\(y={ sin }^{ -1 }(2x),x\in \left[ -\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right] \)

\(y=-\frac { 1 }{ 2 } { cos }^{ 2 }x+\frac { { cos }^{ 5 }x }{ 5 } +{ xe }^{ x }-{ e }^{ x }+c\)

c=-1

\( \frac { \partial w }{ \partial u } =\frac { 2u }{ \sqrt { 1-\left( { u }^{ 2 }-{ v }^{ 2 } \right) } } ;\frac { \partial w }{ \partial v } =\frac { -2v }{ \sqrt { 1-\left( { u }^{ 2 }-{ v }^{ 2 } \right) } } \)

(−∞,−2)concave downward
(5,∞)concave upward
Point of inflection is (5, 75)

\(\frac { 1 }{ 48\pi } cm/sec\)

Given r² = x² +y² + z²
log r² = log (x² + y² + z²)
⇒ 2 log r = log (x² + y² + z²)
∴ 2V = log (x² + y² + z²) [∵ V = log r ]
⇒ V = \(\frac12\) log (x² + y² + z²)
\(\frac { \partial V }{ \partial x } =\frac { 1 }{ 2 } \frac { 2x }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } =\frac { x }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } \)
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } =\frac { ({ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 })(1)-x(2x) }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }-{ x }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
III^ty \(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } =\frac { { x }^{ 2 }+{ y }^{ 2 }-{ y }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
∴ \(\frac { { \partial }^{ 2 }V }{ \partial { z }^{ 2 } } =\frac { { x }^{ 2 }+{ y }^{ 2 }-{ z }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
\(\therefore \frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } +\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } +\frac { { \partial }^{ 2 }V }{ \partial { z }^{ 2 } } =\frac { { y }^{ 2 }+{ z }^{ 2 }-{ x }^{ 2 }+{ z }^{ 2 }+{ x }^{ 2 }-{ y }^{ 2 }+{ x }^{ 2 }+{ y }^{ 2 }-{ z }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 }) }^{ 2 } } \)
= \(\frac { { x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 }) }^{ 2 } } =\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 } } \)
= \(\frac { 1 }{ { r }^{ 2 } } \)
Hence proved.

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 *.

The entries in the columns corresponding to p ↔ q and ( p ⟶ q) ∧ (q ⟶ p) are identical and hence they are equivalent

Given (x,y) = x³ - 2x²y + 3xy² + y³ ...(1)
f(tx, ty) = (tx)³ - 2(tx)² (ty) + 3 (tx) (ty)² + (ty)³
= t³ x³ - 2t² x²ty + 3txt²y²+ t³y³
= t³ (x³ - 2x²y + 3xy² +y³)
f(tx, ty) = t³.f(x, y)
∴ f is a homogeneous function and its degree is 3.
Differentiate (1) partially with respect to 'x' and 'y' we get
\(\frac { \partial f }{ \partial x } { =3 }^{ 2 }-4xy+3{ y }^{ 2 }\)
\(\Rightarrow x\frac { \partial f }{ \partial x } ={ 3x }^{ 2 }-4xy+{ 3xy }^{ 2 }\)  ...(2)
\(\frac { \partial f }{ \partial y } =-{ 2x }^{ 2 }+6xy+3{ xy }^{ 2 }\)
\(\Rightarrow y\frac { \partial f }{ \partial y } =-2{ x }^{ 2 }+{ 6xy }^{ 2 }+{ 3y }^{ 3 }\) ...(3)
Adding (2) and (3) we get,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 3x² - 4x²y + 3x²y- 2x²y + 6xy² + 3y³
= 3x² - 6x²y + 9xy² + 3y³
= 3 (x³ - 2x²y - 3xy² +y³)
= 3f [using (1)]
∴ \(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 3f = nf where 3 is the degree of (x, y)
Hence Euler's theorem is verified.

Given U (x, y) = e^x sin y ; x = st² ; y = s²t
\(\frac { \partial U }{ \partial x } \) = e^x sin y ; \(\frac { \partial U }{ \partial y } \) = e^x cos y
\(\frac { \partial U }{ \partial x } \) = \({ e }^{ { st }^{ 2 } }\) sin (s²t)
\(\frac { \partial U }{ \partial y } \) = \({ e }^{ { st }^{ 2 } }\) cos (s²t)
\(\frac{dx}{dt}\) = 2st; \(\frac{dy}{dt}\) = s²
\(\frac{dx}{ds}\) = t²; \(\frac{dy}{ds}\) = 2 st
By chain rule
\(\frac { dU }{ ds } =\frac { \partial U }{ \partial x } .\frac { dx }{ ds } +\frac { \partial U }{ \partial y } .\frac { dy }{ ds } \)
= \({ e }^{ { st }^{ 2 } }\). sin (s²t) (t²) + \({ e }^{ { st }^{ 2 } }\) cos(s^2t).(2^st)
∴ \({ \left( \frac { \partial U }{ \partial s } \right) }_{ (s=t=1) }\) = e¹ sin (1) + 2e¹ cos (1)
= e [sin (1) + 2 cos (1)] and
\(\frac { dU }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { \partial u }{ \partial y } .\frac { dy }{ dt } \)
=  \({ e }^{ { st }^{ 2 } }\) . sin (s²t)(2^st) + \({ e }^{ { st }^{ 2 } }\) cos (s²t). (s²)
∴ \({ \left( \frac { \partial U }{ \partial t } \right) }_{ (s=t=1) }\) = 2e¹ sin (1) + e¹ cos (1)
= e [2 sin (1) + cos (1)]

