Class	Mid value x	f	c.f.	D=|x-28|	f|D|
0-10	5	5	5	23	115
10-20	15	8	13	13	104
20-30	25	15	28	3	45
30-40	35	16	44	7	112
40-50	45	6	50	17	102
			N = 50		\(\sum { f|D|=478 } \)

\(\frac { N }{ 2 } =\frac { 50 }{ 2 } =25\)
\(\therefore\) Median lies in the interval (20, -30) and its corresponding values are
L = 20, f = 15, pcf = 13 and c = 10
\(\therefore Median=L+\left( \frac { \frac { N }{ 2 } -pcf }{ f } \right) \times c=20+\frac { 25-13 }{ 15 } \times 10\)
\(=20+\frac { 120 }{ 15 } =20+8=28\)
Now, mean deviation from median
\(=\frac { \sum { f|D| } }{ N } =\frac { 478 }{ 50 } =9.56\)

Given b_xy =2 and b_yx=0.45
We know r= \(\sqrt { { b }_{ xy }.{ b }_{ yx } } =\sqrt { 2(0.45) } =\sqrt { 0.9 } \)=0.949
\(\therefore\)r=0.949

Here A = 1,67,160, n = 16 i = \(\frac{15}{100}=0.15\) a = ?
To find: a
Now A = \(\frac{a}{i}[(1+i)^n-1]\)
1,67,160 = \(\frac{a}{0.15}[(1+0.15)^{16}-1]\)
= \(\frac{a}{15}[(1.15)^{16}-1]\)
⇒ a = \(\frac{1,67,160\times 0.15}{(1.15)^{16}-1}\)
a = \(\frac{1,67,160\times 0.15}{9.358-1}\)
= \(\frac{1,67,160\times 0.15}{8.358}\)
= 3,000
Therefore a = Rs.3,000

Demand : R = 1000 per month
Setup cost : C₃ = Rs.200 per order
Carrying cost: C₁= Rs. 20 per item per month

Run size q	Set up cost \(\frac { q }{ q } \times { C }_{ 3 }\)	Carrying cost \(\frac { q }{ 2 } \times { C }_{ 1 }\)	Total cost (Set up cost + Carrying cost)
500	\(\frac { 1000 }{ 500 } \times 200=400\)	\(\frac { 500 }{ 2 } \times 20=5000\)	5400
600	\(\frac { 1000 }{ 600 } \times 200=333.3\)	\(\frac { 600 }{ 2 } \times20= 6000\)	6333.3
700	\(\frac { 1000 }{ 700 } \times 200=285.7\)	\(\frac { 700 }{ 20 } \times20= 7000\)	7285.7
800	\(\frac { 1000 }{ 800 } \times 200=250\)	\(\frac { 800 }{ 2 } \times20= 8000\)	8250

EOQ = \(\sqrt { \frac { 2{ RC }_{ 3 } }{ { C }_{ 1 } } } =\sqrt { \frac { 2\times 1000\times 200 }{ 20 } } \)
= \(\sqrt { 20000 } \) = 141 units (app)

(i) variables: Let x₁ and x₂ represents the types of pen A and B respectively.
(ii) Objective function:
Profit on x₁ pens = 5x₁
Profit on x₂ pens = 3x₂
Total profit = 5x₁ + 3x₂
Let Z = 5x₁ + 3x₂ which is the objective function.
Since the total is to be maximized, we have to maximize Z = 5x₁ + 3x₂
(iii) Constraints:
The supply of raw material is sufficient only for 1000 pens per day.
2x₁ + x₂ ≤ 1000 [\(\because\) pen A is twice as that of pen B]
clips are available per day, for pen A: x₁ ≤ 400
clips are available per day, for pen B: x₂ ≤ 700
(iv) Non-negative restrictions: Since the number of pen A and pen B cannot be negative, we have x₁ ≥ 0, x₂ ≥ 0.
Thus, the mathematical formulation of the LPP is
Max Z = 5x₁ + 3x₂
Subject to the constraints
2x₁ + x₂ ≤ 1000
x₁ ≤ 400
x₂ ≤ 700
x₁, x₂ ≥ 0 (non-negative constraints)

x + y = 6 .....(1)  
x + 2y = 4  .....(2)
On solving (1) and (2) we get the centre of the circle
C(-h, - k) = (8, - 2)
Equation of circle is \((x-h)^2+(y-k)^2=r^2\)
\((x-8)^2+(y+2)^2=r^2\)
It passes through (2, 6)
\((2-8)^2+(6+2)^2=r^2\)
\(36+64=r^2 \Rightarrow r^2=100\)
Required equation is \((x-8)^2+(y+2)^2=100\)
\(x^2-16 x+64+y^2+4 y+4=100\)
\(x^2+y^2-16 x+4 y-32=0\)

