\(\underset { x\rightarrow 0 }{ lim } \cfrac { log\left( 1+{ x }^{ 3 } \right) }{ { sin }^{ 3 }x } \) 
\(=\underset { x\rightarrow 0 }{ lim } \left\{ \cfrac { log\left( 1+{ x }^{ 3 } \right) }{ { x }^{ 3 } } \times \cfrac { { x }^{ 3 } }{ { sin }^{ 3 }x } \right\} \)
\(=\lim _{x \rightarrow 0} \frac{\log \left(1+x^3\right)}{x^3} \times \frac{1}{\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^3}\)
\(=1\times \cfrac { 1 }{ 1 } =1\)

Let A=69, B=112, dx=x-69, dy=y-112.
Given \(\sum\)dxdy=2176, \(\bar{X}\)=74.5, \(\bar{Y}\)=125.5
\({ \sigma }_{ x }\)=13.07, \({ \sigma }_{ y }\)=15.85, n=8, A=69, B=112
we know, \(\bar{x}\)=A+\(\frac { \sum { dx } }{ n } \)
\(\Rightarrow\)74.5=69+\(\frac { \sum { dx } }{ 8 } \) \(\sum\)dx=(74.5-69)(8)
\(\Rightarrow\)\(\sum\)dx =(5.5)(8)=4
Also, \(\bar{y}\)=B+\(\frac { \sum { dy } }{ n } \)
\(\Rightarrow\)125.5=112+\(\frac { \sum { dy } }{ 8 } \)
\(\Rightarrow\)(125.5-112)8=\(\sum\)dy
\(\Rightarrow\)\(\sum\)dy=108
The correlation co-efficient between x and y
r(x, y)=\(\frac { n\sum { dxdy } -(\sum { dx } )(\sum { dy } ) }{ \sqrt { n\sum { d{ x }^{ 2 } } }-(\Sigma dx)^{ 2 } \sqrt { n\sum { d{ y }^{ 2 }-{ (\sum { dy } ) }^{ 2 } } } } \) ...(1) 
\({ \sigma }_{ x }=\sqrt { \frac { { \sum { dx } }^{ 2 } }{ n } -{ \left( \frac { \sum { dx } }{ n } \right) }^{ 2 } } =\sqrt { \frac { { \sum { dx } }^{ 2 } }{ n } -\frac { { (\sum { dx } ) }^{ 2 } }{ { n }^{ 2 } } } \)
13.07= \(\frac { \sqrt { n\sum { d{ x }^{ 2 }-{ (\sum { dx } ) }^{ 2 } } } }{ n } \)  
13.07(8)=\(\sqrt { n\sum { { dx }^{ 2 }-({ \sum { dx } ) }^{ 2 } } } \) 
\(\sqrt { n\sum { { dx }^{ 2 }-({ \sum { dx } ) }^{ 2 } } } \)=104.56 .....(2)
SImilarly, \(\sqrt { n\sum { { dy }^{ 2 }-({ \sum { dy } ) }^{ 2 } } } \)=8(15.85)-126.8   ..(3)
Substuting (2) and (3) in (1) we get,
r=\(\frac { 8(2176)-44(108) }{ (104.56)(126.8) } =\frac { 12656 }{ 13258.208 } \)
=0.9545

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

FV = 100
Market Price = Rs.80
Dividend Rate = 20
Yield = \(\frac { FV }{ \text {Market Price } }\) x 20
=\(\frac { 100 }{ 80 } \) x 20 = 25%

Let n(s) denote the number of men psychologist.
\(\therefore\) n(s) = 50
Let n(A) denote the number of men psychologist
n(A) = 15
P(A) = \(\frac { 15 }{ 50 } =\frac { 3 }{ 10 } \)