Equation of the given curves are y = sin x  ..(1)
Y = cos x ...(2)
from (1) and (2), sin x = cos x
y = sin x

y = cos x

\(\Rightarrow x=\frac { \pi }{ 4 } \)

\(\therefore \) Required area = \(2\int _{ \frac { \pi }{ 4 } }^{ \frac { 3\pi }{ 4 } }{ (sinx-cosx)dx } \)
[\(\because\) the area is symmetrical about X - axis]
\(=2{ \left[ -cosx-sinx \right] }_{ \frac { \pi }{ 4 } }^{ \frac { 3\pi }{ 4 } }\)
\(=-2\left[ \left( cos\frac { 3\pi }{ 4 } +sin\frac { 3\pi }{ 4 } \right) -\left( cos\frac { \pi }{ 4 } +sin\frac { \pi }{ 4 } \right) \right] \)
= -2\(\left[ \left( -\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \right) -\left( \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \right) \right] \)
[cos 135^o = cos(180^o- 45) = -cos 45^o sin135^o = sin(180^o- 45) = -sin 45^o]
\(=-2\left[ \frac { -2 }{ \sqrt { 2 } } \right] =\frac { 4 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 4\sqrt { 2 } }{ 2 } =2\sqrt { 2 } \) 

\(Let\ I=\int _{ 0 }^{ \pi }{ x\left[ { sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx) \right] } dx\quad ...(1)\)
\(\left[ \because \int _{ 0 }^{ a }{ f(x)dx=\int _{ 0 }^{ a }{ f(a-x)dx } } \right] \)
\(I=\int _{ 0 }^{ \pi }{ (\pi -x) } [{ sin }^{ 2 }(sin(\pi -x))+{ cos }^{ 2 }(cos(\pi -x)]dx\)
\(=\int _{ 0 }^{ \pi }{ (\pi -x)[{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(-cosx)]dx } \)
\(=\int _{ 0 }^{ \pi }{ (\pi -x)[{ sin }^{ 2 }(sin\quad x)+{ cos }^{ 2 })(cosx)]dx } \)
\([\because cos(-x)=cos\quad x]\)
\(=\int _{ 0 }^{ \pi }{ \pi [{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)] } dx\)
\(-\int _{ 0 }^{ \pi }{ x[{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)] } dx\)
\(=\int _{ 0 }^{ \pi }{ \pi [{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)]dx-I } [from\quad (1)]\)
\(2I=\pi \int _{ 0 }^{ \pi }{ [{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)] } dx\)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ \left[ { sin }^{ 2 }(sin\quad x)+{ cos }^{ 2 }(cos\quad x) \right] } dx\quad ...(2)\)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ \left[ { sin }^{ 2 }(sin\frac { \pi }{ 2 } -x)+{ cos }^{ 2 }(cos\frac { \pi }{ 2 } -x) \right] dx } \)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ [{ sin }^{ 2 }(cosx)+{ cos }^{ 2 }(sin\quad x)]dx] } ....(3)\)
Adding (2) and (3) we get,
\(2I=\pi \int _{ 0 }^{ \pi /2 }{ \left[ { sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx) \right] dx } +{ sin }^{ 2 }(cosx)+{ cos }^{ 2 }(sin\quad x)]dx\)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ (1+1)dx } \ [\because { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1]\)
\(=\pi (2){ [x] }_{ 0 }^{ \frac { \pi }{ 2 } }=2\pi \left[ \frac { \pi }{ 2 } -0 \right] ={ \pi }^{ 2 }\ \)
\( \therefore I=\frac { { \pi }^{ 2 } }{ 2 } \)

