² + 5C₃ \({ \left( \sqrt { 2 } \right) }^{ 2 }\) . (1)³+ 5C₄ \({ \left( \sqrt { 2 } \right) }^{ 1}\) .(1)⁴ + (1)⁵] +[\({ \left( \sqrt { 2 } \right) }^{ 5 }\) - 5C_I \({ \left( \sqrt { 2 } \right) }^{ 4}\) (l)¹ + 5C₂ \({ \left( \sqrt { 2 } \right) }^{ 3 }\) (1)² - 5C₃ \({ \left( \sqrt { 2 } \right) }^{ 2 }\) (1)³+ 5C₄ \({ \left( \sqrt { 2 } \right) }^{ }\) (1)⁴ -1⁵ ]
=2[\({ \left( \sqrt { 2 } \right) }^{ 5 }\) + 10\({ \left( \sqrt { 2 } \right) }^{ 3 }\) +5 \({ \left( \sqrt { 2 } \right) }^{ }\)] = 2[ 4\(\sqrt { 2 } \) + 20\(\sqrt 2\) + 5.\(\sqrt 2\)] = 2[29.\( \sqrt { 2 } \)] = 58\( \sqrt 2\)

Since the numerator degree is equal to the degree of the denominator, let us divide.

\(\therefore\)   \({{x^2+x+1}\over{x^2+2x+1}}=1-{{x}\over{x^2+2x+1}}\)  ...(1)
Consider    \({{x}\over{x^2+2x+1}}={{x}\over{(x+1)^2}}={{A}\over{x+1}}+{{B}\over{{(x+1)}^{2}}}\)
\(\therefore\ {{x}\over{x^2+2x+11}}={{A(x+1)+B}\over{{(x+1)}^{2}}}\)
\(\Rightarrow\)      x = A(x+ 1) + B  ..(2)
Putting x = - 1 in (2) we get
-1 = 0 + B   \(\Rightarrow\ \ \boxed{B=-1}\)
Putting x = 0 in (2) we get,
0 =A+B \(\Rightarrow\) A - 1 = 0  \(\Rightarrow\)  \(\boxed{A=1}\)
\(\therefore\)   \({{x}\over{{(x+1)}^{2}}}={{1}\over{x+1}}-{{1}\over{{(x+1)}^{2}}}\) ..(2)
Substituting (2) in (1) we get,
\({{x^2+x+1}\over{x^2+2x+1}}=1-{{1}\over{x+1}}+{{1}\over{{(x+1)}^{2}}}\)

\(\frac { \sin { \left( { 180 }^{ o }+A \right) \cos { \left( { 90 }^{ o }-A \right) \tan { \left( { 270 }^{ o }-A \right) } } } \quad \quad }{ \sec { \left( { 540 }^{ o }-A \right) \cos { \left( { 360 }^{ o }+A \right) \csc { \left( { 270 }^{ o }+A \right) } } } } \)
\(=\frac{(-\sin A)(\sin A)(\cot A)}{\sec (360+180-A) \cos A(-\sec A)} \)
\(=\frac{(-\sin A)(\sin A) \frac{\cos A}{\sin A}}{(-\sec A) \cos A(-\sec A)} \)
\(=-\frac{\sin A \cos A}{\frac{1}{\cos A} \cos A \frac{1}{\cos A}} \)
\(=-\sin A \cos ^2 A\)
= R.H.S
Hence proved.

Let P (n) denote the statement 1.2 + 2.3 + 3.4 + ..... + (n + 1) = \(\frac { n(n+1)(n+2) }{ 3 } \)
Put n = 1
LHS = 1 (2) = 2
\(=\frac { 1(2)(3) }{ 3 } \Rightarrow 2=2\)
LHS = RHs
\(\therefore\) P(1) is true.
Let us assume that P(k) IS true i.e, to P.T. 1.2 + 2.3 + 3.4 + ...... + k(k + 1)
= \(\frac { k(k+1)(k+2) }{ 3 } \) ......(1)
To prove that P(k+1) is true i.e to P.T. 1.2 + 2.3 + 3.4 + .... + k(k + 1) + (k + 1)(k + 2)
= \(\frac { k(k+2)(k+2) }{ 3 } \) + (k + 1)(k + 2)
\(=\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}\)
\(=\frac { (k+1)(k+2)(k+3) }{ 3 } \) = RHS
\(\therefore\) P(k + 1) is true whenever P(k) is true.
\(\therefore\) p(n) is true for all \(n\in N\).