S = {(1, 1), (1, 2) ....... (1, 6)
(2, 1). (2, 2) ........ (2, 6)
(6, 1), (6, 2) ....... (6, 6)}
\(\therefore\) n(S) = 36
A = {(1, 1),(1, 2) .. (1, 6)
(3, 1),(3, 2) ........ (3, 6)
(5, 1), (5, 2) ....... (5, 6)}
\(P(A) =\frac { 18 }{ 36 } =\frac { 1 }{ 2 } \)
B = {(1, 1),(2, 1) ....... (6, 1)
(1, 3),(2, 3)  ....... (6, 3)
(1, 5), (2, 5) ... (6, 5)}
\(P(B)=\frac { 18 }{ 36 } =\frac { 1 }{ 2 } \)
(A\(\cap \) B) = {(1, 1) (1, 3) ...... (1, 5)
(3, 1) (3, 3) ...... (3, 5)
(5, 1) (5, 3) ....... (5, 5)}
\(P(A\cap B)=\frac { 9 }{ 36 } =\frac { 1 }{ 4 } \) 
\(P(A).P(B)=\frac { 1 }{ 2 } . \frac { 1 }{ 2 } =\frac { 1 }{ 4 } =P(A\cap B)\)
\(\therefore\) A and B are independent events.

Weight (x)	f	log x	flog x
130	3	2.1139	6.3417
135	4	2.1303	8.5212
140	6	2.1461	12.8766
145	6	2.1614	12.9684
146	3	2.1644	6.4932
148	5	2.1703	10.8515
149	2	2.1732	4.3464
150	1	2.1761	2.1761
157	1	2.1959	2.1959
	N = 31		\(\sqrt{\sum flogx}\) = 66.771

GM = Antilog\(\left( \frac { \sum { f\ log\ x } }{ N } \right) \)
\(=Anitlog\left( \frac { 66.771 }{ 31 } \right) \)
= 142.5 Ibs

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 p(x,y) be any point on the parabola whose focus is F(-3, 2) and the directrix is x + y - 4 = 0.
Draw pm perpendicular to x + y - 4 = 0
Then FP = pm \(\Rightarrow\) FP² = pm²
\(\Rightarrow { (x+3) }^{ 2 }+{( y-2) }^{ 2 }={ \left[ \frac { x+y-4 }{ \sqrt { 1+1 } } \right] }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+6x+9+{ y }^{ 2 }-4y+4=\frac { { x }^{ 2 }+{ y }^{ 2 }+16+2xy-8x-8y }{ 2 } \)
\(\Rightarrow\) 2(x² + y² + 6x - 4y + 13) = x² + y² + 2y - 8x - 8y + 16
\(\Rightarrow\) x² + y² - 2xy + 20x + 10 = 0.

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

\(A^{-1}=\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)\)
\(A^2=\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)\)
\(=\left(\begin{array}{ll} 1-2 & -1-3 \\ 2+6 & -2+9 \end{array}\right)=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)\)
\(\mathrm{LHS}=A^2-4 A+5 l_2\)
\(=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)-4\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)+5\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)\)
\(=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)-\left(\begin{array}{cc} 4 & -4 \\ 8 & 12 \end{array}\right)+\left(\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right)\)
\(=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)+\left(\begin{array}{cc} 1 & 4 \\ -8 & -7 \end{array}\right)=\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right)\)
\(=O=R H S\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A\)
\(|A|=3+2=5 \neq 0\)
\(A^{-1}=\frac{1}{5}\left(\begin{array}{cc} 3 & 1 \\ -2 & 1 \end{array}\right)\)

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

(999)⁵ = (1000 - 1)⁵
\(=5 C_0(1000)^5-5 C_1(1000)^4+5 C_2(1000)^3 -5 C_3(1000)^2+5 C_4(1000)-5 C_5\)
= 1000,000,000,000,000 - 5 (1,000,000,000,000) + 10 (1,000,000,000) -10 (1000,000) + 5 (1000)-1
= 1000000000000000 - 5000000000000 + 10000000000 - 10000000+ 5000 1
= 99 500 999 000 4999

\(P =\frac{\frac{a}{i}}{k} \)
\(=\frac{\frac{2000}{0.1}}{12}=2,40,000\)

\(E O Q=\sqrt{\frac{2 C_3 R}{C_1}}=\sqrt{\frac{2 \times 20 \times 5000}{20}}=1000\)

2x	+	y	=	20		x	+	y	=	20
x		0		10		x		0		20
y		20		0		y		20		0

Corner points	Z = x + 3y
(0, 0)	0
(0, 20)	60
(10, 0)	10

\(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}} =\sin 2\left(30^{\circ}\right)=\sin 60^{\circ}=\frac{\sqrt{3}}{2} \)

\(\sin15^o = \sin(45^o - 30^o) = \sin45^o \cos 30^o - \cos 45^o \sin 30^o\)
\(= \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}\)

