\(\underset { x\rightarrow 1 }{ lim } \frac { { x }^{ 3 }-1 }{ x-1 } \) is of the type \(\frac { 0 }{ 0 } \)
\(\underset { x\rightarrow 1 }{ lim } \cfrac { { x }^{ 3 }-1 }{ x-1 } =\underset { x\rightarrow 1 }{ lim } \cfrac { \left( x-1 \right) \left( { x }^{ 2 }+x+1 \right) }{ \left( x-1 \right) } \)
\(=\underset { x\rightarrow 1 }{ lim } \left( { x }^{ 2 }+x+1 \right) \)
\(=\left( 1 \right) ^{ 2 }+1+1\) 
= 3

LHS = \(\frac { sin(-\theta )tan({ 90 }^{ o }-\theta )sec\left( { 180 }^{ o }-\theta \right) }{ sin(180+\theta )cot(360-\theta )cosec({ 90 }^{ o }-\theta ) } =1\)
= \(\cfrac { \left( -sin\theta \right) cot\theta \left( -sec\theta \right) }{ \left( -sin\theta \right) \left( -cot\theta \right) sec\theta } =1\)

Given line is x - y + 5 = 0
Perpendicular distance of the given line from P(2, 2) is
= \(\left| \cfrac { 2-2+5 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } } \right| \) 
= \(\left| \cfrac { 5 }{ \sqrt { 2 } } \right| =\cfrac { 5 }{ \sqrt { 2 } } \) 
Distance of (0,0) from the given line
= \(\left| \cfrac { 5 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } } \right| =\left| \cfrac { 5 }{ \sqrt { 2 } } \right| =\cfrac { 5 }{ \sqrt { 2 } } \) 
The given line is equidistance from origin and (2, 2).

(i) Variables:
Let x₁, x₂ represent the number of units of X and Y. 
(ii) Constraints:

	Labour	Capital
X	2	3
Y	3	1

 

∴ 2x₁ + 3x₂ ≤ 30 and 3x₁ + x₂ ≤ 17
(iii) Non-negative restrictions:
Since the number of units of X and Y cannot be negative,x₁, x₂ ≥ 0.
Hence, the mathematical formulation of the LPP is maximize \(Z=100{ x }_{ 1 }+120{ x }_{ 2 }\)
Subject to the constraints
\( { 2x }_{ 1 }+3{ x }_{ 2 }\le 30\)
\({ 3x }_{ 1 }+{ x }_{ 2 }\le 17\)
and x₁, x₂ ≥ 0.

Given f(x, y) =3x²+4y³+6xy-x²y³+6
Differentiating partially w.r.t. 'x' we get,
fx(x, y)=6x + 0 + 6y(1) -y³(2x) + 0
=6x + 6y- 2xy³
\(\therefore f_x(1,-1)=\) 6(1) +6(-1)-2(1)(-1)³
= 6-6+2
= 2

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 \(\cos^{-1}\left(\frac{-1}{\sqrt2}\right)=\theta\)
\(\Rightarrow\cos\theta=-\frac{1}{\sqrt2}\)
We know that the range of principal value of \(\cos^{-1}\)is \([0,\pi]\)
\(\therefore\cos\theta=-\frac{1}{\sqrt2}\Rightarrow-\cos\frac{\pi}{4}=\cos\left(\pi-\frac{\pi}{4}\right)=\cos\frac{3\pi}{4}\)
\(\therefore\cos\theta=\cos\frac{3\pi}{4}\)
\(\Rightarrow\theta=\frac{3\pi}{4}\epsilon[0,\pi]\)
Thus the principal value of \(\cos^{-1}\left(-\frac{1}{\sqrt2}\right)\)is \(\frac{3\pi}{4}\)

B = \(\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}\) 
I - B = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}=\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
|I - B| =\(\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
= 0.08 - 0.18 = - 0.1 < 0
Since |I - B| is negative, Hawkins - Simon conditions are not satisfied.
Therefore the given system is not viable.