Let the 3 numbers be x, y and z.
Given x + y + z = 20
2x + y - z = 23
3x+ y + z  = 46
It can be rewritten as 
\(\begin{bmatrix} 1&1&1\\2&1&-1\\3&1&1 \end{bmatrix}\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 20\\23\\46 \end{bmatrix}\)
\(\Rightarrow\)   \(A X=B \Rightarrow X=A^{-1} B\) 
Where \(A=\begin{bmatrix} 1&1& 1\\2&1&-1\\3&1&1 \end{bmatrix},X=\begin{bmatrix} x\\y\\z \end{bmatrix},B=\begin{bmatrix} 20\\23\\46 \end{bmatrix}\)
= 1 ( 1 + 1 ) - 1 ( 2 + 3 ) + 1 ( 2 - 3)
= 2 - 5 - 1 = 2 - 6 = - 4 \(\neq \) 0 
\(\therefore\) A^-1 exists.
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 2 & -5 & -1 \\ 0 & -2 & 2 \\ -2 & 3 & -1 \end{array}\right)\)
\(\therefore{A}^{-1}={{1}\over{|A|}}adj\ A={{-1}\over{4}}\begin{bmatrix} 2&0&-2\\-5&-2&3\\-1&2&-1 \end{bmatrix}\)
\(X={A}^{-1}B={{-1}\over{4}}\begin{bmatrix}2&0&-2\\5&-2&3\\-1&2&-1 \end{bmatrix}\begin{bmatrix} 20\\23\\46 \end{bmatrix}\)
\(= \begin{bmatrix} x\\y\\z \end{bmatrix},={{-1}\over{4}}\begin{bmatrix} 40+0-92\\-100-46+138\\-20+46-46 \end{bmatrix}={{-1}\over{4}}\begin{bmatrix} -5\\-8\\-20 \end{bmatrix}\)
\(X=\begin{bmatrix} 13\\2\\5 \end{bmatrix}\)
\(x=13, y=2, z=5\)
\(\therefore\) The 3 numbers are 13, 2 and 5.

Now AB = \(\begin{bmatrix} 2&2&1\\1&3&1\\1&2&2 \end{bmatrix} \begin{bmatrix} {{4}\over{5}} &{{-2}\over{5}}&-{{1}\over{5}}\\-{{1}\over{5}}&{{3}\over{5}} &-{{1}\over{5}} \\ -{{1}\over{5}}&-{{2}\over{5}} &{{4}\over{5}}\end{bmatrix}\)
\(=\frac{1}{5}\left(\begin{array}{lll} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{array}\right)\left(\begin{array}{ccc} 4 & -2 & -1 \\ -1 & 3 & -1 \\ -1 & -2 & 4 \end{array}\right)\)
\(=\frac{1}{5}\left(\begin{array}{lll} 8-2-1 & -4+6-2 & -2-2+4 \\ 4-3-1 & -2+9-2 & -1-3+4 \\ 4-2-2 & -2+6-4 & -1-2+8 \end{array}\right)\)
\(=\frac{1}{5}\left(\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right)=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=I\)
\(\mathrm{BA}=\frac{1}{5}\left(\begin{array}{ccc} 4 & -2 & -1 \\ -1 & 3 & -1 \\ -1 & -2 & 4 \end{array}\right)\left(\begin{array}{lll} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{array}\right)\)
\(=\frac{1}{5}\left(\begin{array}{ccc} 8-2-1 & 8-6-2 & 4-2-2 \\ -2+3-1 & -2+9-2 & -1+3-2 \\ -2-2+4 & -2-6+8 & -1-2+8 \end{array}\right)\)
\(=\frac{1}{5}\left(\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right)=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=I\)
\(\mathrm{AB}=\mathrm{BA}=\mathrm{I}\)
\(\therefore\) A and B are inverse of each other

