\(f(x)= \begin{cases}\frac{1}{30} & 0
Mean =\(E(X)=\int _{ 0 }^{ 30 }{ x3f(x)dx } \) 
= \(\int _{ 0 }^{ 30 }{ x.\frac { 1 }{ 30 } dx } \) 
\(E(X)=\frac { 1 }{ 30 } \left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 30 }\) 
= \(\frac { 1 }{ 30 } [ \frac{30\times 30}{2}-0]\)  
E(X) = 15 minutes
The average waiting time for the student is 15| minutes.

First let us find w_x, w_y, and w_z
Now w_x = 2xy + z², w_y = 2yz +x² and w_z = 2zx + y².
Thus,by (15), the differential is
dw = (2xy + z² )dx + (2yz + x² )dy+ (2zx + y² )dz.

Observe that, f (2) = 0 = f (3) and f (x) is continuous in the interval [2, 3] and differentiable in (2, 3). Now,
\(f'(x)=\frac{x^{2}-6}{x(x^{2}+6)} \)
Therefore, \(f'{(c)}=0\) gives
\(\frac{c^{2}-6}{c(c^{2}+6)}=0\)
which implies \( c=\pm\sqrt{6}\)
Now c = \(\pm\sqrt{6} \in (2,3).\)
Observe that \(-\sqrt{6}\notin (2,3)\) and hence \(c=\pm\sqrt{6}\) satisfies the Rolle’s theorem.
Rolle’s theorem can also be used to compute the number of roots of an algebraic equation in an interval without actually solving the equation.

\(\overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } \)= \(\overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } =\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } \right) -\left( \overset { \wedge }{ j } -\overset { \wedge }{ k } \right) =\left( \overset { \wedge }{ i } +\overset { \wedge }{ k } \right) \)
\(\overset { \rightarrow }{ b } -\overset { \rightarrow }{ c } =\left( \overset { \wedge }{ j } -\overset { \wedge }{ k } \right) -\left( \overset { \wedge }{ k } -\overset { \wedge }{ i } \right) =\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } =\left( \overset { \wedge }{ k } -\overset { \wedge }{ i } \right) -\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } \right) =-2\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
\(\therefore \left[ \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ b } -\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right] =\left| \begin{matrix} 1 \\ 1 \\ -2 \end{matrix}\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right| \)
= 1 ( 1 + 2) + 0 + 1( 1+ 2)
= 3 + 3 = 6

\(\Delta = \left| \begin{matrix} t & 3 & -1 \\ 1 & 2 & 1 \\ -t & 1 & 2 \end{matrix} \right| =t\left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ -t & 2 \end{matrix} \right| -\left| \begin{matrix} 1 & 2 \\ -t & 1 \end{matrix} \right| \)
= t(4 - 1) -3 (2 + t) -1(1 + 2t)
= 3t - 6i - 3t - 1 - 2t = - 7 - 2t
The system will fail to have unique solution if
Δ = 0 ⇒ -7-2t = 0 ⇒ -2t = 7 ⇒ t = \(\frac { -7 }{ 2 } \)
∴ t = \(\frac { -7 }{ 2 } \).

Put y = \(\frac { x-1+x-5 }{ 2 } =-3\)
⇒ x = y + 3
∴ (x-1)⁴+(x-5)⁴ = 82
⇒ (y+3-1)⁴+(y+3-5)⁴ = 82
⇒ (y+2)⁴+(y-2)⁴ = 82
⇒ 2(y⁴+24y²+16) = 82
⇒ y⁴+24y²+16 = 41
⇒ y⁴+24y²-25 = 0
⇒ (y²+25)(y²-1) = 0
⇒ y = 土5i, y = 士1
∴ x = 3土5i, 4, 2.

