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

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

Since there is no common area between the lines x₁ + x₂ ≤ 1 and 5x₁ + 5x₂ ≥ 0

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

\(\cos \left(-480^{\circ}\right) =\cos 480^{\circ}=\cos \left(360^{\circ}+120^{\circ}\right) \)
\(=\cos 120^{\circ}=\cos \left(180^{\circ}-60^{\circ}\right) \)
\(=-\cos 60^{\circ}=-1 / 2 \)

\(a+b=0 \Rightarrow k-2=0\)

\((\sqrt{2})^8=2^4=16\)

nP₂ = 20
n(n - 1) = 5 x 4
n = 5

cov(x, y)=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ { n }^{ 2 } } =\frac { 10(-115)-50(-30) }{ { 10 }^{ 2 } } \)
=\(\frac { -1150+1500 }{ 100 } =\frac { 350 }{ 100 } \)=3.5
Again, co-efficient correlation
r(x,y)=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ \sqrt { n{ \sum { x } }^{ 2 }-{ (\sum { x } ) }^{ 2 } } \sqrt { n{ \sum { y } }^{ 2 }-{ (\sum { y } ) }^{ 2 } } } \)
=\(\frac { 10(-115)-50(-30) }{ \sqrt { 10(290)-{ (50) }^{ 2 } } .\sqrt { 10(300)-{ (-30) }^{ 2 } } } \)
=\(\frac { -1150+1500 }{ \sqrt { 2900-2500 } .\sqrt { 3000-900 } } =\frac { 350 }{ 20\times 10\sqrt { 21 } } \)=0.3819
∴ r(x,y)=0.3819

Since the decision variables are non-negative, the solution lies in the I quadrant of the plane.
Consider the equation 
\({ x }_{ 1 }+{ x }_{ 2 }= 4\)

The feasible region is OAB and its co-ordinates are O(0, 0), A(4, 0) and B(0, 4)

Maximum of Z occurs at B(0, 4)
Hence, the solution is x₁= 0, x₂ = 4 and Z_max = 16

Given a = Rs.1000,r = \(\cfrac { 8 }{ 100 } \times \cfrac { 1 }{ 4 } \) = 0.02,n =3 X 4 = 12
A =\(\cfrac { a }{ i } \left( 1+i \right) \left[ \left( 1+i \right) ^{ n }-1 \right] \)
=\(\cfrac { 1000 }{ 0.02 } \left( 1.02 \right) \left[ \left( 1.02 \right) ^{ 12 }-1 \right] \)
= 51,000[1.269-1]
= 51,000[0.269]=Rs.13,719
(1.02)¹² = 12log(1.02)
=12(0.0086)
= 0.1032
Antilog of 0.1032 is 1.269

Calculation of Harmonic Mean

\(HM=\frac { n }{ \sum { \left( \frac { f }{ X } \right) } } =\frac {120 }{ 5.975 } =20.08\)

Calculation of Geometric Mean

GM = Anti \(\log { \left( \frac { \sum { flogX } }{ N } \right) } \)
= Anti \(\log { \left( \frac { 173.7907 }{ 93 } \right) } \)
= Anti log(1.8687)
GM = 73.91

Slope of the line 3x-y+5=0 is
\(\Rightarrow \quad { m }_{ 2 }=-\frac { co-efficient\quad of\quad x }{ co-efficient\quad of\quad y } =\frac { -3 }{ -1 } =3\)
Let m₁ = m and \(\theta =45\)
\(\therefore \quad tan\theta =\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \)
\(\Rightarrow \quad tan\quad 45=\left| \frac { m-3 }{ 1+3m } \right| \)
\(\Rightarrow \quad 1=\left| \frac { m-3 }{ 1+3m } \right| \)
\(\Rightarrow \quad 1|1+3m|=|m-3|\)
\(\Rightarrow \quad 1+3m=\pm (m-3)\)
\(\Rightarrow \quad 1+3m=m-3\quad or\quad 1+3m=-m+3\)
\(\Rightarrow\)   2m = - 4 or  4m = 2
\(\Rightarrow \)  m = - 2  or  \(m=\frac { 2 }{ 4 } =\frac { 1 }{ 2 }​​​​​​​\)
\(\therefore \)   m =  - 2 or \(\frac { 1 }{ 2 } \)

