0.677

(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.

Given (x, y, z) = log (x³ + y³ + z³)
\(\frac { \partial U }{ \partial x } =\frac { 1 }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } { (3x }^{ 2 });\)
\(\frac { \partial U }{ \partial y } =\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \) and
\(\frac { \partial U }{ \partial z } =\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(\therefore \frac { \partial U }{ \partial x } +\frac { \partial U }{ \partial y } +\frac { \partial U }{ \partial z } =\frac { { 3x }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(=\frac { { 3({ x }^{ 2 }+y }^{ 2 }+{ z }^{ 2 }) }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)

Taking u = x² and v = sin nx, and applying the Bernoulli’s formula, we get
\(I=\int _{ 0 }^{ 2\pi }{ { x }^{ 2 } } sin\quad nx\quad dx={ \left[ \left( { x }^{ 2 } \right) \left( -\frac { cos\quad nx }{ n } \right) -(2x)\left( -\frac { sin\quad nx }{ { n }^{ 2 } } \right) +(2)\left( \frac { cos\quad nx }{ { n }^{ 3 } } \right) \right] }_{ 0 }^{ 2x }\)
\(=\left[ (4{ \pi }^{ 2 })\left( -\frac { 1 }{ n } \right) -0+(2)\left( \frac { 1 }{ { n }^{ 3 } } \right) \right] -\left[ 0-0+(2)\left( \frac { 1 }{ { n }^{ 3 } } \right) \right] \) since cos 2n\(\pi\) = 1 and sin 2n \(\pi\) = 0
\(=-\frac { 4{ \pi }^{ 2 } }{ n } +\frac { 2 }{ { n }^{ 3 } } -\frac { 2 }{ { n }^{ 3 } } =\frac { { 4\pi }^{ 2 } }{ n } \)

By the mean value theorem we have, there exists 'c'∈(2, 7) such that,
\(\frac { f(7)-f(2) }{ 7-2 } \) = f'(c) ≤ 29
Hence, f(7) ≤ 5× 29 +17 = 162
Therefore, the maximum value of f (7) is 162.

Let z = x + iy be a complex number
∴ \(\left| \frac { x+iy-3 }{ x+iy+3 } \right| \)  = 2
⇒ \(\left| \frac { x+iy-3 }{ x+iy+3 } \right| \) = 2
⇒ |(x-3) + iy| = 2|(x+3)+iy|
⇒ \(\sqrt { (x-3)^{ 2 }+{ y }^{ 2 } } =2\sqrt { (x+3)^{ 2 }+{ y }^{ 2 } } \)
Squaring both sides we get,
(x - 3)² + y² = 4[(x + 3)² + y²]
⇒ x² + 9 - 6x + y² = 4 [x²+ 9 + 6x + y²]
x² + 9 - 6x + 1 = 4x² + 36 + 24x + 4y²
⇒ 3x² + 3y² + 30x + 27 = which represent a circle.

Let \(\hat { d } =6\hat { i } +2\hat { j } -3\hat { k } \) and p = 12
If \(\hat { d } \) is the unit normal vector in the direction of the vector \(6\hat { i } +2\hat { j } -3\hat { k } \)
then \(\hat { d } =\frac { \hat { d } }{ \left| \hat { d } \right| } =\frac { 1 }{ 7 } (6\hat { i } +2\hat { j } -3\hat { k } )\)
If \(\hat { r } \) is the position vector of an arbitrary point (x, y, z) on the plane, then using \(\vec { r } .\hat { d } =p\), the vector equation of the plane in normal form is \(\vec { r } .\frac { 1 }{ 7 } (6\hat { i } +2\hat { j } -3\hat { k } )=12\)
Substituting \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) in the above equation, we get \((x\hat { i } +y\hat { j } +z\hat { k } ).\frac { 1 }{ 7 } (6\hat { i } +2\hat { j } -3\hat { k } )=12\)
Applying dot product in the above equation and simplifying, we get 6x + 2y - 3z = 84,  which is the Cartesian equation of the required plane.

AS = 94.5 × 10⁶ km,
SA' = 152 × 10⁶ km
a+c = 152 × 10⁶
a-c = 94.5 × 10⁶
Subtracting 2c = 57.5 × 10⁶ = 575 × 10⁵ km
Distance of the Sun from the other focus is SS' = 575 × 105 km.