\(3 x-y+2 z=13 ; 2 x+y-z=3\)
\(x+3 y-5 z=-8\)
The given system can be written as
\(\left(\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & -1 \\ 1 & 3 & -5 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(A X=B \Rightarrow X=A^{-1} B\)
\(\text {Where } A=\left(\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & -1 \\ 1 & 3 & -5 \end{array}\right), X=\left(\begin{array}{l} x \\ y \\ z \end{array}\right)\)
\(B=\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(|A|=3(-5+3)+1(-10+1)+2(6-1)\)
\(=-6-9+10=-5 \neq 0\)
\(\therefore A^{-1} exists\)
\(\mathrm{A}_{11}=\text {Co-factor of } 3=(-5+3)=-2 \)
\(\mathrm{A}_{12}=\text {Co-factor of }-1=-(-10+1)=9 \)
\(\mathrm{A}_{13}=\text {Co-factor of } 2=(6-1)=5\)
\(\mathrm{A}_{21}=\text { Co-factor of } 2=-(5-6)=1\)
\(\mathrm{A}_{22}=\text { Co-factor of } 1=-15-2=-17 \)
\(\mathrm{A}_{23}=\text { Co-factor of }-1=-(9+1)=-10\)
\(\mathrm{A}_{31}=\text {Co-factor of } 1=1-2=-1\)
\(\mathrm{A}_{32}=\text {Co-factor of } 3=-(-3-4)=7 \)
\(\mathrm{A}_{33}=\text {Co-factor of }-5=3+2=5\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -2 & 9 & 5 \\ 1 & -17 & -10 \\ -1 & 7 & 5 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{-5}\left(\begin{array}{ccc} -2 & 1 & -1 \\ 9 & -17 & 7 \\ 5 & -10 & 5 \end{array}\right)\)
\(X=A^{-1} B=\frac{-1}{5}\left(\begin{array}{ccc} -2 & 1 & -1 \\ 9 & -17 & 7 \\ 5 & -10 & 5 \end{array}\right)\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(=\frac{-1}{5}\left(\begin{array}{ccc} -26 & 3 & 8 \\ 117 & -51 & -56 \\ 65 & -30 & -40 \end{array}\right)\)
\(=\frac{-1}{5}\left(\begin{array}{c} -15 \\ 10 \\ -5 \end{array}\right)\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 3 \\ -2 \\ 1 \end{array}\right)\)
\(x=3, \mathrm{y}=-2, \mathrm{z}=1\)

\(|A|=\left| \begin{matrix}1 & tan\quad x \\ -tan\quad x & 1 \end{matrix} \right| =1+{ tan }^{ 2 }x={ sec }^{ 2 }x\neq 0\)
\(\Rightarrow\) A^-1 exists
Let C_ij be the cofactor of a_ij  in A
C₁₁ = (-1)¹⁺¹ M₁₁ = (-1)²(1) = 1
C₁₂ = (-1)¹⁺² (-tan x) = tan x
C₂₁ = (-1)²⁺²(1) = 1
\(\therefore \quad adj\quad A={ \left[ \begin{matrix} 1 & tan\quad x \\ -tan\quad x & 1 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ |A| } adjA=\frac { 1 }{ 1+{ tan }^{ 2 }x } \left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] \)
\(\therefore \quad { A }^{ T }{ A }^{ -1 }=\left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] \left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } +\frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] \)
\(=\left[ \begin{matrix} \frac { 1-tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -2tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { 2tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1-{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] =\left[ \begin{matrix} cos\quad 2x & -sin2x \\ sin\quad 2x & cos\quad 2x \end{matrix} \right] \)  (Using multiple angle formula)

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]  

