A = {(2, 2)}, n(S) = 36; P(A) = 1/36

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

\(\text { M.V }=100-9 \frac{1}{2}+\frac{1}{2}=Rs.91\)

\(r =\frac{N \Sigma X Y-\Sigma X \Sigma Y}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{25(520)-(125)(100)}{\sqrt{25(650)-(125)^2} \sqrt{25(436)-(100)^2}} \)
= 0.667

\(\eta_d=-\frac{p}{x} \frac{d x}{d p}=\frac{-p}{\frac{1}{p}}\left(\frac{-1}{p^2}\right)=1\)

\(\frac{d y}{d x}=1, \frac{d z}{d x}=\frac{-1}{x^2} \quad \frac{d y}{d z}=-x^2\)

\(\tan 2 A= \frac{2 \tan A}{1-\tan ^2 A}=\frac{2(1 / 2)}{1-1 / 4}=\frac{1}{3 / 4}=4 / 3 \)
\(\tan (2 A+B)= \frac{\tan 2 A+\tan B}{1-\tan 2 A \cdot \tan B}=\frac{\frac{4}{3}+\frac{1}{3}}{1-4 / 9} \)
\(=\frac{5 / 3}{5 / 9}=3 \)

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

x + y = n
3 + 2 = 5 = n
nC₄ = 5C₄ = 5C₁ = 5

\(\text {Since }|A|=4-4=0\)

\(L\left[ f(x) \right] _{ x=1 }=\underset { x\rightarrow { 1 }^{ - } }{ lim } f\left( x \right) \\ =\underset { h\rightarrow 0 }{ lim } f\left( 1-h \right) ,x=1-h\\ =\underset { h\rightarrow 0 }{ lim } \left[ 5\left( 1-h \right) -4 \right] \\ =\underset { h\rightarrow 0 }{ lim } \left( 1-5h \right) =1\) 
\(R\left[ f\left( x \right) \right] _{ x=1 }=\underset { x\rightarrow { 1 }^{ + } }{ lim } f(x)\\ =\underset { h\rightarrow 0 }{ lim } f\left( 1+h \right) ,x=1+h\\ =\underset { h\rightarrow 0 }{ lim } \left[ 4\left( 1+h \right) ^{ 3 }-3\left( 1+h \right) \right] \\ =4\left( 1 \right) ^{ 3 }-3\left( 1 \right) =1\) 
Clearly,\(L\left[ f(1) \right] =R\left[ f(1) \right] \) 
\(\therefore \underset { x\rightarrow 1 }{ lim } f\left( x \right) \) exists and equal to 1

\(\cos 55^{\circ}+\sin 55^{\circ}=\cos 55^{\circ}+\cos \left(90^{\circ}-55^{\circ}\right)\)
\(=\cos 55^{\circ}+\cos 35^{\circ}\)
\(=2 \cos \left(\frac{55^{\circ}+35^{\circ}}{2}\right) \cos \left(\frac{55^{\circ}-35^{\circ}}{2}\right)\)
\(=2 \cos 45^{\circ} \cos 10^{\circ}=2\left(\frac{1}{\sqrt{2}}\right) \cdot \cos 10^{\circ}\)
\(=\sqrt{2} \cos 10^{\circ}\)

B = \(\left[ \begin{matrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{matrix} \right] \)
I - B=\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)-\(\left[ \begin{matrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 0.2 & -0.2 \\ - 0.9 & 0.3 \end{matrix} \right] \)
|I - B|=  \(\left[ \begin{matrix} 0.2 & -0.2 \\ - 0.9 & 0.3 \end{matrix} \right] \)
= (0.2)(0.3) - (-0.2)(-0.9)
= 0.06 - 0.18
= 0.12 < 0
Since |I - B| is negative, Hawkins – Simon conditions are not satisfied.
Therefore, the given system is not viable.