\(\frac { 3x+7 }{ { x }^{ 2 }-3x+2 } =\frac { 3x+7 }{ (x-1)(x-2) } =\frac { A }{ x-2 } +\frac { B }{ x-1 } \)
\(\frac{3 x+7}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}\)
\(3 x+7=A(x-2)+B(x-1)\)  ...(1)
If x = 1
\(3+7=\mathrm{A}(1-2) \Rightarrow \mathrm{A}=-10\)
If x = 2
\(6+7=\mathrm{B}(2-1) \Rightarrow \mathrm{B}=13\) 
\(\therefore\)\(\frac { 3x+7 }{ { x }^{ 2 }-3x+2 } \)=\(\frac { -10 }{ x-1 } + \frac { 13 }{ x-2 } \)

\(\frac{13}{52}\)=1/4

Brokerage = 100 x 400 x \(\frac{1}{100}=400\)

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

Since x₁ = 2.5, x₂ = 35 satisfies all the Constraints

cos x sin x - cos x sin x = 0

\(37^{\circ} 30^{\prime}=37 \frac{1}{2}^{\circ}=\frac{75}{2} \times \frac{\pi}{180}=\frac{5 \pi}{24}\)

\(4 a=1 \Rightarrow a=\frac{1}{4}\)
Equation x = a
\(x=\frac{1}{4}\)

\(r=\sqrt{g^2+f^2-c}=\sqrt{1+4+4}=3\)

Since sum of all binomial coefficients is 2n
5C₀ + 5C₁ + 5C₂ + 5C₃ + 5C₄ + 5C₅ + (5C₁ + 5C₂ + 5C₃ + 5C₄)
= 2⁵ + (2⁵ - (5C₀ + 5C₅)
= 32 + 32 - (1 + 1)
= 64 - 2 = 2⁶ - 2

No. of ways = 5C₄ = 5C₁ = 5

\(\left|\begin{array}{ccc} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array}\right|+\left|\begin{array}{ccc} x & -y z & 1 \\ y & -z x & 1 \\ z & -x y & 1 \end{array}\right|\)
Multiply R₁, R₂, R₃, of second determinant by x, y, z respectively
\(\left|\begin{array}{ccc} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array}\right|-\frac{1}{x y z}\left|\begin{array}{ccc} x^2 & x y z & x \\ y^2 & x y z & y \\ z^2 & x y z & z \end{array}\right|\)
\(=\left|\begin{array}{lll} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array}\right|-\frac{x y z}{x y z}\left|\begin{array}{ccc} x^2 & 1 & x \\ y^2 & 1 & y \\ z^2 & 1 & z \end{array}\right|\)
\(=\left|\begin{array}{lll} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array}\right|-\left|\begin{array}{lll} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array}\right|=0\)

\(|A|=4-2=2\)
\(A(\operatorname{adj} A)=|A| I=\left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right)\)

Given lines 3x - 4y - 13 = 0...(1)
8x - 11y = 33  ...(2)
2x - 3y - 7 = 0 ...(3)
Conditon for concurrent lines is
\(\left| \begin{matrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{matrix} \right| =0\) 
i.e.,\(\left| \begin{matrix} 3 & -4 & -13 \\ 8 & -11 & -33 \\ 2 & -3 & -7 \end{matrix} \right| =3\left( 77-99 \right) +4\left( 56+44 \right) -13\left( -24+22 \right) \) 
= -66 + 40 + 26 = 0
\(\Rightarrow \) Given lines are concurrent. To get the point of concurrency solve the equations (1) and (3)

Equation (1) \(\times\) 2	\(\Rightarrow \)	6x	- 8y	= 26
Equation (3) \(\times\) 3	\(\Rightarrow \)	6x	- 9y	= 21
			y	= 5

When y = 5 from (2) 8x = 88
x = 11
Point of concurrency is (11, 5)

Since the decision variables x₁,x₂ are non-negative, the solution lies in the I quadrant.
Consider the equations
\(3{ x }_{ 1 }+5{ x }_{ 2 }=15\)                                             