(0, a), a = 4

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

(Since \(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}\) is the correct answer)

\(\cfrac { d }{ dx } \left( { x }^{ \frac { 3 }{ 2 } } \right) =\cfrac { 3 }{ 2 } { x }^{ \frac { 3 }{ 2 } -1 }\) 
\(=\cfrac { 3 }{ 2 } { x }^{ \frac { 1 }{ 2 } }=\cfrac { 3 }{ 2 } \sqrt { x } \)

(i) Variables:
Let x₁, x₂ represent the dolls of A and B produced in a week.
(ii) Objective function:
Let Z be the total profit in a week.
\(\therefore Z={ 12x }_{ 1 }+16{ x }_{ 2 }\)
Since we have to maximize the profit, we have maximize \(Z={ 12x }_{ 1 }+16{ x }_{ 2 }\)
(iii) Constraints:
\({ x }_{ 1 }+{ x }_{ 2 } \le 2000\)
\({ x }_{ 1 }-{ 2x }_{ 2 }\ge 0\)
\( { x }_{ 1 }-{ 3x }_{ 2 }\le 600\)
(iv) Non-negative restictions:
 Since the number of dolls on type A and B cannot be negative, we have \({ x }_{ 1 },{ x }_{ 2 }\ge 0\)
Hence, the mathematical formation of LPP is
Maximize \(Z={ 12x }_{ 1 }+16{ x }_{ 2 }\)
Subject to the constraints
\({ x }_{ 1 }+{ x }_{ 2 } \le 2000\)
\({ x }_{ 1 }-{ 2x }_{ 2 }\ge 0\)
\( { x }_{ 1 }-{ 3x }_{ 2 }\le 600\)
and x₁, x₂ ≥ 0.

We know that correlation coefficient
r= \(\frac { N\Sigma XY-(\Sigma X)(\Sigma Y) }{ \sqrt { N\Sigma X^{ 2 }-(\Sigma X)^{ 2 }\sqrt { N\Sigma { Y }^{ 2 }-(\Sigma Y)^{ 2 } } } } \)
= \(\frac { 9(597)-(45\times 108) }{ \sqrt { 9(285)-(45)^{ 2 }\times \sqrt { 9(1356)-(108^{ 2 }) } } } \)
r = +0.95

Given p = 10e^{\(\frac { -x }{ 2 } \)}
\({dp\over dx}=10e^{-x/2}(-1/2)=-5e^{-x\over 2}\)
Elasticity of demand \((\eta_d)\)=\({-p\over x}.{dx\over dp}\) 
\(\eta_d=\frac{-10 e^{-\frac{x}{2}}}{x} \cdot \frac{1}{-5 e^{-\frac{x}{2}}}=\frac{2}{x}\)

\(\frac{15\pi}{4}=15\times45^o=675^o\)
\(\cot\left(\frac{-15\pi}{4}\right)=\)cot (-675°)= - cot 675° = - cot (720 -45°) = -cot (2 x 360° - 45°)
= -(-cot 45°)(\(\therefore\) 675° is in the IV quadrant) =-(-1) = 1.

sin (-105)° = - sin (105°) 
= - sin (60° + 45°)
= - [sin 60° cos 45° + cos 60° sin 45°]
= \(-\left[ \frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } .\frac { 1 }{ \sqrt { 2 } } \right] \)
= \(-\left[ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right] \)

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

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

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 \(cosA=\cfrac { 4 }{ 5 } \ cosB=\cfrac { 12 }{ 13 } \)
Therefore  \(\sin A=-\sqrt{1-\cos ^2 A}\)
\(=-\sqrt{1-\frac{16}{25}}\)
\(=-\sqrt{\frac{25-16}{25}}\)
\(=-\frac{3}{5}\)
\(\operatorname{Sin} B =-\sqrt{1-\cos ^2 B} \)
\(=-\sqrt{1-\frac{144}{169}} \)
\(=-\sqrt{\frac{169-144}{169}} \)
\(=-\frac{5}{13}\)
\(sin(A-B)=sinAcosB-cosAsinB\\ =\left( \cfrac { -3 }{ 5 } \right) \left( \cfrac { 12 }{ 13 } \right) -\left( \cfrac { 4 }{ 5 } \right) \left( \cfrac { -5 }{ 13 } \right) \\ =\cfrac { -36 }{ 65 } +\cfrac { 20 }{ 65 } =\cfrac { -6 }{ 65 } \)

