Mean deviation about median = co-eficient of mean deviation about median x Median
= 0.25 x 45
= 11.25

GM = \(\sqrt{8 \times 18} = \sqrt{144} = 12\)

Investment = 80, Income = 7
Investment = 100,
Income \(= \frac{ 7 \times 100}{80} = 8.75 \%\)

If FV 100, Investment = 120
FV = 20,000,
Investment = \(\frac{120\times 20,000}{100} \) = Rs. 24000

R = px = 20x- 3x²
M.R = dR/dx = 20 - 6x

\(p =\frac{\tan 50^{\circ}}{\sec 50^{\circ}}=\frac{\sin 50^{\circ} / \cos 50^{\circ}}{1 / \cos 50^{\circ}} =\sin 50^{\circ}\)

\(\frac{1}{2}\left(2 \sin 15^{\circ} \cos 15^{\circ}\right) =\frac{1}{2} \sin 2\left(15^{\circ}\right)=\frac{1}{2} \sin 30^{\circ} =\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)

4a = 5

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

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

5x - 16 = 0
5x = 16
\(x=\frac{16}{5}\)

Given R = - 90 + 6x² - x³
For Revenue Function: \({dR\over dx}=12x-3x^2\)
\({dR\over dx}=0⇒12x-3x^2=0\)
⇒ 4x-x² = 0 ⇒ x(4 - x) = 0

The possible intervals are (-∞,0), (0, 4) and (4, ∞)

∴ Revenue function is increasing in (0, 4) and decreasing in (-∞, 0) and (4, ∞).
For Marginal Revenue Function:
\(MR={dR\over dx}={d\over dx}(-90+6x^2-3x^3)=12x-3x^2\)
Let y = 12x - 3x²
\({dy\over dx}=0⇒12-6x=0⇒x=2\)

The possible intervals are (-∞, 2) and (2, ∞).

 Hence, Marginal Revenue function is increasing in (-∞, 2) and decreasing in (2, ∞).

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

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.

Shares of tea company
No.of shares = 2000
FV = Rs.10
MV = Rs.33
Rate of dividend = 25%
S.P of a share = Rs.33
S.P of 2000 shares = 2000 x 33 = Rs. 66,000
Shares of cotton textiles
Investment = Rs. 66,000
FV = Rs. 25
MV = Rs. 44
Dividend =15%
(i) Number of cotton textile shares
= \(\frac {\text{ Investment} }{ M.V } \)
=  \(\frac { 66,000 }{ 44 } =1500\)
(ii) Income from tea company shares
= 2000 x 10 x \(\frac { 25 }{ 100 } \) = Rs. 5000
Income from cotton textiles shares
= 1 500 x 25 x \(\frac { 15 }{ 100 } \) = Rs. 5625
Change in his dividend income = 5625 - 5000 = Rs. 625

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

First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant write all the inequalities of the constraints in the form of equations.
∴ We have the lines \(x_1+x_2 \leq 50; 3 x_1+x_2 \leq 90\)
x₁ + x₂ = 50 is a line passing through the points (0,50) and (50,0).
[(0,50) is obtained by taking x₁ = 0 in x₁ + x₂ = 50, (50,0) is obtained by taking x₂ = 0 in \(\left.x_1+x_2=50\right]\)
Any point lying on or below the line x₁ + x₂ = 50. Satisfies the constraint \(x_1+x_2 \leq 50\)
We follow the same steps for the following
x₁ + x₂ = 50

\({ 3x }_{ 1 }+{ x }_{ 2 }=90\)

Now we draw the graph

The feasible region satisfying all the conditions is OABC. The co-ordinates of the points are O(0,0), A(30, 0), B(20,30), C(0,50).

Optimal solution is at A(30,0)
x₁ = 30, x₂ = 0 and Z_max = 120.
Verification
\(3 x_1+x_2 =90 \)
\(x_1+x_2 =50 \)
\(2 x_1 =40 \)
\(x_1 =20 x_2 =50-20=30 \)
B(20,30)

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.

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.