\({ x }_{ 1 }\)	0	5
\({ x }_{ 2 }\)	3	0

\(5{ x }_{ 1 }+2{ x }_{ 2 }=10\)

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

The feasible region is OABC and its co-ordinates are O(0, 0) A(2, 0) C(0, 3) and B is the point of intersection of the lines
\(3{ x }_{ 1 }+5{ x }_{ 2 }=15\) ----(1)
and \(5{ x }_{ 1 }+2{ x }_{ 2 }=10\)   ... (2)
Verification of B:
\((1)\times 5\Rightarrow 15{ x }_{ 1 }+25{ x }_{ 2 }=75\\ \quad \quad (-)\quad (-)\quad \quad (-) \\ (2)\times 3\Rightarrow 15{ x }_{ 1 }+6{ x }_{ 2 }=30\\ --------------\\ 19{ x }_{ 2 }=45 \Rightarrow { x }_{ 2 }=\frac { 45 }{ 19 } \)
\(From(1), 3{ x }_{ 1 }+5\left( \frac { 45 }{ 19 } \right) =15\)
\(\Rightarrow 3{ x }_{ 1 }=15-\frac { 225 }{ 19 } \)
\(\Rightarrow { x }_{ 1 }=\frac { 20 }{ 19 } \)
\(\therefore \ B \ is\ \left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)

Corner Points	\(Z=6{ x }_{ 1 }+5{ x }_{ 2 }\)
O(0,0)	0
A(2,0)	12
B\(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)	\(6\times \frac { 20 }{ 19 } +5\times \frac { 45 }{ 19 } =\frac { 345 }{ 19 } \)
C(0,3)	15

Maximum of Z occurs at B \(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)
Hence, the solution is \({ x }_{ 1 }=\frac { 20 }{ 19 } ,{ x }_{ 2 }=\frac { 45 }{ 19 } \ and\ { Z }_{ max }=\frac { 345 }{ 19 } \) 

Investment = Rs.12,000
FV = Rs.100
Gain in selling one share = 96-92 = 4
\(\therefore\)Gain in the transaction = \(\cfrac { 12000 }{ 100 } \) x 4 = Rs.480

The two lines of regression are
3X-2Y =-1    ..(1)
2X-Y =2       ..(2)
Since these lines meet in (\(\bar {X}\), \(\bar{Y}\)), let us solve (1) and (2)
(1)     3X-2Y =-1
         (-)  (+)  (-)
(2)x2   4X-2Y=4
_________________
Subtracting, -X=-5        \(\Rightarrow\)\(\bar{X}\)=5
Substitutuing \(\bar{X}\)=5 in (1) we gwt,
\(\Rightarrow\)3(5)-2Y=-1 15-2Y=-1
\(\Rightarrow\)16=2Y
\(\Rightarrow\)Y=8
\(\bar{Y}\)=8
Hence, \(\bar{X}\)=5 and \(\bar{Y}\)=8

(i) S = {1, 2, 3, 4, 5, 6}
n(S) = 6
(i) Let A be the event of getting a prime number
A = {2, 3, 5}
n(A) = 3
\(P(A)=\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let B be the event that the number is greater than or equal to 3.
B = {3, 4, 5, 6}    
n(B) = 4
\(P(B) =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)

Market value of one share = 25 - 7 = 18
Market value of 125 shares = 125 x 18
= Rs. 2250

In averaging ratios and percentages, geometric mean is more appropriate. Let us consider X represents Expenses at the end of the year.