\(A B=\left(\begin{array}{cc} 2 & 3 \\ 1 & -6 \end{array}\right)\left(\begin{array}{cc} -1 & 4 \\ 1 & -2 \end{array}\right)\)
\(=\left(\begin{array}{cc} -2+3 & 8-6 \\ -1-6 & 4+12 \end{array}\right)=\left(\begin{array}{cc} 1 & 2 \\ -7 & 16 \end{array}\right)\)
\(\mathrm{LHS}=\operatorname{adj}(\mathrm{AB})=\left(\begin{array}{cc} 16 & -2 \\ 7 & 1 \end{array}\right)\)
\(\mathrm{RHS}=(\operatorname{adj} B)(\operatorname{adj} \mathrm{A})\)
\(=\left(\begin{array}{cc} -2 & -1 \\ -1 & -1 \end{array}\right)\left(\begin{array}{cc} -6 & -3 \\ -1 & 2 \end{array}\right)\)
\(=\left(\begin{array}{cc} 12+4 & 6-5 \\ 6+1 & 3-2 \end{array}\right)=\left(\begin{array}{cc} 16 & -2 \\ 7 & 1 \end{array}\right)\)
\(a d j(A B)=(a d j\ B)(a d j\ A)\)

x = a (\(\theta\) - sin \(\theta\)), y = a (1 - cos \(\theta\))
Differentiating with respect to 'x' we get,
\(\frac { dx }{ d\theta } =a\left( 1-\cos { \theta } \right) \)
\(=a\left( 2\sin ^{ 2 }{ \frac { \theta }{ 2 } } \right) \)
Now,    \(\frac { dy }{ dx } =\frac { \frac { dy }{ d\theta } }{ \frac { dx }{ d\theta } } \)
\(=\frac { 2a\sin { \frac { \theta }{ 2 } } \cos { \frac { \theta }{ 2 } } }{ 2a\sin ^{ 2 }{ \frac { \theta }{ 2 } } } =\frac { \cos { \frac { \theta }{ 2 } } }{ \sin { \frac { \theta }{ 2 } } } =\cot { \left( \frac { \theta }{ 2 } \right) } \)

Given equations are 2x + 5y = 1; 3x + 2y = 7
This system of equations can be written in matrix form as \(\begin{pmatrix}2 & 5 \\3 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} 1\\7 \end{pmatrix}\Rightarrow Ax=B\)
where A = \(\begin{pmatrix} 2 & 5 \\3 & 2 \end{pmatrix},X=\begin{pmatrix} x\\y \end{pmatrix}\) and B = \(\begin{pmatrix} 1 \\ 7 \end{pmatrix}\)
\(\therefore\)  X = A^-1 B.
\(|A|=\begin{vmatrix} 2 &5 \\3 & 2 \end{vmatrix}=4-15=-11\)
A₁₁ = 2, A₁₂ = -3, A₂₁ = -5, A₂₂ = 2
\(\therefore\)   adj A = \(\begin{bmatrix} 2 & -3 \\ -5 & 2 \end{bmatrix}^{T}=\begin{bmatrix} 2 & -5 \\-3 & 2 \end{bmatrix}\)
\(\therefore\)  \({A}^{-1}={{1}\over{|A|}}\) . adj A = \({{-1}\over{11}}\begin{bmatrix} 2 & -5\\ -3& 2 \end{bmatrix}\)
X = A^{- 1} B \(={{-1}\over{11}}\begin{bmatrix} 2 & -5\\ -3&2 \end{bmatrix}\begin{bmatrix} 1\\7 \end{bmatrix}\)
\(=\frac { 1 }{ 11 } \left[ \begin{matrix} +2-35 \\ -3+14 \end{matrix} \right] ={{-1}\over{11}}\begin{bmatrix} -33\\11 \end{bmatrix}=\begin{bmatrix} 3\\-1 \end{bmatrix}\)
\(\therefore\)  x = 3.and y = -1.

