\(L\left| f(x) \right| _{ x=1 }=\underset { x\rightarrow { 1 }^{ - } }{ lim } f(x)\\ =\underset { h\rightarrow 0 }{ lim } f\left( 1-h \right) ,x=1-h\\ =\underset { h\rightarrow 0 }{ lim } \left[ 5\left( 1-h \right) -4 \right] \\ =5(1)-4=1\) 
\(R\left[ f\left( x \right) \right] _{ x=-1 }=\underset { x\rightarrow { 1 }^{ + } }{ lim } f(x)\\ =\underset { h\rightarrow 0 }{ lim } f\left( 1+h \right) ,x=1+h\\ =\underset { h\rightarrow 0 }{ lim } \left[ 4(1+h)^{ 2 }-3\left( 1+h \right) \right] \\ =4\left( 1 \right) ^{ 3 }-3(1)\\ =4-3=1\)
Now.\(f(1)=5(1)-4=5-4=1\) 
\(\therefore \underset { x\rightarrow { 1 }^{ - } }{ lim } f(x)=\underset { x\rightarrow { 1 }^{ + } }{ lim } f(x)=f\left( 1 \right) \) 
f(x) is  continuous at x = 1

Given f(x)= x³-27x+108
f'(x) = = 3x² - 27
f'(x) = 0 ⇒ 3x² - 27 = 0
⇒ x² - 9 = 0 [Divided by 3]
⇒ x²=9
⇒ x= 3, - 3
Also,f"(x) = 6x
when x = 3, f"(x) = 6(3) = 18 > 0
∴ f is minimum at x = 3.
∴ Minimum Value = f(3) = 3³ - 27(3) + 108
=27-81+108=54 ....(1)
when x = -3,f"(x) = 6(-3) = -18 < 0
∴ f is maximum at x = - 3
Maximum value = 1(-3) = (-3)³ - 27(-3) + 108
= -27 + 81 + 108 = 162   ...(2)
From (1) and (2)
Maximum value - Minimum value = 162 - 54 = 108.
Hence, the maximum value of f(x) is 108 more than the minimum value

x	y	x2	y2	xy
4	2	16	4	8
2	3	4	9	6
3	2	9	4	6
4	4	16	16	16
2	4	4	16	8
15	15	49	49	44

Here n=5,
b_yx=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ n\sum { { x }^{ 2 }-{ (\sum { x } ) }^{ 2 } } } \)
=\(\frac { 5(44)-(15)(15) }{ 5(49)-{ (15) }^{ 2 } } \)
=\(\frac { 220-225 }{ 245-225 } =\frac { -5 }{ 20 } =\frac { -1 }{ 4 } \)
b_xy=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ n\sum { { y }^{ 2 }-{ (\sum { y } ) }^{ 2 } } } \)
=\(\frac { 5(44)-15(15) }{ 5(49)-{ (15) }^{ 2 } } \)
\(\frac { 220-225 }{ 245-225 } =\frac { -5 }{ 20 } =\frac { -1 }{ 4 } \)
b_xy and b_yx are negative, r is also negative.
\(\therefore\)Correlation co-efficient
r(x,y)=-\(\sqrt { { b }_{ xy }.{ b }_{ yx } } \)
=-\(\sqrt { \left( \frac { -1 }{ 4 } \right) \left( \frac { -1 }{ 4 } \right) } =\frac { -1 }{ 4 } \)
r=-0.25

We know that the lines of regression intersect at the mean values of x and y.
Given lines are 4x+3y =-7  ....(i)
and      3x+4y =-8  ....(ii)
\(\Rightarrow\)(1)x3    12x+9y =-21
                  (-)   (+)    (-)
\(\Rightarrow\)(2)x4    12x+16y=-32
             _______________
Subtracting,  -7y =+11   \(\Rightarrow\) y =-\(\frac{11}{7}\)
Subtarcting y=-\(\frac{11}{7}\) in (1) we get,
4x+3(\(\frac{11}{7}\)) =-7
4x-\(\frac{33}{7}\) =-7
4x=-7+\(\frac{33}{7}\) =\(\frac{-49+33}{7}\)
4x=\(\frac{-16}{7}\)
x=\(\frac{-16}{4\times7}=\frac{-4}{7}\)
\(\bar{X}\)=\(\frac{-4}{7}\) and \(\bar{Y}\)=\(\frac{11}{7}\)
(ii) Let us take the equation 4x+3y+7=0 as the line of regression of X on Y and 3x+4y+8=0 as the line of regression of Y on X.
Then, \(\Rightarrow\)4x+3y+7 =-3y-7
x=\(\frac{-3}{4}\) y -\(\frac{7}{4}\)
\(\therefore\)b_xy=\(\frac{-3}{4}\)
and 3x+4y+8=0  \(\Rightarrow\)4y=-3x-8
\(\Rightarrow\)y=\(\frac{-3}{4}x-\frac{8}{4}\)
\(\Rightarrow\)b_yx=\(\frac{-3}{4}\)
(iii) The correlation co-efficient is
r=\(\sqrt { { b }_{ xy }-{ b }_{ yx } } =\sqrt { \left( \frac { -3 }{ 4 } \right) \left( \frac { -3 }{ 4 } \right) } =\frac { 3 }{ 4 } \)=0.75
As b_xy and b_yx are both negative,
r=-0.75

