given A = {Q\{1}}
A is defined on A by x*y = x+y-xy
Let x,y ≠ 1
∴ x*y = x + y - xy
Now to prove that x + y - xy ≠ 1
Let us assume that x + y - xy = 1
x+y-xy-1 = 0
(x-1)-y(x-1) = 0
(x-1)(1-y) = 0
x =1 or y = 1 which is a false [∵x, y ≠ 1]
∴ is a binary operation on A.
Commutative property:
Let x,y ∈A ⇒ x, y≠1
∴x*y = x+y-xy
and y*x = y+x-yx
⇒x+y = y*x∀x, y∈A
A has commutative property under *
Associative property:
Let x,y,z ∊A ⇒x,y,z≠1
Consider (x*y)*z = (x+y-xy)*z
= x +y~xy +z- (x +y-xy)z
= x +y-xy+ z-xz- yz + xyz
= x +y+z-xy-yz-zx +xyz ...(1)
= x +y+z - yz - x (y + z - yz)
= x +y +z-yz-zy-xz +xyz ..(2)
From (1) & (2), (x*y)*z = x*(y*z)
A has associative property under *.

Since the given function is a probability density function

\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
That is \(\int _{ -\infty }^{ 1 }{ f(x) } dx+\int _{ 1 }^{ 4 }{ f(x) } dx+\int _{ 4 }^{ \infty }{ f(x) } dx=1\) 
From the given information
\(\int _{ -\infty }^{ 1 }{ 0dx } +\int _{ 1 }^{ 4 }{ { Cx }^{ 2 }dx } +\int _{ 4 }^{ \infty }{ 0dx } =1\) 
\(0+C\left[ \cfrac { { x }^{ 3 } }{ 3 } \right] _{ 1 }^{ 4 }+0=1\Rightarrow C\left[ \cfrac { 64-1 }{ 3 } \right] =1\Rightarrow 21C\Rightarrow C=\cfrac { 1 }{ 21 } \) 
Therefore the probability density function is
\(f(x)= \begin{cases}C x^{2} & 1
Since f (x) is continuous, the probability that X is equal to any particular value is zero. Therefore when the random variable is continuous, either or both of the signs < by ≤ and > by ≥ can be interchanged. Thus
(i) P(1.5 < X < 3.5) = P(1.5 ≤ X< 3.5)= P(1.5 < X ≤3.5) = P(1.5 ≤X ≤ 3.5)
Therefore
\(P(1.5
 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) =\cfrac { 1 }{ 21 } \left( \cfrac { \left( 3.5 \right) ^{ 3 }-\left( 1.5 \right) ^{ 3 } }{ 3 } \right) \) 
= \(\cfrac { 79 }{ 126 } \) 
(ii) \(P(X\le 2)=\int _{ -\infty }^{ 2 }{ f(x) } dx=\int _{ -\infty }^{ 1 }{ f(x)dx } +\int _{ 1 }^{ 2 }{ f(x)dx } \)
 
Therefore
\(P(X\le 2)=0+\cfrac { 1 }{ 21 } \int _{ 1 }^{ 2 }{ { x }^{ 2 }dx=\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) } ^{ 2 }_{ 1 }\) 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { 2 }^{ 3 }-{ 1 }^{ 3 } }{ 3 } \right) =\cfrac { 7 }{ 63 } \) 
(iii) \(P(3
 
= \(\cfrac { 1 }{ 21 } \int _{ 3 }^{ 4 }{ { x }^{ 2 }dx+0=\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) } _{ 3 }^{ 4 }\) 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { 4 }^{ 3 }-{ 3 }^{ 3 } }{ 3 } \right) =\cfrac { 37 }{ 63 } \)

Since X denotes the total score in two throws, it takes on the values 2, 3, 4, 5 and 6. From the Sample space S, we have

\(P(X=2)=\frac { 1 }{ 36 } \), \(P(X=3)=\frac { 4 }{ 36 } \) 
\(P\left( X=4 \right) =\frac { 10 }{ 36 } \) , \(P(X=5)=\frac { 12 }{ 36 } \) and 
\(P(X=6)=\frac { 9 }{ 36 } \)
(i) Probability mass function is