Given A = Rs.3783,i = 0.05,n = 3
A = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
3783 = \(\cfrac { a }{ i } \) [(1.05)³-1]
3783 x 0.05 = a[1.1576-1]
a = \(\cfrac { 189.15 }{ 0.1576 } \) = 1200.19
\(\therefore\) a =Rs.1200
(1.05)³ = 3 log (1.05)
= 3(0.212)
= 0.636
Antilog of 0.636 is 1.1576

r(x,y) =\(\frac { N\sum { XY-(\sum { X)(\sum { Y) } } } }{ \sqrt { N.{ \sum { X } }^{ 2 }-({ \sum { X) } }^{ 2 } } .\sqrt { N.{ \sum { Y } }^{ 2 }-{ (\sum { Y) } }^{ 2 } } } \)
=\(\frac { 9(597)-45(108) }{ \sqrt { 9(285)-{ (45) }^{ 2 } } .\sqrt { 9(1356)-({ 108) }^{ 2 } } } \)
=0.95
\(\therefore\)X and Y are highly positively correlated.

Q₁ = Size of \({ \left( \frac { N+1 }{ 4 } \right) }^{ th }\) value
= Size of (165.25)^th value = 10
Q₃ = Size of 3\({ \left( \frac { N+1 }{ 4 } \right) }^{ th }\)value
= size of (495.75)^th value = 30
Q.D = \(\frac { 1 }{ 2 } ({ Q }_{ 3 }-{ Q }_{ 1 })=\frac {30-10}{2}=\frac { 20 }{ 2 } =10\)
Co-efficient of Q.D = \(\frac { { Q }_{ 3 }-{ Q }_{ 1 } }{ { Q }_{ 3 }+{ Q }_{ 1 } } =\frac { 20 }{ 40 } =\frac { 1 }{ 2 } \) = 0.5

Given the production is P = 10L - 0.1L² +15K - 0.2K² + 2KL
\(\frac { \partial P }{ \partial L } \) = 10 - 0.2L + 2K
\(\frac { \partial P }{ \partial K } \) = 15 - 0.4 k + 2L
When L = K = 10 units,
Marginal productivity of labour 
\(\left( \frac { \partial P }{ \partial L } \right) _{ 10,10 }\) = 10 - 2 +20 = 28
Marginal productivity of capital :
\(\left( \frac { \partial P }{ \partial K } \right) _{ 10,10 }\) = 15 - 4  + 20 =31
(ii) Upper limit for use of labour when K=10 is given by \(\left( \frac { \partial P }{ \partial L } \right) \)\(\ge\)0
10 − 0.2L+20 \(\ge\)0
30 \(\ge\) 0.2L
i.e., L \(\le\) 150
Hence the upper limit for the use of labour will be 150 units.

Given that f(x) = 2x³ + 9x² + 12x + 1.
f'(x) = 6x² + 18x + 12
= 6(x² + 3x + 2)
= 6(x + 2)(x + 1)
f'(x) = 0 \(\Rightarrow\) 6 (x + 2)(x + 1) = 0
\(\Rightarrow\) x + 2 = 0 (or) x + 1 = 0.
x = –2 (or) x = –1
f(x) has stationary points at x = – 2 and x = – 1
Stationary values are obtained by putting x = – 2 and x = – 1
When x = – 2, f(–2) = 2(–8)  +9(4) + 12(–2) + 1 = –3
When x = – 1, f(–1) = 2(–1) + 9(1) + 12(–1) + 1 = –4
The stationary points are (–2, –3) and (–1,–4).

Let P(x₁, y₁) be any point on the locus and let A(0, 2) B(0, -2) be the fixed points.
By the given condition, PA + PB =6
\(\Rightarrow \sqrt { { \left( { x }_{ 1 }-0 \right) }^{ 2 }+{ \left( { y }_{ 1 }-2 \right) }^{ 2 } } +\sqrt { { ({ x }_{ 1 }-0 })^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } =6\)
\(\Rightarrow \sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }-2) }^{ 2 } } =6-\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
Squaring both sides we get,
\({ x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }-2) }^{ 2 }=36-12\sqrt { { x }_{ 1 }^{ 2 }+({ y }_{ 1 }+2)^{ 2 } } +{ x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 }\)
\(\Rightarrow \quad { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4-4{ y }_{ 1 }=36-12\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } +{ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4+4{ y }_{ 1 }\)
\(\Rightarrow \quad { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4-4{ y }_{ 1 }-36-{ x }_{ 1 }^{ 2 }-{ y }_{ 1 }^{ 2 }-4-4{ y }_{ 1 }=-12\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
= -8y₁ - 36 = -12 \(\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
= 2y₁ + 9 = 3\(\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
Squaring again we get,
(2y₁+9)² = 9[\({ x }_{ 1 }^{ 2 }+({ y }_{ 1 }+2)^{ 2 }]\)
\(\Rightarrow 4{ y }_{ 1 }^{ 2 }+81+36{ y }_{ 1 }=9[{ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4+4{ y }_{ 1 }]\)
\(\Rightarrow 4{ y }_{ 1 }^{ 2 }+81+36{ y }_{ 1 }-9{ x }_{ 1 }^{ 2 }-9{ y }_{ 1 }^{ 2 }-36-36{ y }_{ 1 }=0\)
\(\Rightarrow -9{ x }_{ 1 }^{ 2 }-5{ y }_{ 1 }^{ 2 }+45=0\)
\(\Rightarrow 9{ x }_{ 1 }^{ 2 }+5{ y }_{ 1 }^{ 2 }=45\)
\(\therefore \)  Locus of (x₁ , y₁) is 9x² + 5y² = 45