Compare the equation
12x² - 10xy + 2y² + 14x - 5y + 2 = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c = 0
We get a = 12, 2h = -10, b = 2, 2g = 14, 2f = -5
\(h=-5\quad g=7\quad f=-\frac { 5 }{ 2 } ,c=2\)
\(\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=\left|\begin{array}{ccc} 12 & -5 & 7 \\ -5 & 2 & \frac{-5}{2} \\ 7 & \frac{-5}{2} & 2 \end{array}\right|\)
\(=12\left(4-\frac{25}{4}\right)+5\left(-10+\frac{35}{2}\right)+7\left(\frac{25}{2}-14\right)\)
\(=48-75-50+\frac{175}{2}+\frac{175}{2}-98\)
= -175 + 175 = 0
Hence the given equations represent a pair of straight lines.
To find separate equation
\(12 x^2-10 x y+2 y^2 =12 x^2-6 x y-4 x y+2 y^2 \)
\(=6 x(2 x-y)-2 y(2 x-y) \)
\(=(6 x-2 y)(2 x-y)\)
\(12 x^2-10 x y+2 y^2+ 14 x-5 y+2 =(6 x-2 y+l)(2 x-y+\mathrm{m})\)
Comparing the coefficient of x and y
14 = 6m + 2l
divided by 2
7 = 3m + l .........(1)
-5 = -2m - l .......(2)
Solving (1) and (2) we get m = 2, 1 = 1
The separate equations are
6x - 2y + 1 = 0
2x - y + 2 = 0

\(x-y+z=2\)
\(2 x-y=0 \)
\(2 y-z=1\)
The given equations can be written in matrix form as
\(\begin{bmatrix} 1&-1&1\\2&-1&0\\0&2&-1 \end{bmatrix}\begin{bmatrix} x \\ y\\z \end{bmatrix}=\begin{bmatrix} 2\\0\\1 \end{bmatrix} \)
\(A X=B \Rightarrow X=A^{-1} B\)
Where \(A=\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 0 & 2 & -1 \end{matrix} \right] ,\ X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,\ B=\left[ \begin{matrix} 2 \\ 0 \\ 1 \end{matrix} \right] \)
 \(|A|=\begin{vmatrix} 1&-1&1\\2&-1&0\\0&2&-1 \end{vmatrix}=-1\begin{vmatrix} -1&0\\2&-1\end{vmatrix}+1\begin{vmatrix} 2&0\\0&-1 \end{vmatrix}+1\begin{vmatrix} 2&-1\\0&2 \end{vmatrix}\)
\(|A|=1(1-0)+1(-2-0)+1(4-0)\)
\(=1-2+4=3\neq0\)
\(\therefore\) A^-1 exists.
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 1 & 2 & 4 \\ 1 & -1 & -2 \\ 1 & 2 & 1 \end{array}\right)\)
\(\therefore\ {A}^{-1}={{1}\over{|A|}}adj\ A={{1}\over{3}}\begin{bmatrix} 1&1&1\\2&-2&2\\4&-2&1 \end{bmatrix}\)
Now \(X={A}^{-1}B={{1}\over{3}}\begin{bmatrix} 1&1&1\\2&-1&2\\4&-2&1 \end{bmatrix}\begin{bmatrix} 2\\0\\1 \end{bmatrix}=\frac{1}{3}\left(\begin{array}{l} 2+0+1 \\ 4+0+2 \\ 8+0+1 \end{array}\right)=\frac{1}{3}\left(\begin{array}{l} 3 \\ 6 \\ 9 \end{array}\right)\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right)\)
\(x=1, y=2, z=3\)

\({ \left( x+\frac { 1 }{ x } \right) }^{ 2n }\)
t_{r+ 1} = nC_r x^n-r a^r
\(\therefore\) t_r+1 = 2nC_r x^2n-r \({\left( {{1}\over{x}} \right)}^{r}\) = 2nC_r x^2n-r \({\left( {{1}\over{x}} \right)}^{r}\) = 2nC_r x^2n-2r
To find the term independent of x, equate the power of x as 0
 2n = 2r \(\Rightarrow\) r = n
\(\therefore\) t_r+1 = 2nC_n.x^2n-2n = 2nC_n
\(={{2n!}\over{n!n!}}={{(2n)(2n-1)(2n-2)(2n-3)......4,3,2,1}\over{n!n!}}\)
\(={{[(2n)(2n-2)(2n-4)...4.2][(2n-1)(2n-3)......3.1]}\over{n!n!}}\)
\(={{2^n(n)(n-1)(n-2).2.1.[(2n-1)(2n-3)....3.1]}\over{n!n!}}\) 
\(=\frac{2^n . n !(2 n-1) \ldots 5.3 .1}{n ! n !}\)
\(=\frac{1.3 .5 \ldots(2 n-1) 2^n}{n !}\) 
Hence proved.