(ii) Cumulative distribution function By definition of the cumulative distribution function for discrete random variable we have
\(f(x)=P(X\le x)=\underset { x_{ 1 }\le x }{ \Sigma } P(X={ x }_{ 1 })\)
\(P(X for -∞
\(F(2)=P(X\le 2)=\sum _{ -\infty }^{ 2 }{ P(X=x)=P\left( X \right) <2)+P(X=2) } =0+\frac { 1 }{ 36 } =\frac { 1 }{ 36 } \)
\(F(3)=P(X\le 3)=\sum _{ -\infty }^{ 3 }{ P(X=x) } =P\left( x<2 \right) +P(X=2)+P(X=3)+P(X=4)=0+\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } =\frac { 15 }{ 36 } \)
\(F(4)=P(X\le 3)=\sum _{ -\infty }^{ 3 }{ P(X=x) } =P\left( x<2 \right) +P(X=2)+P(X=3)+P(X=4)\)
\(0+\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } =\frac { 15 }{ 36 } \)
\(F(5)=P\left( X\le 5 \right) =\sum _{ -\infty }^{ 5 }{ P(X=x) } =P\left( X<2 \right) +P(X=3)+P\left( X=4 \right) +P\left( X=5 \right) \)
= \(0+\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } =\frac { 27 }{ 36 } \)
\(F(6)=P(X\le 6)=\sum _{ -\infty }^{ 6 }{ P(X=x) } \) 
= \(P(X<2)+P(X=2)+P(X=3)+P(x=4)+P(x=5)P(X=6)\)
 \(0+\frac { 1 }{ 36 } +{ \frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =1 }\) 
Therefore the cumulative distribution function is


(iii) \(P(3\le X\le 6)=\sum _{ x=3 }^{ 5 }{ P(X={ { x }_{ 1 })=P(X=3) }+P(X=4) } +P(X=5)\)
\(=\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 26 }{ 36 } \)
(iv) \(X\ge 4)=\sum _{ x=4 }^{ 5 }{ P(X={ x }_{ 1 }) } \)
 = \(\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =\frac { 31 }{ 36 } \)

Let X be the random variablc denotes number of red balls.
Then X take the values 0, 1, 2
Sample space = ⁷C₂ = 21
Let X denote the drawing the red ball.
Then X take the values 0, 1, 2
P(X = 0), X^-1 (BB) = ³C₂ = 3
P(X = 1), X^-1 (BR) = ³C₁ x ⁴C₁ = 12
P(X = 2), X^-1 (BR) = ³C₂ = 6

The probability mass function is

Mean :
\(E(x)=\Sigma x.f\left( x \right) \) 
= \(0(\frac { 1 }{ 7 } )+1\left( \frac { 4 }{ 7 } \right) +2\left( \frac { 2 }{ 7 } \right) \) 
= \(\frac { 4 }{ 7 } +\frac { 4 }{ 7 } =\frac { 8 }{ 7 }\)

\(Let\quad I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 4x{ sin }^{ 2 }x+5{ cos }^{ 2 }x } } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x }{ 4{ tan }^{ 2 }x+5 } } dx\)
(Dividing both numerator and denominator by cos² x).
Let u = tan x 
Then du = sec² x dx
When x = 0, u = tan 0 = 0
When x = \(\frac{\pi}{2}\), u = tan\(\frac{\pi}{2}\) = \(\infty\)
\(\therefore I=\int _{ 0 }^{ \infty }{ \frac { du }{ 4{ u }^{ 2 }+5 } } \) (This is an improper integral)
\(\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ \frac { du }{ \left[ { u }^{ 2 }+\left( \frac { \sqrt { 5 } }{ 2 } \right) \right] } =\frac { 1 }{ 4 } \times \frac { 2 }{ \sqrt { 5 } } { \left[ { tan }^{ -1 }\left( \frac { u }{ \frac { \sqrt { 5 } }{ 2 } } \right) \right] }_{ 0 }^{ \infty } } =\frac { 1 }{ 2\sqrt { 5 } } \left( { tan }^{ -1 }\infty -{ tan }^{ -1 }0 \right) =\frac { 1 }{ 2\sqrt { 5 } } \left( \frac { \pi }{ 2 } \right) =\frac { \pi }{ 4\sqrt { 5 } } \)

Factorising the given function we have
\(y=f(x)\frac { { x(x }-3x) }{ (x-1) } \)
(1) The domain and the range of f (x) are respectively R \{1} and the entire real line.
(2) Putting y = 0 we get the x = 0, 3. Therefore the x -intercept is (3,0). Putting x = 0, we get y = 0. Therefore the curve passes through the origin.
(3) \(f'(x)\frac { { x }^{ 2 }-2x+3 }{ { (x-1) }^{ 2 } } \) and hence the critical point of the curve occurs at x =1 as f′(1) does not exist. But x² -x − 2 + 3 = 0 has no real solution. Hence the only critical point occurs at x = 1.
(4) x =1 is not in the domain of the function and f'(x) ≠0 ∀x ∈R \{1}, there is no local maximum or local minimum.
(5) \(f"(x)=-\frac { 4 }{ { (x-1) }^{ 3 } } \)∀x∈R\{1}. Therefore when x <1, f"(x)>0the curve is concave upwards in (-∞,1) and x>1, f"(x)<0 the curve is concave downwards in (1,∞). Since f"(x} ≠ 0 ∈ R\{1} there is no point of infection for f(x).
(6) Since, \(\underset { x\rightarrow { 1 }^{ - } }{ lim } \frac { { x }^{ 2 }-3x }{ (x-1) } =+\infty \ and\ \underset { x\rightarrow { 1 }^{ + } }{ lim } \frac { { x }^{ 2 }-3x }{ (x-1) } =-\infty ,\ x=1\) is a vertical asymptote.
The rough sketch is shown.

