Given \(\frac { dy }{ dx } =-\left( \frac { y+3 }{ x+2 } \right) \)
⇒ \(\frac { dy }{ y+3 } =-\frac { dx }{ x+2 } \)
⇒ \(\int { \frac { dy }{ y+3 } } =-\int { \frac { dx }{ x+2 } } \)
⇒ log(y + 3) = -log (x + 2) + log c
⇒ log(y + 3) + log (x + 2) = log c
⇒ log(x + 2) (y + 3) = log c
⇒ (x+2)(y+3) = c ..(1)
Since the curve passes through (0, 0)
(0 + 2)(0 + 3) = c ⇒ c = 6
(1) becomes,
(x + 2)(y + 3) = 6
⇒ xy + 3x + 2y + 6 = 6
⇒ xy + 3x + 2y = 0

Since the entries in the last column are F, (p ∧ ~p) ∧ (~q ∧ P) is a contradiction

Given that n = 6
Probability for selecting a defective product is \(\frac { 20 }{ 100 } \) that is \(p=\frac { 1 }{ 5 } \) 
Since X denotes the number defective products, X can take on the values 0,1,2,...,6
The probability for defective (success) is \(p-\frac { 1 }{ 5 } \) and for failure \(q=1-p=\frac { 4 }{ 5 } \), and n = 6 
Therefore X follows a binomial distribution denoted by \(X\sim B\left( 6,\frac { 1 }{ 5 } \right) \) 
This gives \(f(x)=\left( \begin{matrix} 6 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 5 } \right) ^{ x }\left( \cfrac { 4 }{ 5 } \right) ^{ 6-x }\), x = 0,1,2,...,6,
(i) Probability for two defective products is
\(P(X=2)=f(2)=\left( \begin{matrix} 6 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 5 } \right) ^{ x }\left( \cfrac { 4 }{ 5 } \right) ^{ 6-x }=15\left( \cfrac { { 4 }^{ 4 } }{ { 5 }^{ 6 } } \right) \)
(ii) Probability for at most one defective products is 
P(X ≤1) = P(X = 0) + P(X = 1)
\(-\left( \begin{matrix} 6 \\ 0 \end{matrix} \right) \left( \cfrac { 1 }{ 5 } \right) ^{ 0 }\left( \cfrac { 4 }{ 5 } \right) ^{ 6-0 }+\left( \begin{matrix} 6 \\ 1 \end{matrix} \right) \left( \frac { 1 }{ 5 } \right) ^{ 1 }\left( \cfrac { 4 }{ 5 } \right) ^{ 6-1 }\) 
\(-\left( \cfrac { 4 }{ 5 } \right) ^{ 6 }+\left( 6 \right) \left( \cfrac { { 4 }^{ 2 } }{ { 5 }^{ 2 } } \right) =2\left( \cfrac { 4 }{ 5 } \right) ^{ 2 }\)
Probability for at most one defective products is \(2\left( \frac { 4 }{ 5 } \right) ^{ 5 }\) 
(iii) Probability for at least two defective products is 
P(X≥2)-1-P(X<2) = 1-P(X≤1) = \(1-2\left( \frac { 4 }{ 5 } \right) ^{ 5 }\) 
Probability for at least two defective products is \(1-2\left( \frac { 4 }{ 5 } \right) ^{ 5 }\)

Given f(x, y) = \(\frac { 3x }{ y+sinx \ } \)
\({ f }_{ x }=\frac { \partial f }{ \partial x } \)
= \(\frac { (y+sin \ x)(3)-3x(cos \ x) }{ { (y+sin \ x) }^{ 2 } } \)
\(=\frac { 3y+3sin \ x-3x \ cos \ x }{ (y+sin{ x })^{ 2 } } \)
\({ f }_{ yx }=\frac { \partial ^{ 2 }f }{ \partial y\partial x } \)
\(=\frac { (y+{ sinx) }^{ 2 }[3]-(3y+3sinx-3xcosx)(2)(y+sinx)(1) }{ { (y+sinx) }^{ 3 } } \)
= \(\frac { (y+{ sinx) }-[3y+3sinx-6y-6sinx+6xcosx] }{ { (y+sinx) }^{ 4 } } \)
\({ f }_{ yx }=\frac { -3y-3sinx+6xcosx }{ { (y+sinx) }^{ 3 } } \) ......(1)
\({ f }_{ y }={ [3x[-1][y+sinx] }^{ -2 }\)
\(=\frac { -3x }{ (y+sin{ x) }^{ 2 } } \)
\(\therefore { f }_{ xy }=-3\left[ \frac { ({ y+sinx) }^{ 2 }(1)-x(2)(y+sinx)(cosx) }{ { (y+sinx) }^{ 4 } } \right] \)
\({ f }_{ xy }=\frac { -3(y+sinx)[y+sinx-2xcosx] }{ { (y+sinx) }^{ 4 } } \)
\({ f }_{ xy }=\frac { -3(y+sinx-2xcosx) }{ ({ y+sinx })^{ 3 } } \) ...(2)
From (1) and (3)
f_xy = f^yx