Let A be the matrix. Performing elementary row operations, we get
A = \(\left[ \begin{matrix} 2 \\ \begin{matrix} -3 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -1 \\ 7 \end{matrix} \end{matrix} \right] \)\(\overset { { R }_{ 2 }\longrightarrow 2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} -6 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 8 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -4 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -2 \\ 7 \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -13 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -2 \end{matrix} \end{matrix} \right] \).
\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -45 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -30 \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -15 \right) }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ 3 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ 2 \end{matrix} \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has three non-zero rows. So, ρ(A) = 3.

cos x = 0
\(\Rightarrow x=\left( 2n+1 \right) \frac { \pi }{ 2 } ,n\varepsilon Z\)
But \(-6\pi \le x\le 6\pi \)
\(\therefore \) n can take values from  
\(x=(2n+1)\frac { \pi }{ 2 } ,n=0\pm 1,\pm 2,...\pm 5\), -6

As the circle touches both the axes, the distance of the centre from both the axes is 3 units, centre can be (±3, ±3) and hence there are four circles with radius 3, and the required equations of the four circles are
x² + y² ± 6x ± 6y + 9 = 0.

Equation of family of straight lines in my plane is y = mx - c where m and c are arbitrary constraints.
Differentiating, y' = m
Differentiating again, y" = 0, is the required differential equation.

x = 2 and dx = 0.1
Taking differentials,
df = (2x + 3) dx
Whenx = 2, dx = 0.1
df = (2(2) + 3)(0.1) = 7(0.1) = 0.7

Given f(x) = tan x, x ∈ [0, π]
Rolle's theorem is not applicable since tan x is not continuous at x = \(\frac{\pi}{2}\) [∵ tan \(\frac{\pi}{2}\) = ∞]

z-iw
= (5-2i) - i(-1+3i)
= (5-2i) + (+1-3i²)
5 - 2i + i - 3(-1) = 5 - i + 3 = 8 - i

\(\left[ \begin{matrix} 1 \\ 3 \end{matrix}\begin{matrix} -2 \\ -6 \end{matrix}\begin{matrix} -1 \\ -3 \end{matrix}\begin{matrix} 0 \\ 1 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 \\ 3 \end{matrix}\begin{matrix} -2 \\ -6 \end{matrix}\begin{matrix} -1 \\ -3 \end{matrix}\begin{matrix} 0 \\ 1 \end{matrix} \right] \)
A is a matrix of order (2 \(\times\) 4)
∴ \(\rho \)(A) ≤ min(2, 4) = 2
The highest order of minor of A is 2
It is \(\left| \begin{matrix} 1 & -2 \\ 3 & -6 \end{matrix} \right| \) = -6 + 6 = 0
Also, \(\left| \begin{matrix} -1 & 0 \\ -3 & 1 \end{matrix} \right| \) = -1 + 0 = -1 ≠ 0
∴ \(\rho \)(A) = 2

Here A = 3, C = 2 and F = -14
A ≠ C and A and C are of the same sign.
Hence, the given equation represents an ellipse.

Let cosec⁻¹(−1) = y. Then, cosec y = −1
Since the range of principal value branch of y = cosec^-1x is\(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)\{0} and
cosec \((-\frac{\pi}{2})=-1,\) 
We have y = \(-\frac{\pi}{2}\).
Note that \(-\frac{\pi}{2}\in\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)\{0}.
Thus, the principal value of cosec⁻¹(−1) is −\(\frac{\pi}{2}\)

~(p V q) V (~p ∧ q) and ~p

The entries in column (5) and column (7) are identical.
∴ ~(p V q) V (~p ∧ q) and ~p are logically equivalent.

Consider the set Z_o of all odd integers Z_o = {2k+1 : k∈Z} = {.., -5,-3,-1,1,3,5,...}. + is not a binary operation on Z_o because when x = 2m+1, y = 2n +1, x + y = 2(m+ n) + 2 is even for all m and n. For instance, consider the two odd numbers 3, 7 ∈Z_o. Their sum 3 + 7 = 10 is an even number. In general, if x, y∈Z₀, then (x+y)∉Z_o. Other properties need not be checked as it is not a binary operation.