Let T represent the temperature of the boiling water and Tm represents the temperature of the kitchen.
By Newton's law of cooling
\(\Rightarrow \int { \frac { dT }{ T-{ T }_{ m } } =K\int { dt } } \)
\(\Rightarrow log(T-{ T }_{ m })=Kt+logC\)
\(\Rightarrow log(T-{ T }_{ m })-logC=Kt\)
\(\Rightarrow log\left( \frac { T-{ T }_{ m } }{ C } \right) =Kt\)
\(\Rightarrow T-{ T }_{ m }={ Ce }^{ Kt } ...(1)\)
when t=0,T=100
\(\therefore 100-{ T }_{ m }={ Ce }^{ 0 }\)
\(\Rightarrow C=100-{ T }_{ m }\)
\(\Rightarrow becomes,\ T-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ Kt }\)
Also when t = 5, T = 80
\(\therefore 80-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ 5K }\)
\(\Rightarrow { e }^{ 5K }=\frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } ..(2)\)
When t = 10, T = 65
(2) \(\Rightarrow\) 65 - T = (100-T_m)e^10K
= (100-T_m)(e^5K)²
\(=(100-{ T }_{ m }){ \left( \frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } \right) }^{ 2 }\)
[using(2)]
\(\Rightarrow 65-{ T }_{ m }=\frac { { (80-{ T }_{ m } })^{ 2 } }{ 100-{ T }_{ m } } \)
\(\Rightarrow\) 6500-65T_m-100T_m+T_m² = 6400+T_m²-160T_m
\(\Rightarrow\) 6500-6400 = 165Tm-160Tm
\(\Rightarrow\) 100 = 5Tm
\(\\ \Rightarrow { T }_{ m }=\frac { 100 }{ 5 } ={ 20 }^{ o }C\)
Hence the temperature of the kitchen is 20^oC

Given \(v\frac { dv }{ dx } =g\left( 1-\frac { { v }^{ 2 } }{ { k }^{ 2 } } \right) =g\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ { k }^{ 2 } } \right) \)
On separating the variables we get,
\(\frac { vdv }{ { k }^{ 2 }-{ v }^{ 2 } } =\frac { g }{ { k }^{ 2 } } .dx\)
Multiplying by -2 both sides we get,
\(v\frac { dv }{ dx } =g\left( 1-\frac { { v }^{ 2 } }{ { k }^{ 2 } } \right) =g\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ { k }^{ 2 } } \right) \)
\(\frac { -2vdv }{ { k }^{ 2 }-{ v }^{ 2 } } =\frac { -2g }{ { k }^{ 2 } } dx\)
Taking integrating on both sides, we get
\(\int { \frac { -2v }{ { k }^{ 2 }-{ v }^{ 2 } } } =\frac { -2g }{ { k }^{ 2 } } .x+log\quad c\)
\(\Rightarrow log({ k }^{ 2 }-{ v }^{ 2 })=\frac { -2g }{ { k }^{ 2 } } .x+log\quad c\)
\(\Rightarrow log({ k }^{ 2 }-{ v }^{ 2 })-log\quad c=\frac { -2g }{ { k }^{ 2 } } .x\)
\(\Rightarrow log\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ c } \right) -\frac { -2gx }{ { k }^{ 2 } } \)
\(\Rightarrow \frac { { k }^{ 2 }-{ v }^{ 2 } }{ e } ={ e }^{ -\frac { -2gx }{ { k }^{ 2 } } }\)
\(\Rightarrow { k }^{ 2 }-{ v }^{ 2 }{ ce }^{ \frac { -2gx }{ { k }^{ 2 } } }...(1)\)
Initial condition:
when v = 0, x = 0 we get
\( k^2-(0)^2 =C e^{\frac{-2g(0)}{k^2}} \)
\(k^2 =C e^0 \Rightarrow k^2=C \)
\((1) \Rightarrow k^2-v^2 =k^2 e^{\frac{-2 x^2}{k^2}} \)
\(k^2-k^2 e^{\frac{-2 s x}{k^1}} =\mathrm{v}^2 \)
\(k^2\left[1-e^{\frac{-2 s x}{k^2}}\right] =\mathrm{v}^2\)

Let AB be the position of the ladder at any time t such that OA = x and OB = y
Then OA² + OB² = AB²
⇒ x² + y² = 17²
Given \(\frac { dx }{ dt } \) = 5 and x = 8
When x = 8, 8² + y² = 17²
 ⇒ y² = 289 - 64 = 225
 ⇒ y = 15
Differentiating (1) with respect to 't' we get,
\(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } =0\)
⇒ 8(5) + 15 \(\frac { dy }{ dt } \) = 0 [∵ x = 8, \(\frac { dx }{ dt } \) = 15, y = 15]
⇒ 40 + 15\(\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dt } =\frac { -40 }{ 15 } =\frac { -8 }{ 3 } \) m/sec
∴ The rate of top of the ladder moving down the wall is \(\frac{-8}{3}\) m/sec
(ii) The ladder, the wall and the floor forms a right angled triangle.
 Area = \(\frac12\)xy
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } =\frac { 1 }{ 2 } \left[ x\frac { dy }{ dx } +y\frac { dx }{ dt } \right] \)
\(=\frac { 1 }{ 2 } \left[ 8\left( -\frac { 8 }{ 3 } \right) +15(5) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64 }{ 3 } +75 \right] =\frac { 1 }{ 2 } \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64+225 }{ 3 } \right] =\frac { 1 }{ 2 } \left( \frac { 161 }{ 3 } \right) \)
= \(\frac { dA }{ dt } \) = 26.83 sq.m/sec