(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

\(P=\frac{a}{i}=\frac{5000}{0.1}=50,000\)

If F.V = 100, Investment = 90
FV 10,000,
Investment \(= \frac{ 90 \times 10,000}{100} = 9000\)

\(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)=4 / 5\)

\(\frac{3 \tan 10^{\circ}-\tan ^3 10^{\circ}}{1-3 \tan ^2 10^{\circ}} =\tan 3(10) =\tan 30^{\circ}=\frac{1}{\sqrt{3}} \)

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

\(n P_r =720(\mathrm{nCr}) \)
\(\frac{n !}{(n-r) !} =720\left(\frac{n !}{r !(n-r) !}\right) \)
\(r ! =720 \)
\(r ! =6 ! \)
r = 6

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

(Corresponding co-factor)

(Since \(\left|A^{-1}\right|=\frac{1}{|A|}\))

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

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

Let the events E₁, E₂ and A be defined as
E₁ - event that male subscriber is under 30
E₂ - event that male subscriber is above 30
A is event that subscriber is male
P(E₁) = 0.80 and P(E₂) = 1 - 0.8 = 0.2
P(B/E₁) = 0.20, P(B/E₂) = 0.30
\(P({ E }_{ 1 }/B)=\frac { P({ E }_{ 1 }).P\left( \frac { B }{ { E }_{ 1 } } \right) }{ P({ E }_{ 1 }).P\left( \frac { B }{ { E }_{ 1 } } \right) +P({ E }_{ 2 }).P\left( \frac { B }{ { E }_{ 2 } } \right) } \)
\(=\frac { 0.8\times 0.2 }{ 0.8\times 0.2+0.2\times 0.3 } \)
\(=\frac { 0.16 }{ 0.16+0.06 } =\frac { 16 }{ 22 } =0.727\)

Let x be the amount invested in each stock
Income on 10% stock at 89:
F.V. = Rs. 100, M.V. = Rs. 90
Number of shares = \(\frac{Investments}{M.V}\) = \(\frac{x}{90}\)
Annual income = \(\frac{x}{90}\times \frac{10}{100}\times 100\)
= \(\frac{x}{9}\) .......(1)
Income on 7% stock at 90:
M.V. = Rs. 91
Number of shares = \(\frac{Investments}{M.V}\) = \(\frac{x}{91}\)
Annual income = \(\frac{x}{91}\times \frac{7}{100}\times 100\)
= \(\frac{x}{13}\) .....(2)
Given: \(\frac{x}{9}-\frac{x}{13}=100\) ⇒ x = Rs. 2925
\(\therefore\) The amount invested in each stock ≈ Rs. 2,925

Given
Ordering cost per order : C₃ = Rs. 150 per order.
Number of units per order: q = 500 units
Annual demand = 500 × 4 = 2000 units
∴ Demand rate : R = 2000 per year
Carrying cost : C₁ = 20% of unit value
C₁ = \(\frac { 20 }{ 100 } \times 125=Rs25\)
(i) Total annual cost of due existing inventory policy
= \(\frac { R }{ q } \times { C }_{ 3 }+\frac { q }{ 2 } { C }_{ 1 }\) = \(\frac { 2000 }{ 500 } \times 150+\frac { 500 }{ 2 } \times 25\)
= Rs. 6850
(ii) EOQ =  \(\sqrt { \frac { 2Rc_{ 3 } }{ { c }_{ 1 } } } \)
= \(\sqrt { \frac { 2\times 2000\times 150 }{ 25 } } \)
= \(\sqrt { 12\times 2000 } \)
= 155 units (app.)
(iii) Minimum annual cost \(= \sqrt { { 2Rc_{ 3 } }{ { c }_{ 1 } } } \)
= \(\sqrt { 2\times 2000\times 150\times 25 } \)
= Rs. 3873.
By applying the economic order quantity, money saved by a company = 6850 – 3873  = Rs. 2977.

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.