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
 

Compare the equation ax² + 5xy - 6y² + 12x + 5y + c = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c = 0
We get a = a, 2h = 5, b = -6, 2g = 12, 2f = 5 c = c
\(h=\frac { 5 }{ 2 } \quad g=6\quad f=\frac { 5 }{ 2 } \)
since the lines are perpendicular
a + b = 0
 \(\Rightarrow\) a - 6 = 0
\(\Rightarrow\) a = 6
since it represents a pair of straight lines
\(\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc} 6 & \frac{5}{2} & 6 \\ \frac{5}{2} & -6 & \frac{5}{2} \\ 6 & \frac{5}{2} & c \end{array}\right|=0\)
\(6\left(-6 c-\frac{25}{4}\right)-\frac{5}{2}\left(\frac{5 c}{2}-\frac{30}{2}\right)+6\left(\frac{25}{4}+36\right)=0 \)
\(=-36 c-\frac{150}{4}-\frac{25 c}{4}+\frac{150}{4}+\frac{150}{4}+216=0 \)
\(-36 c-\frac{25 c}{4}=\frac{-150}{4}-216 \)
\(\frac{-169 c}{4}=\frac{-1014}{4} \Rightarrow c=6 \)

\(\left(2 x^2-\frac{3}{x^3}\right)^{10}\)
n = 10
middle term is \(t_{\frac{n}{2}+1}=t_{\frac{10}{2}+1}=t_6\)
r = 5
\(t_{r+1}=n C_{r x}^{n-r} a^r\)
\(t_6=10 C_5\left(2 x^2\right)^5\left(\frac{-3}{x^3}\right)^5\)
\(=-10 C_5 \frac{x^{10}}{x^{15}} 2^5 \cdot 3^5\)
\(=-10 C_5 \frac{(6)^5}{x^5}\)

(101)⁴ = (100 + 1)⁴
= 4C₀ (100)⁴ + 4C₁ (100)³ (1)¹ + 4C₂ (100)² (1)² + 4 C₃ (100)¹ (1)³ + 4 C₄(1)⁴
= 100,000,000 + 4 (1,000,000) + 6 (10000) + 4 (100) + 1
= 10,40,60,401

Given \(TR=12x+{x^2\over2}-{x^3\over 3}\)
Average Revenue \(AR={TR\over x}={12x+x^2/2-x^3/3\over x}\)
\(AR=12+{x\over2}-{x^2\over3}\)     ...(1)
Let y \(=12+{x\over2}-{x^2\over3}\)
\({dy\over dx}={1\over2}-{2x\over 3}\)
Condition for maximum is
\({dy\over dx}=0\ and\ {d^2 y\over dx^2}<0\)
\(∴\ {1\over2}-{2x\over3}=0⇒{1\over2}={2x\over 3}\)
\(x={1\over2}\times{3\over 2}={3\over 4}\)
\({d^2y\over dx^2}={-2\over3}<0\)
∴ AR is maximum at x=3/4
when x=3/4, \(AR=12+{3/4\over2}-{\left(3/4\right)^2\over 3}\)  [From )1)]
\(=12+{3\over8}-{9\over 48}=12.1875\)     ....(2)
\(MR={{dR\over dx}}={d\over dx}\left(12x+{x^2\over 2}-{x^3\over 3}\right)=12+x-x^2\)
when x = 3/4, \(MR=12+{3\over 4}-\left(3\over4\right)^2=12+{3\over4}-{9\over 16}=12.1875 \)    ..(3)
From (2) and (3), at the highest point of AR,
AR = MR = 12.1875