Observe that the given distribution is continuous
By definition \(\mu =E(X)=\int _{ -\infty }^{ \infty }{ xf(x) } dx\) (We can also use integration by parts or Bernoulli’s formula) 
= \(\int _{ -\infty }^{ 0 }{ 0\left( \lambda { e }^{ -2x } \right) dx } +\int _{ 0 }^{ \infty }{ x\left( { \lambda e }^{ -\lambda x } \right) } dx\) 
= \(0+\lambda \int _{ 0 }^{ \infty }{ x\left( { e }^{ -\lambda x } \right) dx } \) 
=  \(0+\lambda \left( \frac { 1 }{ { \lambda }^{ 2 } } \right) \) (using Gamma integral for positive integer n,\(\int _{ 0 }^{ \infty }{ { x }^{ n } } { e }^{ -ax }dx=\cfrac { n }{ { a }^{ n+1 } } \))
= \(\frac { 1 }{ \lambda } \) 
Variance :
By definition,\(E\left( { X }^{ 2 } \right) =\int _{ -\infty }^{ \infty }{ { x }^{ 2 }f(x) } dx\) (We can also use integration by parts or Bernoulli’s formula) 
= \(\int _{ -\infty }^{ 0 }{ 0\left( \lambda { e }^{ -\lambda x } \right) } dx+\int _{ 0 }^{ \infty }{ { x }^{ 2 }\left( \lambda { e }^{ -2x } \right) } dx\) 
= \(0+\lambda \int _{ 0 }^{ \infty }{ { x }^{ 2 }\left( { e }^{ -2x } \right) dx } \)
 (using Gamma integral for positive integer)
Therefore Var(X ) = E(X² )- E(X )²
= \(\frac { 2 }{ { \lambda }^{ 2 } } -\left( \frac { 1 }{ \lambda } \right) ^{ 2 }=\frac { 1 }{ { \lambda }^{ 2 } } \) 
Hence the mean and variance are respectively \(\frac { 1 }{ \lambda } \)  and \(\frac { 1 }{ { \lambda }^{ 2 } } \)

Let I = \(\int ^{\pi}_{-\pi} \frac{cos ^2 x}{1+ a^x} dx\)-- (1) 
Using \(\int ^{b}_{a}\) f(x) dx =\(\int ^{b}_{a}\) f(a+b -x)dx  we get,
I = \(\int ^{\pi}_{-\pi} \frac{cos ^2 (\pi -\pi - x)}{1+ a^{\pi -\pi - x}} dx\)
= \(\int ^{\pi}_{-\pi} \frac{cos ^2 (-x)}{1+ a^{-x}} dx\)
= \(\int ^{\pi}_{-\pi} a^x (\frac{cos ^2 x }{a^x+ 1} )dx\) --- (2)
Adding (1) and (2) we get
2I =  \(\int ^{\pi}_{-\pi} \frac{cos ^2 x }{a^x+ 1}(a^x+ 1) dx\) =  \(\int ^{\pi}_{-\pi} cos ^2 x dx\)
= 2 \(x =\int ^{\pi}_{-\pi} cos ^2 x dx\)  (since cos² x is an even function)
Hence, I =  \(\int ^{\pi}_{0} (\frac{1 + cos2x }{2} )dx\) 
=  \(\frac {1}{2} [ x + \frac {sin 2x}{2}]^{x}_{0}\) 
= \(\frac {1}{2} [\pi]\)
= \(\frac {\pi}{2}\)

Given probability mass function is

(i) Since f(x) is a probability mass function.
\(\sum _{ i=1 }^{ 5 }{ f({ x }_{ i }) } =1\) 
⇒ k² + 2k² + 3k² + 2k + 3k = 1
⇒ 6k² + 5k = 1
⇒ 6k² + 5k - 1 = 0 
⇒ (k + 1) (6k - 1) = 0
⇒ k = -1 or  ⇒ \(k=\frac { 1 }{ 6 } \) 
⇒ \(k=\frac { 1 }{ 6 } \)
(ii) p(2 ≤ x < 5)
= p(x = 2) + p(x = 3) + p(x = 4)
= 2k² + 3k² + 2k = 5k² + 2k
= \(5\left( \frac { 1 }{ 36 } \right) +2\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 36 } +\frac { 1 }{ 3 } =\frac { 5+12 }{ 36 } \) 
= \(\frac { 17 }{ 36 } \) 
(iii) p(3 < x) = p(x > 3)
= p(x = 4) + p(x = 5)
= 2k + 3k = 5k
= \(5\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 6 } \)