f'(x) = -3.x⁵ + 5x³
f'(x) = -3x⁵ + 5x
f'(x) = - 15x⁴ + 15x²
f''(x) = 0
⇒ - 15x⁴ + 15x² = 0
⇒ - 15x² (1 - x²) = 0
⇒ x² = 0, 1-x² = 0
⇒ x = 0, x = 1, x = -1
ஃ The critical numbers are 0, 1,-1.
f"(x) - 60x³ + 30 x
f"(0) = 0
f"(1) - 60(1)³ + 30 (1)
-60 + 30 = -30
- 60(-1)³ + 30 (-1)
60- 30 = -30
Since f"(-1) < 0, it has a local maximum at
x = 1.
ஃ f(1) = -3(-1)⁵ + 5(-1)³
= -3 + 5 = 2
Since f"(-1) > 0, it has a local maximum at
x = -1.
ஃ f(-1) -3(1)⁵ + 5(1)³
3 - 5 = -2
∴ Local maximum is 2 which occurs at,
x = 1 and local minimum is -2 which occurs at 
x = -1.

Given that the equation is \(\frac{dv}{dx}+2y\ cot\ x=3x^2 cosec^2x\)
This is a linear differential equation. Here, P = 2 cot x ; Q = 3x²cosec²x.
\(\int { Pdx=\int { 2cot\quad xdx=2log|sin\quad x|=log|sin\quad x{ | }^{ 2 } } =log{ sin }^{ 2 }x } \)
Thus, \(I.F={ e }^{ \int { Pdx } }={ e }^{ log{ sin }^{ 2 } }x={ sin }^{ 2 }x\)
Hence, the solution is \({ ye }^{ \int { Pdx } }={ \int { Qe } }^{ \int { Pdx } }dx+C\)
That is, \(y{ sin }^{ 2 }x=\int { { 3x }^{ 2 }cose{ c }^{ 2 }x.{ sin }^{ 2 }xdx+C=\int { { 3x }^{ 2 }dx+C={ x }^{ 3 }+C } } \)
Hence, \(y{ sin }^{ 2 }x={ x }^{ 3 }+C\) is the required solution

Let us now find the point of intersection of the two given curves. Equating x² = (x - 3)² we get, x = \(\frac32\). Therefore, the point of intersection is \({(\frac{3}{2},\frac{9}{4})}\). Let θ be the acute angle between the curves. The slopes of the curves are as follows:

For the curve y = x² 
\(\frac{dy}{dx}=2x\) 
\(m_{1}=(\frac{dy}{dx})\ at\ {(\frac{3}{2},\frac{9}{4})}=3\).
For the curve y = (x-3)² 
\(\frac{dy}{dx}=2(x-3)\) 
\(m_{2}=(\frac{dy}{dx}) \ at \ {(\frac{3}{2},\frac{9}{4})}= -3\)
Using (3), we get
\(tan \theta=|\frac{3-(-3)}{1-9}|=\frac{3}{4}\) 
Hence,  \(\theta\) =  \(tan^{-1}(\frac{3}{4})\).

arg(1+i) + arg(1-i) = arg[(1+i) (1-i)]
LHS = arg (1+i) + arg(1-i)
1+i = \(\sqrt { 2 } \left( \frac { 1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \)
= \(\sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
∴ arg (1+i) = π/4
-1+i =\(\sqrt { 2 } \left( \frac { -1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \)
= \(\sqrt { 2 } \left( cos3\frac { \pi }{ 4 } +isin3\frac { \pi }{ 4 } \right) \)
∴ (-1+i) = 3\(\frac { \pi }{ 4 } \)
∴ LHS = \(\frac { \pi }{ 4 } +\frac { 3\pi }{ 4 } =\frac { 4\pi }{ 4 } =\pi \)
RHS = arg[(1+i) (-1+i)]
= arg[-1-i + i + i²]
= (-1-i + i-1) = arg(-2)
= arg(2) - (1) = 2 arg(-1)
= 2 (cos π + isin π) = π
∴ LHS = RHS