E1= 0	L5= 48
E2= 0+21=21	L4= 48 -11 = 37
E3 =(21 + 11) or (0 + 26)                            Whichever is maximum  =32	L3= 37 - 5 = 32
E4= (32 + 5) or (21 + 13)                                   =Whichever is maximum  = 37	L2= (37 - 13) or (32 - 11) Whichever is minimum =21
E5=37 + 11 = 48	L1=(21 - 21) or (32 - 26) Whichever is minimum = 0

Activity	Duration	EST	EFT = EST + tij	EFT = EST - tij	LFT
1-2	21	0	21	21-21=0	21
1-3	26	0	26	32-26=6	32
2-3	11	21	32	32-11=21	32
2-4	13	21	34	37-13=24	37
3-4	5	32	37	37-5=32	37
4-5	11	37	48	48-11=37	48

EFT and LFT are same in the activities.
1 - 2, 2 - 3, 3 - 4 and 4 - 5
Hence, the critical path is 1 - 2 - 3 - 4 - 5 and the duration of project completion is 48 days .

Given Z = 4x⁶ - 8x³ - 7x + 6xy + 8y + x³y⁵
(i) Differentiating partially w.r.t. 'y' we get,
\({\partial z\over \partial y^2}=0-0-0+6x(1)+8+x^3(5y^4)\)
= 6x + 8 + 5x³y⁴
Differentiating again partially w.r.t. 'y' we get
\({\partial^2z\over \partial y^2}=.0 + 0 + 5x^3( 4y^3)\)
=20x³,y³
(ii) We know that \({\partial z\over \partial y}=6x+8+5x^3y^4\)
Differentiating partially w.r.t. 'x' we get,
\({\partial^2z\over \partial x \partial y}=6(1)+0+5y^4(3x^2)\)
= 6+15x²y⁴
(iii) \({\partial z\over \partial x}=4( 6x^5) - 8 (3x^2) - 7 + 6y(1) + 0 + y^5(3x2^)\)
= 24x⁵ - 24x² - 7 + 6y + 3x²y⁵
Differentiating again partially w.r.t. 'y' we get,
\({\partial^2z\over \partial y \partial x}=0+ 0 - 0 + 6(1) + 3x^2 (5y^4)\)
= 6 + 15x²y⁴

Given \(P(A)=\frac { 1 }{ 2 } \Rightarrow p(\bar { A } )=1-P(A)=1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
\(P(B)=\frac { 2 }{ 3 } \Rightarrow P(\bar { B } )=1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
\(P(C)=\frac { 3 }{ 4 } \Rightarrow P(\bar { C } )=1-P(C)=1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } \)
P(Only one of them hits the target)
\(=P(A\cap \bar { B } \cap \bar { C } )+P(\bar { A } \cap B\cap \bar { C } )+P(\bar { A } \cap \bar { B } \cap C)\)
\(=P(A).P(\bar { B) } .P(\bar { C) } +P(\bar { A } ).P(B).P(\bar { C } )+P(\bar { A } ).P(\bar { B } ).P(C)\)
\(=\frac { 1 }{ 2 } \times \frac { 1 }{ 3 } \times \frac { 1 }{ 4 } +\frac { 1 }{ 2 } \times \frac { 2 }{ 3 } \times \frac { 1 }{ 4 } +\frac { 1 }{ 2 } \times \frac { 1 }{ 3 } \times \frac { 3 }{ 4 } \)
\(=\frac { 1 }{ 24 } +\frac { 2 }{ 24 } +\frac { 3 }{ 24 } =\frac { 6 }{ 24 } =\frac { 1 }{ 4 } \)

Given observations arranged in ascending order are
38, 42, 57, 58, 61, 65, 66, 72 and n = 8
D₆ = Size of 6\({ \left( \frac { n+1 }{ 10 } \right) }^{ th }\) value = Size of \(6{ \left( \frac { 8+1 }{ 10 } \right) }^{ th }\) value
= Size of 5.4^th value \(\simeq \) Size of 5^th value
= 61
D₈ = Size of \(8{ \left( \frac { n+1 }{ 10 } \right) }^{ th }\) value = Size of \(8{ \left( \frac { 8+1 }{ 10 } \right) }^{ th }\) value
=size of 7.2^th value \(\simeq \) size of 7^th value
= 66
P₇ = Size of \(7{ \left( \frac { n+1 }{ 100 } \right) }^{ th }\) value = size of \(7{ \left( \frac { 8+1 }{ 100 } \right) }^{ th }\) value
= size of (0.6)^th value \(\simeq \) size of first value
=38
P₂₀=Size of 20\({ \left( \frac { n+1 }{ 100 } \right) }^{ th }\) value = Size of \({ 20\left( \frac { 9 }{ 100 } \right) }^{ th }\) value
= Size of (1.8^th value) \(\simeq \) size of 2^nd value
= 42