\(\frac { dy }{ dx } +\frac { 3y }{ x } =\frac { 1 }{ { x }^{ 2 } } \), given that y = 2 when x = 1 
This is a linear differential equation.
\(\therefore P=\frac { 3 }{ x } ;Q=\frac { 1 }{ { x }^{ 2 } } \)
\(\int { pdx } =3\int { \frac { 1 }{ x } dx=3logx=log{ x }^{ 3 } } \)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log{ x }^{ 3 } }={ x }^{ 3 }\)
\(\therefore\) The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 3 }=\int { \frac { 1 }{ { x }^{ 2 } } .{ x }^{ 3 } } dx+c\)
\(\Rightarrow { yx }^{ 3 }=\int { xdx+c } \)
\(\Rightarrow { yx }^{ 3 }=\frac { { x }^{ 2 } }{ 2 } +c...(1)\)
When x = 1, y = 2
\(\Rightarrow 2{ (1) }^{ 3 }=\frac { 1 }{ 2 } +c\Rightarrow 2-\frac { 1 }{ 2 } =\frac { 3 }{ 2 } \)
\({ yx }^{ 3 }=\frac { { x }^{ 2 } }{ 2 } +\frac { 3 }{ 2 } \)
\({ 2x }^{ 3 }y={ x }^{ 2 }+3\)

Let \(f(x)=\frac{1}{x}\). then the Taylor series of f (x) is
\(f(x)=\sum^{n=\infty}_{n=0}a_{n}(x-2)^{n},\) where \(a_{n}=\frac{f^{(n)}(2)}{n!}\)
Various derivatives of the function f (x) evaluated at x = 2 are given below.

Substituting these values, we get the required expansion of the function as:
\(\frac{1}{x}=\frac{1}{2}-\frac{1}{4}\frac{(x-2)}{1!}+\frac{1}{4}\frac{(x-2)^{2}}{2!}-\frac{3}{8}\frac{3(x-2)^{3}}{3!}+...\)
which is, \(\frac{1}{x}=\frac{1}{2}-\frac{(x-2)}{4}+\frac{(x-2)^{2}}{8}-\frac{(x-2)^{3}}{16}+...\)

From the line  \(\frac { x-3 }{ 3 } =\frac { y-8 }{ -1 } =\frac { z-3 }{ 1 } \), we get
\(\overset { \rightarrow }{ a } =3\overset { \wedge }{ i } +8\overset { \wedge }{ j } +3\overset { \wedge }{ k } \), \(\overset { \rightarrow }{ b } =3\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
From the line \(\frac { x+3 }{ -3 } =\frac { y+7 }{ 2 } =\frac { z-6 }{ 4 } \) we get
\(\overset { \rightarrow }{ c } =-3\overset { \wedge }{ i } -7\overset { \wedge }{ j } +6\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ d } =-3\overset { \wedge }{ i } +2\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
Since the given lines are not parallel, the shortest distance between the line is 
\(d=\left| \frac { \left( \overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right) }{ \left| \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right| } \right| \)
\(\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } =-6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } =\left| \begin{matrix} \overset { \wedge }{ i } \\ 3 \\ -3 \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ -1 \\ 2 \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ 1 \\ 4 \end{matrix} \right| \)
\(=\overset { \wedge }{ i } (-4-2)-\overset { \wedge }{ j } (12+3)+\overset { \wedge }{ k } (6-3)\\ \)
\(=-6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \)
\(\therefore \left| \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right| =\sqrt { 36+225+9 } \)
\(=\sqrt { 270 } \)
\(\therefore \left( \overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right) =\left( -6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) .\left( -6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) \)
= -6 (-6) + 15 _(15) + 3(3)
= 270 ≠ 0
Since the given lines are neither intersecting, nor parallel they are skew lines
\(\therefore d=\frac { 270 }{ \sqrt { 270 } } =\sqrt { 270 } units\)

