\(\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\)
\(\Rightarrow \ \tan ^{-1}\left(\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}\right)=\frac{\pi}{4}\)
\(\Rightarrow\left(\frac{\frac{(x-1)(x+2)+(x+1)(x-2)}{(\not x-\not 2)(\not x+\not 2)}}{\frac{(x-2)(x+2)-(x-1)(x+1)}{(\not x-\not 2)(\not x+\not 2)}}\right)=\tan \frac{\pi}{4}=1\)
\(\Rightarrow \ \frac{x^{2}+x-2+x^{2}-x-2}{\left(x^{2}-4\right)-\left(x^{2}-1\right)}=1\)
\(\Rightarrow \ \frac{2 x^{2}-4}{x^{2}-4-x^{2}+1}=1\)
\( \Rightarrow 2 x^{2}-4=-3 \)
\(\Rightarrow 2 x^{2}=-3+4=1 \)
\(\Rightarrow x^{2}=\frac{1}{2} \)
\(\Rightarrow x=\frac{1}{\sqrt{2}}\)

Let the roots are in H.P. Then their reciprocals are in A.P.
\( \frac{15}{x^{3}}-\frac{23}{x^{2}}+\frac{9}{x}-1=0 \)
\( 15-23 x+9 x^{2}-x^{3}=0 \)
\(x^{3}-9 x^{2}+23 x-15=0\)
Since the roots are in A.P. Assume that the roots are in the form a-d, a, a + d
\( \Sigma_{1}=a-d+a+a+d =9 \)
\(3 a =9 \)
\(a =3 \)
\(\Sigma_{3}=(a-d)(a)(a+d) =15 \)
\(a\left(a^{2}-d^{2}\right) =15 \)
\(3\left(9-d^{2}\right) =15 \)
\(9-d^{2} =\frac{15}{3} \)
\(9-d^{2} =5 \\d^{2} =4 \)
\(d =\pm 2 \)
a = 3, d = 2 then roots are 1, 3, 5
a = 3, d = -2 then roots are 5, 3, 1

(i) The sample space S = {H,T}\(\times\){H,T}
That is S = {TT,TH,HT,HH}
(ii)  Let X : S ⟶R be the number of tails
Then X (TT) = 2 (2 Tails)
X (TH ) = 1 (1 Tail)
X (HT) = 1 (1 Tail)
and X (HH) = 0 (0 Tails).
Then X is a random variable that takes on the values 0, 1 and 2.
Let X (ω) denotes the number of tails, this gives
\(\\ \\ \\ \\ \\ \\ X\left( \omega \right) =\begin{cases} \begin{matrix} 2 & if\omega =TT \end{matrix} \\ \begin{matrix} 1 & if\omega =HT,TH \end{matrix} \\ \begin{matrix} 0 & if\omega =HH \end{matrix} \end{cases}\) 
The inverse images of 1 {TH, HT} . That is X^-1{1} = {TH, HT}.
(iii) Number of elements in inverse images are shown in the table.

Let f (t) be the distance travelled by the trucker in 't' hours. This is a continuous function in [0, 2] and differentiable in (0, 2). Now, f (0) = 0 and f (2) =164. By an application of the Mean Value Theorem, there exists a time c such that, \(f'(c)=\frac{164-0}{2-0}=82>80\)
Therefore at some point of time, during the travel in 2 hours the trucker must have travelled with a speed more than 80 km which justifies the issuance of a speed violation ticket.

Let \(\vec { b } =5\hat { i } +6\hat { j } +7\hat { k } \) and \(\vec { a } =7\hat { i } +9\hat { j } +13\hat { k } \) 
The parametric form of vector equation of a straight line passing through two points \(\vec { a } \) and \(\vec { b } \) is
\(\vec { r } =\vec { a } +t(\vec { b } -\vec { a } )\)
∴ \(\vec { r } =(7\hat { i } +9\hat { j } +13\hat { k } )+t(7-5)\hat { i } +(9-6)\hat { j } +(13-7)\hat { k } \)
\(\vec { r } =(7\hat { i } +9\hat { j } +13\hat { k } )+t(2\hat { i } +3\hat { j } +6\hat { k } ),t\in R\)
The Cartesian equation of a straight line passing through two points as
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \) [(x₁,y₁,z₁) is (7,9,13) (x₂,y₂,z₂) is (5,6,7))
\(\frac { x-7 }{ 5-7 } =\frac { y-9 }{ 6-9 } =\frac { z-13 }{ 7-13 } \)
⇒ \(\frac { x-7 }{ -2 } =\frac { y-9 }{ -3 } =\frac { z-13 }{ -6 } \)
= \(\frac { x-7 }{ 2 } =\frac { y-9 }{ 3 } =\frac { z-13 }{ 6 } \)