Let \(\overset { \rightarrow }{ B } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
Given \(\overset { \rightarrow }{ A } \times \overset { \rightarrow }{ B } =\overset { \rightarrow }{ C } \Rightarrow \left| \begin{matrix} \overset { \wedge }{ i } \\ 1 \\ x \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ 1 \\ y \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ 1 \\ z \end{matrix} \right| =\overset { \wedge }{ j } -\overset { \wedge }{ k } \)
\(\Rightarrow \overset { \wedge }{ i } (z-y)-\overset { \wedge }{ j } (z-x)+\overset { \wedge }{ k } (y-x)\quad \overset { \wedge }{ j } -\overset { \wedge }{ k } \)
Equating the like components on both sides, we get
z - y = 0   .....(1)
x - y = 1   .....(2)
y - x = -1  .....(3)
Also, \(\overset { \rightarrow }{ A } .\overset { \rightarrow }{ B } =3\Rightarrow \left( \overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) .\left( x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \right) =3\)
⇒ x + y + z = 3   ....(4)
Solving (1), (2), (3) and (4), we get \(x=\frac { 5 }{ 3 } ,y=\frac { 2 }{ 3 } \)and \(z=\frac { 2 }{ 3 } \)
\(\therefore \overset { \rightarrow }{ B } =\frac { 5 }{ 3 } \overset { \wedge }{ i } +\frac { 2 }{ 3 } \overset { \wedge }{ j } +\frac { 2 }{ 3 } \overset { \wedge }{ k } \)

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 = 2 arg (-1)
= 2 arg (cos π + i sin π) = 2π
RHS = arg (-1)² = arg (1)
= arg (cos θ + isin θ) = 0
∴ LHS ≠ RHS

Equation of the ellipse is \(\frac { { x }^{ 2 } }{ 25 } +\frac { y^{ 2 } }{ 9 } =1\)
∴ a² = 25, b² = 9
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
Focus is (ae, 0) = \(\left( 5\times \frac { 4 }{ 5 } \right) \) = (4, 0)
Since the focus of the hyperbola coincides with the focus of the ellipse, foci of the hyperbola are (±4,0).
Let A be the length of the semi-transverse axis
∴ Ae - 4 ⇒ 2A = \(\frac { 4 }{ e } =\frac { 4 }{ 2 } =2\) [∵ e = 2]
Let B b th length of the semi conjugate axis
B² = A²(e² - 1) = 4(4 - 1) = 12
Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { A }^{ 2 } } -\frac { { y }^{ 2 } }{ { B }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1\)

Let us take the axis AX as the and the tangent AY at A as y-axis.
Equation of the parabola is y² = 4ax
CA 15, FG = 120
CF = CG= 60
F is (15, 60)
Since F lies on (1), 602 = Aa(15) ∴ ⇒ a = 60
y² = 240x
When y = 24, 24² = 240(x)
⇒ x = \(\frac { 24\times 24 }{ 240 } \)
x = \(\frac{24}{10}\) = \(\frac{12}{5}\) = 2.4
From the diagram
BD = BE - ED = 15-2.4 = 12.6m.
Hence the required height is 12.6 m.

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

Put \(x=a\ cos\theta \)
\(f(x)={ tan }^{ -1 }\sqrt { \frac { a-acos\theta }{ a+acos\theta } } ={ tan }^{ -1 }\sqrt { \frac { 1-cos\theta }{ 1+cos\theta } } \)
= \({ tan }^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\frac { \theta }{ 2 } }{ 2{ cos }^{ 2 }\frac { \theta }{ 2 } } } =tan|tan\frac { \theta }{ 2 } |={ tan }^{ -1 }\left( tan\frac { \theta }{ 2 } \right) \)
= \(\frac { \theta }{ 2 } \) \([\because-a \(\left.\Rightarrow 0<\frac{\theta}{2}<\frac{\pi}{2} \therefore\left|\tan \frac{\theta}{2}\right|=\tan \frac{\theta}{2}\right]\)
= \(\frac { 1 }{ 2 } .{ cos }^{ -1 }\left( \frac { x }{ a } \right) \)\(\left[ \because x=acos\theta \Rightarrow cos\theta =\frac { x }{ a } \Rightarrow { cos }^{ -1 }\left( \frac { x }{ a } \right) \right] \)

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]
\(\lambda =\frac { 5 }{ 2 } \)

The augmented matrix [A|B] is \(\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 1 & \lambda \\ 1 & -7 & 8 \end{matrix}|\begin{matrix} 12 \\ \mu \\ 17 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 3 & 1 & \lambda \\ 2 & -3 & 5 \end{matrix}|\begin{matrix} 17 \\ \mu \\ 12 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2R_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 22 & \lambda -51 \\ 0 & 11 & -11 \end{matrix}|\begin{matrix} 17 \\ \mu -51 \\ -22 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 3 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 11 }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 0 & \lambda -2 \\ 0 & 1 & -1 \end{matrix}|\begin{matrix} 17 \\ \mu -7 \\ -2 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & \lambda -2 \end{matrix}|\begin{matrix} 17 \\ -2 \\ \mu -7 \end{matrix} \right] \)
Case (i) : when λ ≠ 2,
\(\rho\) ([A|B]) = 3 and \(\rho\)(A) = 3
∴ \(\rho\)([AIB])= \(\rho\)(A) = 3 = the number of unknowns
∴ The system has unique solution
Case (ii) : when λ = 2, μ =7
\(\left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 17 \\ -2 \\ 0 \end{matrix} \right] \)
Here \(\rho\)(A) = 2 and \(\rho\)([A|B]) = 2
∴ \(\rho\)(A) = \(\rho\)([A|B]) = 2 < number of unknowns
Thus the system is consistent with infinitely many solutions.
Case (iii) : When λ= 2 and μ ≠ 7
\(\rho\) (A) = 2 and \(\rho\) ([A|B]) = 3
∴ \(\rho\) (A) ≠ \(\rho\) ([A|B])
Thus, the given system of equations is inconsistent.