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

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
\(3{ x }_{ 1 }+4{ x }_{ 2 }=8\)

\(5{ x }_{ 1 }+2{ x }_{ 2 }=11\)

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:
\((1) \Rightarrow 3{ x }_{ 1 }+4{ x }_{ 2 }=8\)
\( (-)\quad (-)\quad \quad (-)\)
\((2)\times 2\Rightarrow 10{ x }_{ 1 }+4{ x }_{ 2 }=22\)
\(--------------\)
\( -7x_{ 1 }=-14 \Rightarrow { x }_{ 1 }=2\)
\(From(1), 3(2)+4{ x }_{ 2 }=8\)
\(4{ x }_{ 2 }=8-6=2\Rightarrow { x }_{ 2 }=\frac { 1 }{ 2 } \)
\( \therefore \ B\ is\ \left( 2,\frac { 1 }{ 2 } \right) \) 

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.3000,r =\(\cfrac { 8 }{ 100 } \times \cfrac { 1 }{ 4 } \) = 0.02,n = 3 x 4 =12
A=\(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
3000=\(\cfrac { a }{ 0.02 } \left[ \left( 1.02 \right) ^{ 12 }-1 \right] \)
3000 x 0.02 = a[1.2690-1]
60 = a[0.2690]
a=\(\cfrac { 60 }{ 0.2690 } =223.04\)
a = Rs.223 (app)
(1.02)¹² = 12 log (1.02)
= 12 (0.0086)
= 0.1032
Antilog of 0.1032 is 1.2690

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.

Q₁ = Size of \(\left( \frac { N }{ 4 } \right) ^{th}\) value = \(\frac{67}{4}\) = 16.75^th value.
Thus Q₁ lies in the class (20 – 30) and its corresponding values are L = 20;
\(\frac{N}{4}\) = 16.75; pcf =12; f = 19 ; c = 10
Q₁ = L + \(\left( \frac { \frac { N }{ 4 } -pcf }{ f } \right) \times c\)
Q₁ = 20 +\(\left( \frac { 16.75-12 }{ 19 } \right) \times 10\) = 20 + 2.5 = 22.5
Q₃ = Size of \(\left( \frac { 3N }{ 4 } \right) ^{th}\)value = 50.25^th value
So Q₃ lies in the class (50-60) corresponding values are L = 50,\(\left( \frac { 3N }{ 4 } \right) \)= 50.25;
pcf = 46, f = 9, c = 10 
\({ Q }_{ 3 }=L+\left( \frac { \frac { 3N }{ 4 } -pcf }{ f } \right) \times c\)
\({ Q }_{ 3 }=50+\left( \frac { 50.25-25 }{ 9 } \right) \times 10=54.72\)
\({ D }_{ 4 }=L+\left( \frac { \frac { 4N }{ 4 } -pcf }{ f } \right) \times c\)
D₄ = Size of \(\left( \frac { 4N }{ 10 } \right) ^{th}\) value = 26.8^th value.Thus D₄ lies in the class (20 – 30) and its corresponding values are
L = 20,\( \frac { 4N }{ 10 } \) = 26.8; pcf = 12, f = 19, c = 10.
\({ D }_{ 4 }=20+\left( \frac { 26.8-12 }{ 19 } \right) \times 10\) = 27.79
P₇₅ = Size of \(\left( \frac { 75N }{ 100 } \right) ^{th}\) value = 50.25^th value.Thus P₇₅ lies in the class ( 50 – 60) and its corresponding values are L = 50; \( \frac { 75N }{ 100 } \) = 50.25 ; pcf = 46, f = 9, c= 10.
\({ P }_{ 75 }=L+\left( \frac { \frac { 75N }{ 100 } -pcf }{ f } \right) \times C\)
\(=50+\left( \frac { 50.25-46 }{ 9 } \right) \times 10\) = 54.72