LHS =\({ tan }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) -{ tan }^{ -1 }\left( \frac { 1-y }{ 1+y } \right) \)
= tan^-1(1) - tan^-1 (x) - (tan^-1(1) - tan^-1(y)
\(\left[ \because { tan }^{ -1 }(\frac { x-y }{ 1+xy } )={ tan }^{ -1 }x-{ tan }^{ -1 }y \right] \)
= tan^-1(1) - tan^-1 (x) - tan^-1(1) + tan^-1(y)
= tan^-1(y) - tan^-1(x)
= \({ tan }^{ -1 }\left( \frac { y-x }{ 1+xy } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { 1+\left( yx \right) ^{ 2 }+\left( y-x \right) ^{ 2 } } } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { (1+{ x }^{ 2 })(1+{ x }^{ 2 }) } } \right) \)
RHS

Given \(\overset { \rightarrow }{ AB } =\overset { \rightarrow }{ \alpha } \), \(\overset { \rightarrow }{ AD } =\overset { \rightarrow }{ \beta } \) and \(\overset { \rightarrow }{ AC } =2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \)
Area of the quadrilateral ABCD 
∴ = are of ∆ ABC + area of ∆ ACD
\(=\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ AB } \times \overset { \rightarrow }{ AC } \right| +\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ AC } \times \overset { \rightarrow }{ AD } \right| \)
\(=\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ \alpha } \times \left( 2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| \left( 2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \right) \times \overset { \rightarrow }{ \beta } \right| \)
\(=\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \alpha } \right) +3\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) +3\left( \overset { \rightarrow }{ \beta } \times \overset { \rightarrow }{ \beta } \right) \right| \)
\(=\frac { 1 }{ 2 } \left| 3\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| \quad \quad \quad \left[ \because \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \alpha } =\overset { \rightarrow }{ \beta } \times \overset { \rightarrow }{ \beta } =0 \right] \)
\(=\left( \frac { 3 }{ 2 } +\frac { 2 }{ 2 } \right) \left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) =\left( \frac { 5 }{ 2 } \right) \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| \quad \quad (1)\)
Now, Area of the parallelogram with \(\overset { \rightarrow }{ AB } \) and \(\overset { \rightarrow }{ AD } \) as
adjacent sides = \(\left| \overset { \rightarrow }{ AB } \times \overset { \rightarrow }{ AD } \right| =\left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| .... (2)\)
From (1) & (2), \(\frac { 5 }{ 2 } \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| =\lambda \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| \) [Given]
\(\lambda =\frac { 5 }{ 2 } \)

Augmented matrix [A|B] is \(\left[ \begin{matrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{matrix}|\begin{matrix} a \\ b \\ c \end{matrix} \right] \)
[A|B]\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ -2 & 1 & 1 \\ 1 & 1 & -2 \end{matrix}|\begin{matrix} b \\ a \\ c \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & -3 & 3 \\ 0 & 3 & -3 \end{matrix}|\begin{matrix} b \\ a+2b \\ c-b \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} b \\ c+2b \\ a+b+c \end{matrix} \right] \)
Here \(\rho\) (A) = 2
The given system is consistent only when \(\rho\)([A|B]) = 2\(\rho\)([A|B]) = 2 only if a + b + c = 0 Hence proved.

\(\alpha +\beta =\frac { 1 }{ { \alpha }^{ 2 } } +\frac { 1 }{ { \beta }^{ 2 } } =\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 } }{ { \alpha }^{ 2 }{ \beta }^{ 2 } } =\frac { { (\alpha +\beta ) }^{ 2 }-2\alpha \beta }{ { \alpha }^{ 2 }{ \beta }^{ 2 } } \)
\(\Rightarrow \frac { -b }{ a } =\frac { \frac { { b }^{ 2 } }{ { a }^{ 2 } } -2\frac { c }{ a } }{ \frac { { c }^{ 2 } }{ { a }^{ 2 } } } =\frac { { b }^{ 2 }-2ac }{ { c }^{ 2 } } \)
\(\Rightarrow \frac { 2a }{ c } =\frac { { b }^{ 2 } }{ { c }^{ 2 } } +\frac { b }{ a } \)
\(\frac { 2a }{ c } =\frac { { ab }^{ 2 }+{ bc }^{ 2 } }{ { ac }^{ 2 } } \)
\(2{ a }^{ 2 }c={ ab }^{ 2 }+{ b }^{ 2 }\)
\(\frac { 2a }{ b } =\frac { b }{ c } +\frac { c }{ a } \) [Dividing by abc]
\(\Rightarrow \frac { c }{ a } ,\frac { a }{ b } ,\frac { b }{ c } \) are in A.P \(\Rightarrow \frac { a }{ c } ,\frac { b }{ a } ,\frac { c }{ b } \) are in H.P