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.

Given equation is
(2x² - 3x + 1) (2x² + 5x + 1) = 9x² ....(1)
Clearly x = 0 does not satisfy (1),
∴ (1) can be rewritten as  
\(\left( 2x+\frac { 1 }{ x } -3 \right) \left( 2x+\frac { 1 }{ x } +5 \right) =9...(2)\)
put  \(2x+\frac { 1 }{ x } =y\)
∴ (2)⇒ (y - 3) (y + 5) = 9
⇒ y²+ 2y-15 = 9
or y²+ 2y - 24 = 0
⇒ (y + 6) (y - 4) = 0
⇒  y = -6, -4
Case (i)
When \(y=-6,2x+\frac { 1 }{ x } =-6\)
2x²+6x+1 = 0
\(\Rightarrow =\frac { -6\pm \sqrt { 36-8 } }{ 4 } \)
\(x=-3\pm \frac { \sqrt { 7 } }{ 2 } \)
Case(ii)
When \(y=4,2x+\frac { 1 }{ x } =4\)
⇒ 2x²- 4x+1 = 0
\(x=\frac{4 \pm \sqrt{16-8}}{4}\)
\(x=\frac { 2\pm \sqrt { 2 } }{ 2 } \)
This, the roots are
\(\\ \frac { -3+\sqrt { 7 } }{ 2 } ,\frac { -3-\sqrt { 7 } }{ 2 } ,\frac { 2+\sqrt { 2 } }{ 2 } ,\frac { 2-\sqrt { 2 } }{ 2 } \)

Given equation is (a²+b²+c²) p²-2
(ab+bc+cd) p+(b²+c²+d²) ≤ 0...(1)
(1) can be rewritten as
(a²p² - 2abp + b²) + (b²p² - 2bcp + c²) + (c²p2 - 2cdp + d2) ≤ 0
Since a, b, c, d, p∈ R
(ap-b)² ≥ 0, (bp-c)² ≥ 0 and (cp-d)² ≥ 0
∴ (2) will be satisfied only if
ap - b = 0, bp - c = 0, cp - d = 0
\(\Rightarrow \frac { b }{ a } =\frac { c }{ b } =\frac { d }{ c } =p\)
⇒ a, b, c, d are in G.P and ad = bc

2x + 3y - z = 9, x + y + z = 9, 3x - y - z = -1
The matrix form of the system is
\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
⇒ AX = B where A =\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right] \)
X =\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
⇒ X = A^-1N
|A| = \(\left| \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right| =2\left| \begin{matrix} 1 & 1 \\ -1 & -1 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| \)
= 2(-1+1)-3(-1-3)-1(-1-3)
= 0-3(-4) = 12+4 = 16
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 1 & 1 \\ -1 & -1 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & -1 \\ -1 & -1 \end{matrix} \right| & +\left| \begin{matrix} 2 & -1 \\ 3 & -1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & -1 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & -1 \\ 1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \end{matrix} \right] \)
=\(\left[ \begin{matrix} +(-1+1) & -(-1-3) & +(-1-3) \\ -(-3-1) & +(2+3) & -(-2-9) \\ +(3+1) & -(2+1) & +(2-3) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 0 & 4 & -4 \\ 4 & 1 & 11 \\ 4 & -3 & -1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA\frac { 1 }{ 16 } \left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 16 } \left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
= \(\frac { 1 }{ 16 } \left[ \begin{matrix} 0+36-4 \\ 36+9+3 \\ -36+99+1 \end{matrix} \right] =\frac { 1 }{ 16 } \left[ \begin{matrix} 32 \\ 48 \\ 64 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 3 \\ 4 \end{matrix} \right] \)
∴ x = 2, y = 3, z = 4
∴ Solution set is {2, 3, 4}

Here the number of unknowns is 3. So, if the system is consistent and has a non-trivial solution, then the rank of the coefficient matrix is equal to the rank of the augmented matrix and is less than 3.
So the determinant of the coefficient matrix should be 0.
Hence we get
\(\left| \begin{matrix} 3\lambda -8 & 3 & 3 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 or \(\left| \begin{matrix} 3\lambda -2 & 3\lambda -2 & 3\lambda -2 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by applying R₁ ➝ R₁ + R₂ + R₃)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by taking out (3λ − 2) from R₁)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -11 & 3 \\ 3 & 3 & 3\lambda -11 \end{matrix} \right| \) = 0 (by applying R₂ ➝ R₂ - 3R₁, R₃ ➝ R₃ - 3R₁)
or (3λ - 2)(3λ - 11)² 0. So λ = \(\frac { 2 }{ 3 } \) and λ = \(\frac { 11 }{ 3 } \).
We now give an application of system of linear homogeneous equations to chemistry. You are already aware of balancing chemical reaction equations by inspecting the number of atoms present on both sides.