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

Given \(f(x)=\left\{ \begin{matrix} 6ax+3b & if\quad x>1 \\ ax-2b & if\quad x<1 \\ 15 & if\quad x=1 \end{matrix} \right\} \)
\(L[{ f(x)] }_{ x=1 }=\lim _{ x\rightarrow { 1 }^{ - } }{ f(x) =\lim _{ h\rightarrow 0 }{ f(1-h)=\lim _{ h\rightarrow 0 }{ a(1-h)-2b\quad [\because f(x)=ax-2b\quad if\quad x<1] } } } \)
\(=\lim _{ h\rightarrow 0 }{ a-ah-2b=a-2b } \) .....(1)
R[f(x)]_x=1=\(\lim _{ x\rightarrow { 1 }^{ + } }{ f(x)=\lim _{ h\rightarrow 0 }{ f(1+h) } } \)
\(=\lim _{ x\rightarrow { 1 }^{ + } }{ 6a(1+h)+3b\quad [\because f(x)=6ax+3b\quad if\quad x>1] } \)
= 6(91+0) + 3b = 6a + 3b   ....(2)
Also, f(1) = 15 ....(3)
Since f(X) is continuous at x =1,
L[f(x)]_x=1=R[f(x)]_x=1=f(1)
\(\Rightarrow\) a - 2b = 6a + 3b = 15 [using (1), (2) and (3)]
\(\Rightarrow\) a - 2b = 15 ....(4)
and 6a+3b=15 \(\Rightarrow\) 2a + b =5  ..(5)
(4) \(\times\) 2 \(\rightarrow \) 2a - 4b = 30
       -             -      -
(5)        \(\rightarrow \)  2a + b = 5
____________________
              -5b = 25  \(\Rightarrow\) b = -5
Substituting b = -5 in (4) we get,
a - 2(-5) = 15
\(\Rightarrow\) a + 10 = 15
\(\Rightarrow\) a = 15 - 10 = 5
\(\therefore\)    a = 5 and b = -5.   

\(L H L=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} 1-x=0\)
\(R H L=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(1-x)(2-x)=0\)
LHL = RHL
\(\therefore\) f is continuous at x = 1
\(L H D=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{1-h-1}\)
\(=\lim _{h \rightarrow 0} \frac{(1-1-h)(2-1-h)}{h}\)
\(R H D=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{1+h-1}\)
\(=\lim _{h \rightarrow 0} \frac{-h(1-h)}{h}=-1\)
LHD = RHD
\(\therefore\) f is differentiable at x = 1
\(L H L=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}(1-x)(2-x)=0\)
\(R H L=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2} 3-2=1\)
\(L H L \neq R H L\)
\(\therefore\) f is not continuous at x = 2 and hence not differentiable at x = 2

Let the number of radios of type P be x and the radios of type Q be y.
Given 2x + 3y = 130 and 3x + 4y = 180
\(\Rightarrow \left( \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 130 \\ 180 \end{matrix} \right) \)
AX = B where A = \(\left( \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right) ,X=\left( \begin{matrix} x \\ y \end{matrix} \right) and\quad B=\left( \begin{matrix} 130 \\ 180 \end{matrix} \right) \)
|A|=\(\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| =8-9=-1\neq 0\Rightarrow { A }^{ -1 }\) existsadj A=\(\left( \begin{matrix} 4 & -3 \\ -3 & 2 \end{matrix} \right) \)    [\(\because\) A₁₁ = 4, A₁₂ = -3 A₂₁ = -3, A₂₂ = 2]
\(\therefore \ { A }^{ -1 }=\frac { 1 }{ |A| } adj\quad A=\frac { 1 }{ -1 } \left( \begin{matrix} 4 & -3 \\ -3 & 2 \end{matrix} \right) =\left( \begin{matrix} -4 & 3 \\ 3 & -2 \end{matrix} \right) \)
\(\therefore \ X={ A }^{ -1 }B=\left( \begin{matrix} -4 & 3 \\ 3 & -2 \end{matrix} \right) \left( \begin{matrix} 130 \\ 180 \end{matrix} \right) =\left( \begin{matrix} -520 & +540 \\ 390 & -360 \end{matrix} \right) =\left( \begin{matrix} 20 \\ 30 \end{matrix} \right) \)
\(\therefore\) Number of radios of Type P is 20.
Number of radios of Type Q is 30.