LHS = \(\left( \frac { i+\sqrt { 3 } }{ -i+\sqrt { 3 } } \right) ^{ 2\omega }+\left( \frac { i-\sqrt { 3 } }{ i+\sqrt { 3 } } \right) ^{ 2\omega }\)
= \(\left( \frac { \sqrt { 3 } +i }{ \sqrt { 3 } -i } \times \frac { \sqrt { 3 } +i }{ \sqrt { 3 } +i } \right) ^{ 2\omega }+\left( \frac { -\sqrt { 3 } +i }{ \sqrt { 3 } +i } \times \frac { \sqrt { 3 } -i }{ \sqrt { 3 } -1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 3-1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }+\left( \frac { -3+1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }\)
= \(\left( \frac { 1+\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }\)
=\(\left[ -\left( \frac { -1-\sqrt { 3 } i }{ 2 } \right) \right] ^{ 2\omega }+\left[ \frac { -1+\sqrt { 3 } i }{ 2 } \right] ^{ 2\omega }\)
= (-ω²)^2ω+(ω)^2ω
[∴ ω = \(\frac { -1+i\sqrt { 3 } }{ 2 } \), ω² = \(\frac { -1-i\sqrt { 3 } }{ 2 } \)]
= ω^4ω+ω^2ω
= (ω³)¹³³. ω¹ + (ω³)⁶⁶.ω²
= 1.ω+1.ω²  [∴ 1+ω+ω² = 0 & ω³ = 1]
= ω + ω²
= -1 = RHS

Given F₁P + F₂P = 8
By the focal property of ellipse
F₁P + F₂P = 2a
∴ 2a = 8 ⇒ a = 4
and distance between the foci = F₁F₂ = 6
2ae = 6 ⇒ ae = 3
∴ 4(e) = 3 ⇒ e \(\frac34\)

∴ b² = a²(1- e²)
= \(16\left( { 1-\left( \frac { 3 }{ 4 } \right) }^{ 2 } \right) =16\left( 1-\frac { 9 }{ 10 } \right) =16\left( \frac { 7 }{ 16 } \right) =7\)
∴ The path traced by him is an ellipse and its equation is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 7 } \) = 1

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

Let the required numbers be x, y and z
By the given data,
x + y + z = 20  ....(1)
2z + x = 23 ⇒ x + 2z = 23...(2)
y + z + 3x = 46 ⇒  3x + y + z = 46..(3)
Δ = \(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{matrix} \right| =1\left| \begin{matrix} 0 & 2 \\ 0 & 1 \end{matrix} \right| -1\left| \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix} \right| \)
= -2 + 5 + 1 =4
Δ₁ = \(\left| \begin{matrix} 20 & 1 & 1 \\ 23 & 0 & 2 \\ 46 & 1 & 1 \end{matrix} \right| =20\left| \begin{matrix} 0 & 2 \\ 1 & 1 \end{matrix} \right| -1\left| \begin{matrix} 23 & 2 \\ 46 & 1 \end{matrix} \right| +1\left| \begin{matrix} 23 & 0 \\ 46 & 1 \end{matrix} \right| \)
= -40 + 69 + 23 = 52
Δ₂ = \(\left| \begin{matrix} 1 & 20 & 1 \\ 1 & 23 & 2 \\ 3 & 46 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 23 & 2 \\ 46 & 1 \end{matrix} \right| -20\left| \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right| +1\left| \begin{matrix} 1 & 23 \\ 3 & 46 \end{matrix} \right| \)
= -69 + 100 - 23 = 8
Δ₃ = \(\left| \begin{matrix} 1 & 1 & 20 \\ 1 & 2 & 23 \\ 3 & 1 & 46 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 0 & 23 \\ 1 & 46 \end{matrix} \right| -1\left| \begin{matrix} 1 & 23 \\ 3 & 46 \end{matrix} \right| +20\left| \begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix} \right| \)
= -23 + 23 + 20 = 20
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 52 }{ 4 } \) = 13
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 8 }{ 4 } \) = 2 and z =\(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 20 }{ 4 } \) = 5
Hence the required numbers are 13, 2 and 5.