Y = -x² + 10 x - 15
\(\Rightarrow\) x² - 10x = - y - 15
\(\Rightarrow\) (x- 5)² = -y - 15 + 25
\(\Rightarrow\) (x- 5)² = -y + 10 
\(\Rightarrow\) (x - 5)² = (-y - 10)
The best time to end the project is when x = 5 months.

y = x³e^x
\(\frac{d y}{d x}=x^x\left(e^x\right)+e^x\left(3 x^2\right)=x^2 e^x(x+3)\)

\(2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2 \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}=\frac{3}{2} \)
\(\operatorname{cosec}^2 \frac{7 \pi}{6} =\operatorname{cosec}^2\left(\pi+\frac{\pi}{6}\right) \)
\(=\operatorname{cosec}^2 \frac{\pi}{6}=\operatorname{cosec}^2 30^{\circ}=2 \)
\(\text { LHS }= 2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2 \frac{\pi}{6} \cos ^2 \frac{\pi}{3} \)
\(= 2 \sin ^2 30^{\circ}+2^2\left(\cos ^2 60^{\circ}\right) \)
\(= 2\left(\frac{1}{2}\right)^2+4\left(\frac{1}{2}\right)^2 \)
\(= \frac{2}{4}+\frac{4}{4}=\frac{6}{4}=3 / 2=\mathrm{RHS} \)
Hence proved.

\(\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right] \)
\(A=\left(\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right)\)
\(|A|=1(10-0)=10 \neq 0(\text {expanding along} \ C_1 )\)
\(\therefore \mathrm{A}^{-1} \text { exists }\)
\(A_{11}=\text {Cofactor of } 1=10-0=10 \)
\(A_{12}=\text {Cofactor of } 2=-(0-0)=0 \)
\(A_{13}=\text {Cofactor of } 3=0-0=0 \)
\(A_{21}=\text {Co-factor of } 0=-(10-0)=-10 \)
\(A_{22}=\text {Co-factor of } 2=5-0=5 \)
\(A_{23}=\text {Co-factor of } 4=-(0-0)=0\)
\(A_{31}=\text {Co-factor of } 0=8-6=2 \)
\(A_{32}=\text {Co-factor of } 0=-(4-0)=-4 \)
\(A_{33}=\text {Co-factor of } 5=2-0=2\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \end{array}\right)\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{10}\left(\begin{array}{ccc} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right)\)

\(\left(x-\frac{2}{x^2}\right)^{15}\)
\(t_{r+1}=n C_r x^{n-r} a^r\)
\(=15 C_r(x)^{15-r}\left(\frac{-2}{x^2}\right)^r\)
\(=(-1)^r 15 C_r 2^r x^{15-r}\left(\frac{1}{x^{2 r}}\right)\)
\(=(-1)^r 15 C_r 2^r x^{15-r-2 r}\)
\(=(-1)^r 15 C_r 2^r x^{15-3 r}\)
To find the term independent of x, equate the power of x to 0
15 - 3r = 0
\(3 r=15 \Rightarrow r=\frac{15}{3}=5\)
\(t_{r+1}=(-1)^5 15 C_5 2^5\)
\(=-32\left(15 C_5\right)\)

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

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, \(\sin A=-\sqrt{1-\cos ^2 A}\)
\(=-\sqrt{1-\frac{16}{25}}\)
\(=-\sqrt{\frac{25-16}{25}}\)
\(=-\frac{3}{5}\)
\(\operatorname{Sin} \mathrm{B}=-\sqrt{1-\cos ^2 B}\)
\(=-\sqrt{1-\frac{144}{169}}\)
\(=-\sqrt{\frac{169-144}{169}}\)
\(=-\frac{5}{13}\)
\( \cos (A+B)= \cos A \cos B- \sin A \sin B \)
\(=\frac{4}{5} \times \frac{12}{13}-\left(\frac{-3}{5}\right) \times\left(\frac{-5}{13}\right)\)
\(=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}\)