% Rise	X	logX
32	132	2.1206
40	140	2.1461
50	150	2.1761
		\(\sum\)log X = 6.4428

GM = Anti \(\log { \left( \frac { \sum { logX } }{ N } \right) } \)
= Anti \(\log { \left( \frac { 6.4428 }{ 3 } \right) } \)
= Anti log(2.1476)
GM = 140.5
Average rate of increase in overhead expenses
140.5 – 100 = 40.5 %

(i) Variables: Let x₁, x₂ and x₃ be the number of units of products P₁, P₂ and P₃ to be produced.
(ii) Objective function: Profit on x₁ units of the product P₁ = 20 x₁
Profit on x₂ units of the product P₂ = 25 x₂
Profit on x₃ units of the product P₃ = 15 x₃
Total profit = 20 x₁ + 25 x₂ + 15 x₃
Since the total profit is to be maximized, we have to maximize Z = 20 x₁ + 25 x₂ + 15 x₃
Constraints: 6x₁ + 3x₂ + 12x₃ ≤ 200
2x₁ + 5x₂ + 4x₃ ≤ 350
x₁ + 2x₂ + x₃ ≤ 100
Non-negative restrictions: Since the number of units of the products A, B and C cannot be negative, we have x₁, x₂, x₃ ≥ 0
Thus, we have the following linear programming model.
Maximize Z = 20 x₁ + 25 x₂ + 15 x₃
Subject to 6 x₁ + 3 x₂ + 12 x₃ ≤ 200
2x₁ + 5x₂ + 4x₃ ≤ 350
x₁ + 2x₂ + x₃ ≤ 100
x₁, x₂, x₃ ≥ 0

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 \(\frac { 1 }{ 2 } \)

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

y = 2 sin x + 3cos x
y₁ = 2 cos x - 3 sin x
y₂ = 2 (-sin x) -3 cos x
y₂ = -2 sin x -3 cos x = -y
y₂ + y = 0.

There are 13 letters in the given word of which 4 are S's, 2 are I's , 3 A's, 2N's 1 - 0 and 1 - T.
No. of arrangements \(=\frac{13 !}{4 ! 3 ! 2 ! 2 !}\)

u(x, y) = (x²+y²)^-1/2
u(tx, ty) = (t²x²+t²y²)^-1/2 = t^-1(x²+y²)^-1/2
∴ u is a homogeneous function of degree –1
By Euler’s theorem \(x.\frac { \partial u }{ \partial x } +y.\frac { \partial u }{ \partial y } =\left( -1 \right) u=-u\)
Verification:
\(u =\left(x^2+y^2\right)^{-\frac{1}{2}} \)
\(\frac{\partial u}{\partial x} =-\frac{1}{2}\left(x^2+y^2\right)^{-\frac{3}{2}} \cdot 2 x=\frac{-x}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(x \cdot \frac{\partial u}{\partial x} =\frac{-x^2}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(\frac{\partial u}{\partial y} =-\frac{1}{2}\left(x^2+y^2\right)^{-\frac{3}{2}} \cdot 2 y=\frac{-y}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(y \cdot \frac{\partial u}{\partial y} =\frac{-y^2}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(\therefore x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y}=\frac{-\left(x^2+y^2\right)}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(=(-1) \frac{1}{\sqrt{x^2+y^2}}=(-1) u=-u \)
Hence Euler’s theorem verified

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, ∞)

Intervals	Sign of \({dy\over dx}\)	Nature of Function
In (-∞, 0) say x =-1	12(-1) - 3(-1)2 =.-15 (Negative)	Decreasing
In (0,4) say x = 1	12(1) - 3(1)2 = 9 (Positive)	Increasing
In (4, ∞) say x = 5	12(5) - 3(5)2 = -15 (Negative)	Decreasing

∴ 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, ∞).

Intervals	Sign of \({dy\over dx}\)	Nature of Function
In (-∞,2) say x =0	12- 6(0) = 12 (Positive)	Increasing
In (2, ∞) say x = 3	12- 6(3) = - 6 (Negative)	Decreasing

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

Let x₁ g of wheat and x₂ g of rice be mixed in the daily diet. Let Z be the minimum cost of diet.