Let \(\vec { { F }_{ 1 } } \) and \(\vec { { F }_{ 2 } } \) be the two forces given
Given \(|\vec { { F }_{ 1 } } |=5\sqrt { 2 } \) and its direction is along \(3\hat { i } +4\hat { j } +5\hat { k } \)
∴ \(\vec { { F }_{ 1 } } =5\sqrt { 2 } \) (unit vector of \(3\hat { i } +4\hat { j } +5\hat { k } \))
= \(5\sqrt { 2 } \frac { (3\hat { i } +4\hat { j } +5\hat { k } ) }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 } } } \)   \(\left[ \because \hat { n } =\frac { \vec { n } }{ |\vec { n } | } \right] \)
= \(5\sqrt { 2 } \frac { (3\hat { i } +4\hat { j } +5\hat { k } ) }{ \sqrt { 9+16+25 } } =\frac { 5\sqrt { 2 } (3\hat { i } +4\hat { j } +5\hat { k } ) }{ 5\sqrt { 2 } } \)
= \(3\hat { i } +4\hat { j } +5\hat { k } \)
and  \(\vec { { F }_{ 2 } } =10\sqrt { 2 } \) (unit vector of \(10\hat { i } +6\hat { j } -8\hat { k } \))
= \(10\sqrt { 2 } \frac { (10\hat { i } +6\hat { j } -8\hat { k } ) }{ \sqrt { { 10 }^{ 2 }+{ 6 }^{ 2 }+(-8)^{ 2 } } } \)
\(=\frac{10 \sqrt{\not 2}(10 \hat{i}+6 \hat{j}-8 \hat{k})}{10 \sqrt\not {2}}\)
= \(10\hat { i } +6\hat { j } -8\hat { k } \)
∴ Resistant force \(\vec { F } =\vec { { F }_{ 1 } } +\vec { { F }_{ 2 } } \)
=\((3\hat { i } +4\hat { j } +5\hat { k } )+(10\hat { i } +6\hat { j } -8\hat { k } )\)
\(\vec { F } =13\hat { i } +10\hat { j } -3\hat { k } \)
\(\hat { d } \) = displacement to the point - displacement from the point
= \((6\hat { i } +\hat { j } -3\hat { k } )-(4\hat { i } -3\hat { j } -2\hat { k } )\)
= \(2\hat { i } +4\hat { j } -\hat { k } \)
∴ Work done
w = \(\vec { F } .\vec { d } =(13\hat { i } +10\hat { j } -3\hat { k } ).(2\hat { i } +4\hat { j } -\hat { k } )\)
w = 13(2) + 10(4) - 3(-1) = 26 + 40 + 3
w = 69 units.

Given vertex A(5, -2) and focus S(2, -2) and the focal distance
AS = a = 3  
Parabola is open left and symmetric about the line parallel to x -axis.
Then, the equation of the required parabola is
(y + 2)² = −4(3)(x − 5)
y² + 4y + 4 = −12x + 60
y² + 4y +12x − 56 = 0  

Equation of the parabola is y² = 4ax
Latus rectum LL′ passes through the focus (a, 0)  
Hence the point L is (a, y₁) 
Therefore y₁² = 4a²
Hence y₁ = ±2a
The end points of latus rectum are (a, 2a) and (a, -2a)  

\(={ sin }^{ -1 }\left( sin\frac { 5\pi }{ 9 } cos\frac { \pi }{ 9 } +cos\frac { 5\pi }{ 9 } sin\frac { \pi }{ 9 } \right) \)
= \({ sin }^{ - }\left( sin\left( \frac { 5\pi }{ 9 } +\frac { \pi }{ 9 } \right) \right) \)
(\(\because \) sin A cos B + cos A sin B = sin (A + B))
= \({ sin }^{ -1 }\left( sin\left( \frac { 6\pi }{ 9 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \frac { 2\pi }{ 3 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \pi -\frac { \pi }{ 3 } \right) \right) \) \(\left[ \because \frac { 2\pi }{ 3 } \notin \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
= \({ sin }^{ -1 }\left( sin\frac { \pi }{ 3 } \right) \) \(\left( \because sin\left( \pi -\theta \right) =sin\theta \right) \)
= \(\frac { \pi }{ 3 } \) \(\left[ \because \frac { \pi }{ 3 } \quad \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \) 