Let V be the velocity and the retardation (negative acceleration) be -\(\frac{dv}{dt}\)
Given \(\frac{dv}{dt}\) = -V
Separating the variables,
\(\frac{dv}{v}=-dt\)
\(\Rightarrow \int { \frac { dv }{ v } } =-\int { dt } \)
\(\Rightarrow log\quad v=-t+logC\)
\(\Rightarrow logv-logC=-t\)
\(\Rightarrow log\left( \frac { v }{ { C }_{ v } } \right) =-t\)
\(\Rightarrow ={ e }^{ -t }\)
\(\Rightarrow \frac { v }{ { C }_{ v } } ={ Ce }^{ -t }...(1)\)
Given when t = 0, v = m/sec
\(\therefore\) (1) become 10 = C_e0 \(\Rightarrow\) C = 10
\(\therefore\) (1)  v = 10e^-t
When t = 2, v = 10e^-2
\(\Rightarrow v=\frac { 10 }{ { e }^{ 2 } } \)

Let us take the semi circle with center (0, 0) an radius r

x = r cos θ,
y = r sin θ
ஃ Length of the rectangle = 2x = 2r cos θ
Breadth of the rectangle = y = r sin θ
ஃ Area = 2xy = 2r² cos θ sin θ
= r² sin 2θ
Letf'(θ) = r² 2 cos 2θ
f(θ) = r²cosθ sinθ
⇒ 2r² cos 2θ = 0
\(\Rightarrow cos2\theta =0=cos\frac { \pi }{ 2 } \)
\(\Rightarrow 2\theta =\frac { \pi }{ 2 } \)
\(\Rightarrow \theta =\frac { \pi }{ 4 } \) 
\(f''={ 2r }^{ 2 }(2)-(-sin2\theta )\) 
= \(-4{ r }^{ 2 }sin2\theta \) 
\(\therefore f''\left( \frac { 4\pi }{ 4 } \right) =-4{ r }^{ 2 }sin2\left( \frac { \pi }{ 4 } \right) \) 
= \(-4{ r }^{ 2 }sin\frac { \pi }{ 2 } ={ 4r }^{ 2 }<0\) 
ஃ f(θ) is maximum when \(\theta =\frac { \pi }{ 4 } \)
ஃ Length of the rectangle \((x)=2ros\frac { \pi }{ 4 } \)
= \(2r\times \frac { 1 }{ \sqrt { 2 } } =\sqrt { 2 } r\) 
ஃ Breadth of the rectangle = y = r sin θ
= \(rsin\frac { \pi }{ 4 } =\frac { r }{ \sqrt { 2 } } \)

ydx + (1 + x²) tan^-1 xdy = 0
\(\mathrm{yd} x=-\left(1+x^2\right) \tan ^{-1} x \mathrm{dy} \)
\(\frac{d x}{\left(1+x^2\right) \tan ^{-1} x}=-\frac{d y}{y} \)
Take \(\mathrm{t}=\tan ^{-1} x \)
\(\mathrm{dt}=\frac{1}{1+x^2} d x\)
The equation can be written as
\(\frac{d t}{t}=-\frac{d y}{y}\)
Taking Integration on both sides, we get
\(\int \frac{d t}{t}=-\int \frac{d y}{y}\)
log t = - log y + log C
log (tan^-1 x) = -log y+ log C
log (tan^-1 x) + log y = log C
log y(tan^-1 x) = log c
y tan^-1 x = c

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.

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

Given lines are
\(\vec { r } =(6\hat { i } +\hat { j } +2\hat { k } )+s(\hat { i } +2\hat { j } -3\hat { k } )\)
\(\vec { a } =6\hat { i } +\hat { j } +2\hat { k } ,\vec { b } =\hat { i } +2\hat { j } -3\hat { k } \)
and \(\vec { r } =(3\hat { i } +2\hat { j } -2\hat { k } )+t(2\hat { i } +4\hat { j } -5\hat { k } )\)
\(\vec { c } =3\hat { i } +2\hat { j } -2\hat { k } \quad and\quad \vec d = 2\hat { i } +4\hat { j } -5\hat { k } \)
Since \(\vec { b } \neq \vec { d } \), they are not parallel and they do not intersect.
Hence the given lines are skew lines.
Shortest distance between the two skew lines
\(\delta =\frac { |(\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )| }{ |(\vec { b } \times \vec { d } )| } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{matrix} \right| \)
\(=\hat { i } (-10+12)-\hat { j } (-5+6)+\hat { k } (4-4)\)
\(=2\hat { i } -\hat { j } \Rightarrow |\vec { b } \times \vec { d } |\quad \sqrt { { 2 }^{ 2 }++(-1)^{ 2 } } =\sqrt { 5 } \)
\(\vec { c } =\vec { a } =(3\hat { i } +2\hat { j } -2\hat { k } )-(6\hat { i } +\hat { j } +2\hat { k } )\)
\(=-3\hat { i } +\hat { j } -4\hat { k } \)
\(\therefore (\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=(-3\hat { i } +\hat { j } -4\hat { k } ).(2\hat { i } -\hat { j } )\)
= -6-1 = -7
\(\therefore \delta =\frac { |-7| }{ \sqrt { 5 } } =\frac { 7 }{ \sqrt { 5 } } units\)