	Proteins	Carbohydrates	Cost
1 g of Wheat	0.1 g	0.25 g	Rs.4/kg
1 g of Rice	0.05 g	0.5 g	Rs.6/kg
Minimum Requirement	50 g	200 g

Thus, the mathematical formulation of the LPP
Minimize  \(Z=\frac { 4{ x }_{ 1 } }{ 1000 } +\frac { 6{ x }_{ 2 } }{ 1000 } \quad \Rightarrow \quad Z=\frac { { x }_{ 1 } }{ 250 } +\frac { 3{ x }_{ 2 } }{ 500 } \)
Subject to the constraints
\(0.1{ x }_{ 1 }+0.05{ x }_{ 2}\ge 50 \Rightarrow 2{ x }_{ 1 }+{ x }_{ 2 }\ge 1000\)
\( 0.25{ x }_{ 1 }+0.5{ x }_{ 2\quad }\ge 200 \Rightarrow { x }_{ 1 }+2{ x }_{ 2 }\ge 800\)
\(and\ { x }_{ 1 },{ x }_{ 2 }\ge 0\)
Consider the equation
\(2{ x }_{ 1 }+{ x }_{ 2 }= 1000\)

\({ x }_{ 1 }\)	0	500
\({ x }_{ 2 }\)	1000	0

\({ x }_{ 1 }+2{ x }_{ 2 }= 800\)

\({ x }_{ 1 }\)	0	500
\({ x }_{ 2 }\)	1000	0

 

 

The feasible region is ABC an its co-ordinates are A(800, 0), C(0, 1000) and B is the point of intersection of the lines 2x₁+ x₂ = 1000 ... (1) x₁+ 2x₂ = 800 ... (2)
Verification of B:
\(\Rightarrow (1)\times 2 4{ x }_{ 1 }+2{ x }_{ 2 }=2000\)
\(\quad (-)\quad (-)\quad \quad (-)\)
\(\Rightarrow (2) 15{ x }_{ 1 }+6{ x }_{ 2 }=30 \Rightarrow { x }_{ 1 }=400\)
\(--------------\)
Substracting,     \(3{ x }_{ 1 }=2000\)
From (2),    \(\quad 400+2{ x }_{ 2 }=800\)
\(2{ x }_{ 2 }=400\quad \Rightarrow \quad { x }_{ 2 }=200\)  

Corner Points	\( Z=\frac { { x }_{ 1 } }{ 250 } +\frac { 3{ x }_{ 2 } }{ 500 } \)
A(800,0)	\(\frac { 800 }{ 250 } =3.2\)
B(400, 200)	\(\frac { 400 }{ 250 } +\frac { 600 }{ 500 } =2.8\)
C(0,1000)	\(\frac { 3000 }{ 500 } =6\)

Minimum of Z occurs at B(400, 200). Hence, the solution is x₁= 400, x₂ = 200 and Z_rnin = 2.8

Stock = Rs.9000
FV =Rs.100
Dividend Rate = 7%
Income on 7% stock =\(\frac { \text {stock} }{ 100 } \)x FV x Rate percentage
=\(\frac { 9000 }{ 100 } \times 100\times \frac { 7 }{ 100 } \) = Rs.630    ...(1)
Cost of price of one share = 80
Sale proceeds = \(\frac { \text {stock} }{ 100 } \) x Cost Price
=\(\cfrac { 9000 }{ 100 } \)x 80 =7200
Investment = Rs.7200
Market Price = Rs.120
Dividend Rate =15%
Income =\(\frac {\text { Investment} }{ \text {Market Price } }\)x Dividend Rate
=\(\frac { 7200 }{ 120 } \)x15=Rs.900 .....(2)
Change in income = 930 - 630 = Rs.270

Already the values are arranged in ascending order then Median is obtained by the following.