The roots of x³+2x2+3x+4 = 0 are ∝, β, ૪
∴ ∝+β+૪ = -co-efficient of x² = -2 ...(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3 ....(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4 ...(3)
Form a cubic equation whose roots are 2∝, 2β, 2૪
2∝+2β+2૪ = 2(∝+β+૪) = 2(-2) = -4 [from (1)]
4∝β+4β૪+4૪∝ = 4(∝β+β૪+୪∝) = 4(3) = 12 [from (2)]
(2∝)(2β)(2૪) = 8(∝β૪) = 8(-4) = -32 [from (3)]
∴ The required cubic equation is
x³-(2∝+2β+2૪)x² + (2∝β+2β૪+2୪∝)x - (2∝)(2β)(2૪) = 0
⇒ x³+(-4)x²+12x+32 = 0
⇒ x³+4x²+12x+32 = 0

\(y=\frac { { x }^{ 2 } }{ 2 } +2x-13log|x+2|+c,x\varepsilon R-\{ 2\} \)

\(\frac { 4 }{ 2 } \left( 4\pi +\sqrt { 3 } \right) \)

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 

(i) The probability density function. Differentiating F(x) with respect to 'x' at continuity points  of F(x), we get
\(F(x)=\begin{cases} \begin{matrix} 0 & x<0 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 2 } ({ 2x }+1) & 0\le x<1 \end{matrix} \\ \begin{matrix} 1 & x\ge 1 \end{matrix} \end{cases}\) 
(ii) \(p(0.3\le X\le 0.6)=\int _{ 0.3 }^{ 0.6 }{ f(x)dx } \) 
= \(\int _{ 0.3 }^{ 0.6 }{ \frac { 1 }{ 2 } \left( 2x+1 \right) dx } =\frac { 1 }{ 2 } \left[ \frac { { 2x }^{ 2 } }{ 2 } +x \right] _{ 0.3 }^{ 0.6 }\) 
= \(\frac { 1 }{ 2 } \left( { x }^{ 2 }+x \right) _{ 0.3 }^{ 0.6 }=\frac { 1 }{ 2 } \left[ \left( { 0.6 }^{ 2 }+0.6 \right) -\left( { 0.3 }^{ 2 }+0.3 \right) \right] \) 
= \(\frac { 1 }{ 2 } \left[ \left( .36.6 \right) \right] -\left( .09+0.3 \right) ]\) 
= \(\frac { 1 }{ 2 } \left[ 0.96-.39 \right] =\frac { 0.57 }{ 2 } =0.285\) 
= 0.285

Given E = Ri + L \(\frac{di}{dt}\)
\(\frac { E }{ L } =\frac { Ri }{ L } +\frac { di }{ dt } \)
\(\Rightarrow \frac { Ri }{ L } +\frac { di }{ dt } =\frac { E }{ L } \)
This is a linear differential equation
\(Here\quad P=\frac { R }{ L } and\quad Q=\frac { E }{ L } \)
\(\therefore \int { pdt } =\int { \frac { R }{ L } dt } =\frac { R }{ L } t\)
\(\therefore I.F={ e }^{ \int { pdt } }={ e }^{ \frac { Rt }{ L } }\)
\(\therefore\) Solution is i\({ e }^{ \int { pdt } }=\int { Q{ e }^{ \int { pdt } }dt+C } \)
\(\Rightarrow i{ e }^{ \frac { Rt }{ L } }=\int { \frac { E }{ L } . } { e }^{ \frac { Rt }{ L } }dt+C\)
\(\therefore i{ e }^{ \frac { Rt }{ L } }=\frac { E }{ L } \frac { { e }^{ \frac { Rt }{ L } } }{ \frac { R }{ L } } dt+C\)
\(i=\frac { E }{ R } { e }^{ \frac { Rt }{ L } }+C\)
\(i=\frac { E }{ R } +c{ e }^{ -\frac { Rt }{ L } }\)
When E = 0,
\(i=0+c{ e }^{ -\frac { Rt }{ L } }\)
\(\Rightarrow i=c{ e }^{ -\frac { Rt }{ L } }\)