Given y = - x³ + 3x² +  9x - 27    ...(1)
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=-3x^2 + 6x + 9\)
∴ Slope of the tangent is - 3x² + 6x + 9
Let M = - 3x² + 6x + 9
\({dM\over dx}=-6x+6\)    ..(2)
Slope is maximum when \({dM\over dx}=0\) and \({d^2M\over dx^2}<0\)
\({dM\over dx}=0⇒-6x+6=0⇒ -6x =- 6⇒x=1\)
\({d^2M\over dx^2}=-6<0\)
∴ Slope is maximum when x = 1
∴ Maximum slope= -3(1)² + 6(1) + 9 = -3 + 15 = 12 [From (2)]
when x = 1, y = - (1)³ + 3(1)² + 9(1) - 27
-1+3+9-27=-16    [From (1)]
∴ Maximum slope is 12 and the required point is (1, -16)

Let the manufacturer produce x₁ trunks of first type and x₂ trunks of second type each day.
Let Z be the total profit of the manufacturer.

Thus, the mathematical formulation of the LPP is maximize Z = 30x₁+ 40x₂
Subject to the constraints
3x₁ + 3x₂ ≤ 18,2x₁ + 3x₂ ≤ 14 and x₁ + x₂ ≥ 0
Consider the equations
\(3{ x }_{ 1 }+3{ x }_{ 2 }= 18\) 

\(2{ x }_{ 1 }+3{ x }_{ 2 }=14\)

The feasible region is OABC and its co-ordinates are O(0, 0), A(6, 0), c(0, 14/3) and B is the point of  intersection of 3x₁ + 3x₂ = 18 and 2x₁ + 3x₂ = 14
Verification of B:
\(3{ x }_{ 1 }+3{ x }_{ 2 }=18 ...(1)\)
\( (-)\quad (-)\quad \quad (-)\)
\(2{ x }_{ 1 }+3{ x }_{ 2 }=14...(2)\)
\(-------------\)
x₁ = 4
From 2x₁ + 3x₂ = 14, we get 8 + 3x₂ = 14 ⇒ 3x₂ = 6 ⇒ x₂ = 2
\(\therefore \text {B is} (4,2)\)

Maximum of Z occurs at B( 4, 2)
Hence, the solution is x₁ = 4, x₂ = 2 and Z_max = 200.

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.

\(\bar{X} =\frac{\Sigma X}{N}=\frac{1686}{10}=168.6 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{1690}{10}=169 \)
\(b_{x y} =\frac{N \Sigma d x d y-(\Sigma d x)(\Sigma d x)}{N \Sigma d y^2-(\Sigma d y)^2} \)
\(=\frac{10(248)-0}{10(446)-0}=\frac{248}{446}=0.556 \)
\(b_{y x} =\frac{N \Sigma d x d y-(\Sigma d x)(\Sigma d y)}{N \Sigma d x^2-(\Sigma d x)^2} \)
\(=\frac{2480}{4100-36}=\frac{2480}{4064}=0.6102\)
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

\(p = 4L^{ \frac { 3 }{ 4 } }K^{ \frac { 1 }{ 4 } }\) is a homogeneous function of degree 1.
Marginal productivity of labour is
\(\frac { \partial P }{ \partial L } \) = \(4 \times { \frac { 3 }{ 4 } }L^{ \frac { -1 }{ 4 } } K^{ \frac { 1 }{ 4 } }\) = 3 \(\left( \frac { K }{ L } \right) ^{ \frac { 1 }{ 4 } }\)
Marginal productivity of capital is
\(\frac { \partial P }{ \partial L } \) = \(4L^{ \frac { 3 }{ 4 } }\times \frac { 1 }{ 4 } K^{ \frac { -3 }{ 4 } }\) = \(\left( \frac { L }{ K } \right) ^{ \frac { 3 }{ 4 } }\)
\(L\frac { \partial P }{ \partial L } +K\frac { \partial P }{ \partial K } =3L\left( \frac { K }{ L } \right) ^{ \frac { 1 }{ 4 } }+k\left( \frac { L }{ K } \right) ^{ \frac { 3 }{ 4 } }\)
\(= 3L^{ \frac { 3 }{ 4 } }K^{ \frac { 1 }{ 4 } }+L^{ \frac { 3 }{ 4 } }K^{ \frac { 1 }{ 4 } }\)
\(= 4L^{ \frac { 3 }{ 4 } }K^{ \frac { 1 }{ 4 } }\) = P
Hence Euler’s theorem is verified.