\(\frac{9}{(x-1)(x+2)^2} =\frac{A}{(x-1)}+\frac{B}{(x+2)}+\frac{C}{(x+2)^2} \) ...(1)
Multiplying both sides by \((x-1)(x+2)^{ 2 }\) 
\(\therefore 9=A(x+2)^{ 2 }+B\left( x-1 \right) (x+2)+C(x-1)\) ..(2)
Put x = –2 in (2)
\(\\ 9=A(0)+B(0)+C(-2-1)\) 
\(\therefore C=-3\) 
Put x = 1 in (2)
\(9=A\left( 3 \right) ^{ 2 }+B(0)+C(0)\\ \therefore A=1\) 
Equating the coefficient of x² on both the sides of (2), we get 
\(0=A+b\\ B=-A\\ \therefore B=-1\) 
Substituting the values of A, B and C in (1), we get 
\(\cfrac { 9 }{ (x-1)(x+2)^{ 2 } } =\cfrac { 1 }{ x-1 } -\cfrac { 1 }{ x+2 } -\cfrac { 3 }{ \left( x+2 \right) ^{ 2 } } \)

Consider the equations
\(36{ x }_{ 1 }+6{ x }_{ 2 }= 108\)

\({ x }_{ 1 }\)	0	3
\({ x }_{ 2 }\)	18	0

\(3{ x }_{ 1 }+12{ x }_{ 2 }= 36\)

\({ x }_{ 1 }\)	0	12
\({ x }_{ 2 }\)	3	0

\(20{ x }_{ 1 }+10{ x }_{ 2 }= 100\)

\({ x }_{ 1 }\)	0	5
\({ x }_{ 2 }\)	10	0

 

The feasible region is ABCD and its co-ordinates are A(12, 0), D(0, 18), B is the point of intersection of the lines
Verification of B:
\(3{ x }_{ 1 }+12{ x }_{ 2 }=36 \Rightarrow { x }_{ 1 }+4{ x }_{ 2 }=12 ...(1)\)
\(20{ x }_{ 1 }+10{ x }_{ 2 }=100 \Rightarrow { 2x }_{ 1 }+{ x }_{ 2 }=10...(2)\)
\( (1)\times 2\Rightarrow 2{ x }_{ 1 }+8{ x }_{ 2 }=24\)
\( (-)\quad (-)\quad \quad (-)\)
\((2)\Rightarrow 2{ x }_{ 1 }+{ x }_{ 2 }=10\)
\(--------------\)
\(7{ x }_{ 2 }=14 \Rightarrow { x }_{ 2 }=2\)
\(From(1), { x }_{ 1 }+8=12 \Rightarrow { x }_{ 1 }=4\)
∴ B is (4,2) 
Also C is the point of intersection of the lines
Verification for C:
\(3{ 6x }_{ 1 }+6{ x }_{ 2 }=108\Rightarrow 6{ x }_{ 1 }+{ x }_{ 2 }=18 ...(3)\)
\( (-)\quad (-)\quad \quad (-)\)
\(20{ x }_{ 1 }+10{ x }_{ 2 }=100\Rightarrow 2{ x }_{ 1 }+{ x }_{ 2 }=10 ...(4)\)
\( --------------\)
\(4{ x }_{ 1 }=8 \Rightarrow { x }_{ 1 }=2\)
\( From(4), 4+{ x }_{ 2 }=10 \Rightarrow { x }_{ 2 }=6\)
∴ C is (2,6) 

Corner Points	z = 20x1+ 40x2
A(12,0)	240
B(4,2)	160
C (2, 6)	280
D(0, 18)	720

Minimum of B occurs at B(4, 2)
Hence, the solution is x₁ = 4, x₂ = 2 and Z_min = 160.