(i) 
Rearrange the terms as,
(x-5) (x+4) (x-7) (x+6) = 504
{-2, 3, -7, 8}
⇒ (x² -x - 20) (x² -x - 42) = 504
Put x²- x = y
⇒ (y-20)(y-42) = 504
⇒ y²-62y+840-504 = 0
⇒ y²-62y+336 = 0
⇒ (y - 56) (y - 6) = 0
⇒ y = 56, 6
Case (i)
When y = 56, x² - x = 56

⇒ x² - x - 56 = 0
⇒ (x - 8)(x + 7) = 0
⇒ x = 8, -7
Case (ii) 
When y = 6,
x²- x = 6
x²- x - 6 = 0
⇒ (x - 3)(x + 2) = 0
⇒ x = 3, - 2

Hence the roots are -2, 3, 8, -7

Since two roots are in the ratio p : q : r, we can assume the roots as pλ, qλ and rλ .
Then, we get
Σ₁ = pλ + qλ + rλ = -a .....(1)
Σ₂ = (pλ)(qλ)+(qλ)(rλ)+(rλ)(pλ) ..........(2)
Σ₃ = (pλ)(qλ)(rλ) = -c .....(3)
Now, we get
(1) ⇒ λ = -\(\frac { a }{ p+q+r } \) ..........(4)
(3) ⇒ λ³ = \(\frac { c }{ pqr } \) ...........(5)
Substituting (4) in (5), we get
\(\left( \frac { a }{ p+q+r } \right) ^{ 3 }=-\frac { c }{ pqr } \) ⇒ pqra³ = c(p+q+ r)³.

Rolle’s theorem is not valid.

Given curve is y = \(\sqrt x\) + 1

Required area
\(\int _{ 0 }^{ 4 }{ ydx } =\int _{ 0 }^{ 4 }{ (\sqrt { x } +1)dx } \)
\({ \left[ \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+x \right] }_{ 0 }^{ 4 }=\frac { 2 }{ 3 } { (4) }^{ \frac { 3 }{ 2 } }+4\)
\(\frac { 2 }{ 3 } (4)\sqrt { 4 } +4=\frac { 16 }{ 3 } +4\)
\(\frac { 16+12 }{ 3 } =\frac { 28 }{ 3 } \) sq.units

Given \(f(x)=\frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } \)
\(\quad \underset { x\rightarrow 1^{ + } }{ lim } \frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } =\underset { x\rightarrow 1^{ + } }{ lim } \frac { { x }^{ 2 } }{ \left( x+1 \right) \left( x-1 \right) } \) 
= \(\underset { h\rightarrow { 0 }^{ + } }{ lim } \frac { \left( 1+h \right) ^{ 2 } }{ \left( 1+h+1 \right) \left( 1+h-1 \right) } \)
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { (1+h)^{ 2 } }{ (2+h)(h) } \infty \) 
Also  \(\underset { x\rightarrow 1^{ - } }{ lim } \frac { { x }^{ 2 } }{ (x+1)/(x-1) } =\underset { h\rightarrow 0^{ + } }{ lim } \frac { \left( 1-h \right) ^{ 2 } }{ \left( 1-h+1 \right) \left( 1-h-1 \right) } \)
= \(\underset { h\rightarrow { 0 }^{ + } }{ lim } \frac { \left( 1-h \right) ^{ 2 } }{ (2-h)(-h) } =-\infty \) 
ஃ x = -1 and x = 1are vertical asymptotes,
Also
\(\underset { x-\rightarrow \infty }{ lim } \frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } =\underset { \frac { 1 }{ x } \rightarrow 0 }{ lim } \frac { 1 }{ 1-\frac { 1 }{ { x }^{ 2 } } } =1\)
[Divide numerator and denominator by x²]
ஃ y = 1 is a horizontal asymptote,

The given differential equation is \(3\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 3 }{ 2 } }\)Squaring both sides, we get 
\(9{ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 2 }={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ 3 }\)
In this equation, the highest order derivative is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) whose power is 2.
Therefore, the given differential equation is of order 2 and degree 2.

Let AOB be the vertical section of the reflector and m is the mid-point of AB. Let the equation of the parabola be y² = 4ax A(6, 12) lies on (1)

∴ 12² = 4a(6) ⇒ a = 6
∴ Focus is (a, 0) = (b, 0)
Hence focus coincides with m, the mid-point of AB.