X	f	cf
15	12	12
25	11	23
35	10	33
45	15	48
55	22	70
65	13	83
75	18	101
85	19	120
	N = 120

Median = size of \(\left( \frac { (n+1) }{ 2 } \right) ^{ th }\)value
= size of \(\left( \frac { (120+1) }{ 2 } \right) ^{ th }\) value
= size of 60.5^th item = 55
MD about Median = \(\frac { \Sigma f|X-Median| }{ N } =\frac { \Sigma f|D| }{ N } \)
Mean deviation about Median

X	f	|D|=|X-55|	f|D|
15	12	40	480
25	11	30	330
35	10	20	200
45	15	10	150
55	22	0	0
65	13	10	130
75	18	20	360
85	19	30	570
	N = 120		Σf|D| = 2220

MD about Median = \(\frac { 2220 }{ 120 } \) = 18.5
Coefficient of mean deviation about median = \(\frac { MD \ about \ Median }{ Median } \)
= \(\frac { 18.5}{ 55 } \) = 0.34

X	Y	X2	Y2	XY
1	4	1	16	4
2	8	4	64	16
3	2	9	4	6
4	12	16	144	48
5	10	25	100	50
6	14	36	196	84
7	16	49	256	112
8	6	64	36	48
9	18	81	324	162
​\(\Sigma X\)​ = 45	​\(\Sigma Y\)​ = 90	​\(\Sigma X^2\)​ = 285	​\(\Sigma Y^2\)​ = 1140	​\(\Sigma XY\)​ = 530

\(\bar{X} =\frac{45}{9}=5 \quad \bar{Y}=\frac{90}{9}=10 \)
\(b_{y x} =\frac{N \Sigma X Y-\Sigma X \Sigma Y}{N \Sigma X^2-(\Sigma X)^2} \)
\(=\frac{9(530)-4050}{9(285)-2025}=\frac{720}{540}=1.33\)
Regression equation of Y on X
\(Y-\overset{-}{Y}=b_{yx}(X-\overset{-}{X})\)
Y - 10 = 1.33(X - 5)
Y = 1.33X - 6.65 + 10
Y = 1.33X + 3.35

\(\frac { \partial p }{ \partial { P }_{ 1 } } =2-{ p }_{ 1 }{ p }_{ 2 }\)
\(\frac { \partial p }{ \partial { P }_{ 2 } } =1-{ p }_{ 1 }^{ 2 }\)
(i) \(\frac { Eq }{ { EP }_{ 1 } } =\frac { { p }_{ 1 } }{ q } \frac { \partial p }{ \partial { P }_{ 1 } } =\frac { -p }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } (-2-{ p }_{ 1 }{ p }_{ 2 })\)
= \(\frac { { 2p }_{ 1 }+2{ p }_{ 1 }^{ 2 }{ p }_{ 2 } }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } \)
when p₁ = 3 and p₂ = 7
\(\frac { { E }q }{ { Ep_{ 1 } } } =\frac { 2(3)+2(9)(7) }{ 5-6+7-(9)(7) } =\frac { 132 }{ -57 } =\frac { -132 }{ 57 } \)
\(\frac { Eq }{ { EP }_{ 1 } } =\frac { { p }_{ 2 } }{ q } \frac { \partial p }{ \partial { P }_{ 2 } } =\frac { -p\left( 1-{ p }_{ 1 }^{ 2 } \right) }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } \)
\(\frac { { -p }_{ 2 }+{ p }_{ 2 }{ p }_{ 1 }^{ 2 } }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } \)
when p₁ - 3 and p₂ = 7
\(\frac { { E }q }{ { Ep_{ 2 } } } =\frac { -7+7(9) }{ 5-6+7-(9)(7) } =\frac { 56 }{ -57 } =\frac { -56 }{ 57 } \)

\(\tan \frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{1+\cos A}}\)
Put \(A=\frac{\pi}{4}\)
\(\tan \frac{\pi}{8}=\sqrt{\frac{1-\cos \pi / 4}{1+\cos \pi / 4}}\left[\because \tan \frac{\pi}{8} \text { is }+v e\right]\)
\(=\sqrt{\frac{1-1 / \sqrt{2}}{1+1 / \sqrt{2}}}\)
\(=\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}}\)
\(=\sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}=\sqrt{2}-1\)