\(\vec { r } =(5\hat { i } +7\hat { j } -3\hat { k } )+s(-4\hat { i } +4\hat { j } -5\hat { k } )\) and \(\vec { r } =(8\hat { i } +4\hat { j } +5\hat { k } )+t(7\hat { i } +\hat { j } +3\hat { k } )\)
Let \(\vec { a } =5\hat { i } +7\hat { j } -3\hat { k } ,\vec { b } =4\hat { i } +4\hat { j } -5\hat { k } \)
\(\vec { c } =8\hat { i } +4\hat { j } +5\hat { k } \ and\ \vec { d } =7\hat { i } +\hat { j } +3\hat { k } \)
We know that the two given lines are co-planar
if \(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0\)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & 4 & -5 \\ 7 & 1 & 3 \end{matrix} \right| \)
= \(\hat { i } \left( 12+5 \right) -\hat { j } \left( 12+35 \right) +\hat { k } \left( 4-28 \right) \)
= \(17\hat { i } -47\hat { j } -24\hat { k } \)
\(\left( \vec { c } -\vec { a } \right) =\left( 8-5 \right) \hat { i } +\left( 4-7 \right) \hat { j } +\left( 5+3 \right) \hat { k } \)
= \(3\hat { i } -3\hat { j } +8\hat { k } \)
Now,\(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =\left( 3\hat { i } -3\hat { j } +8\hat { k } \right) \)
\(\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) \)
= \(51+141-192=192-192=0\)
\(\therefore\) The two given lines are co-planar.
The plane containing the two given co-planar lines is
\(\left( \vec { r } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0\)
\(\Rightarrow \left( \vec { r } -5\hat { i } +7\hat { j } -3\hat { k } \right) \times \left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =0\)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) -\left[ \left( 5\hat { i } +7\hat { j } -3\hat { k } \right) .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) \right] =0\)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =\left[ 85-329+72 \right] \)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =-172 \)
which is the required vector equation of the plane.

Given that \(\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =r\Rightarrow { z }_{ 1 }\bar { { z }_{ 1 } } ={ z }_{ 2 }\bar { { z }_{ 2 } } ={ r }^{ 2 }\)
\(\Rightarrow { z }_{ 1 }=\frac { { r }^{ 2 } }{ \bar { { z }_{ 1 } } } ,{ z }_{ 2 }=\frac { { r }^{ 2 } }{ \bar { { z }_{ 2 } } } ,{ z }_{ 3 }=\frac { { r }^{ 2 } }{ { \bar { z } }_{ 3 } } \)
Therefore \({ z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 }=\frac { { r }^{ 2 } }{ { \bar { z } }_{ 1 } } +\frac { { r }^{ 2 } }{ \bar { { z }_{ 2 } } } +\frac { { r }^{ 2} }{ \bar { { z }_{ 3 } } } \)
= \({ r }^{ 2 }\left( \frac { \bar { { z }_{ 2 } } \bar { { z }_{ 3 } } +\bar { { z }_{ 1 } } \bar { { z }_{ 3 } } +\bar { { z }_{ 1 } } { \overline { z } }_{ 2 } }{ \overline { { z }_{ 1 } } \bar { { z }_{ 2 } } \bar { { z }_{ 3 } } } \right) \) 
\(\left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| =\left| { r }^{ 2 } \right| \left| \frac { \overline { { z }_{ 2}{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } } }{ \overline { { z }_{ 1 }{ z }_{ 2 }{ z }_{ 3 } } } \right| \) \(\left(\because \bar{z}_{1}+\bar{z}_{2}=\overline{z_{1}+z_{2}}\right)\)
= \({ r }^{ 2 }\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ \left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \left| { z }_{ 3 } \right| } \)  \(\left( \because |z|=|\bar { z } |and\ \left| { z }_{ 1 }{ z }_{ 2 }{ z }_{ 3 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \left| { z }_{ 3 } \right| \right) \)
= \(\left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| ={ r }^{ 2 }\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ { r }^{ 3 } } =\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ r } \)
\(\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ \left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| } \) = r   (given that \(z_{1}+z_{2}+z_{3} \neq 0\))
Thus,\(\left| \frac { { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } }{ { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } \right| \) = r