Given x = 106 - 2p; 2p = 106 - x
P = \(\frac { 106-x }{ 2 } \)
(R) = px = \(\left( \frac { 106-x }{ 2 } \right) = 56x -\frac { { x }{ 2 } }{ 2 } \)
\(A C=\frac{C}{x} =5+\frac{x}{50} \)
\(C =5 x+\frac{x^2}{50} \)
\(p=R-C =53 x-\frac{x^2}{2}-5 x-\frac{x^2}{50} \)
\(=48 x-\frac{26 x^2}{50} \)
\(\frac{d p}{d x} =48-\frac{52 x}{50} \)
\(\text {For maximum profit } \frac{d P}{d x} =0 \)
\(48-\frac{52 x}{50} =0 \)
\(x =\frac{48 \times 50}{52} \approx 46 \)
\(\frac{d^2 p}{d x^2} =-\frac{52}{50}<0\)
\(\therefore\)  At x  = 46, Profit is maximum
Maximum profit  = 48(46) - \(\frac { 26\left( 46 \right) ^{ 2 } }{ 50 } \) [From (1)]
= 2208 - 1100.32
= Rs.1107.68
P is maximum when x ≈ 46, maximum  profit =  Rs.1107.68

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

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,
\({{x^2+x+1}\over{x^2+2x+1}}=1-{{1}\over{x+1}}+{{1}\over{{(x+1)}^{2}}}\)

y = \({ \left( x+\sqrt { 1+{ x }^{ 2 } } \right) }^{ m }\)  
\(\frac{d y}{d x}=m\left(x+\sqrt{1+x^2}\right)^{m-1}\left(1+\frac{1}{2 \sqrt{1+x^2}}(2 x)\right) \)
\(y_1=m\left(x+\sqrt{1+x^2}\right)^{m-1}\left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right) \)
\(\sqrt { 1+{ x }^{ 2 } } y_{ 1 } = { m{ \left( x+\sqrt { 1+{ x }^{ 2 } } \right) }^{ m } }\)
\(\Rightarrow \sqrt { 1+{ x }^{ 2 } } y_{ 1 } =my\)
Squaring both sides
\(\left(1+x^2\right) y_1^2=m^2 y^2\)
\(\left(1+x^2\right) 2 y_1 y_2+y_1^2(2 x)=2 m^2 y y_1\)
\(\div 2 y_1 \ \left(1+x^2\right) y_2+x y_1-m^2 y=0\)

Let P (n) denote the statement. 4 + 8 + 12 + ...... + 4n = 2n(n + 1)
Put n = 1
LHS = 4
RHS ⇒ 2(2) = 4
LHS = RHS
∴ P(1) is true.
Let us assume that P(k) is true.
p(k) : 4 + 8 + 12 + ..... + 4 = 2k(k + 1)
To prove that P(k + 1) is true
4 + 8 + .... + 4k + 4(k + 1) = 2(k + 1) = 2(k + 1)(k + 2)
P(k) + 4(k + 1)
= 2k(k + 1) + 4(k + 1)
= (k + 1)(2k + 4) [using (1)]
= 2(k + 1)(k + 2) = RHS
∴ P (k + 1) is true whenever P(k) is true.
∴ p(n) is true for all \(n\in N\).

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

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

nP _r= 360 = 36 \(\times\) 10
= 3 \(\times\) 3 \(\times\) 4 \(\times\) 5 \(\times\) 2
= 6 \(\times\) 5 \(\times\) 4 \(\times\) 3 = 6P₄
Therefore n = 6 and r = 4

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

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

Since A, B, C, D are angles of a quadrilateral, A + B + C + D =\(2\pi\)
A + B + C + D =\(2\pi\)
\(\Rightarrow A+B=2\pi-(C+D)\)
\(\Rightarrow\sin(A+B)=\sin[2\pi-(C+D)]\)
=-sin(C+D)[\(\therefore2\pi-(C+D)\)is in the IV quadrant]
\(\Rightarrow\sin(A+B)+\sin(C+D)=0\)

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 \(\theta \) =\(\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \)
\(tan\quad \theta =\left| \frac { 2-(-1) }{ 1+2(-1) } \right| =3\)
\(\Rightarrow \theta ={ tan }^{ -1 }(3)\)