\(r(x, y)=\frac{N \Sigma X Y-\left(\sum X\right)\left(\sum Y\right)}{\sqrt{N \Sigma X^2-(\Sigma Y)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{10(10718)-(310)(340)}{\sqrt{10(10480)-(310)^2} \sqrt{10(12280)-(340)^2}} \)
\(=\frac{107180-105400}{\sqrt{104800-96100} \sqrt{122800-115600}} \)
\(=\frac{1780}{\sqrt{8700 \times 7200}}=\frac{1780}{7914.54}=0.2249\)

Since EFT and LFT is same on 1 - 3, 3 - 5, 5 -7 and 7 - 8 and 8 -10 the critical path is 1- 3 - 5 -7 - 8 - 10 and the duration is 22 time units.

LHS = (sin 3x + sin x) sin x + (cos 3x - cos x) cos x
= sin 3x sin x + sin²x + cos 3x cos x - cos²x
= sin 3x sin x + cos 3x cos x - (cos²x - sin²x)
= cos (3x - x) - cos 2x [∴ cosA cosB + sinA sinB = cos(A - B) and cos²A - sin²A = cos²A]
= 0 = RHS. Hence proved

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.

LHS\(=\frac{\cos4x+\cos2x+\cos3x}{\sin4x+\sin2x+\sin3x}\)
\(=\frac{2\cos\left(\frac{4x+2x}{2}\right).\cos\left(\frac{4x-2x}{2}\right)+\cos3x}{2\sin\left(\frac{4x+2x}{2}\right).\cos\left(\frac{4x-2x}{2}\right)+\sin3x}\)
\(=\frac{2 \cos 3 x \cos x+\cos 3 x}{2 \sin 3 x \cos x+\sin 3 x} \)
\(=\frac{\cos 3 x(2 \cos x+1)}{\sin 3 x(2 \cos x+1)}=\frac{\cos 3 x}{\sin 3 x}=\cot3x=RHS\)
Hence Proved

The technology matrix is given under.

B =\(\begin{bmatrix} 0.4 & 0.1 \\ 0.7 & 0.7 \end{bmatrix}\)
I - B =\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.4 & 0.1 \\ 0.7 & 0.7 \end{bmatrix}=\begin{bmatrix} 0.6 & -0.1 \\ -0.7 & 0.3 \end{bmatrix}\)
|I - B| = \(\begin{bmatrix} 0.6 & -0.1 \\ -0.7 & 0.3 \end{bmatrix}\)
|I - B| = 0.18 - 0.07 = 0.11 > 0
Since the diagonal elements of (I - B) are positive and |I - B| is positive, the system is viable.
\(\therefore \) ( I - B)^-1 = \(\frac { 1 }{ |I-B| } adj(I-B)=\frac { 1 }{ 0.11 } \begin{bmatrix} 0.3 & 0.1 \\ 0.7 & 0.6 \end{bmatrix}\)
Now X = (I - B)^-1 D where D = \(\left[ \frac { 6.8 }{ 10.2 } \right] \)
X = \(\frac { 1 }{ 0.11 } \begin{bmatrix} 0.3 & 0.1 \\ 0.7 & 0.6 \end{bmatrix}\left[ \frac { 6.8 }{ 10.2 } \right] =\frac { 1 }{ 0.11 } \left[ \frac { 0.3\times6.8+0.1\times10.2 }{ 0.7\times6.8+0.6\times10.2 } \right] \)
=\(\frac { 1 }{ 0.11 } \left[ \frac { 2.04+1.02 }{ 4.76+6.12 } \right] =\frac { 1 }{ 0.11 } \left[ \frac { 3.06 }{ 10.88 } \right] =\left[ \frac { 27.81 }{ 98.90 } \right] \)
Gross production of commodity A and B are 27.81
Gross production of B is 98.91 tonnes