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 1 \\ \begin{matrix} 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 3 \\ \begin{matrix} 1 \\ \begin{matrix} 3 \\ -1 \end{matrix} \end{matrix} \end{matrix} \right] \).
The augmented matrix is [A | B] = \(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 3 & -1 & 2 & 1 \\ 1 & -2 & 3 & 3 \\ 1 & -1 & 1 & -1 \end{array}\right]\).
Applying Gaussian elimination method on [A | B], we get
[A | B] \(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 1 }, \\ { R }_{ 4 }\longrightarrow { R }_{ 4 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -7 \\ \begin{matrix} -4 \\ -3 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 5 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} -8 \\ \begin{matrix} 0 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow \left( -1 \right) { R }_{ 2 }, \\ { R }_{ 3 }\longrightarrow \left( -1 \right) { R }_{ 3 } \\ { R }_{ 4 }\longrightarrow \left( -1 \right) { R }_{ 4 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 4 \\ 3 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} -4 \\ -2 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} 0 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 3 }\longrightarrow 7{ R }_{ 3 }-4{ R }_{ 2 } \\ { R }_{ 4 }\longrightarrow 7{ R }_{ 4 }-3{ R }_{ 2 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} -8 \\ 1 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} -32 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -8 \right) }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} 4 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \overset { { R }_{ 4 }\longrightarrow { R }_{ 4 }-{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \)
There are three non-zero rows in the row-echelon form of [A | B]. So, ρ([A | B]) = 3.
So, the row-echelon form of A is \(\left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \). There are three non-zero rows in it. So ρ(A) = 3.
Hence, ρ(A) = ρ([A | B]) = 3.
From the echelon form, we write the equivalent system of equations
x + 2y − z = 3,7y − 5z = 8, z = 4, 0 = 0.
The last equation 0=0 is meaningful. By the method of back substitution, we get
z = 4
7y − 20 = 8 ⇒ y = 4 ,
x = 3 − 8 + 4 ⇒ x = −1.
So, the solution is (x = −1, y = 4, z = 4).(Note that A is not a square matrix.)
Here the given system is consistent and the solution is unique.

z³ = -27 = (-1 \(\times\) 3)³ = -1 \(\times\) 33
z = \((-1)^{ \frac { 1 }{ 3 } }\times 3^{ 3\times \frac { 1 }{ 3 } }=(-1)^{ \frac { 1 }{ 3 } }\)\(\times\) 3
∴ z = 3\(\left[ cos\pi +isin\pi \right] ^{ \frac { 1 }{ 3 } }\)
[∵ cos π = -1 and sin π = 0]
= 3\(\left[ cos\frac { 1 }{ 3 } (2k\pi +\pi )isin\frac { 1 }{ 3 } (2k\pi +\pi ) \right] \)
k = 0, 1, 2
When k = 0,
z = 3\(\left[ cos\frac { 1 }{ 3 } (\pi )isin\frac { 1 }{ 3 } (\pi ) \right] =3cos\frac { \pi }{ 3 } \)
When k = 1
z = 3\(\left[ cos\frac { 1 }{ 3 } (3\pi )isin\frac { 1 }{ 3 } (3\pi ) \right] \)
= 3[cos π + i sin π] = 3(-1+0) 
When k = 2
z = 3\(\left[ cos\frac { 1 }{ 3 } (5\pi )isin\frac { 1 }{ 3 } (5\pi ) \right] =3\left[ cos5\frac { \pi }{ 3 } \right] \)
Hence, the roots are 3 cis\(\frac { \pi }{ 3 } \), -3, 3 c is 5\(\frac { \pi }{ 3 } \)