We have to find \((\sqrt{3}+1)^{\frac{1}{3}}\). Let \(z=(\sqrt{3}+i)^{\frac{1}{3}}\). Then \({ z }^{ 3 }=\sqrt { 3 } +i=r\left( cos\theta +isin\theta \right) \)
Then, \(r=\sqrt { 3+1 } =2\) and \(\alpha =\theta =\frac { \pi }{ 6 } \) (\(\because \sqrt{3}+i\) lies in the first quadrant)
Therefore, \({ z }^{ 3 }=\sqrt { 3 } +i=2\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)
\(\Rightarrow z=\sqrt [ 3 ]{ 2 } \left( cos\left( \frac { \pi +12k\pi }{ 18 } \right) +isin\left( \frac { \pi +12k\pi }{ 18 } \right) \right) \), k = 0, 1, 2.
Taking k = 0, 1, 2, we get
k = 0, z \(={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 1, \(z={ z }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 2, \(z={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { 25\pi }{ 18 } +sin\frac { 25\pi }{ 18 } \right) ={ 2 }^{ \frac { 1 }{ 3 } }\left( -cos\frac { 7\pi }{ 18 } -sin\frac { 7\pi }{ 18 } \right) \)

Let P(x, y) be the location of explosion
Given PB - PA = 6

Using distance formula,
\(\sqrt { { (x-5) }^{ 2 }+({ y-0) }^{ 2 } } -\sqrt { ({ x+5 })^{ 2 }+({ y-0) }^{ 2 } } =6\)
\(\Rightarrow \sqrt { { (x-5) }^{ 2 }+({ y })^{ 2 } } -\sqrt { (x+{ 5) }^{ 2 }+({ y) }^{ 2 } } =6\)
Squaring both sides, we get,
(x - 5)² + y² + (x + 5)² + y²
\(-2\sqrt { [(x-{ 5 })^{ 2 }+{ y }^{ 2 }][(x+{ 5 })^{ 2 }+{ y }^{ 2 }] } \) = 36

\(2\sqrt { [(x-{ 5 })^{ 2 }+{ y }^{ 2 }][(x+{ 5 })^{ 2 }+{ y }^{ 2 }] } \)
⇒ 2x² + 2y²+14
= \(2\sqrt { ({ x }^{ 2 }-10x+25+{ y }^{ 2 })({ x }^{ 2 }+10x+25+{ y }^{ 2 }) } \)
⇒ x² + y² + 7
= \(2\sqrt { ({ x }^{ 2 }-10x+25+{ y }^{ 2 })({ x }^{ 2 }+10x+25+{ y }^{ 2 }) } \)
Squaring again,
x⁴+ y⁴ + 49 +2x²y² + 14y² + 14x²
= (x² - 10x + 25 + y²) (x² + 10x + 25 + y²)

⇒ 14y² + 14x² = -50x² + 50y² + 625 - 49
⇒ 64x² - 36y² = 576
\(\div \) 4 we get,
16x² - 9y = 144
\(\frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\)
Hence the location of explo ion is restricted to a hyperbola whose equation is \(\frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\)

x²+3y² = 12
\(\div 12\) we get, \(\frac { { x }^{ 2 } }{ 12 } +\frac { { y }^{ 2 } }{ 4 } =1\)
∴ a² = 12, b² = 4
The line x-y+ 4 = 0 can be rewritten as y = x+4.
∴ m = 1, c = 4
The condition for y = mx + 4 to be a tangent to the ellipse is c² = a²m² + b²
∴ (4)² = 12(1)² + 4
⇒ 16 = 12+4
⇒ 16 = 16
Since the condition is satisfied, the line x - y + 4 = 0 is a tangent to the ellipse x² + 3y² = 12.
Also, the point of contact is \(\left( -\frac { { a }^{ 2 }m }{ c } ,\frac { { b }^{ 2 } }{ c } \right) \)
\(\Rightarrow \left( -\frac { 12(1) }{ 4 } ,\frac { 4 }{ 4 } \right) \Rightarrow (-3,1)\)
∴The point of contact is (-3, 1).