Rearranging the terms, the equation of ellipse is 4x² + 24x + y²− 2y + 21 = 0

That is, 4(x² + 6x + 9 − 9) + (y² − 2y + 1 − 1) + 21 = 0,
 4(x + 3)² − 36 + (y−1)² −1 + 21 = 0,
 4(x + 3)² + (y − 1)² = 16,
\(\frac { { \left( x+3 \right) }^{ 2 } }{ 4 } +\frac { { \left( y+1 \right) }^{ 2 } }{ 16 } =1\)
Centre is (-3, 1) a = 4, b = 2, and the major axis is parallel to y-axis c² = 16−4 = 12
c = ±2\(\sqrt { 3 } \)
Therefore, the foci are (−3, 2\(\sqrt { 3 } \) +1) and (−3, −2\(\sqrt { 3 } \) +1).
Vertices are (3, ±4 +1).
That is the vertices are (3, 5) and (3, -3) and the length of Latus rectum = \(\frac { { 2b }^{ 2 } }{ a } \) = 2 units.

Given cubic equation 3x²-16x²+23x-6 = 0
Let ∝, \(\frac{1}{\alpha}\) and ૪ be the roots of the equation
[∵ product of two roots is 1]
\((1)\rightarrow { x }^{ 2 }-\frac { 16 }{ 3 } { x }^{ 2 }+\frac { 23 }{ 3 } -2=0 \) ......(1)
comparing  (1) with 
\({ x }^{ 3 }-\left( \frac { \alpha +\beta +\gamma }{ \alpha } \right) +\left( \alpha \frac { 1 }{ \alpha } +\frac { 1 }{ \alpha } .\gamma +\gamma \alpha \right) \)
\(-\alpha \frac { 1 }{ \alpha } .\gamma =0\) ..........(2)
We get, 
\(\alpha +\frac { 1 }{ \alpha } +\gamma =\frac { 16 }{ 3 } \) .......(3)
\(1+\frac { \gamma }{ \alpha } +\gamma \alpha =\frac { 23 }{ 3 } \)
\(\alpha .\frac { 1 }{ \alpha } .\gamma =2\Rightarrow \gamma =2\) ............(4)
Substituting ૪ = 2 in (3)
 \(\alpha +\frac { 1 }{ \alpha } +2=\frac { 16 }{ 3 } \)
\(\Rightarrow \alpha +\frac { 1 }{ \alpha } =\frac { 16 }{ 3 } -2=\frac { 16-6 }{ 3 } =\frac { 10 }{ 3 } \)
\(\Rightarrow \frac { { \alpha }^{ 2 }+1 }{ \alpha } =\frac { 10 }{ 3 } \)

\({ 3x }^{ 2 }+3=10\alpha \)
\({ 3\alpha }^{ 2 }-10\alpha +3=0\)
\(\alpha =\frac { -10 }{ 3 } or\ \alpha =\frac { 1 }{ 3 } \)
\((3\alpha +10)(3\alpha -1)=0\)
\(\alpha =\frac { -10 }{ 3 } \) is not possible \(\Rightarrow \alpha =\frac { 1 }{ 3 } \)
[\(\because \alpha =\frac { -10 }{ 3 } \) will not satisfy(5)]
∴ The roots are 3, \(\frac{1}{3}\), 2.

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1\vee 0 & 0\vee 1 \\ 0\vee 1 & 1\vee 0 \\ 1\vee 1 & 0\vee 0 \end{matrix}\begin{matrix} 1\vee 0 & 0\vee 1 \\ 0\vee 1 & 1\vee 0 \\ 0\vee 0 & 1\vee 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 1 \\ 1 & 1 \\ 0 & 1 \end{matrix} \right) \) [∵ a∨b=max(a,b)]

Since × is binary operation on Z, a,b ∈ Z⇒ a × b = ab∈Z and b × b = b²∈Z ...(1)
The fact that + is binary operation on Z and (1) ⇒ 3ab = (ab + ab + ab) ∈Z and 5b²= (b²+b²+b²+b²+b²)∈Z     ...(2)
Also a∈Z and 3ab ∈Z implies a+3ab∈Z   ...(3)
(2),(3), the closure property of -on Z yield a * b = (a+3ab-5b²)∈Z. Since a * b belongs to Z, * is a binary operation on Z.

(cos θ + i sin θ )² = cos 2θ + isin 2θ
[By De moivre's theorem]
⇒ cos 2θ + isin 2θ = x + iy
Equating the real and imaginary parts we get,
x = cos 2θ, y = sin 2θ
∴ x² + y² = cos²2θ + sin²2θ = 1
Hence proved

We have \(x=\sqrt { 2+x } \)
\(\Rightarrow { x }^{ 2 }=2+x \Rightarrow { x }^{ 2 }-x-2=0\)
\(\Rightarrow x=\frac { 1\pm \sqrt { 1+8 } }{ 2 } \Rightarrow x=\frac { 1\pm 3 }{ 2 } \)
\(\Rightarrow x=\frac { 1+3 }{ 2 } ,\frac { 1-3 }{ 2 } \Rightarrow x=2,-1\)
Also x>0, we get x = 2