Since x lies in the II quadrant ,cos x is negative
∴ cos x = \(-\left( \frac { 1 }{ 1+\sqrt { { tan }^{ 2 }x } } \right) =\frac { -1 }{ \sqrt { 1+\frac { 16 }{ 9 } } } =\frac { -3 }{ 5 } \)
Now \(\frac { \pi }{ 2 }     lies in I quadrant \(sin\frac { x }{ 2 } ,cos\frac { x }{ 2 } \) \(tan\frac { x }{ 2 } \) are positive
\(\therefore cos\frac { x }{ 2 } =\sqrt { \frac { 1+cos\quad x }{ 2 } } =\sqrt { \frac { 1-\frac { 3 }{ 5 } }{ 2 } } =\sqrt { \frac { 5-3 }{ 2\times 5 } } =\sqrt { \frac { 1 }{ 5 } } =\frac { 1 }{ \sqrt { 5 } } \)
\(sin\frac { x }{ 2 } =\sqrt { \frac { 1-cos\quad x }{ 2 } } =\sqrt { \frac { 1+\frac { 3 }{ 5 } }{ 2 } } =\sqrt { \frac { 5+3 }{ 2\times 5 } } =\sqrt { \frac { 4 }{ 5 } } =\frac { 2 }{ \sqrt { 5 } } \)
\(tan\frac { x }{ 2 } =tan\frac { x }{ 2 } =\sqrt { \frac { 1-cos\quad x }{ 1+cos\quad x } } =\sqrt { \frac { 1+\frac { 3 }{ 5 } }{ 1-\frac { 3 }{ 5 } } } =\sqrt { \frac { 8 }{ 2 } } =\sqrt { 4 } =2\)

Any number greater than a million will contain all the seven digits.
Now, we have to arrange these seven digits, out of which 2 occur twice, 3 occurs twice and the rest are distinct.
The number of such arrangements =\(\frac { 7! }{ 2!3! } =\frac { 7\times 6\times 5\times 4\times 3! }{ 2\times 3! } =420\)
These arrangements also include those numbers which contain 0 at the million's place.
Keeping 0 fixed at the million place, we have 6 digits out of which 2 occurs twice, 3 occurs thrice and the rest are distinct.
These 6 digits can be arranged in \(\frac { 6! }{ 2!3! } =\frac { 6\times 5\times 4\times 3! }{ 2\times 1\times 3! } =60\quad ways\)
Hence, the number of required numbers = 420 - 60 = 360.

Let P (n) denote the statement 1³ + 2³ + 3³ + ....... + n³ = \(\frac { { n }^{ 2 }(n+1)^{ 2 } }{ 4 } \)
Put n = 1
LHS = 1³ = 1
\(=\frac { 1^2(2)^2 }{ 4 } \Rightarrow 1\)
LHS = RHS
\(\therefore\) P (1) is true
Let us assume that P(k) is true
p(k) : 1³ + 2³ + ..... + k³ = \(\frac { { k }^{ 2 }(k+1)^{ 2 } }{ 4 } \)
To prove that P(k+1) IS true
p(k) : 1³ + 2³ + ..... + k³ + (k + 1)³
\(=P(k)+(k+1)^3\)
= \(\frac { { k }^{ 2 }(k+1)^{ 2 } }{ 4 } +(k+1)^{ 3 }\)
\(=\frac{k^2(k+1)^2+4(k+1)^3}{4}\)
\(=\frac{(k+1)^2\left(k^2+4(k+1)\right)}{4}\)
\(=\frac{(k+1)^2\left(k^2+4 k+4\right)}{4}=\frac{(k+1)^2(k+2)^2}{4}\)
∴ p(k + 1) is true if P(k) is true.
∴ p(n) is true for all \(n\in N\)