We know that \(sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) =cos^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
Thus, \(cos\left( sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right) =\frac { 1 }{ \sqrt { 1+x^{ 2 } } } \) ...(1)
Let \(\cot ^{-1}\left(\frac{3}{4}\right)=\theta\). Then \(\cot \theta=\frac{3}{4}\) and so \(\theta\) is cute. 
From the diagram, we get,
Hence \(\sin \left\{\cot ^{-1}\left(\frac{3}{4}\right)\right\}=\sin \theta=\frac{4}{5}\) ................ (2)
Using (1) and (2) in the given equation, we \(\frac { 1 }{ \sqrt { 1+x^{ 2 } } } =\frac { 4 }{ 5 } \) \(\sqrt{1+x^2}=\frac{5}{4}\)
Thus, x = \(\pm\frac{3}{4}\)

Now, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } =tan^{ -1 }{ a }_{ 2 }-tan^{ -1 }{ a }_{ 1 }\)
Similarly, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) =tan^{ -1 }{ a }_{ 3 }-tan^{ -1 }{ a }_{ 2 }\)
Continuing inductively, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ n-1 } }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }{ a }_{ n }-tan^{ -1 }{ a }_{ n-1 }\)
Adding vertically, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +....+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) tan[tan^{ -1 }{ a }_{ n }-{ tan }^{ -1 }{ a }_{ 1 }]\\ \)
\(tan\left[ tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +...+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) \right] =tan\left[ tan^{ -1 }{ a }_{ n }-tan^{ -1 }a_{ 1 } \right] \)\(=\left[ tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) \right] =\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \)

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

Let f(x) x⁶-3x⁵-5x⁴+22x³-39x²-39x+135
Given (1+2i) is a root \(\Rightarrow\)(-2i) is also a root
Also \(\sqrt3\) is a root \(\Rightarrow\)-\(\sqrt3\) is also a root.
Hence, the factors of f(x) are [x - (1 + 2i)]
[x-(1-2i)] [x\(\sqrt3\)] [x+\(\sqrt3\)]
[(x-1)-2i] [(x-1)+2i] [x-\(\sqrt3\)][x+\(\sqrt3\)]
((x-1)²+2²)(x²-3) = (x²-2x+1+4)(x²-3)
\(\Rightarrow\) factor of f(x) is (x²-2x+5)(x²-3)
\(\Rightarrow\)x⁴-3x²-2x³+6x+5x²-15
\(\Rightarrow\)(x⁴-3x²-2x³+6x-15) is a factor of f(x) 
To find the other factor, let us divide f(x) by
x⁴ - 2x³ + 2x² + 6x - 15

The other factor is x² - x - 9
\(\Rightarrow x=\frac { 1\pm \sqrt { { (-1) }^{ 2 }-4(1)(-9) } }{ 2 } \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
\(\Rightarrow x=\frac { 1\pm \sqrt { 37 } }{ 2 } \)
Hence the roots are
1 - 2i, 1 + 2i, \(\sqrt { 3 }, -\sqrt { 3 }, \frac { 1+\sqrt { 37 } }{ 2 } ,\frac { 1-\sqrt { 37 } }{ 2 } \).

The given equation is same as
(2x-3)(3x-2)(6x-1)(x-2)-5 = 0
After a computation, the above equation becomes
(6x²-13x+6)(6x²-13x+12)-5 = 0
By taking y = 6x²-13x, the above equation becomes
(y+6)(y+12)-5 = 0
which is same as
y²+18y+7 = 0
Solving this equation, we get y = −1 and y = −7.
Substituting the values of y in y = −6x²-13x, we get
6x²-13x+1 = 0
6x²-13x+7 = 0
Solving these two equations, we get
x = 1, x = \(\frac { 7 }{ 6 } \), x = \(\frac { 13 + \sqrt { 145 } }{ 12 } \) and x = \(\frac { 13-\sqrt { 145 } }{ 12 } \)
as the roots of the given equation.

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α and β, respectively, with positive x-axis, where A and B are as in the diagram.
Draw AL and BM perpendicular to the x-axis. Then \(\left| \vec { OL } \right| =\left| \vec { OA } \right| \) cos α = cos α, \(\left| \vec { LA } \right| =\left| \vec { OA } \right| \) sin α =  sin α
So, \(\vec { OL } =\left| \vec { OL } \right| \)\(\hat { i } \) =  cos,α \(\hat { i } \), \(\overrightarrow { LA } \) = sin α (-\(\hat { j } \))
Therefore, \(\hat { a } =\overrightarrow { OA} = \overrightarrow { OL } +\overrightarrow { LA } \) = cos α \(\hat { i } \) - sin α \(\hat { j } \) ..(1)
Similarly \(\hat { b } \) = cos β \(\hat { i } \)+ sin β \(\hat { j } \) ....(2)
The angle between \(\hat { a } \) and \(\hat{b}\) is α + β and so,
\(\hat { a } .\hat { b } =\left| \hat { a } \right| \left| \hat { b } \right| \) cos (α + β) = cos (α + β) ... (3)

On the other hand, from (1) and (2)
\(\hat { a } .\hat { b } =(cos\alpha \hat { i } -sina\hat { j } )(cos\beta \hat { i } -sin\beta \hat { j } )\) = cos α cos β - sin α sin β....(4)
From (3) and (4), we get cos(α + β) = cos α cos β - sin α sin β