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

Number of preference shares = 1000
Preferential stock = 1000 X 100 = Rs.1,00,000
Ordinary shares = 4000
Ordinary shares = 4000 x 100 = Rs.4,00,000
Total dividend = Rs.40,000
Dividend on preference share=\(\frac { 1,00,000 }{ 100 } \) x 6 = Rs.6,000
Dividend on ordinary shares=Total dividend-dividend on preference shares
= 40,000-6000
=Rs.34,000
Sundar's income on preference shares
= No.of shares X FV XRate percentage
=100 x 100 x \(\frac { 6 }{ 100 } \)
=Rs.600         ...(1)
Sundar's income on ordinary shares
=\(\frac { 2500 \times 100 }{ 4,00,000 } \) x 34,000=Rs.1700   ..(2)
\(\therefore\)Sundar's total income = 600+1700
= Rs.2300

Let the amount invested in each stock be Rs.x
For 12% Stock
Investment =Rs.x
Purchased Price = 89+1 = 90
Income=\(\frac {\text { Investment} }{ \text {Purchase Price } }\) x Dividend Rate
=\(\frac { x }{ 90 } \times 12=\frac { 2x }{ 15 } \)   ...(1)
For 8% Stock
Investment = Rs.x
market Price = 95+1= 96
Dividend Rate = 8%
Income=\(\cfrac { x }{ 96 } \times 8 =\cfrac { x }{ 12 } \)  ...(2)
Difference in income = Rs.120
From (1) and (2),  \(\cfrac { 2x }{ 15 } -\cfrac { x }{ 12 } \) = 120
\(x\left( \cfrac { 2 }{ 15 } -\cfrac { 1 }{ 12 } \right) \) = 120
\(x\left( \cfrac { 8-5 }{ 60 } \right) \) = 120
\(\cfrac { 3x }{ 60 } \) = 120
\(\Rightarrow\) x = \(\cfrac { 120\times 60 }{ 3 } \) = Rs.2400
Hence, investment in each stock is Rs.2400

Here the sample space of the experiment is
S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}
A = {Three heads or Three tails}
= {HHH, TTT}
B = {at least two heads}
= {HHH, HHT, HTH, THH} and
C = {at most two heads} = {HHT, HTH, HTT, THH, TTH, THT, TTT}
Also (A∩B) = {HHH}; (A∩C) = {TTT} and (B∩C) ={HHT, HTH, THH}
∴ P(A) = \(\frac { 2 }{ 8 } =\frac { 1 }{ 4 } \); P(B) =\(\frac{1}{2}\); P(C) = \(\frac{7}{8}\) and
P(A∩B)= \(\frac{1}{8}\), P(A∩C)=\(\frac{1}{8}\), P(B∩C)=\(\frac{3}{8}\)
Also P(A). P(B)=\(\frac { 1 }{ 4 } .\frac { 1 }{ 2 } =\frac { 1 }{ 8 } \)
P(A). P(C) =\(\frac { 1 }{ 4 } .\frac { 7 }{ 8 } =\frac { 7 }{ 32 } \)
and P(B). P(C) =\(\frac { 1 }{ 2 } .\frac { 7 }{ 8 } =\frac { 7 }{ 16 } \)
Thus, P(A∩B) = P(A). P(B)
P(A∩C) ≠ P(A) P(C) and
P(B∩C) ≠ P(B). P(C)
Hence, the events (A and B) are independent, and the events (A and C) and (B and C) are dependent.

Calculation for median follows by the following table

X	f	cf
0-10	20	20
10-20	25	45
20-30	32	77
30-40	40	117
40-50	42	159
50-60	35	194
60-70	10	204
70-80	8	N = 212

\(\frac { N }{ 2 } =\frac { 212 }{ 2 } =106\) Class interval corresponding to cumulative frequency 106 is (30 – 40). So, the corresponding values from the median class are L = 30, pcf = 77, f = 40 and c = 10.
Median = L+\(\left( \frac { \left( \frac { N }{ 2 } \right) -pcf }{ f } \right) \times c\)
Median = \(30+\left( \frac { 106-77 }{ 40 } \right) \times 10\)
∴ Median = 37.25 (corrected to two places of decimals)
Calculations proceeded for mean deviation about the median.

X	f	M	|D|=|X-37.25|	f|D|
0-10	20	5	32.25	645
10-20	25	15	22.25	556.25
20.-30	32	25	12.25	392
30-40	40	35	2.25	90
40-50	42	45	7.75	325.5
50-60	35	55	17.75	321.25
60-70	10	65	27.75	277.5
70-80	8	75	37.75	302
	N = 212			Σf|D| = 3209.5

Then the mean deviation about median is to be computed by the following
MD about Median = \(\frac { \Sigma f|D| }{ N } =\frac { 3209.5 }{ 212 } \) = 15.14
Coefficient of MD about Median = \(\frac { M.D \ about \ median }{ Median } =\frac { 15.14 }{ 37.25 } \) = 0.4064 = 0.41 (corrected to two decimal places).