From the given data the equation is of the form
\(\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1\)
Centre C (h, k) is (2, 1)
CF = ae = 6
(Draw CZ perpendicular to x = 4)
The distance between the centre and directrix
\( C Z=\frac{a}{e}=2 \)
(a e)\( \frac{a}{e}=6 \times 2 \Rightarrow a^{2} =12 \)
\(\frac{a e}{a / e} =\frac{e}{2} \Rightarrow e^{2}=3 \)
\(b^{2} =a^{2}\left(e^{2}-1\right) \)
\(\therefore b^{2} =12(3-1)=24\)
The required equation is
\(\frac{(x-2)^{2}}{12}-\frac{(y-1)^{2}}{24}=1\)

Y = x² is continuous on [-2, 2] and differentiable on [-2, 2]
f(a) = f(-2) = (-2)² = 4
f(b) = f(2) = 2² = 4
∴ f(a) = f(b)
Since the tangent is parallel to X - axis, f'(c) = 0
⇒ 2c = 0
⇒ c = 0
∴ When c = 0, y = 0
∴ AE (0, 0) the tangent is parallel to X- axis.

Given \(p=\frac { 3 }{ 4 } \) 
n = 5
P(X = x) = nC_xp^x (1-p)^n-x
\(P(X=3)={ 5C }_{ 3 }\left( \frac { 3 }{ 4 } \right) ^{ 3 }\left( 1-\frac { 3 }{ 4 } \right) ^{ 2 }\) 
\(P(X=3)={ 5C }_{ 2 }\left( \frac { 3 }{ 4 } \right) ^{ 3 }\left( \frac { 1 }{ 4 } \right) ^{ 2 }\) [∵nCr = nC_n-r]

= \(\frac { 135 }{ 512 } \)

If we directly substitute x = 0 we get an indeterminate form \(\frac{0}{0}\) and hence we apply the l’Hôpital’s rule to evaluate the limit as
\(\underset{x\rightarrow 0}{lim}(\frac{sin \ mx}{x})\)=\(\underset{x\rightarrow 0}{lim}(\frac{m\times cos \ mx}{1})\)
= m
The next example tells that the limit does not exist.

Given v = x³
Differentiating with respect to x we get,
\(\frac { dv }{ dt } \) = 3x²
When x = 5, \(\frac { dv }{ dt } \) = 3(52) = 75
∴ \(\frac { dv }{ dt } \) when x = 5 is 75 units.

Equation of the circle is
2x² + 2y² - 6x + 4y + 2 = 0
Dividing by 2, we get
x² + y² - 3x + 2y + 1 = 0
Here 2g = -3 ⇒ g = \(\frac { -3 }{ 2 } \)
2f = 2 ⇒ f = 1
and c = 1
∴ Centre is (-g, -f) = \(\left( \frac { 3 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { \left( \frac { 3 }{ 2 } \right) }^{ 2 }+{ 1 }^{ 2 }-1 } \)
= \(\sqrt { \frac { 9 }{ 4 } } =\frac { 3 }{ 2 } \) units.

Let \(\vec { a } =-6\hat { i } +14\hat { j } +10\hat { k } \), \(\vec { b } =14\hat { i } -10\hat { j } -6\hat { k } \) and \(\vec { c } =2\hat { i } +4\hat { j } -2\hat { k } \)
Volume of the parallelepiped having \(\vec { a } ,\vec { b } \) and \(\vec { c } \) as its co-terminus edges is \(\vec { a } .(\vec { b } \times \vec { c } )\).
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{matrix} \right| \)
= \(-6\left| \begin{matrix} -10 & -6 \\ 4 & -2 \end{matrix} \right| -14\left| \begin{matrix} 14 & -6 \\ 2 & -2 \end{matrix} \right| +10\left| \begin{matrix} 14 & -10 \\ 2 & 4 \end{matrix} \right| \)
= -6(20 + 24) - 14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760 = 720.
∴ Volume of the required parallelepiped = 720 cubic units.