Given P = 10K - K² + KL
Differentiating partially w.r.t. 'K' we get,
\({\partial P\over \partial K}=10(1)-2K+L(1)\)
=10-2K+L
When K = 2 and L = 6.
Marginal Productivity of Capital
\({\partial P\over \partial K}=10 - 2(2) + 6 = 10 - 4 + 6 = 12\)
Differentiating 'P' partially w.r.t. 'L' we get,
\({\partial P\over \partial L}=0-0+K=K\)
When K = 2 and L = 6,
Marginal Productivity of Labour
\({\partial P\over \partial L}=2\)

Stock = Rs.8000
FV = Rs.100
Dividend Rate = 7%
Income = \(\frac { Stock }{ FV } \times Dividend \ Rate=\frac { 8000 }{ 100 } \times 7\) = Rs.560  ....(1)
Selling price of one share=Rs.96
Sale proceeds = \(\frac { Stock }{ FV } \)x S.P of one share = \(\frac { 8000 }{ 100 } \) x 96 = Rs.7680
Given that the income on 10% stock
= 560+80 = 640
FV = Rs.10
Income = \(\frac { \text {Investment }}{ \text {Purchase Price } }\) x Dividend Rate
640 = \(\frac { 7680 }{\text { Purchase Price } }\) x 10
Purchase Price = \(\frac { 7680 }{ 640 } \) x 10 = Rs.120
Hence, purchase price of 10% stock=Rs.120

Let y=x³-6x²+7
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=3x^2-12x\)
\({dy\over dx}=0\)
\(\Rightarrow3x^2-12x=0\)
\(\Rightarrow3x(x-4)=0\)
\(\Rightarrow x=0 \ or \ x=4\)
\({d^2y\over dx^2}=6x-12\)
when x=0 \({d^2y\over dx^2}=-12<0\)
\(\therefore \) y is maximum at x=0
\(\therefore \) maximum value=0³-6(0)²+7=7
when x=4, \({d^2y\over dx^2}=-6(4)-12=12>0\)
\(\therefore \) y is minimum at x=4
\(\therefore \) Minimum value =4⁴-6(4)²+7=64-96+7=-25
Hence maximum value is 7 and minimum value is -25.

Let us consider Height of father (in inches) is represented as X and Height of son (in inches) is represented as Y

X	dx = (X-67)	dx2	Y	dy = (Y-68)	dy2	dxdy
65	-2	4	67	-1	1	2
66	-1	1	68	0	0	0
67	0	0	64	-4	16	0
67	0	0	68	0	0	0
68	1	1	72	4	16	4
69	2	4	70	2	4	4
71	4	16	69	1	1	4
73	6	36	70	2	4	12
ΣX = 546	Σdx = 10	Σdx2 = 62	ΣY = 548	Σdy = 4	Σdy2 = 42	Σdxdy = 26

r = \(\frac { N\Sigma dxdy-(\Sigma dx)(\Sigma dy) }{ \sqrt { N\Sigma dx^{ 2 }-(\Sigma dx)^{ 2 }\times \sqrt { N\Sigma dy^{ 2 }-(\Sigma dy)^{ 2 } } } } \)
Here Σdx = 10, Σdx² = 62, Σdy = 4, Σdy² = 42 and Σdxdy = 26
r = \(\frac { (8\times 26)-(10\times 4) }{ \sqrt { (8\times 62)-(10)^{ 2 }\times \sqrt { (8\times 42)-(4)^{ 2 } } } } \)
r = \(\frac { 168 }{ \sqrt { 396 } \times \sqrt { 320 } } \)
r =  \(\frac { 168 }{ { 355.98 } } \) =0.472
i.e., r = +0.472
Heights of fathers and their respective sons are positively correlated.

Activity	Duration (in week)	EST	EFT	LST	LFT
1-2	8	0	8	0	8
1-3	7	0	7	8	15
1-5	12	0	12	9	21
2-3	4	8	12	11	15
2-4	10	8	18	8	18
3-4	3	12	15	15	18
3-5	5	12	17	16	21
3-6	10	12	22	15	25
4-6	7	18	25	18	25
5-6	4	17	21	21	25

Here the critical path is 1–2–4–6
The project completion time is 25 weeks.

f(x) = 3x⁵ - 25x³ + 60x + 1 … (1)
f ' (x) = 15x⁴ – 75x² + 60 = 15(x⁴ – 5x² + 4)
f '(x) = 0 \(\Rightarrow\) 15(x⁴ – 5x² + 4) = 0
\(\Rightarrow\)  (x² – 4)(x² – 1) = 0
x = \(\pm\)2 (or) x = \(\pm\)1
of these four points −2,\(\pm\) ∈ |-2,1| and 2∉[−2,1]
From (1)
f(-2) = 3(-2)⁵ - 25(-2)³+60(-2)+1 =-15
When x = 1
f(1) = 3(1)⁵ -25(1)³+ 60(1) +1 = 39
when x = -1
f(–1) = 3(-1)5 - 25(-1)³ + 60(-1) +1 =-37
Absolute maximum is 39 and
Absolute minimum is - 37.