(1 + x)²ⁿ
2n is even
Middle term is \(t_{\frac{n}{2}+1}=t_{\frac{2 n}{2}+1}=t_{n+1}\)
\(r=n\)
\(t_{r+1}=n C_r x^{n-r} a^r\)
\(t_{n+1}=2 n C_n(1)^{2 n-n} x^n\)
\(=2 n C_n \cdot x^n\)
\(n C_r=\frac{n !}{r !(n-r) !}\)
\(=\frac{(2 n) !}{n !(2 n-n) !} x^n\)
\(=\frac{(2 n)(2 n-1)(2 n-2)(2 n-3) \ldots 5 \cdot 4 \cdot 3 \cdot 2.1}{n ! n !} x^n\)
\(=\frac{(2 n)(2 n-2) \ldots 4.2(2 n-1)(2 n-3) \ldots 5.3 .1}{n ! n !} x^n\)
\(=\frac{2^n(n(n-1) \ldots 2.1)(2 n-1)(2 n-3) \ldots 5.3 .1}{n ! n !} x^n\)
\(=\frac{2^n \cdot n !(2 n-1)(2 n-3) \ldots 5 \cdot 3 \cdot 1}{n ! n !} x^n\)
\(=\frac{1.3 .5 \ldots(2 n-3)(2 n-1) 2^n x^n}{n !}\)

\(A=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
\(|A|=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
=16 ≠ 0
Since A is a nonsingular matrix, A ^-1 exists
Now  adj \(A=\left[ \begin{matrix} 2 & -4 \\3 & 2 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } adjA\)
\(=\frac { 1 }{ 16 } \left[ \begin{matrix} 2 & -4 \\ 3 & 2 \end{matrix} \right] \)

Given data in ascending order are
21, 27, 28, 30, 31, 32, 34, 36, 37, 39, 45, 46, 57 and n = 13
Q₂ = size of \({ 2\left( \frac { N+1 }{ 4 } \right) }^{ th }\) value = Size of \(2{ \left( \frac { 13+1 }{ 4 } \right) }^{ th }\) value
= Size of 7^th value
Q₂ = 34 

The regression co-efficient of X on Y is
b_xy= \(\frac { N\sum { XY-(\sum { X)(\sum { Y) } } } }{ N.\sum { { Y }^{ 2 }-(\sum { { Y) }^{ 2 } } } } =\frac { 5(300)-(20)(40) }{ 5(345)-({ 40) }^{ 2 } } \)=5.6
Also, \(\bar {X}\) =\(\frac { \sum { X } }{ N } =\frac { 20 }{ 5 } \)=4
and \(\bar {Y}\)=\(\frac { \sum { Y } }{ N } =\frac { 40 }{ 5 } \)=8
\(\therefore\)The regression equation of X on Y is
X- \(\bar {X}\)=b_xy(Y-\(\bar {Y}\) )
X-4=5.6(Y-8)
X-4=5.6Y-44.8
X=5.6Y-40.8
When Y=6, X=5.6(6)-40.8=33.6-40.8
\(\therefore\)X=-7.2

Let Investment in each stock be 120 x 135
At 7% stock
Investment = 120, Income = 7
Investment = 120 x 135
Income = \(\cfrac { 7\times120\times135 }{ 120 }=Rs.945\)
At 8% stock
Investment = 135, Income = 8
Investment = 120 x 135
Income =\(\cfrac { 8 \times 120 \times135 }{ 135 }=Rs.960\)
\(\therefore\) 8% of Rs.100 shares at Rs.135 is a better investment .