This is an indeterminate of the form \(1^{\infty}\).
Let \(g(x)=(1+x)^{\frac{1}{x}}\). Taking the logarithm, we get
\(log \ g(x)=\frac{log(1+x)}{x}\)
\(\underset{x\rightarrow 0^{+}}{lim} log (g(x))=\underset{x\rightarrow 0^{+}}{lim}(\frac{log(1+x)}{x})\) \((\frac{0}{0})\) 
=\(\underset{x\rightarrow0^{+}}{lim}(\frac{\frac{1}{1+x}}{1})\)  (by 1’Hôpital Rule)
= 1.
But, \(\underset{x\rightarrow0^{+}}{lim}log g(x)=log(\underset{x\rightarrow 0^{+}}{lim} g(x))\)
Therefore, log\((\underset{x\rightarrow 0^{+}}{lim} g(x))=1\).
Hence by exponentiating, we get, \(\underset{x\rightarrow 0^{+}}{lim}g(x)=e.\)

\(\frac { dy }{ dx } +\frac { { 3x }^{ 2 }y }{ 1+{ x }^{ 3 } } =\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
This is a linear differential equation 
\(\therefore P=\frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } ;Q=\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
\(\therefore \int { pdx } =\int { \frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } =log(1+{ x }^{ 3 })\)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log(1+{ x }^{ 3 }) }=(1+{ x }^{ 3 })\)
\(\therefore\)The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } (1+{ x }^{ 3 })dx+c } \)
\(cos2x=1-2{ sin }^{ 2 }x\)
\(sin2x=\frac { 1-cos2x }{ 2 } =\int { { sin }^{ 2 }xdx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { 1-cos2x }{ 2 } } dx+c\)
\(\Rightarrow y(1+{ x }^{ 3 })=\frac { x }{ 2 } -\frac { sin2x }{ 4 } +c\)

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 \(z=\frac { 2 }{ 3 } \)
\(\therefore \overset { \rightarrow }{ B } =\frac { 5 }{ 3 } \overset { \wedge }{ i } +\frac { 2 }{ 3 } \overset { \wedge }{ j } +\frac { 2 }{ 3 } \overset { \wedge }{ k } \)

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

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 \(\frac { { x }^{ 2 } }{ { A }^{ 2 } } -\frac { { y }^{ 2 } }{ { B }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1\)

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

\({ tan }^{ -1 }\left( { \frac { x-1 }{ x-2 } } \right) +{ tan }^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =\frac { \pi }{ 4 } \)
\(\Rightarrow { tan }^{ -1 }\left( \cfrac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\left( \frac { x-1 }{ x-2 } \right) \left( \frac { x+1 }{ x+2 } \right) } \right) =\frac { \pi }{ 4 } \)

\(\Rightarrow \frac { 2{ x }^{ 2 }-4 }{ { x }^{ 2 }-4-{ x }^{ 2 }+1 } =1\)
\(\Rightarrow\) 2x²- 4 = -3
\(\Rightarrow \) 2x²- 4 = -3
\(\Rightarrow\) 2x² = -3 + 4 = 1
\(\Rightarrow\) \({ x }^{ 2 }=\frac { 1 }{ 2 } \)
\(\Rightarrow\) \(x=\frac { 1 }{ \sqrt { 2 } } \)

Augmented matrix = [A|B] = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{matrix}|\begin{matrix} 1 \\ \lambda \\ { \lambda }^{ 2 } \end{matrix} \right] \)
[A|B] =  \(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 3 & 9 \end{matrix}|\begin{matrix} 1 \\ \lambda -1 \\ { \lambda }^{ 2 }-1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ \lambda -1 \\ { \lambda }^{ 2 }-1-3\lambda +3 \end{matrix} \right] \)
⟶ \(\left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ \lambda -1 \\ { \lambda }^{ 2 }-3\lambda +2 \end{matrix} \right] \)
Here \(\rho\)(A) = 2
The given system of equations is consistent only when \(\rho\)([AIB]) = 2
\(\rho\)([AIB]) = 2 only when λ²-3λ + 2 = 0
⇒ (λ-1) (λ-2) = 0 ⇒ λ = 1 or λ = 2
∴ The given system is consistent when the values of A are 1 and 2.