Given that 2a = 36.18, 2b = 9.12 , we get
e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\frac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \) = \(\frac { \sqrt { { \left( \frac { 36.18 }{ 2 } \right) }^{ 2 }{ \left( \frac { 9.12 }{ 2 } \right) }^{ 2 } } }{ \frac { 36.18 }{ 2 } } \)
\(\frac { \sqrt { { \left( 18.09 \right) }^{ 2 }-{ \left( 4.56 \right) }^{ 2 } } }{ \left( 8.09 \right) } \approx0.97\)

Here Δ = b²−4ac = 0 for equal roots. This implies 4(k + 2)² = 4(9)k. This implies k = 4 or 1.

cos^-1(-x) = \(\pi\)-cos⁻¹(x)
Let cos^-1(-x) = \(\theta \)   ..(1)
\(\Rightarrow -x=cos\theta \)
\(\Rightarrow x=-cos\theta =cos\theta =cos\left( \pi -\theta \right) \)
\(\Rightarrow \pi -\theta ={ cos }^{ -1 }\left( x \right) \)
\(\Rightarrow \theta =\pi -{ cos }^{ -1 }x\) ...(2)
From (1) & (2) \({ cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }\left( x \right) \)
\({ cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }\left( x \right) \) is true.

\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 0 & 1 \end{array}\right|=\hat{i}+\hat{j}-2 \hat{k}\)
\(\vec{c} \times \vec{d}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{array}\right|=\hat{i}-3 \hat{j}+\hat{k}\)
\((\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \end{array}\right|=-5 \hat{i}-3 \hat{j}-4 \hat{k}\) .......(1)
\([\vec{a}, \vec{b}, \vec{c}]=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 2 & 1 & 1 \end{array}\right|=1\)
\([\vec{a}, \vec{b}, \vec{c}]=\left|\begin{array}{lll} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 1 & 1 & 2 \end{array}\right|=-2\)
\( {[\vec{a}, \vec{b}, \vec{d}] \vec{c}-[\vec{a}, \vec{b}, \vec{c}] \vec{d} } =(-4 \hat{i}-2 \hat{j}-2 \hat{k})-(\hat{i}+\hat{j}+2 \hat{k}) \)
\( =-5 \hat{i}-3 \hat{j}-4 \hat{k}\) .........(2)
\( (1), (2) \Rightarrow(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=[\vec{a}, \vec{b}, \vec{d}] \vec{c}-[\vec{a}, \vec{b}, \vec{c}] \vec{d}\)

\( \mathrm{x} =a \cos ^{3} \theta \mathrm{y} =a \sin ^{3} \theta \)
\(\frac{d x}{d \theta} =-3 a \cos ^{2} \theta \sin \theta \frac{d y}{d \theta} =3 a \sin ^{2} \theta \cos \theta\)
Slope of the tangent \( \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta} \)
\(=-\frac{\sin \theta}{\cos \theta}=\mathrm{m} \)
Slope of the normal = \(\frac{\cos \theta}{\sin \theta}=-\frac{1}{m}\)
Equation of normal is \(y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right)\)
\( \Rightarrow \quad y-a \sin ^{3} \theta =\frac{\cos \theta}{\sin \theta}\left(x-a \cos ^{3} \theta\right) \)
\(y \sin \theta-a \sin ^{4} \theta =x \cos \theta-a \cos ^{4} \theta \)
\(x \cos \theta-y \sin \theta =a\left(\cos ^{4} \theta-\sin ^{4} \theta\right) \)
\( =a\left(\cos ^{2} \theta+\sin ^{2} \theta\right)\left(\cos ^{2} \theta-\sin ^{2} \theta\right) \)
\(\therefore x \cos \theta -y \sin \theta=a \cos 2 \theta \)
Hence Proved.

Let X be random variable of denoting number of red balls
X = {0, 1, 2, 3}
n(S) = 7 \(\times\)7 \(\times\)7 = 343

Probability mass function

Mean:
\( \mu=\mathrm{E}(\mathrm{X})= 0 \times \frac{64}{343}+1 \times \frac{144}{343} +2 \times \frac{108}{343}+3 \times \frac{27}{343} \\ \)
\(= \frac{9}{7} \)
Variance:
\( \mathrm{E}\left(\mathrm{X}^{2}\right)= 0 \times \frac{64}{343}+1^{2} \times \frac{144}{343}+2^{2} \times \frac{108}{343} +3^{2} \times \frac{27}{343}=\frac{117}{49} \)
Variance \( =E\left(X^{2}\right)-[E(X)]^{2} =\frac{117}{49}-\left(\frac{9}{7}\right)^{2} \)
\( =\frac{36}{49}\)