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 22 P_{r + 1} :20 P_{r+ 2} = 11 : 52
\(\Rightarrow\)52.22P_r+1 = 11.20 P_r+2 
\(\Rightarrow\)    \(52.\frac { 22! }{ (22-r-1)! } =11.\frac { 20! }{ (20-r)-2! } \quad \left[ \therefore npr=\frac { n! }{ n-r! } \right] \)
\(\Rightarrow\)    \(\frac { 52\times 2\times 21 }{ (21-r)! } =11.\frac { 20! }{ (18-r)! } \)
\(\Rightarrow\)    \(\frac { 52\times 2\times 21 }{ (21-r)(20-r)(19-r)(18-r)! } =\frac { 1 }{ (18-)! } \)
\(\Rightarrow\)    52 x 2 x 21 = (21- r) (20 - r) (19- r)
\(\Rightarrow\)    2 x 3 x 7 x 4 x 13 = (21- r) (20 - r) (19- r)
\(\Rightarrow\)    12 x 13 x 14 = (21- r)(20 - r)(19- r)
\(\Rightarrow\)    12 x 13 x 14 = (19- r) (20 - r) (21- r)
\(\Rightarrow\)    19-r=12
\(\Rightarrow\)r= 19-12
∴  r =7

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

a₁₁ = 30, a₁₂ = 40, x₁ = 120
a₂₁ = 20, a₂₂ = 10, x₂ = 60
\({b}_{11}={{{a}_{11}}\over{x_1}}={{30}\over{120}}={{1}\over{4}}\)
\({b}_{12}={{{a}_{12}}\over{{x}_{2}}}={{40}\over{60}}={{2}\over{3}}\)
\({b}_{21}={{{a}_{21}}\over{x_1}}={{20}\over{120}}={{1}\over{6}}\)
\({b}_{22}={{{a}_{22}}\over{{x}_{1}}}={{10}\over{60}}={{1}\over{6}}\)
The technology matrix is   B =  \(\begin{bmatrix}{{1}\over{4}}&{{2}\over{3}}\\ {{1}\over{6}}&{{1}\over{6}} \end{bmatrix}\)
I - B = \(\begin{bmatrix} 0&0\\0&1 \end{bmatrix}-\begin{bmatrix} {{1}\over{4}} &{{2}\over{3}}\\{{1}\over{6}}&{{1}\over{6}} \end{bmatrix}=\begin{bmatrix} {{3}\over{4}}&{{-2}\over{3}}\\ {{-1}\over{6}}&{{5}\over{6}} \end{bmatrix}\)
\(|I-B|=\frac{3}{4} \times \frac{5}{6}-\frac{2}{3} \times \frac{1}{6}=\frac{5}{8}-\frac{1}{9}=\frac{37}{72}=0\)
Since diagonals of I - B are positive and | I - B | is positive, the system is viable
\({(I-B)}^{-1}={{1}\over{|I-B|}}adj\ (I-B)={{72}\over{37}}\begin{bmatrix}{{5}\over{6}}&{{2}\over{3}}\\{{1}\over{6}}&{{3}\over{4}} \end{bmatrix}\)
X = (I - B)^-1 D where D = \(\begin{bmatrix} 80\\40 \end{bmatrix}\)
\(={{72}\over{37}}\begin{bmatrix}{{5}\over{6}}&{{2}\over{3}}\\{{1}\over{6}}&{{3}\over{4}} \end{bmatrix}\begin{bmatrix} 80 \\ 40 \end{bmatrix}\)
\(=\frac{72}{37}\left(\begin{array}{cc} 66.67 & +26.67 \\ 13.33 & +30 \end{array}\right)=\frac{72}{37}\left(\begin{array}{l} 93.34 \\ 43.33 \end{array}\right)\)
\(=\left(\begin{array}{c} 181.63 \\ 84.32 \end{array}\right)\)
The output for production section X and Y are 181.63 and 84.32 respectively.