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 = 2000 + 1800x -75 x² + x³
\({dC\over dx}=1800-150x+3x^2\)
\({dC\over dx}=0\)
\(\Rightarrow 3(x^2-50x+600)=0\)
\(\Rightarrow x^2-50x+600=0\) (Divided by 3)
\(\Rightarrow x=30,20\)
The intervals are (0, 20) (20, 30) and (30, \(\infty\))

Intervals	Sign of \({dC\over dx}\)	Nature of Function
(0,20)	+ ve	Increasing
(20,30)	- ve	Decreasing
(30,\(\infty\))	+ ve	Increasing

X	Y	dx = X-168	dy = Y-169	dx2	dy2	dxdy
158	163	-10	-6	100	36	60
166	158	-2	-11	4	121	22
163	167	-5	-2	25	4	10
165	170	-3	1	9	1	-3
167	160	-1	9	1	81	-9
170	180	2	11	4	121	22
167	170	-1	1	1	1	-1
172	175	4	6	16	36	24
177	172	9	3	25	9	27
181	175	13	6	169	36	78
\(\Sigma X\) = 1686	\(\Sigma Y\) = 1690	\(\Sigma dx\) = 6	\(\Sigma dy\) = 0	\(\Sigma dx^2\) = 410	\(\Sigma dy^2\)= 446	\(\Sigma dxdy\) = 248

\(\bar{X} =\frac{\Sigma X}{N}=\frac{1686}{10}=168.6 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{1690}{10}=169 \)
\(b_{x y} =\frac{N \Sigma d x d y-(\Sigma d x)(\Sigma d x)}{N \Sigma d y^2-(\Sigma d y)^2} \)
\(=\frac{10(248)-0}{10(446)-0}=\frac{248}{446}=0.556 \)
\(b_{y x} =\frac{N \Sigma d x d y-(\Sigma d x)(\Sigma d y)}{N \Sigma d x^2-(\Sigma d x)^2} \)
\(=\frac{2480}{4100-36}=\frac{2480}{4064}=0.6102\)
Regression equation of X on Y
\(X-\bar{X}=b_{x y}(Y-\bar{Y}) \)
X - 168.6 = 0.556(Y - 169)
X = 0.556 Y + 168.6-93.964
X = 0.556 Y + 74.64
Regression equation of Y on X
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 169 = 0.6102(X - 168.6)
Y = 0.6102 X - 102.8 + 169
Y = 0.6102X + 66.12
If X = 164
Y= 100.07 + 66.12
= 166.19
Height of son is 166.19

a = 2000 + 2000 = Rs. 4000
i = \(\cfrac { 8 }{ 12 \times 100} \)= 0.0067, n = 20 x 12 = 240
A = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
\(=\cfrac { 4000 }{ 0.0067 } \left[ \left(1.0067 \right) ^{ 240 }-1 \right] \)
= 597014.93 [4.9661 - 1]
= 597014.93 x 3.9661
= 2367820.91.

(i) First convert the given equations Y on X and X on Y in standard form and find their regression coefficients respectively.
Given regression lines are
3X–2Y = 5      ... (1)
X–4Y = 7        ... (2)
Let the line of regression of X on Y is
3X–2Y = 5
3X = 2Y+5
X = \(\frac{1}{3}\)(2Y+5)
X = \(\frac{1}{3}\)(2Y+5)
X = \(\frac { 2 }{ 3 } Y+\frac { 5 }{ 3 } \)
∴ Regression coefficient of X on Y is
b_xy =\(\frac{2}{3}\)(<1)
Let the line of regression of Y on X is
X–4Y = 7
–4Y = –X+7
4Y = X–7
Y = \(\frac{1}{4}\)(X-7)
Y = \(\frac{1}{4}\)X-\(\frac{7}{4}\)
∴ Regression coefficient of Y on X is
b_yx = \(\frac{1}{4}\)(<1)
(ii) Coefficient of correlation
Since the two regression coefficients are positive then the correlation coefficient is also positive and it is given by
r = \(\sqrt { { b }_{ yx }.{ b }_{ xy } } \)
= \(\sqrt { \frac { 2 }{ 3 } .\frac { 1 }{ 4 } } \)
= \(\sqrt { \frac { 1 }{ 6 } } \)
= 0.4082
∴ r = 0.4082

q = 80 - \({ p }_{ 1 }^{ 2 }\) + 5p₂ - p₁p₂
\({\partial q\over \partial p_1}=-2p_1-p_2\)
\({\partial q\over \partial p_2}=5 - p_1\)
\(\frac{E q}{E p_1}=\frac{-p_1}{q} \frac{\partial q}{\partial p_1}=-\frac{p_1\left(-2 p_1-p_2\right)}{80-p_1^2+5 p_2-p_1 p_2}=\frac{2 p_1^2+p_1 p_2}{80-p_1^2+5 p_2-p_1 p_2}\)
\(\frac{E q}{E p_2}=\frac{-p_2}{q} \frac{\partial q}{\partial p_2}\)
\(=\frac{-p_2\left(5-p_1\right)}{80-p_1^2+5 p_2-p_1 p_2} \)
\(={-5p_2+p_1p_2\over 80-p^2_1+5p_2-p_1p_2}\)
\(\frac { { E }q }{ { E }p_{ 1 } } ={8+2\over 80-4+5-2}={10\over 79}\)
\(\frac { { E }q }{ { E }p_{ 2 } } ={-5+2\over 80-4+5-2}={-3\over 79}\)