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α,β respectively with positive x-axis
Draw AL and BM 丄 to x-axis
Then \(|\vec { OL } |=|\vec { OA } |cos\alpha \Rightarrow \vec { OL } =\vec { |OL| } \hat { i } =cos\alpha \hat { i } \)
\(|\vec { LA } |=|\vec { OB } |\) sin α
⇒ \(\vec { LA } =|\vec { OB } |\hat { j } =sin\alpha (-\hat { j } )=-sin\alpha \hat { j } \)
[\(\vec { LA } \) is in the opp direction of y axis]
\(\hat { a } =\vec { OA } =\vec { OL } +\vec { LA } =cos\alpha \hat { i } -sin\alpha \check { j } \) ..(1)
Similarly \(\hat { b } =\vec { OB } =\vec { OM } +\vec { MB } =cos\beta \hat { i } +sin\beta \hat { j } \) ...(2)
Now \(\hat { a } \times \hat { b } =|\hat { a } ||\hat { b } |sin(\alpha +\beta )\hat { k } =sin(\alpha +\beta )\hat { k } \) ....(3)
[\(|\hat { a } |=|\hat { b } |\) = 1]
Also \(\hat { a } \times \hat { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ cos\alpha & -sin\alpha & 0 \\ cos\beta & cos\beta & 0 \end{matrix} \right| \)
= \(\hat { i } (0)-\hat { j } (0)+\hat { k } \)(cosα sinβ + sinα cosβ)
= (sin α cos β + cos α sin β)\(\hat { k } \) .(4)
using (3) and (4), sin(α+β) = sin α cos β + cos α sin β

Given cubic equation is 2x³-6x²+3x+k = 0
Here, a = 2, b = -6, c = 3, d = k
Let ∝, β, ૪ be the roots
Given ∝ = 2(β+૪) ⇒ \(\frac{\alpha}{2}\) = β+૪ ...(1)
Now, \(\alpha +\beta +\gamma =\frac { -b }{ a } =-\frac { (-6) }{ 2 } =3\)
\(\frac { \alpha }{ 2 } +\alpha =3\Rightarrow \frac { \alpha +2\alpha }{ 2 } =3\Rightarrow \frac { 3\alpha }{ 2 } =3\)
\(\Rightarrow \alpha =2\)
\(\alpha \beta \gamma =\frac { -d }{ a } =\frac { -k }{ 2 } \Rightarrow 2.\beta \gamma =\frac { -k }{ 2 } \)
\(\beta \gamma =\frac { -k }{ 4 } ...(2)\)
Also, \(\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \) 
\(2\beta +\beta \gamma +2\gamma =\frac { 3 }{ 2 } \)
\(2(\beta +\gamma )+\beta \gamma =\frac { 3 }{ 2 } \)
\(\alpha \frac { -k }{ 4 } =\frac { 3 }{ 2 } \quad [from(1)\& (2)]\)
Also, \(2-\frac { k }{ 4 } =\frac { 3 }{ 2 } [\because \alpha =2]\)
\(2-\frac { 3 }{ 2 } =\frac { k }{ 4 } \Rightarrow \frac { 1 }{ 2 } =\frac { k }{ 4 } \)
\(\\ k=\frac { 4 }{ 2 } \Rightarrow k=2\)
From(2), \(\beta \gamma =\frac { -k }{ 4 } =\frac { -2 }{ 4 } =\frac { -1 }{ 2 } \)\(\Rightarrow \gamma =\frac { -1 }{ 2\beta } \)
From \((1),\beta +\gamma =\frac { \alpha }{ 2 } =\frac { 2 }{ 2 } =1\)
Substituting \(\gamma =\frac { -1 }{ 2\beta } \) We get
\(\beta -\frac { 1 }{ 2\beta } =1\Rightarrow 2{ \beta }^{ 2 }-1=2\beta \Rightarrow 2\beta -2\beta -1=0\)
\(\beta =\frac { 2\pm \sqrt { 4-4(2)(-1) } }{ 4 } =\frac { 2\pm \sqrt { 4+8 } }{ 4 } \)
\(=\frac { 2\pm \sqrt { 12 } }{ 4 } =\frac { 2\pm 2\sqrt { 3 } }{ 4 } \)
\(\beta =\frac { 1\pm \sqrt { 3 } }{ 2 } \)
Hence the roots are \(2,\frac { 1+\sqrt { 3 } }{ 2 } ,\frac { 1-\sqrt { 3 } }{ 2 } \) and k = 2