Given A \(=\begin{bmatrix} 1&1&1\\3&4&7\\1&-1&1 \end{bmatrix}\)
\(=(4+7)-1(3-7)+1(-3-4)\)
\(=11+4-7=8\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 11 & 4 & -7 \\ -2 & 0 & +2 \\ 3 & -4 & 1 \end{array}\right)\)
\(\operatorname{adj} A=\left(\begin{array}{ccc} 11 & -2 & 3 \\ 4 & 0 & -4 \\ -7 & 2 & 1 \end{array}\right)\)
\(\mathrm{A}(\operatorname{adj} A)=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 3 & 4 & 7 \\ 1 & -1 & 1 \end{array}\right)\left(\begin{array}{ccc} 11 & -2 & 3 \\ 4 & 0 & -4 \\ -7 & 2 & 1 \end{array}\right)\)
\(=\left(\begin{array}{ccc} 11+4-7 & -2+0+2 & 3-4+1 \\ 33+16-49 & -6+0+14 & 9-16+7 \\ 11-4-7 & -2+0+2 & 3+4+1 \end{array}\right)\)
\(=\left(\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}\right)=8\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=|A| I_3\)
\((adj\ A)\ A=\begin{bmatrix}11&-2&3\\4&0&-4\\-7&2&1\end{bmatrix}\begin{bmatrix} 1&1&1\\3&4&7\\1&-1&1 \end{bmatrix}\)
\(=\begin{bmatrix} 11-6+3&11-8-3&11-14+3\\4+0-4&4+0+4&4+0-4\\-7+6+1&-7+8-1&-7+14+1 \end{bmatrix}=\begin{bmatrix} 8&0&0\\0&8&0\\0&0&8 \end{bmatrix}\) ...(2)
\(|A|.{I}_{3}=8\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}=\begin{bmatrix} 8&0&0\\0&8&0\\0&0&8\end{bmatrix}\)  ...(3)
From (1), (2) and (3)
A (adj A) = (adj A) A = |A|I₃.

Let A \(=\begin{vmatrix} 1 & a&a^2&-bc \\1 &b&{b}^{2}&-ca\\1&c&c^2&-ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|+\left|\begin{array}{ccc} 1 & a & -b c \\ 1 & b & -c a \\ 1 & c & -a b \end{array}\right|\)
\(A=\begin{vmatrix} 1 & a&{a}^{2} \\ 1 &b&b^2\\1&c&c^2 \end{vmatrix}-\begin{vmatrix} 1 & a&bc \\1 &b&ca\\1&c&ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{1}{a b c}\left|\begin{array}{ccc} a & a^2 & a b c \\ b & b^2 & a b c \\ c & c^2 & a b c \end{array}\right|\)
(Multiplying R₁, R₂ and R₃ of II det by a, b, c respectively)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array}\right|\)
\(\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|=0\)

\(L\left[ f\left( x \right) \right] _{ x=3 }=\underset { x-{ 3 }^{ - } }{ lim } f\left( x \right) \)
\(=\lim _{h \rightarrow 0} f(3-h), x=3-h\)
\( =\underset { h\rightarrow 0 }{ lim } \cfrac { \left| (3-h)-3 \right| }{ \left( 3-h \right) -3 }\)
\(=\underset { h\rightarrow 0 }{ lim } \cfrac { \left| -h \right| }{ h } \)
\(=\underset { h\rightarrow 0 }{ lim } \cfrac { h }{ -h } \)
\(=\underset { h\rightarrow 0 }{ lim } −1 =−1 \) 
\(R\left[ f(x) \right] _{ x=3 }=\underset { x\rightarrow { 3 }^{ + } }{ lim } f(x)\)
\(=\underset { h\rightarrow 0 }{ lim } f\left( 3+h \right) \)
\(=\underset { h\rightarrow 0 }{ him } \cfrac { \left| \left( 3+h \right) -3 \right| }{ \left( 3+h \right) -3 } \)
\(=\underset { h\rightarrow 0 }{ lim } \cfrac { \left| h \right| }{ h } \)
\(=\underset { h\rightarrow 0 }{ lim } 1 = 1 \)