A =\(\left[ \begin{matrix} 4 \\ 7 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 0 \end{matrix}\begin{matrix} 1 \\ 8 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ -1 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 12 \end{matrix}\begin{matrix} 1 \\ 6 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 4 \end{matrix}\begin{matrix} -3 \\ 5 \end{matrix}\begin{matrix} 12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }+(-1){ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 4 \end{matrix}\begin{matrix} 13 \\ 5 \end{matrix}\begin{matrix} -12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \end{matrix}\begin{matrix} 13 \\ -47 \end{matrix}\begin{matrix} -12 \\ 42 \end{matrix}\begin{matrix} -6 \\ 25 \end{matrix} \right] \)
The equivalent row-echelon matrix hats two non zero rows.
∴ \(\rho\) (A) = 2

The quadratic equation 3x² + 2(a² + 1)x + (a² - 3a + 2)
Will have two roots of opposite sign if it has real roots and the product of the roots is negative.
⇒ 4(a²+1)²-12(a²-3a+2)\(\ge\) 0 and \(\frac { { a }^{ 2 }-3a+2 }{ 3 } <0\)
Both of these conditions are true if 
⇒ a²-3a+2< 0
⇒ (a-1)(a-2)< 0
⇒ 1<a<2

Let \(\vec { a } \) = \(2\hat { i } +3\hat { j } +\hat { k } \), \(\vec { b } \)= \(\hat { i } -2\hat { j } +2\hat { k } \) and \(\vec { c } \) = \(\hat { 3i } +\hat { j } +3\hat { k } \)
\(\vec { a } ,\vec { b } \) and \(\vec { c } \) are coplanar if  \(\vec { a } .(\vec { b } \times \vec { c } )\)
Consider \(\vec { a } .(\vec { b } \times \vec { c } )\)
= \(\left| \begin{matrix} 2 & 3 & 1 \\ 1 & -2 & 2 \\ 3 & 1 & 3 \end{matrix} \right| =2\left| \begin{matrix} -2 & 2 \\ 1 & 3 \end{matrix} \right| -3\left| \begin{matrix} 1 & 2 \\ 3 & 3 \end{matrix} \right| +1\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| \)
= 2 (-6- 2) -3 (3 - 6) + 1(1 + 6)
= 2(-8) - 3(-3) + 1(7)
= -16 + 9 + 7
= -16+16
= 0.
Hence, the given vectors are co-planar.

Let cos⁻¹\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \) = y. Then, cos y = \( \frac { \sqrt { 3 } }{ 2 } \)
The range of the principal values of y = cos⁻¹x is [0, \(\pi\)].
So, let us find y in [0, \(\pi\)] such that cos y =\( \frac { \sqrt { 3 } }{ 2 } \)
But, cos\(\frac{\pi}{6}=\frac{\sqrt3}{2} and \frac{\pi}{6}\in[0,\pi]\). Therefore, y = \(\frac{\pi}{6}\)
Thus, the principal value of cos^-1 \(\left( \frac { \sqrt { 3 } }{ 2 } \right) is\frac { \pi }{ 6 } \)

\(\frac { 17 }{ 12 } \)

(i) Given that five fair coins are tossed once. Since the coins are fair coins the probability of getting an head in a single coin is
 \(p=\frac { 1 }{ 2 } \) and \(q=1-p=\frac { 1 }{ 2 } \) 
Let X denote the number of heads that appear in five coins. X is binomial random variable that takes on the values 0, 1, 2,  3, 4 and 5 and \(p=\frac { 1 }{ 2 } \) That is \(X\sim B\left( 5,\cfrac { 1 }{ 2 } \right) \) 
Therefore the binomial distribution is
\(f(x)=\left( \begin{matrix} n \\ x \end{matrix} \right) p*\left( 1-p \right) ^{ n-x }\), x = 0, 1, 2,..,n
becomes
\(f(x)=\left( \begin{matrix} 5 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 2 } \right) ^{ x }\left( \cfrac { 1 }{ 2 } \right) ^{ n-x }\), x = 0, 1, 2,..,5
That is 
\(f(x)=\left( \begin{matrix} 5 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 2 } \right) ^{ n }\), x = 0, 1, 2...,n
(ii) A fair die is rolled ten times and X denotes the number of times 4 appeared. X is binomial
random variable that takes on the values 0, 1, 2, 3,...10 , with n = 10 and \(p=\cfrac { 1 }{ 6 } \). That is \(X\sim B\left( 10,\cfrac { 1 }{ 6 } \right) \) 
Probability of getting a four in a die is \(p=\frac { 1 }{ 2 } \) and \(q=1-p=\frac { 5 }{ 6 } \)
Therefore the binomial distribution is 
\(f(x)=\left( \begin{matrix} 10 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 6 } \right) ^{ x }\left( \cfrac { 5 }{ 6 } \right) ^{ 10-x }\) x = 0, 1, 2,...,10