The parametric form of vector equation of a line passing through a point \(\left( \overset { \rightarrow }{ a } \right) \) and parallel to \(\overset { \rightarrow }{ b } \) is
\({ \overset { \rightarrow }{ r } }=\overset { \rightarrow }{ a } +t\overset { \rightarrow }{ b } \), t ∈ R
⇒ Here \(\overset { \rightarrow }{ a } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ b } =2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \)
\(\therefore { \overset { \rightarrow }{ r } }=\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) +t\left( 2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \right) \), t ∈ R which is the required equation of a line.

Let \(x={ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(\Rightarrow cosx=\frac { 1 }{ 2 } \)
\(\Rightarrow cosx=\frac { 1 }{ 2 } \)    \(\left[ \because \frac { \pi }{ 3 } \varepsilon \left[ 0,\pi \right] \right] \)
\(\Rightarrow x=\frac { \pi }{ 3 } \)
\(\therefore \) The principal value of \(cos^{ -1 }\left( \frac { 1 }{ 2 } \right) \) is \(\frac { \pi }{ 3 } \)

It is known that cos^-1 x : [-1, 1]\(\rightarrow\)[0, \(\pi\)] is given by
cos⁻¹x = y if and only if x = cos y for -1\(\le x\le1 and 0\le y \le\pi\) 
Thus, we have
cos^-1 \((-\frac{1}{\sqrt2})\) = \(\frac{3\pi}{4}\), since \(\frac{3\pi}{4}\)\(\in[0,\pi]\)cos\(\frac{3\pi}{4}\) = cos\((\pi=\frac{\pi}{4})=-cos \frac{\pi}{4}=-\frac{1}{\sqrt2}\)

Suppose A is symmetric. Then, A^T = A and so, by theorem (vi), we get
adj(A^T) = (adj A)^T ⇒ adj A = (adj A)^T ⇒ adj A is symmetric.

Given g(x, y) = \(\frac { { e }^{ y }sinx }{ x } \) for x ≠ 0 and g(0, 0) = 1
g(0, 0) = 1
The function g is defined for all (x, y) ∈ R²
To check if g has a limit L at (0,0) and if L= g(0,0) = 1
Consider \(\left| g(x,y)-g(0,y) \right| =\left| \frac { { e }^{ y }sinx }{ x } -0 \right| \)
= \(\left| \frac { { e }^{ y }sinx }{ x } \right| =\left| \frac { \left| { e }^{ y } \right| \left| sinx \right| }{ x } \right| =\left| { e }^{ x } \right| \left| \frac { sinx }{ x } \right| =1\)
\(\left[ \because (x,y)\longrightarrow (0,0)\Rightarrow \left| { e }^{ y } \right| =1and\left| \frac { sinx }{ x } \right| =1 \right] \)
\(\because \begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}\frac { { e }^{ y }sinx }{ x } =1=g(0,0)\) Which proves that is continuous at (0, 0)
∴ g(x, y) is continuous at (0, 0)

Observe that the function g is defined for all (x, y)∈R² It is easy to check, as in the above examples, that g is continuous at all point (x, y) ≠ (0, 0). Next, we shall check the continuity of g at (0, 0). For that we see if g has a limit L at (0, 0) and if L = g(0, 0) = 0. So we consider
\(\left| g\left( x,y \right) -g\left( 0,0 \right) \right| =\left| \frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } -0 \right| =\frac { 2\left| { x }^{ 2 }y \right| }{ \left| { x }^{ 2 }+{ y }^{ 2 } \right| } =\frac { 2\left| xy \right| \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \left| x \right| \)   ...(9)
Note that in the final step above we have used 2 \(\left| xy \right| \) \(\le \) x² + y² (which follows by considering 0\(\le \) (x - y)² for all  x, y∈ R . Note that (x, y)→(0, 0) implies |x| → 0. Then from (9) it follows that \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } \) = 0 = g (0, 0) which proves that g is continuous at (0, 0). So g is continuous at every point of R²

Given y = ax + \(\frac { b }{ x } \) .......(1)
Differentiating with respect to x
y' = ax - \(\frac { b }{ x ^2} \) ......(2)
Differentiating again with respect to x
\(y'' = \frac{-b(-2)}{x^3}= \frac{2b}{x^3} \)
\(Now, x^2y'' + xy'-y \)
\( = x^2 \times \frac{2b}{x^3}+x(a- \frac{b}{x^2})-(ax+\frac{b}{x}) \)
\(= 2\times (\frac{b}{x})+ax-(\frac{b}{x})-ax-(\frac{b}{x})\)
= 0
Hence, y = ax + b is the solution of the differential equation x²y"+xy'-y = 0.