(i) Since f (x) is a probability mass function, f (x) ≥ 0 for all x , and d \(\sum_{x} f(x)=1\)
Thus, \(\sum_{x} f(x)=1\)
\(c^{2}+2 c^{2}+3 c^{2}+4 c^{2}+c+2 c=0\)
\(c=\frac{1}{5} \text { or }-\frac{1}{2}\)
Since f x( ) ≥ 0 for all x , the possible value of c is \(\frac{1}{5}\)
Hence, the probability mass function is

 (ii) To find mean and variance, let us use the following table

Mean :  \(E(X)=\Sigma xf(x)=\frac { 115 }{ 25 } =4.6\) 
Variance : \(V(x)=E\left( x \right) ^{ 2 }=\Sigma { x }^{ 2 }f(x)-\left( \Sigma xf(x) \right) ^{ 2 }\) 
= \(\frac { 585 }{ 25 } -\left( \frac { 115 }{ 25 } \right) ^{ 2 }=23.40-21.16=2.24\)
Therefore the mean and variance are 4.6 and 2.24 respectively.

Let p be the probability that indicates the defective rate of an electronic device
n = 10
\(P=5\%=0.05 \) 
q = 1 - p
n = 10, p = 0.05, X ~ B(n, p)
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1,2, .., n
(i) Atleast 1 defective item
P(X ≥ 1) = 1 - P(X < 1)
= 1-P(X = 0)
= 1-¹⁰C₀ (0.05)⁰ (0.95)¹⁰
P(X ≥1) = 1 - (0.95)¹⁰
(ii) Exactly two defective items  
P(X = 2) =¹⁰C₂(0.05)² (0.95)⁸

f(x) = x²e^-2x
f(x) = x² e^-2x
f'(x) = x² (-2) e^-2x + e^-2x(2x)
f''(x) = 2xe^-2x (1 -x)
f'(x) = 0
⇒ 2x e^-2x(1 - x) = 0
⇒ x = 0,1
∴ The critical numbers are x = 0, 1
f"(x) = [x e^-2x (-1) + x (-2)e^-2x(1-x)+1e^-2x(1-x)]
= 2e^-2x(-x^-2x+2x²+ 1-x)
= 2e^-2x(2x²- 4x + 1)
f"(0) = 2(1)(1) = 2 > 0
f"(1) = 2e^-2 (2 - 4 + 1)
= 2e^-2 (-1)
= \(-{ 2e }^{ 2 }=\frac { -2 }{ { e }^{ 2 } } <0\)
Since f"(0) > 0, there is a local minimum at x = 0.
ஃ(0) = 0² e⁰ = 0
Since f"(1) < 0, there is a local maximum at x = 1.
\(\therefore f(1)={ 1 }^{ 2 }e^{ -2(1) }={ e }^{ -2 }=\frac { 1 }{ { e }^{ 2 } } \)

The Cartesian equations of the straight line \(\vec { r } =(\hat { i } +\hat { 3j } -\hat { k } )+t(2\hat { i } +3\hat { j } +2\hat { k } )\) is
\(\frac { x-1 }{ 2 } =\frac { y-3 }{ 3 } =\frac { z+1 }{ 2 } \) = s(say)
Then any point on this line is of the form (2s + 1, 3s + 3, 2s -1) ...............(1)
The Cartesian equation of the second line is \(\frac { x-2 }{ 1 } =\frac { y-4 }{ 2 } =\frac { z+3 }{ 4 } =t\) (say)
Then any point on this line is of the form (t + 2, 2t + 4, 4t - 3)....(2)
If the given lines intersect, then there must be a common point. Therefore, for some s, t ∈ R
we have (2s + 1, 3s + 3, 2s −1 ) = (t + 2, 2t + 4, 4t − 3)
Equating the coordinates of x, y and z we get
2s − t = 1, 3s − 2t = 1 and s − 2t =  −1.
Solving the first two of the above three equations, we get s = 1 and t = 1. These values of s and t satisfy the third equation. So, the lines are intersecting.
Now, using the value of s in (1) or the value of t in (2), the point of intersection (3,6,1) of these two straight lines is obtained.
If we take \(\vec{b}=(2\hat { i } +3\hat { j } +2\hat { k } )\) and \(\vec{d}=\hat { i } +\hat { 3j } -\hat { k } \),
then \(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 2 \\ 1 & 2 & 4 \end{matrix} \right| =8\hat { i } -6\hat { j } +\hat { k } \) is a vector perpendicular to both the given straight lines.
Therefore, the required straight line passing through (3,6,1)
and perpendicular to both the given straight lines is the same as the straight line passing through (3,6,1) and parallel to \(8\hat { i } -6\hat { j } +\hat { k } \). Thus, the equation of the required straight line is
\(\vec { r } =(\hat { 3i } +\hat { 6j } -\hat { k } )+m(8\hat { i } -6\hat { j } +\hat { k } )\), k ∈ R.