Rearranging terms in the equation of hyperbola to bring it to standard form,
We have,
\( 12\left(x^{2}-2 x\right)-4\left(y^{2}-8 y\right)-124 =0 \)
\(12(x-1)^{2}-4(y-4)^{2} =124+12-64 \)
\(12(x-1)^{2}-4(y-4)^{2} =72 \)
\(\frac{(x-1)^{2}}{6}-\frac{(y-4)^{2}}{18} =1\)
Centre (1, 4); a² = 6;
\( b^{2} =18 \)
\(c^{2} =a^{2}+b^{2} \)
\( =6+18 \)
\(c^{2} =24 \)
\(c =\pm 2 \sqrt{6} \)
\(e =\frac{c}{a}=\frac{2 \sqrt{6}}{6}\)
eccentricity, \(e=\frac{\sqrt{6}}{3}\)
centre (h, k) = (1, 4)
foci \( =(h, \pm c+k) \)
\(=(1, \pm 2 \sqrt{6}+4) \)
\( =(1,2 \sqrt{6}+4) 8(1,-2 \sqrt{6}+4)\)

(i) Probability mass function
For a discrete random variable we have
f(x) = p(X = x)
\(\therefore f(0)=F(0)=\frac { 1 }{ 2 } \) 
f(1) = F(1) - F(0)
= \(\frac { 3 }{ 5 } -\frac { 1 }{ 2 } =\frac { 6-5 }{ 10 } =\frac { 1 }{ 10 } \) 
f(2) = F(2)-F(1)
= \(\frac { 4 }{ 5 } -\frac { 3 }{ 5 } =\frac { 1 }{ 5 } \) 
f(3) = F(3) - F(2)
\(\frac { 9 }{ 10 } -\frac { 4 }{ 5 } =\frac { 9-8 }{ 10 } =\frac { 1 }{ 10 } \)
f(4) = F(4)-F(3)
= \(1-\frac { 9 }{ 10 } =\frac { 1 }{ 10 } \) 
ஃThe probability mass function is

(ii) p(x < 3) = p(x = 0) + p(x = 1) + p(x = 2)
= \(\frac { 1 }{ 2 } +\frac { 1 }{ 10 } +\frac { 1 }{ 5 } =\frac { 5+1+2 }{ 10 } =\frac { 8 }{ 10 } \) 
= \(\frac { 4 }{ 5 } \)
(iii) p(x≥2) = p(x = 2) + p(x = 3) + p(x = 4)
= \(\frac { 1 }{ 5 } +\frac { 1 }{ 10 } +\frac { 1 }{ 10 } =\frac { 2+1+1 }{ 10 } =\frac { 4 }{ 10 } \)
= \(\frac { 2 }{ 5 } \)

We have,
\(f(x)=log(1+x)-\frac{x}{1+x}\)
Therefore, \(f'(x)=\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\)
= \(\frac{x}{(1+x)^{2}}\).
Hence, f′(x) is \(\begin{cases} <0 \ when-10 \ whenx>0 \end{cases}\)
Therefore f (x) is strictly increasing for x > 0 and strictly decreasing for x < 0. Since f′(x) changes from negative to positive when passing through x = 0, the first derivative test tells us there is a local minimum at x = 0 which is f (0) = 0. Further, for x > 0, f(x) > f (0) = 0 gives
\(log(1+x)-\frac{x}{1+x}>0 \Rightarrow log(1+x)>\frac{x}{1+x}\).