Let z = x+iy ⇒ \(\bar { z } \) = x+iy
∴ \(\\ \frac { \bar { z } +1 }{ z-1 } =\frac { z-iy+1 }{ x-iy-i } =\frac { (x+1)iy }{ x-i(y+1) }\)
= \(\frac { (x+1)-iy }{ x-i(y+1) } \times \frac { x+i(y+1) }{ x+i(y+1) } \)
Choosing the real part alone we get,
\(\frac { x(x+1)+y(y+1) }{ { x }^{ 2 }+(y+1)^{ 2 } } \) = 0
⇒ x(x+1) + y(y+1) = 0
⇒ x²+x+y²+y = 0 which is the locus of z.

Let \(\overset { \rightarrow }{ r } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
\(\therefore x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } =\left( s-2t \right) \overset { \wedge }{ i } +\left( 3-t \right) \overset { \wedge }{ j } +\left( 2s+t \right) \overset { \wedge }{ k } \)
Equating the co-efficients of like components both sides,
We get, x = s - 2t
y = 3 - t
z = 2s + t
Eliminating x and t using determinates we get
\(\left| \begin{matrix} x \\ y-3 \\ z \end{matrix}\begin{matrix} 1 \\ 0 \\ 2 \end{matrix}\begin{matrix} -2 \\ -1 \\ 1 \end{matrix} \right| =0\)
⇒ x (0+2) -1(y - 3 + z) -2 (2y - 6 - 0) = 0
⇒ 2x - y + 3 - z- 4y + 12 = 0
⇒ 2x - 5y - z + 15 = 0

|A| =\(\left[ \begin{matrix} -2 & -3 \\ 5 & -6 \end{matrix} \right] \) = 12+15 = 27
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & 3 \\ -5 & 2 \end{matrix} \right] \)
(A^-1)^T = \(\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & -5 \\ 3 & 2 \end{matrix} \right] \)...(1)
A^T =\(\left[ \begin{matrix} -2 & 5 \\ -3 & -6 \end{matrix} \right] \)
|A^T| =\(\left[ \begin{matrix} -2 & 5 \\ -3 & -6 \end{matrix} \right] \) = 12+15 = 27
∴ (A^T)^-1 = \(\frac { 1 }{ |A^{ T }| } adj(A^{ T })=\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & -5 \\ 3 & -2 \end{matrix} \right] \)...(2)
From (1) and (2), (A^-1)^T = (A^T)^-1

\({ sin }^{ -1 }(\frac { 3 }{ 5 } )-{ cos }^{ -1 (}\frac { 12 }{ 13 } )={ sin }^{ -1 }(\frac { 16 }{ 65 }) \)
Let \({ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) =x\Rightarrow sinx=\frac { 3 }{ 5 } \)

and \(cosx=\frac { adj }{ hyp } =\frac { 4 }{ 5 } \)
and \({ cos }^{ -1 }\left( \frac { 12 }{ 13 } \right) =y\Rightarrow cosy=\frac { 12 }{ 13 } \)
\(siny=\frac { opp }{ hyp } =\frac { 5 }{ 12 } \)

We know that
sin (x - y) = sin x cos y -cos x siny
= \(\frac { 3 }{ 5 } \times \frac { 12 }{ 13 } -\frac { 4 }{ 5 } \times \frac { 5 }{ 12 } =\frac { 36 }{ 65 } -\frac { 20 }{ 6g } =\frac { 16 }{ 65 } \)
\(\therefore x-y={ sin }^{ -1 }\left( \frac { 16 }{ 65 } \right) \)
\(\Rightarrow { sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) -{ cos }^{ -1 }\left( \frac { 12 }{ 13 } \right) ={ sin }^{ -1 }\left( \frac { 16 }{ 65 } \right) \)
Hence proved

Area of the circle = 9π sq.units

πr² = 9π ⇒ r² ⇒ 9 ⇒ r ⇒ 3
Diameters are x + y = 5 ................(1)
and x - y = 1 .................(2)
We know that centre is the point of intersection of diameters.
∴ To find the centre, solve (1) and (2).

⇒ 2x = 6
⇒ x = 3
∴ (1) ⇒ 3 + y = 5
⇒ y = 5 - 3 = 2
∴ Centre is (3, 2)
Hence, equation of the circle is
(x - h)² + (y - k)² = r2
⇒ (x - 3)² + (y - 2)² = 32
\(\Rightarrow x^{2}-6 x+9+y^{2}-4 y+4=9\)
⇒ x² + y² − 6x − 4y + 4 = 0