We can solve this fourth degree equation by rewriting it suitably and adopting a technique of substitution. Rewriting the equation as
(x−2)(x−3)(x−7)(x+2)+19 = 0
the given equation becomes
(x²−5x+6)(x²−5x−14)+19 = 0 .
If we take x² − 5x as y, then the equation becomes (y+6)(y−14)+19 = 0;
that is, y²-8y-65 = 0
Solving this we get solutions y = 13 and y = −5. Substituting this we get two quadratic equations
x²-5x-13 = 0 and x²-5x+5 = 0
which can be solved by usual techniques. The solutions obtained for these two equations together give solutions as \(\frac { 5\pm \sqrt { 77 } }{ 2 } ,\frac { 5\pm \sqrt { 5 } }{ 2 } \).

concave upward everywhere

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

Let P(x, y) be the movable point.
By focal property of ellipse, PA + PB = 2a
∴ 2a = 10 ⇒ a = 5
Since focus is (4, 0), ae = 4 ⇒ 5e = 4 ⇒ e = \(\frac45\)
Also b² = a²(1 - e²) = 25\(\left( 1-\frac { 16 }{ 25 } \right) =25\left( \frac { 9 }{ 25 } \right) \) = 9
Equation of ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)

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

2x = 3y =  −z \(\Rightarrow \frac{x}{3}=\frac{y}{2}=\frac{-z}{6}\) (Dividing by 6 all)
 \(\Rightarrow \frac{x-0}{3}=\frac{y-0}{2}= \frac{z-0}{-6}\)....(1)
6x = -y = -4z \(\Rightarrow \frac{x}{2}=\frac{-y}{12}=\frac{-z}{3} \) (Dividing by 6 all)
\(\Rightarrow \frac{x-0}{2}=\frac{y-0}{-12}=\frac{z-0}{-3}\) ....(2)
From (1) & (2), we get
\(\vec b = 3\vec i+2\vec j- 6\vec k\)and \( \vec d = 2\vec i-12\vec j- 3\vec k\)
Angle between lines (1) and (2) = Angle between \(\vec b\ and\ \vec d\)
Acute angle between lines cos 0 = \(\frac{|\vec b . \vec d|}{|\vec b||\vec d|} \)
\(\vec b . \vec d \)= (\(\vec b = 3\vec i+2\vec j- 6\vec k\)). (\( 2\vec i-12\vec j- 3\vec k\))
6-24+18 = 0
\( \cos \theta=0 \)
\(\theta=\frac{\pi}{2} \text { or } 90^{\circ}\)

Let y = sec^-1 (-2). Then, sec y = -2
By the definition, the range of the principal value branch of y = sec⁻¹x is [0, \(\pi\)]\{\({{\frac{\pi}{2}}}\)}
Let us find y in [0, \(\pi\)] - {\({{\frac{\pi}{2}}}\)} such that sec y = -2
But, sec y = −2 \(\Rightarrow\) cos y = -\(\frac{1}{2}\)
Now, cos y = -\(\frac { 1 }{ 2 } =-cos\frac { \pi }{ 3 } =cos\left( \pi -\frac { \pi }{ 3 } \right) =cos\frac { 2\pi }{ 3 } \). Therefore, y = \(\frac{2\pi}{3}\)
since \(\frac{2\pi}{3}\in[0,\pi]\)\{\({{\frac{\pi}{2}}}\)}, the principal value of sec^-1(-2) is \(\frac{2\pi}{3}\)

 i¹⁷²⁹ = i¹⁷²⁸ i¹ = i