E1 = 0	L6 = 30
E2 = 0 + 6 = 6	L5 = 30 - 7 = 23
E3 = 6 + 8 = 14	L4 = 30 - 2 = 28
E4 = 6 + 4  = 10	L3 = 23 - 9 = 14
E5 = 14 + 9 = 23	L2 = min of {28 - 4, 14  - 8} = 6
E6 = max of { 23 + 7,10 + 2 } = 30	L1 = 6 - 6 = 0

Activity	Duration ti	EST	EFT=EST+tij	LST=LFT-tij	LFT
1-2	6	0	6	6-6 = 0	6
2-3	8	6	14	14-8 = 6	14
2-4	4	6	10	28-4 = 24	28
3-5	9	14	23	23-9 = 14	23
4-6	2	10	22	30-28 = 28	30
5-6	7	23	30	30-7 = 23	30

Since EFT and LFT. are the same in 1-2,2-3, 3-5, and 5-6, the critical path is 1-2-3-5-6 and time duration is 30 days.

Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant of the plane.
Consider the equations 4x₁ + x₂ = 40 and 2x₁ + 3x₂ = 90
4x₁ + x₂ = 40 is a line passing through the points (0,40) and (10,0). Any point lying on or above the line 4x₁ + x₂ = 40 satisfies the constraint 4x₁ + x₂ ≥ 40.
2x₁ + 3x₂ = 90 is a line passing through the points (0,30) and (45,0). Any point lying on or above the line 2x₁ + 3x₂ = 90 satisfies the constraint 2x₁ + 3x₂ ≥ 90.
Draw the graph using the given constraints.

The feasible region is ABC (since the problem is of minimization type we are moving towards the origin.

Corner points	z = 5x1 + 4x2
A(45,0)	225
B(3,28)	127
C(0,40)	160

The minimum value of Z occurs at B(3,28).
Hence the optimal solution is x₁ = 3, x₂ = 28 and Z_min = 127.

Given \(sin\left( { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x \right) =1\)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) +{ cos }^{ -1 }(x)={ sin }^{ -1 }(1)=\frac { \pi }{ 2 } \)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) +{ cos }^{ -1 }(x)=\frac { \pi }{ 2 } \)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) =\frac { \pi }{ 2 } -{ cos }^{ -1 }(x)\)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) =sin^{ -1 }x\)  \(\left[ \because { sin }^{ -1 }(x)+{ cos }^{ -1 }(x)=\frac { \pi }{ 2 } \right] \)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) =sin^{ -1 }x\)
⇒ \(x=\frac { 1 }{ 5 } \)
Thus, \(x=\frac { 1 }{ 5 } \) is a root of the given equation.

\(\mathrm{LHS} =\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ} \)
\(=\cos 20^{\circ} \cos 40^{\circ}\left(\frac{1}{2}\right) \cos 80^{\circ}\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\cos 40^{\circ} \cos 80^{\circ}\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\cos \left(60^{\circ}-20^{\circ}\right) \cos \left(60^{\circ}+20^{\circ}\right)\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\cos ^2 60^{\circ}-\sin ^2 20\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\frac{1}{4}-\left(1-\cos ^2 20^{\circ}\right)\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\frac{1-4+4 \cos ^2 20^{\circ}}{4}\right)\)
\(=\frac{1}{8}\left(4 \cos ^3 20^{\circ}-3 \cos 20^{\circ}\right)\)
\(=\frac{1}{8} \cos 3\left(20^{\circ}\right)=\frac{1}{8} \cos 60^{\circ}\)
\(=\frac{1}{8}\left(\frac{1}{2}\right)=\frac{1}{16}\)
Hence proved.