Taking vertex of the parabola as reflector at origin, x-axis along the axis of the parabola, equation of the parabola is y² = 4ax
Given depth = 5 cm, diameter = 20 cm
\(\therefore\) (5, 10) lies on the parabola
\(\therefore\) 10² = 4a(5) ⇒ 100 = 20a ⇒ a = \(\frac{100}{20}\) = 5

\(\therefore\) Focus is (a, 0) = (5, 0) which is the mid-point of the given diameter

Since the parabola is symmetric about X-axis and has its vertex at (0,0), its equation will be of the form y² = 4ax or y² = -4ax.
But the parabola passes through (2, 3) which is in the I quadrant, its equation will be of the form y² = 4ax, which is open rightward.
Substituting (2,3) in y² = 4ax, we get
9 = 4a(2) ⇒ 8a = 9
⇒ a = \(\frac{9}{8}\)
\(\therefore\) Equation of the parabola is y² = 4(\(\frac{9}{8}\))x
⇒ y² = \(\frac{9}{2}\)x
⇒ 2y² = 9x ⇒ 2y² - 9x = 0

Let x represent the number of units and y its cost.
By the given data,
x₁(6000) y₁(33,000)
x₂(8000) y₂(40,000)
Using two point form, \(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
⇒ \(\frac { y-33000 }{ 40,000-33000 } =\frac { x-6000 }{ 8000-6000 } \)
⇒ \(\frac { y-33000 }{ 7000 } =\frac { x-6000 }{ 2000 } \)
⇒ \(\frac { y-33000 }{ 7 } =\frac { x-6000 }{ 2 } \)
⇒ 2y - 66000 = 7x - 42000
⇒ 2y = 7x - 42000 + 66000
2y = 7x + 24000, which is the required linear function.

Let y = (sec x -1) (sec x +1)
y = sec² x -1    \(\left[ \therefore (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(\therefore\)y =tan²x
Differentiating with respect to 'x' we get 
\(\frac { dy }{ dx } =2tanx\frac { d }{ dx } (tanx)\) \([\therefore 1+tan^{ 2 }x=sec^{ 2 }x]\)
=2tanx.sec² x

Let y = \(\sqrt { \frac { (x-1)(x-2) }{ (x-3)({ x }^{ 2 }+x+1) } }\)
log y = \(\frac { 1 }{ 2 } \)[log (x - 1) + log (x - 2) -log (x - 3) -log (x² + x + 1]
\( \frac{1}{y} \frac{d y}{d x}=\frac{1}{2}\left[\begin{array}{l} \left.\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}\right] \\ -\frac{1}{x^2+x+1}(2 x+1) \end{array}\right]\)
⇒ \(\frac { dy }{ dx } =\frac { 1 }{ 2 } \sqrt { \frac { (x-1)(x-2) }{ (x-3)(x^{ 2 }+x+1) } } \left[ \frac { 1 }{ x-1 } +\frac { 1 }{ x-2 } -\frac { 1 }{ x-3 } -\frac { (2x+1) }{ { x }^{ 2 }+x+1 } \right] \)

(i) \(\frac { \sin { \left( 180-\theta \right) } \cos { \left( 90+\theta \right) } \tan { \left( 270-\theta \right) \cot { \left( 360-\theta \right) } } }{ \sin { \left( 360-\theta \right) \cot { \left( 360+\theta \right) \sin { \left( 270-\theta \right) \csc { \left( -\theta \right) } } } } } =-1\)
\(=\frac{(\sin \theta)(-\sin \theta)(\cot \theta)(-\cot \theta)}{(-\sin \theta)(\cos \theta)(-\cos \theta)(-\operatorname{cosec} \theta)}\)
\(=\frac{\sin \theta\left(-\cot ^2 \theta\right)}{\cos ^2 \theta(\operatorname{cosec} \theta)}\)
\(=\frac{\sin \theta\left(\frac{-\cos ^2 \theta}{\sin ^2 \theta}\right)}{\cos ^2 \theta\left(\frac{1}{\sin \theta}\right)}\)​​​
= -1 = RHS. 
Hence proved.