\(\frac{x+2}{(x-1)(x+3)^2}=\frac{A}{(x-1)}+\frac{B}{(x+3)}+\frac{C}{(x+3)^2}\)
\(=\frac{A(x+3)^2+B(x-1)(x+3)+C(x-1)}{(x-1)(x+3)^2}\)
\(\Rightarrow\)  x + 2 = A ( x + 3 )² + B ( x - 1 ) ( x + 3 ) + C( x - 1 ) ....(1)
Putting x = 1 in (1) we get,
\(3=A(4)^2 \Rightarrow A=\frac{3}{16}\)
Puttingx = -3 in (1)we get
\(-1=C(-3-1)\)
\(C=\frac{1}{4}\)
Equate co-efficient of x² on both sides of (1)
\(0 =A+B \Rightarrow B=-A=\frac{-3}{16} \)
\(\frac{x+2}{(x-1)(x+3)^2} =\frac{3}{16(x-1)}-\frac{3}{16(x+3)}+\frac{1}{4(x+3)^2}\)

\(x={ at }^{ 2 }\)
\(\cfrac { dx }{ dt } =2at\)
\(y=2at\)
\( \cfrac { dy }{ dt } =2a\) 
\(\therefore \cfrac { dy }{ dx } =\cfrac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \)
\(=\cfrac { 2a }{ 2at } =\cfrac { 1 }{ t } \)

\(\left| \begin{matrix} 1 & 3 & 4 \\ 102 & 18 & 36 \\ 17 & 3 & 6 \end{matrix} \right| =6\left| \begin{matrix} 1 & 3 & 4 \\ 17 & 3 & 6 \\ 17 & 3 & 6 \end{matrix} \right| \)
= 0 (since R₂ ≡ R₃)

(i) Variables:
Let x₁, x₂ represents the first and second type of bonds respectively.
(ii) Objective function:
Let Z be the maximum return
\(\therefore \quad Z=\frac { 8 }{ 100 } { x }_{ 1 }+\frac { 10 }{ 100 } { x }_{ 2 } \Rightarrow Z=0.08{ x }_{ 1 }+0.1{ x }_{ 2 }\)
(iii) Constraints: 
\({ x }_{ 1 }+{ x }_{ 2 } \le 70,000\)
\({ x }_{ 1 } \ge 10,000\)
\( { x }_{ 2 } \le 30,000\)
(iv) Non-negative restrictions:
Since the number of first and second type of bonds cannot be negative,x₁, x₂ ≥ 0.
Hence, the mathematical formulation of the LPP is maximize \(Z=0.08{ x }_{ 1 }+0.1{ x }_{ 2 }\)
Subject to the constraints 
\({ x }_{ 1 }+{ x }_{ 2 } \le 70,000\)
\( { x }_{ 1 }\ge 10,000\)
\({ x }_{ 2 }\le 30,000\)
and x₁, x₂ ≥ 0.

S = {(1, 1) (1, 2) (1,3), ..... (6, 6)}
(2, 1), (2, 2) ...... (2, 6)
(6, 1), (6, 2)  ..... (6, 6)}
n(S) = 36
Let B be the event that sum of numbers appearing is 6
B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
\(P(B)=\frac{5}{36}\)
Let A be the event that 4 has appeared atleast once
A = {(2, 4), (4, 2)}
\(A\cap B\)  = {(2, 4),(4, 2)}
\(P(A\cap B)=\frac{2}{36}\)
\(P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { 2/36 }{ 5/36 } =\frac { 2 }{ 5 } \)

Let ‘x’ be the number of shares.
Face Value = Rs. 100
Face value of ‘x’ shares = Rs. 100x
\(\frac{12}{100}\times 100x=Rs. 3600\)
12 x = 3600 \(\Rightarrow\) x = 300
Hence the number of shares = 300

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

Let \(\cos^{-1}(\frac12)=\theta\)
\(\Rightarrow\frac12=\cos\theta\Rightarrow\cos=\frac{\pi}{3}\cos\theta\)
\(\Rightarrow\theta=\frac{\pi}{3}\)
\(\therefore\cos\left[\frac{\pi}{3}-\cos^{-1}(\frac{1}{2})\right]=\cos\left[\frac{\pi}{3}-\frac{\pi}{3}\right]=\cos(0)=1.\)

Let m₁ and m₂ be the slopes of 2x - y + 3 = 0 and x + y + 2 = 0
Now m₁ = 2, m₂ = –1
Let \(\theta \) be the angle between the given lines
tan \(\theta \) =\(\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \)
\(tan\quad \theta =\left| \frac { 2-(-1) }{ 1+2(-1) } \right| =3\)
\(\Rightarrow \theta ={ tan }^{ -1 }(3)\)