Since the axis of the parabola is vertical, its equation will be x² = 4ay.
Arch is 10m high and 5 m wide at base.
​\(\therefore\)​ Point \((\frac{5}{2},10)\) lies on the parabola
​\(\therefore\)​  \(\frac{25}{4}\) = 4a(10)⇒ 4a = \(\frac{25}{40}\) = \(\frac{5}{8}\)(y)
\(\therefore\) Equation of the parabola becomes x² = \(\frac{5}{8}\)(y)
Let the width of the arch 2m from the vertex is 2b, then point (b, 2) lies on the parabola
\(\therefore\) b² = \(\frac { 5(2) }{ 8 } =\frac { 10 }{ 8 } =\frac { 5 }{ 4 } \Rightarrow b=\frac { \sqrt { 5 } }{ 2 } \)

\(\therefore\) width of arch is  2b = 2.\(\frac{\sqrt{5}}{2}\) = √5m = 2.23m(app)

Given equation of the circle is x² + y² = 169...(1)
Equation of the tangent at (x₁, y₁) to circle (1) is xx₁ + yy₁ = 169
Now, Equation of the tangent at (5,12) to circle (1) is
x(5) + y(12) = 169 ⇒ 5x + 12y - 169 = 0...(2)
and equation of the tangent at (12,-5) to circle (1) is
x(12) + y(-5) = 169 ⇒ 12x - 5y = 169 = 0...(3)
Let m₁ and m₂ be the slopes of the tangents (2) and (3)
ஃ m₁=\(\frac { -Co-efficient\quad of\quad x }{ Co-efficient\quad of\quad y } =\frac { -5 }{ 12 } \)
Similarly m₂ = \(\frac { -12 }{ -5 } =\frac { 12 }{ 5 } \)
Consider m₁m₂ = \(\left( \frac { -5 }{ 12 } \right) \left( \frac { 12 }{ 5 } \right) =-1\)
Since m₁m₂ = -1, the tangents at (5,12) and (12,-5) to the circle x² + y² = 169 are perpendicular to each other.

LHS = sin(n + 1)x sin(n + 2) x + cos(n + 1)xcos(n + 2)x
Let A = (n + 1)x and B = (n + 2)x
= sinAsinB + cosAcosB = cos(A - B)
= cos[(n + 1)x - (n + 2)x] = cos[nx + x - nx - 2x]
= cos[x - 2x] = cos(-x) = cosx[since cos x is an even function]
= RHS Hence proved.

Let f(x)=x³-27
\(\therefore \frac { d }{ dx } (f(x))=\begin{matrix} \underset { h\rightarrow 0 }{ lim } & \frac { f(x+h)-f(x) }{ h } \end{matrix}\)
\(\frac { d }{ dx } (f(x))=\begin{matrix} \underset { h\rightarrow 0 }{ lim } & \frac { \left[ { \left( x+h \right) }^{ 3 }-27 \right] -({ x }^{ 3 }-27) }{ h } \end{matrix}\)
\(=\begin{matrix} \underset { h\rightarrow 0 }{ lim } & \frac { { x }^{ 3 }+3{ x }^{ 2 }h+3x{ h }^{ 2 }+{ h }^{ 3 }-27-{ x }^{ 3 }+27 }{ h } \end{matrix}\)
\(=\begin{matrix} \underset { h\rightarrow 0 }{ lim } & \frac { 3{ x }^{ 2 }h+3x{ h }^{ 2 }+{ h }^{ 3 } }{ h } \end{matrix}\)
\(\begin{matrix} \underset { h\rightarrow 0 }{ lim } & \frac { h(3{ x }^{ 2 }+3x{ h }+{ h }^{ 2 }) }{ h } \end{matrix}\)
\(=\begin{matrix} \underset { h\rightarrow 0 }{ lim } & 3{ x }^{ 2 }+3x{ h }+{ h }^{ 2 }\end{matrix}\)
\(=3{ x }^{ 2 }+3x{ (0) }+0)\)
\(=3{ x }^{ 2 }\)
\(\therefore \frac { d }{ dx } ({ x }^{ 3 }-27)=3{ x }^{ 2 }\)

Given xy + y² = tan x + y
Differentiating with respect to 'x' we get,
\(x.\frac { dy }{ dx } +y\left( 1 \right) +2y\frac { dy }{ dx } =\sec ^{ 2 }{ x } +\frac { dy }{ dx } \)
\(\Rightarrow x.\frac { dy }{ dx } +y+2y\frac { dy }{ dx } =\sec ^{ 2 }{ x } +\frac { dy }{ dx } \)
\(\Rightarrow \frac { dy }{ dx } \left( x+2y-1 \right) =\sec ^{ 2 }{ x-y } \Rightarrow \frac { dy }{ dx } =\frac { \sec ^{ 2 }{ x-y } }{ x+2y-1 } \)

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\( \sqrt 2\)