Given equation of plane is x + 2y + 3z = 2
Length of perpendicular from (4, 3, 2) to the plane is
\(d=\cfrac { 4+2\left( 3 \right) +3\left( 2 \right) }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } } } =\cfrac { 4+6+6 }{ \sqrt { 14 } } \)
= \(\cfrac { 14 }{ \sqrt { 14 } } =\cfrac { \sqrt { 14 } .\sqrt { 14 } }{ \sqrt { 14 } } =\sqrt { 14 } \) units
Let us find the image of the point (4,3,2) to the plane x + 2y + 3z = 2
Here \(\vec { u } =4\hat { i } +3\hat { j } +2\hat { k } ,\vec { n } =\hat { i } +2\hat { j } +3\hat { k } \)
Then the image \(\vec { v } =\vec { u } +\cfrac { 2\left[ p-\left( \vec { u } .\vec { n } \right) \right] }{ \left| \vec { n } \right| ^{ 2 } } \)
\(\vec { v } =\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\cfrac { 2\left[ 2-\left( 4+6+6 \right) \right] }{ \left( \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } } \right) } \left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\frac { 2\left( 2-16 \right) }{ 14 } \left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\cfrac { 2\left( -14 \right) }{ 14 } \left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) -2\left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) -2\left( \hat { i } +4\hat { j } -6\hat { k } \right) \)
= \(2\hat { i } -\hat { j } -4\hat { k } \)
\(\therefore\) The foot of the \(\bot \) from (4, 3, 2) to the plane is
\(\cfrac { \left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\left( 2\hat { i } -\hat { j } -4\hat { k } \right) }{ 2 } \)
= \(\cfrac { 6\hat { i } +2\hat { j } -2\hat { k } }{ 2 } =3\hat { i } +\hat { j } -\hat { k } \)
Hence, the co-ordinates of the foot of the perpendicular is (3, 1, -1)

Since the centre is located at (0, 300), midway between the two foci, which are the coast guard stations, the equation is \(\frac { { \left( y-300 \right) }^{ 2 } }{ { a }^{ 2 } } -\frac { { \left( x-0 \right) }^{ 2 } }{ { b }^{ 2 } } =1\).... (1)
To determine the values of a and b, select two points known to be on the hyperbola and substitute each point in the above equation.
The point(0, 400) lies on the hyperbola, since it is 200 km further from Station A than from station B.
\(\frac { { \left( 400-300 \right) }^{ 2 } }{ { a }^{ 2 } } -\frac { O }{ { b }^{ 2 } } =1\frac { { 100 }^{ 2 } }{ a^{ 2 } } =1,{ a }^{ 2 }=10000.\) There is also a point (x, 600) on the hyperbola such that 600²+ x² = (x + 200)²    
360000 + x² = x²+400x+40000
x = 800
Substituting in (1), we have \(\frac { { \left( 600-300 \right) }^{ 2 } }{ 10000 } -\frac { { \left( 800-0 \right) }^{ 2 } }{ { b }^{ 2 } } =1\)
\(9-\frac { 640000 }{ { b }^{ 2 } } =1\)
b² = 80000
Thus the required equation of the hyperbola is \(\frac { { \left( y-300 \right) }^{ 2 } }{ 10000 } -\frac { { x }^{ 2 } }{ 80000 } =1\)
The ship lies somewhere on this hyperbola. The exact location can be determined using data from a third station.

\(LHS={ tan }^{ -1 }x+{ tan }^{ -1 }\frac { 2x }{ 1-{ x }^{ 2 } } \)
= \({ tan }^{ -1 }\left( \frac { x+\frac { 2x }{ 1-{ x }^{ 2 } } }{ 1-x\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x(1-{ x }^{ 2 })+2x }{ 1-{ x }^{ 2 } } }{ \frac { 1-{ x }^{ 2 }-2{ x }^{ 2 } }{ 1-{ x }^{ 2 } } } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x-{ x }^{ 3 }+2x }{ 1-{ x }^{ 2 } } }{ \frac { 1-3x^{ 2 } }{ 1-{ x }^{ 2 } } } \right) \quad \left[ \because |x|<\frac { 1 }{ \sqrt { 3 } } \right] \)
= \({ tan }^{ -1 }\left( \frac { 3x-{ x }^{ 3 } }{ 1-{ x }^{ 2 } } \times \frac { 1-{ x }^{ 2 } }{ 1-{ 3x }^{ 2 } } \right) \)
If 3x² < 1
⇒ \( |x|<\frac { 1 }{ \sqrt { 3 } }\)

Equation of the ellipse is 2x² + 7y² = 14
\(\div \) we get, \(\frac { { x }^{ 2 } }{ 7 } +\frac { { y }^{ 2 } }{ 2 } =1\)
∴ a² = 7, b² = 2
The condition for the line y = mx + c to be a tangent to the ellipse is
\(y=mx\pm \sqrt { { a }^{ 2 }{ m }^{ 2 }+{ b }^{ 2 } } \) ....(1)
(5, 2) lies on (1) and a² = 7, b² = 2
∴ 2 = \(m(5)\pm \sqrt { 7{ m }^{ 2 }+2 } \)
2 - 5m = \(\pm \sqrt { 7{ m }^{ 2 }+2 } \)