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

At equilibrium, demand = Supply
\(\Rightarrow\)  100 - 2p = 3p - 50
\(\Rightarrow\) 150 = 5p
\(\Rightarrow\) p = 30
Equilibrium price is P_E = 30
x = 100 - 2p
= 100 - 60 = 40
x = 40
\(\therefore\) Equilibrium quantity is x_E = 40 units.

y = \(\sqrt { 1+{ x }^{ 2 } } =(1+{ x }^{ 2 })^{ \frac { 1 }{ 2 } }\)
\(\frac{d y}{d x}=\left(1+x^2\right)^{\frac{1}{2}}=\frac{1}{2}\left(1+x^2\right)^{\frac{1}{2}-1}\)
\(=\frac{1}{2 \sqrt{1+x^2}}(2 x)=\frac{x}{\sqrt{1+x^2}}\)

F is (-1, -2) and directrix is 4x - 3y + 2 =0
Let P (x, y) be any point on the parabola
for parabola, \(\frac { FP }{ Pm } =1\)
FP = PM
\(\Rightarrow\) FP² = PM²
\(\Rightarrow\) (x+1)² + (y+2)² =  \(\left[ \frac { 4x-3y+2 }{ \sqrt { { 4 }^{ 2 }+(-3)^{ 2 } } } \right] ^{ 2 }\)
\(\Rightarrow\) x² + 2x +1 +y² + 4y + 4 = 16x² +9y² + 4 -24xy - 12y + 16x
\(\Rightarrow\) 25x² + 50x + 25y² +50x +100y + 125  - 16x² - 9y² - 4 + 24xy +16x - 12y = 0
\(\Rightarrow\)  9x² + 16y² + 24xy + 34x + 112y + 121 = 0

Let A = \(\left| \begin{matrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1 \\ cos\theta & 1 & x \end{matrix} \right| \)
Expanding along R₁ we get
|A| = x\(\left| \begin{matrix} -x & 1 \\ 1 & x \end{matrix} \right| -sin\theta \left| \begin{matrix} -sin\theta & 1 \\ cos\theta & x \end{matrix} \right| +cos​​\theta \begin{vmatrix} -sin\theta & -x \\ cos\theta & 1 \end{vmatrix}\)
= x(-x² - 1) - sin \(\theta\) (-x sin \(\theta\) - cos \(\theta\)) + cos \(\theta\) (-sin \(\theta\) + x cos \(\theta\))
\(=-x^{ 3 }-x+xsin^{ 2 }\theta +sin\theta cos\theta +xcos^{ 2 }\theta \)
= -x³ - x + x(sin²\(\theta\) + cos²\(\theta\))
= -x³ - x + x(1)    [\(\because\) sin²\(\theta\) + cos²\(\theta\) ] = -1
= -x³ which is independent of \(\theta\)

\(\frac { 1 }{ 6! } +\frac { 1 }{ 7! } =\frac { x }{ 8! } \)
\(\frac { 1 }{ 6! } +\frac { 1 }{ 7\times 6! } =\frac { x }{ 8\times 7\times 6! } \)
\(\frac{1}{6 !}+\left(1+\frac{1}{7}\right)=\frac{x}{8 \times 7 \times 6 !}\)
\(\frac{8}{7}=\frac{6 ! x}{8 \times 7 \times 6 !}\)
\(x=\frac{8 \times 8 \times 7}{7}=64\)

Let \(P\left( { x }_{ 1 },{ y }_{ 1 } \right) \) be any point on the locus and A be the foot of the perpendicular from \(P\left( { x }_{ 1 },{ y }_{ 1 } \right) \) to the y-axis.
Given that  OP = 3 AP
\({ OP }^{ 2 }=9{ AP }^{ 2 }\)
\(\left( { x }_{ 1 }-0 \right) ^{ 2 }+\left( { y }_{ 1 }-0 \right) ^{ 2 }=9x_{ 1 }^{ 2 }\)
\({ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }={ 9x }_{ 1 }^{ 2 }\)
\({ 8x }_{ 1 }^{ 2 }-{ y }_{ 1 }^{ 2 }=0\) 
\(\therefore\) The locus of \(P({ x }_{ 1 },{ y }_{ 1 })\) is \(8 x^2-y^2=0\)

\(\left| \begin{matrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{matrix} \right| \) = 1 (Minor of 1) –2 (Minor of 2) + 4 (Minor of 4)
\(=1\left| \begin{matrix} 3 & 0 \\ 1 & 0 \end{matrix} \right| -2\left| \begin{matrix} -1 & 0 \\ 4 & 0 \end{matrix} \right| +4\left| \begin{matrix} -1 & 3 \\ 4 & 1 \end{matrix} \right| \)
= 0 – 0 – 52 = –52.