Let P (n)be the statement. 1.3+2.3²+3.3 + ... +n.3ⁿ = \({{(2n-1){3}^{n+1}+3}\over{4}} \) for all n \(\in\) N.
Step-1:
Put n = 1  \(\Rightarrow1.3(1){{(2-1){3}^{1+1}+3}\over{4}}={{3^2+3}\over{4}}={{12}\over{4}}\Rightarrow\ 3=3\)
\(\therefore\) P(1) is true.
Step-2:
Let us assume that P(k) is true
\(\therefore\)  1.3 + 2.3² + 3.3³ + ... + k.3k = \({{(2k-1){3}^{k+1}+3}\over{4}}\)  ....(1)
Step-3:
To prove that P (k + 1) is true i.e. to P.T. 1.3 + 2.3² + 3.3³ + ... + k.3^k + (k + 1)3^k+1
\(={{[2(k+1)-1]{3}^{k+2}+3}\over{4}}={{(2k+1){3}^{k+2}+3}\over{4}}\)
LHS = 1.3 + 2.3² + ... + k.3^k + (k+ 1)3^{k+ 1}
\(={{(2k-1){3}^{k+1}+3}\over{4}}+(k+1){3}^{k+1}={{(2k+1){3}^{k+1}+3+(4k+3){3}^{k+1}}\over{4}}\)
\(={{{3}^{k+1}(2k-1+4k+4)+3}\over{4}}={{{3}^{k+1}(6k+3)+3}\over{4}}={{{3}^{k+1}(2k+1)+3}\over{4}}={{{3}^{k+2}(2k+1)+3}\over{4}}\) = RHS
\(\therefore\)  P (k + 1) is true whenever P(k) is true.
\(\therefore\) By mathematical induction, p(n) is true for all values n.

\(\text { LHS }=\sin \theta \cdot \cos \theta\left\{\sin \left(\frac{\pi}{2}-\theta\right) \cdot \operatorname{cosec} \theta\right. \left.+\cos \left(\frac{\pi}{2}-\theta\right) \cdot \sec \theta\right\} \)
\(=\sin \theta \cos \theta\left\{\cos \theta\left(\frac{1}{\sin \theta}\right)+\sin \theta\left(\frac{1}{\cos \theta}\right)\right\} \)
\(=\cos ^2 \theta+\sin ^2 \theta=1=\text { RHS } \)
Hence proved.

Let x and Y represent the number of units produced and the cost of production respectively. By the given data.
x₁ (500)  y₁ (6000)
x₂ (1000)  y₂ (9000)
Equation of straight line is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\Rightarrow \frac { y-6000 }{ 9000-6000 } =\frac { x-500 }{ 1000-500 } \)
\( { y-6000 }{ } =\frac { 3000 }{ 500 } (x - 500)\)
\(\Rightarrow\) y - 6000 = 6(x - 500)
\(\Rightarrow\) y - 6000 = 6x - 3000
\(\Rightarrow\) y = 6x + 3000

Let P (n) denote the statement 1 + 4 + 7 + .............. + 3n - 2 = \({{n(3n-1)}\over{2}}\)
Step-1: Put n = 1
LHS 3 - 2  = 1
RHS \(={{1(2)}\over{2}}\Rightarrow1=1\)
LHS = RHS
\(\therefore\) P(1) is true.
Step-2: Let us assume that P(k) is true.
\(\therefore\) 1 + 4 + 7 + ............+ 3k - 2 = \({{k(3k-1)}\over{2}}\)
Step-3: To prove that P (k+ 1) is true
1 + 4 + 7 + ........... + 3k- 2 + [3 (k+ 1) - 2]
= P (k) + 3k + 3 - 2
\(= \frac{k(3 k-1)}{2}+3 k+1=\frac{3 k^2-k+6 k+2}{2} \) 
\(=\frac{3 k^2+5 k+2}{2}=\frac{3 k^2+3 k+2 k+2}{2} \)
\(=\frac{(3 k+2)(k+1)}{2} \)
= RHS
\(\therefore\)  P (k + 1) is true whenever P(k) is true.
\(\therefore\)  By mathematical induction, pen) is true for all n \(\in\) N.

Equation of circle be x² + y² + 2gx + 2fy + c = 0 ...(1)
It passes through (1, 0)
1 + 0 + 2g + 0 + c = 0   \(\Rightarrow\)  2g + c = -1..(2)
The circle passes through (0, 1)
0 + 1 + 0 + 2f + c = 0 \(\Rightarrow\)  2f + c = -1 ...(3)
centre (-g, -f) lies on x + y = 1
-g - f = 1  ...(4)
Solving (1), (2) and (3) we get
\(g=-\frac { 1 }{ 2 } , f=-\frac { 1 }{ 2 } \) c = 0
Equation of circle is
\(\therefore \)x² + y² + 2 \(\left( -\frac { 1 }{ 2 } \right) x+2\left( -\frac { 1 }{ 2 } \right) y+0=0\)
\(\Rightarrow\) x² + y² - x - y = 0 