\(3 x-y+2 z=13 ; 2 x+y-z=3\)
\(x+3 y-5 z=-8\)
The given system can be written as
\(\left(\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & -1 \\ 1 & 3 & -5 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(A X=B \Rightarrow X=A^{-1} B\)
\(\text {Where } A=\left(\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & -1 \\ 1 & 3 & -5 \end{array}\right), X=\left(\begin{array}{l} x \\ y \\ z \end{array}\right)\)
\(B=\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(|A|=3(-5+3)+1(-10+1)+2(6-1)\)
\(=-6-9+10=-5 \neq 0\)
\(\therefore A^{-1} exists\)
\(\mathrm{A}_{11}=\text {Co-factor of } 3=(-5+3)=-2 \)
\(\mathrm{A}_{12}=\text {Co-factor of }-1=-(-10+1)=9 \)
\(\mathrm{A}_{13}=\text {Co-factor of } 2=(6-1)=5\)
\(\mathrm{A}_{21}=\text { Co-factor of } 2=-(5-6)=1\)
\(\mathrm{A}_{22}=\text { Co-factor of } 1=-15-2=-17 \)
\(\mathrm{A}_{23}=\text { Co-factor of }-1=-(9+1)=-10\)
\(\mathrm{A}_{31}=\text {Co-factor of } 1=1-2=-1\)
\(\mathrm{A}_{32}=\text {Co-factor of } 3=-(-3-4)=7 \)
\(\mathrm{A}_{33}=\text {Co-factor of }-5=3+2=5\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -2 & 9 & 5 \\ 1 & -17 & -10 \\ -1 & 7 & 5 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{-5}\left(\begin{array}{ccc} -2 & 1 & -1 \\ 9 & -17 & 7 \\ 5 & -10 & 5 \end{array}\right)\)
\(X=A^{-1} B=\frac{-1}{5}\left(\begin{array}{ccc} -2 & 1 & -1 \\ 9 & -17 & 7 \\ 5 & -10 & 5 \end{array}\right)\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(=\frac{-1}{5}\left(\begin{array}{ccc} -26 & 3 & 8 \\ 117 & -51 & -56 \\ 65 & -30 & -40 \end{array}\right)\)
\(=\frac{-1}{5}\left(\begin{array}{c} -15 \\ 10 \\ -5 \end{array}\right)\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 3 \\ -2 \\ 1 \end{array}\right)\)
\(x=3, \mathrm{y}=-2, \mathrm{z}=1\)

Given \({ A }^{ -1 }=\left[ \begin{matrix} 1 & 0 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{matrix} \right] \)
\(\left|A^{-1}\right|=1[1-1]-0+3[-2-1]\)
\(=-9 \neq 0\)
\(\therefore\left(\mathrm{A}^{-1}\right)^{-1} \text { exists }\)
\(A_{11}=\text {Co-factor of } 1=1-1=0 \)
\(A_{12}=\text {Co-factor of } 0=-(2+1)=-3 \)
\(A_{13}=\text {Co-factor of } 3=(-2-1)=-3\)
\(A_{21}=\text {Co-factor of } 2=-(0+3)=-3\)
\(\mathrm{A}_{22}= \text {Co-factor of }1=1-3=-2\)
\(A_{23}= \text {Co-factor of }-1=-(-1-0)=1\)
\(A_{31}=\text {Co-factor of }1=0-3=-3\)
\(A_{32}= \text {Co-factor of } -1=-1(-1-6)=7\)
\(\mathrm{A}_{33}= \text {Co-factor of }1=1-0=1\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 0 & -3 & -3 \\ -3 & -2 & 1 \\ -3 & 7 & 1 \end{array}\right)\)
\(\text {Since }\left(A^{-1}\right)^{-1}=A\)
\(\mathrm{A}=\frac{1}{\left|A^{-1}\right|} \text { adj } A^{-1}\)
\(=\frac{1}{-9}\left(\begin{array}{ccc} 0 & -3 & -3 \\ -3 & -2 & 7 \\ -3 & 1 & 1 \end{array}\right)\)
\(A=\frac{1}{9}\left(\begin{array}{ccc} 0 & 3 & 3 \\ 3 & 2 & -7 \\ 3 & -1 & -1 \end{array}\right)\)