\(n(S)=52{ C }_{ 2 }=\frac { 52\times 51 }{ 2\times 1 } =1326\)
Let A be the event that one king and one queen card is drawn
n(A) = 4C₁ \(\times\) 4C₁ = 16
\(P(A)=\frac { 16 }{ 1326 } =0.012\)

a = Rs. 2000; i =10% = 0.1 n = 4
A = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
\(=\frac { 2000 }{ 0.1 } \left[ (1+0.1)^{ 4 }-1 \right] \)
= 20,000[ 1.464-1]
= 20,000 (0.464)
A = Rs. 9280

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

\(p={{4\over x}}\)
\(⇒\ x={4\over p}\)
∴ \({dx\over dp}=-{4\over p^2}\)
At p = 4, \({dx\over dp}=-{1\over 4}=-0.25\)
∴ Rate of change of demand with respect to the price at p = Rs.4 is -0.25
Interpretation:
When the price increases by 1% from the level of p = Rs. 4, the demand decreases (falls) by 0.25%.

We have \(\cos 2x=\frac{1-\tan^2x}{1+\tan^2x}\)
\(\therefore LHS=\cos2x+\sin^2\phi=\frac{1-\tan^2x}{1+\tan^2x}+\sin^2\phi\)
\(=\frac{1-\left(2\tan^\phi+1\right)}{1+(2\tan^2\phi)+1}+\sin^2\phi\)      \([\because tan^{ 2 }x=2tan^{ 2 }\phi +1]\)
\(=\frac{-2\tan^2\phi}{2(1+\tan^2\phi)}+\sin^2\phi=\frac{-\tan^2\phi}{\sec^2\phi}+\sin^2\phi\)
\(=\frac{-\sin^2\phi}{\cos^2\phi.\frac{1}{\cos^2\phi}}+\sin^2\phi=-\sin^2\phi+\sin^2\phi=0=RHS\)
Hence Proved.

Given \(f(x) =\begin{cases} \frac { 1-cos4x }{ 8{ x }^{ 2 } } \quad ifx\neq 0 \\ k\quad \quad \quad ifx=0 \end{cases}\)
\(\neq \underset { x\rightarrow 0 }{ lim } \quad f(x)=\underset { x\rightarrow 0 }{ lim } \frac { 1-cos4x }{ { 8x }^{ 2 } } =\underset { x\rightarrow 0 }{ lim } \frac { 2sin^{ 2 }2x }{ { 8x }^{ 2 } } \)   [ஃ 1-cos²z =sin²x]
= \(\underset { x\rightarrow 0 }{ lim } \frac { sin^{ 2 }2x }{ { 4x }^{ 2 } } =\underset { x\rightarrow 0 }{ lim } \left( \frac { sin2x }{ 2x } \right) ^{ 2 }\)
= (1)²    ....(1)
Given f(0) = k ...(2)
Since f(x) is continous at x = 0,
\(\underset { x\rightarrow 0 }{ lim } \) f(x) = f(0)
1 = k [using (1) and (2)
K = 1

\(\frac { 12x-17 }{ (x-2)(x-1) } =\frac { A }{ x+2 } +\frac { B }{ x-1 } \)
\(\Rightarrow \frac { 12x-17 }{ (x-2)(x-1) } =\frac { A(x-1)+B(x-2) }{ (x-2)(x-1) } \)
\(\Rightarrow 12x-17=A(x-1)+B(x-2)\)
Putting x=1 in (1) we get, 
\(12-17= B(1-2) \Rightarrow -5=-B \Rightarrow \boxed { B=5 } \)
Putting x=2 in (1) we get, 
\(24-17=A(2-1) \Rightarrow 7=A(1) \Rightarrow \boxed { A=7 } \)
\(\therefore \frac { 12x-17 }{ (x-2)(x-1) } =\frac { 7 }{ x+2 } +\frac { 5 }{ x-1 } \)