Let x, y and z be the investments at the rate of interest 6%, 7% and 8% per annum respectively.
Then x + y + z = 5000  ...(1)
Also, \(\frac{6x}{100}+\frac{7y}{100}+\frac{8z}{100}=358\)
\(\Rightarrow\)   6x + 7y + 8z = 35800  ...(2)
And \(\frac{6x}{100}+\frac{7y}{100}=70+\frac{8z}{100}\) (Given)
\(\Rightarrow\)   6x + 7y - 8z = 7000  ...(3)
From (1), (2) and (3),
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5000 \\ 35800 \\ 7000 \end{matrix} \right] \)
AX = B where A = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8 \end{matrix} \right] ,\quad X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,\quad Z=\left[ \begin{matrix} 5000 \\ 35800 \\ 7000 \end{matrix} \right] \) 
 \(\therefore |A|\) \(=\begin{bmatrix} 1&1&1\\6&7&8\\6&7&-8 \end{bmatrix}=1(-56-56)-1(-48-48)1(42-42)\)
\(=-16\neq0\Rightarrow{A}^{-1}\) exists.
A₁₁  = -112, A₁₂ = 96, A₁₃= 0
A₂₁ = 15, A₂₂ = -14  A₂₃ = -1
A₃₁ = 1, A₃₂ = -2, A₃₃ = -1
\(\therefore \ adj\ A=\begin{bmatrix} -112&96&0\\15&-14&-1\\1&-2&1 \end{bmatrix}^{T}=\begin{bmatrix} -112&15&1\\96&-14&-2\\0&-1&1\end{bmatrix}\)
\(\therefore \quad A^{ -1 }=\quad \frac { 1 }{ |A| } adjA=-\frac { 1 }{ 16 } \left[ \begin{matrix} -112 & 15 & 1 \\ 96 & -14 & -2 \\ 0 & -1 & 1 \end{matrix} \right] \)
Hence, the solution is given by
\(X={A}^{-1}B=-\frac{1}{16}\begin{bmatrix} -112&15&1 \\96 &-4&-2\\0&-1&1 \end{bmatrix}\begin{bmatrix} 5000\\35800\\7000 \end{bmatrix}=-{{1}\over{16}}\begin{bmatrix} -560000+537000+7000\\480000-501200-14000\\0-35800+7000 \end{bmatrix}\)
\(X=\begin{bmatrix} 1000\\2200\\1800 \end{bmatrix}\)  [ \(\because\) x = 1000, y = 2200, z = 1800]
Hence, the three investment are of Rs. 1000,Rs. 2200 and Rs. 1800.

Given f(2) = 0
= 2 + 2
\(=4 \neq 0\)
\(\therefore \lim _{x \rightarrow 2} f(x) \neq f(2)\)
The function is not continuous at x = 2

Let \(A=\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] and\quad B=\left[ \begin{matrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{matrix} \right] \)
Now, AB =\(\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] \left[ \begin{matrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{matrix} \right] =\left[ \begin{matrix} -2-9+12 & 0-2+2 & 1+3-4 \\ 0+18-18 & 0+4-3 & 0-6+6 \\ -6-18+24 & 0-4+4 & 3+6-8 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\therefore\)  AB = I₃\(\Rightarrow\) A^-1 = B
\(\therefore { A }^{ -1 }=\left[ \begin{matrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{matrix} \right] \)
Now \(A=\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] ,x=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] and\quad c=\left[ \begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right] \)
\(X={ A }^{ -1 }C=\left[ \begin{matrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right] =\left[ \begin{matrix} -2 & +0 & +2 \\ 9 & +2 & -6 \\ 6 & +1 & -4 \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 5 \\ 3 \end{matrix} \right] \)
\(\because\) x = 0, y = 5 and z = 3

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.

LHS =  \(\begin{vmatrix} {1\over a}&bc&b+c\\{1\over b}&ca&c+a\\{1\over c}&ab&a+b \end{vmatrix}\)
Multiplying R₁ by a, R₂ by b, and R₃ by c respectively and dividing the determinant by abc we get.
\(LHS={1\over abc}\begin{vmatrix} 1&abc&ab+ac\\1&abc&bc+ab\\{1}&abc&ac+bc \end{vmatrix}\)
\(={1\over abc}(abc)\begin{vmatrix} 1&1&ab+ac\\1&1&bc+ab\\{1}&1&ac+bc \end{vmatrix}\) = 0  \([\because C_1\equiv C_2]\) = RHS.
Hence Proved.

\(\frac{1}{\left(x^2+4\right)(x+1)}=\frac{A}{x+1}+\frac{B x+C}{x^2+4}\)
(Since x² + 4 cannot be factorised into linear factors)
\(\frac{1}{\left(x^2+4\right)(x+1)}=\frac{A\left(x^2+4\right)+(B x+C)(x+1)}{x^2+4}\)
\(1=A\left(x^2+4\right)+(B x+C)(x+1)\) ...(1)
putting x = -1 in(1) we get 
1 = c(5) \(\Rightarrow\) \(c=\frac { 1 }{ 5 } \)
Equate co-efficient of x² on both sides of (1)
0 =A + B \(\Rightarrow\) \(B= -A = \frac { -1 }{ 5 } \)
Putting x = 0 in (1) we get,
1 = 4A + C
\(1=\frac{4}{5}+C\)
\(C=1-\frac{4}{5}=\frac{1}{5}\)
\(\frac{1}{\left(x^2+4\right)(x+1)}=\frac{1}{5(x+1)}+\frac{\frac{-1}{5} x+\frac{1}{5}}{x^2+4}\)
\(=\frac{1}{5(x+1)}+\frac{1-x}{5\left(x^2+4\right)}\).