Squaring both sides we get,
(2, - 5m)² = 7m² + 2
4 + 25m²- 20m = 7m² + 2
18m²-2m + 1 = 0 (\(\div \)2)
On factorising we get,
(m-1)(9m-1) = 0
∴ m -1 = 0 or
9m-1 = 0
⇒ m = 1 or \(\frac { 1 }{ 9 } \)
When m = 1, (1) becomes
y = \(1(x)\pm \sqrt { 7{ m }^{ 2 }+2 } \)
⇒ y = x \(x\pm \sqrt { 7+2 } \) ⇒ y = \(\pm \)3
y= x + 3 ⇒ x - y - 3 = 0
When m = \(\frac { 1 }{ 9 } \) (1) becomes
\(y=\frac { 1 }{ 9 } (x)\pm \sqrt { 7\left( \frac { 1 }{ 81 } \right) +2 } \)
⇒ \(y=\frac { x }{ 9 } +\sqrt { \frac { 7+162 }{ 81 } } \)
\(y=\frac { x }{ 9 } +\frac { 13 }{ 9 } \)
⇒ 9y = x + 13 ⇒ x -9y + 13 = 0
Hence the equation of tangents are x - y - 3 = 0 and x −9y + 13 = 0

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
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +....+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) tan[tan^{ -1 }{ a }_{ n }-{ tan }^{ -1 }{ a }_{ 1 }]\\ \)
\(tan\left[ tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +...+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) \right] =tan\left[ tan^{ -1 }{ a }_{ n }-tan^{ -1 }a_{ 1 } \right] \)\(=\left[ tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) \right] =\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \)

In triangle ABC, consider A as the origin. Then the position vectors of D, E, F are given by \(\frac { \vec { AB } +\vec { AC } }{ 2 } ,\frac { \vec { AC } }{ 2 } ,\frac { \vec { AB } }{ 2 } \) respectively.
Since \(​​\left| \vec { AB } \times \vec { AC } \right| \) is the area of the parallelogram formed by the two vectors \(\vec { AB }\), \(\vec { AC } \) as adjacent sides, the area of  ΔABC is \(\frac{1}{2}\) \(​​\left| \vec { AB } \times \vec { AC } \right| \). Similarly, considering ΔDEF, we get

the area of ΔDEF =  \(\frac{1}{2}\)  \(​​\left| \vec { DE } \times \vec { DF } \right| \)
= \(\frac{1}{2}\) \(​​\left| (\vec { AE }-\vec{AD}) \times (\vec { AF }-\vec{AD}) \right|​​\)
= \(\left| \frac { \vec { AB } }{ 2 } \times \frac { \vec { AC } }{ 2 } \right| \)
= \(\frac14\) \(\left( \frac { 1 }{ 2 } \left| \vec { AB } \times \vec { AC } \right| \right) \)
= \(\frac14\)(the area of ΔABC)

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, \(\sqrt { 3 }, -\sqrt { 3 }, \frac { 1+\sqrt { 37 } }{ 2 } ,\frac { 1-\sqrt { 37 } }{ 2 } \).

Given equation of the curve is y = 1 + x³ and the line is x + 12y = 12
Slope of the tangent to the curve
m₁ = \(\frac { dy }{ dx } \) = 3x² and the
Slope of the line = m₂
= \(\frac{-1}{2}\) \(\left[ \because m=\frac { co-efficient\quad of\quad x }{ co-efficient\quad of\quad y } \right] \)
Since the slope of the tangent to the curve and the line are orthogonal, m₁ m₂ = - 1.
∴ 3x²\(\left( \frac { -1 }{ 2 } \right) \) = -1
⇒ \(\frac{x^2}{4}\) = 1
⇒ x² = 4
⇒ x = ±2
When x = 2, y = 1 + 2³ = 9
When x = -2, y = 1+ (-2)³
= 1-8 = -7 
∴ Equation of the tangent at (2, 9) is
y-9 = 12(x-2) [∵ m₁ = 3x² = 3(2)² = 12]
∴ y - 9 = 12x - 24
∴ 12x - y = 15
Equation of the tangent at (-2, -7) is
y + 7= 12(x + 2)
⇒ y + 7 = 12x + 24
⇒ 12x - y = -17