Given lines 3x - 4y - 13 = 0...(1)
8x - 11y = 33  ...(2)
2x - 3y - 7 = 0 ...(3)
Conditon for concurrent lines is
\(\left| \begin{matrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{matrix} \right| =0\) 
i.e.,\(\left| \begin{matrix} 3 & -4 & -13 \\ 8 & -11 & -33 \\ 2 & -3 & -7 \end{matrix} \right| =3\left( 77-99 \right) +4\left( 56+44 \right) -13\left( -24+22 \right) \) 
= -66 + 40 + 26 = 0
\(\Rightarrow \) Given lines are concurrent. To get the point of concurrency solve the equations (1) and (3)

Equation (1) \(\times\) 2	\(\Rightarrow \)	6x	- 8y	= 26
Equation (3) \(\times\) 3	\(\Rightarrow \)	6x	- 9y	= 21
			y	= 5

When y = 5 from (2) 8x = 88
x = 11
Point of concurrency is (11, 5)

A_ij = (–1)^i+jM_ij
A₁₁ = (–1)¹⁺¹M₁₁ = \(\left| \begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix} \right| \) = 0
A₁₂ = (–1)¹⁺²M₁₂ = \(-\left| \begin{matrix} 0 & 0 \\ -4 & 0 \end{matrix} \right| \) = 0
A₁₃ = (–1)¹⁺³M₁₃ = \(\left| \begin{matrix} 0 & 1 \\ -4 & 1 \end{matrix} \right| \)= 0 - (-4) = 4
A₂₁ = (–1)²⁺¹M₂₁ = \(-\left| \begin{matrix} -2 & -3 \\ 1 & 0 \end{matrix} \right| \)= -(0 - (-3)) = -3
A₂₂ = (–1)²⁺²M₂₂ = \(\left| \begin{matrix} 1 & -3 \\ -4 & 0 \end{matrix} \right| \) = 0 - 12 = -12
A₂₃ = (–1)²⁺³M₂₃ = \(-\left| \begin{matrix} 1 & -2 \\ -4 & 1 \end{matrix} \right| \) = -(1-8) = 7
A₃₁ = (–1)³⁺¹M₃₁ = \(\left| \begin{matrix} -2 & -3 \\ 1 & 0 \end{matrix} \right| \) = 0 - (-3) = 3
A₃₂ = (–1)³⁺²M₃₂ = \(-\left| \begin{matrix} 1 & -3 \\ 0 & 0 \end{matrix} \right| \) = 0 - 0 = 0
A₃₃ = (–1)³⁺³M₃₃ = \(\left| \begin{matrix} 1 & -2 \\ 0 & 1 \end{matrix} \right| \) = 1 - 0 = 1
\(\left[ { A }_{ ij } \right] =\left[ \begin{matrix} 0 & 0 & 4 \\ -3 & -12 & 7 \\ 3 & 0 & 1 \end{matrix} \right] \)
Adj A = [A_ij]^T
\(=\left[ \begin{matrix} 0 & -3 & 3 \\ 0 & -12 & 0 \\ 4 & 7 & 1 \end{matrix} \right] \)

Marks	Frequency	Cumulative Frequency
X	f	cf
10	4	4
20	7	11
30	15	26
40	8	34
50	7	41
60	2	43
		N=\(\sum\)f=43

Q₁ = Size of \({ \left( \frac { N+1 }{ 4 } \right) }^{ th }\)value = Size 11^th value = 20
Q₃ = Size of \(\left( \frac { 3\left( N+1 \right) }{ 4 } \right) ^{ th }\)value = Size of 33^rd value = 40
\(QD=\frac { 1 }{ 2 } \left( { Q }_{ 3 }-{ Q }_{ 1 } \right) \frac { 40-20 }{ 2 } =10\)
Coefficient of QD \(=\frac { { Q }_{ 3 }-{ Q }_{ 1 } }{ { Q }_{ 3 }+{ Q }_{ 1 } } =\frac { 40-20 }{ 40+20 } =\frac { 20 }{ 60 } \)
= 0.333

If x = 0, y = 1. The curve cuts the y axis at (0,1)
The curve will not meet the x axis for all real values of x