\(|A|=\left| \begin{matrix}1 & tan\quad x \\ -tan\quad x & 1 \end{matrix} \right| =1+{ tan }^{ 2 }x={ sec }^{ 2 }x\neq 0\)
\(\Rightarrow\) 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)

\(\frac{1}{\left(x^2+4\right)(x+1)}=\frac{A}{x+1}+\frac{B x+C}{x^2+4}\)
(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
\(1=\frac{4}{5}+C\)
\(C=1-\frac{4}{5}=\frac{1}{5}\)
\(\frac{1}{\left(x^2+4\right)(x+1)}=\frac{1}{5(x+1)}+\frac{\frac{-1}{5} x+\frac{1}{5}}{x^2+4}\)
\(=\frac{1}{5(x+1)}+\frac{1-x}{5\left(x^2+4\right)}\).

Equation of circle is
\(\left( x-h \right) ^{ 2 }+\left( y-k \right) ^{ 2 }={ r }^{ 2 }\) 
Here \(\left( h,k \right) =(3,-1)\) and r = 4
Equation of circle is
\(\left( x-3 \right) ^{ 2 }+\left( y+1 \right) ^{ 2 }=16\)
\({ x }^{ 2 }-6x+9+{ y }^{ 2 }+2y+1=16\)
\({ x }^{ 2 }+{ y }^{ 2 }-6x+2y-6=0\)

Let A = \(\left[ \begin{matrix} 8 & 2 \\ 4 & 3 \end{matrix} \right] \)
|A| = \(\left| \begin{matrix} 8 & 2 \\ 4 & 3 \end{matrix} \right| \)
= 24 – 8 = 16 ≠ 0
\(\therefore\) A is a non-singular matrix

Since we are given rate per rupee, harmonic mean will give the correct answer.
\(HM=\frac { n }{ \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\frac { 1 }{ d } } \)
\(=\frac { 4 }{ \frac { 1 }{ 1 } +\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 4 } } \)
\(=\frac { 4\times 12 }{ 25 } \)
= 1.92 kg per rupee.

Here a = 64,000, n = 12 and i = \(\frac{10}{100}=0.1\)
Amount of Ordinary annuity (A) = \(\frac{a}{i}[(1+i)^n-1]\)
= \(\frac{64000}{0.1}[(1+0.1)^{12}-1]\)
= 6,40,000 [(1.1)¹² – 1]
= 6,40,000[3.3184 – 1]
= 6,40,000 [2.3184]
= 64 x 23184
∴ A  = Rs. 14,83,776.

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

Let x be the number of passengers.
\(P=136 -\frac{40}{100}(x-100) \text { if } x \geq 100 \)
\(R=p x =136 x-\frac{2 x^2}{5}+40 x \)
\(=176 x-\frac{2 x^2}{5} \)
\(\frac{d R}{d x} =176-\frac{4 x}{5}\)
For max. revenue \(\frac{d R}{d x}=0\)
\(176 =\frac{4 x}{5} \)
\(x =\frac{176 \times 5}{4}=220 \)
\(\frac{d^2 R}{d x^2} =-\frac{4}{5}<0\)
∴ At  x = 220,
Revenue is maximum.

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

\(\underset { x\rightarrow \frac { 1 }{ 2 } }{ lim } \frac { { 4x }^{ 2 }-1 }{ 2x-1 } =\underset { x\rightarrow \frac { 1 }{ 2 } }{ lim } \frac { \left( { 2x }^{ 2 } \right) -1^{ 2 } }{ 2x-1 } =\underset { x\rightarrow \frac { \pi}{ 2 } }{ lim } \frac { (2x+1)(2x-1) }{ 2x-1 } =\underset { x\rightarrow \frac { \pi }{ 2 } }{ lim } (2x+1)\)
= \(2\left( \frac { 1 }{ 2 } \right) +1=1+1=2\)

Since \(\pi , x lies in III quadrant.
\(\therefore\) sin x is negative and tan x is positive
\(\therefore\sin x=\pm\sqrt{1-\cos^2x}=-\sqrt{1-\frac{1}{4}}=-\frac{\sqrt3}{2}\)
\(\therefore cosec\ x=-\frac{2}{\sqrt{3}}\)
\(\tan x=\frac{\sin x}{\cos x}=\frac{-\frac{\sqrt3}{2}}{-\frac{1}{2}}=\sqrt3\)
Hence, \(4\tan^2x-3cosec^2x=4\times3-3\times\frac43=12-4=8\)

We have P(n): 2³ⁿ-1 is a multiple of 7
 ∴  P (5) : 2¹⁵ -1 = 32767
= 7 (4681) which is multiple of 7
\(\therefore\) P (5) is true