Recall that surface area of a sphere with radius r is given by S(r) = 4\(\pi \)r³. Note that even though we can calculate the exact change using this formula, we shall try to approximate the change using the linear approximation. So, using (4), we have
Change in the surface area = S(5.2) - S(5) ≈ S'(5)(0.2)
= 8\(\pi \)(5)(0.2)
= 8\(\pi \) cm²
Exact calculation of the change in the surface gives
S(5.2) − S(5) = 108.16\(\pi \)-100\(\pi \) = cm².
Percentage error = relative error \(\times\)100 = \(\frac { 8.16\pi -8\pi }{ 8.16\pi } \)\(\times\)100 = 1.9607%

Given  \(sin\left( { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x \right) =1=sin\frac { \pi }{ 2 } \)
\(\left[ \because sin\frac { \pi }{ 2 } =1 \right] \)
\(\Rightarrow { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x={ sin }^{ -1 }\left( sin\left( \frac { \pi }{ 2 } \right) \right) \)
\(\Rightarrow { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x=\frac { \pi }{ 2 } \)
\({ sin }^{ -1 }\frac { 1 }{ 5 } =\frac { \pi }{ 2 } -{ cos }^{ -1 }x{ sin }^{ -1 }x\)
\(\left[ \because { sin }^{ -1 }x+{ cos }^{ -1 }x=\frac { \pi }{ 2 } \right] \)
\(\Rightarrow x=\frac { 1 }{ 5 } \)

Let \(\overset { \rightarrow }{ a } =\overset { \wedge }{ 3i } -5\overset { \wedge }{ k } ,\overset { \rightarrow }{ b } =2\overset { \wedge }{ i } +7\overset { \wedge }{ j } \) and \(\overset { \rightarrow }{ c } =\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
Let the required vector be \(\overset { \rightarrow }{ r } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
Given \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ a } =-1\)
\(\Rightarrow \left( x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ 3i } -5\overset { \wedge }{ k } \right) =-1\)
⇒ 3x - 5z = -1     (1)
\(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ b } =6\)
\(\Rightarrow \left( x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } +7\overset { \wedge }{ j } \right) \)= 2x + 7y = 6       (2)
\(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ i } =5\)
\(\Rightarrow \left( x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) \)= x + y + z = 5   (3)
Solving (1), (2) and (3) we get
x = 3, y = 0 and z = 2.
\(\therefore \overset { \rightarrow }{ r } =\overset { \wedge }{ 3i } +2\overset { \wedge }{ k } \)

Augmented matrix [A|B] =\(\left[ \begin{matrix} 1 & 1 & 3 \\ 2 & 2 & 6 \\ 2 & 1 & 1 \end{matrix}|\begin{matrix} 4 \\ 7 \\ 10 \end{matrix} \right] \)
[A|B] \(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -5 \end{matrix}|\begin{matrix} 4 \\ -1 \\ 2 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -5 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 4 \\ 2 \\ -1 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 [only 2 two-non zero rows]
And \(\rho\) ([A|B]) = 3 [There are 3 non-zero rows]
∴ \(\rho\) (A) ≠ \(\rho\) ([A|B])
∴ The system is inconsistent.

Since x²- y² is even, either both x and y are even or both x and y are odd.
In any case both (x + y) and (x - y) are even integers
∴ x² - y²= (x + y) (x +y) must be divisible by 4 But 4 does not divide 353702
∴ x² - y² = 353702 has no positive integral solutions.

We have |2z-3-i| = 3
|2(x + iy)-3 - i| = 3
Squaring on both sides, we get
|(2x - 3) + (2y - 1)i|² = 9
\(\Rightarrow\) (2x - 3)² + (2y - 1)² = 9
\(\Rightarrow\) 4x² + 4y² -12x - 4y + 1 = 0, the locus of z in Cartesian form

SS′ = 2c and 2c = 4; A'A = 2a = 6
c = 2 and a = 3,
b² = a²−c² = 9−4 = 5.
Major axis is along x-axis, since a > b.
Centre (0, 0) and Foci are (±2, 0) 
Therefore, equation of the ellipse is \(\frac { { x }^{ 2 } }{ 9 } \frac { { y }^{ 2 } }{ 5 } =1\)