Given region is \(\left\{(x, y): y=\sqrt{4-x^{2}}\right\} \text { and } X \text { -axis. }\) 

We have,\(y=\sqrt{4-x^{2}}\) 

We have,\(y=\sqrt{4-x^{2}}\) 
\(\Rightarrow y^{2}=4-x^{2} \Rightarrow x^{2}+y^{2}=4\) 
This represent the equation of circle having centre (0, 0) and radius 2.
But original equation is \(y=\sqrt{4-x^{2}}\). So y is positive. It means that, we have to take a curve above the X-axis.
Thus, only semi-circle is formed above the X-axis
Since, the region is symmetrical about Y-axis
\(\therefore\) Area of shaded region,
\(A=2 \int_{0}^{2} y d x=2 \int_{0}^{2} \sqrt{4-x^{2}} d x=2 \int_{0}^{2} \sqrt{2^{2}-x^{2}} d x\) 
\(=2\left[\frac{x}{2} \sqrt{2^{2}-x^{2}}+\frac{2^{2}}{2} \cdot \sin ^{-1} \frac{x}{2}\right]_{0}^{2} \)
\(=2\left[\frac{2}{2} \cdot 0+2 \cdot \frac{\pi}{2}-\frac{0}{2} \cdot 2-2 \sin ^{-1}(0)\right] \)
\(=2\left[2 \cdot \frac{\pi}{2}+0\right]=2 \pi \text { sq units } \)

Let \(I=\int \tan ^{2} x \sec ^{4} x d x=\int \tan ^{2} x \sec ^{2} x \sec ^{2} x d x\) 
\(=\int \tan ^{2} x\left(1+\tan ^{2} x\right) \sec ^{2} x d x \quad\left[\because \sec ^{2} x=1+\tan ^{2} x\right]\) 
Now, put  \(\tan x=t \Rightarrow \sec ^{2} x d x=d t\) 
\( \therefore I =\int\left(t^{2}-1\right) \cdot t^{2} d t=\int\left(t^{4}-t^{2}\right) d t=\frac{t^{3}}{5}-\frac{t^{3}}{3}+C\)
\( =\frac{1}{5} \sec ^{5} x-\frac{1}{3} \sec ^{3} x+C \quad[\because t=\sec x] \)

\(I =\int _{ 0 }^{ 1 }{ { (\tan ^{ -1 }{ x) } }^{ 2 } } .xdx\)
\(={ \left[ { (\tan ^{ -1 }{ x) } }^{ 2 }.\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ 2 } (\tan ^{ -1 }{ x) } \frac { 1 }{ 1+{ x }^{ 2 } } .\frac { { x }^{ 2 } }{ 2 } dx\)
[Integrating by Parts]
\(=\left( \frac { { \pi }^{ 2 } }{ 32 } -0 \right) -\int _{ 0 }^{ 1 }{ (\tan ^{ -1 }{ x) } } \frac { { x }^{ 2 } }{ 1+{ x }^{ 2 } } dx\)
\(=\frac { { \pi }^{ 2 } }{ 32 } -\int _{ 0 }^{ 1 }{ (\tan ^{ -1 }{ x) } } \frac { { x }^{ 2 } }{ 1+{ x }^{ 2 } } dx....(1)\)
Now \({ I }_{ 1 }=\int _{ 0 }^{ 1 }{ (\tan ^{ -1 }{ x) } } \frac { { x }^{ 2 } }{ 1+{ x }^{ 2 } } dx\)
\(=\int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ x } } \frac { { 1+x }^{ 2 }-1 }{ 1+{ x }^{ 2 } } dx\)
\(=\int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ x } dx- } \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ x } } \frac { 1 }{ 1+{ x }^{ 2 } } dx\)
\(={ I }_{ 2 }-{ \left[ { \frac { 1 }{ 2 } (\tan ^{ -1 }{ x) } }^{ 2 } \right] }_{ 0 }\)
\(={ I }_{ 2 }-\left[ \frac { { \pi }^{ 2 } }{ 32 } -0 \right] ={ I }_{ 2 }-\frac { { \pi }^{ 2 } }{ 32 } ...(2)\)
And  \({ I }_{ 2 }=\int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ x } dx= } \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ x } .1dx } \)
\(={ \left[ { \tan ^{ -1 }{ x } }.x \right] }_{ 0 }^{ 1 }- \int _{ 0 }^{ 1 }{ \frac { 1 }{ 1+{ x }^{ 2 } } } x.dx\)
[Integrating by Parts]
\(=\frac { \pi }{ 4 } -\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { 2x }{ 1+{ x }^{ 2 } } } dx=\frac { \pi }{ 4 } -\frac { 1 }{ 2 } \left( \log { \left| 1+{ x }^{ 2 } \right| } \right) \)
\(=\frac { \pi }{ 4 } -\frac { 1 }{ 2 } \left( \log { 2-0 } \right) =\frac { \pi }{ 4 } -\frac { 1 }{ 2 } \log { 2 } .\)
From(2)  \(I,=\frac { \pi }{ 4 } -\frac { 1 }{ 2 } \log { 2 } -\frac { { \pi }^{ 2 } }{ 32 } \)
From(1), \(I,=\frac { { \pi }^{ 2 } }{ 32 } -\frac { \pi }{ 4 } +\frac { 1 }{ 2 } \log { 2 } -\frac { { \pi }^{ 2 } }{ 32 } \)
\(=\frac { { \pi }^{ 2 } }{ 16 } -\frac { \pi }{ 4 } +\frac { 1 }{ 2 } \log { 2 } .\)
Hence, \(I=\frac { { \pi }^{ 2 }-4\pi }{ 16 } +\log { \sqrt { 2 } } \)

\(I=\int _{ \pi /3 }^{ \pi /2 }{ \sqrt { \frac { 2\cos ^{ 2 }{ \frac { x }{ 2 } } }{ { (2\sin ^{ 2 }{ \frac { x }{ 2 } } ) }^{ 5/2 } } } } dx\)
\(=\frac { 1 }{ 2 } \int _{ \pi /3 }^{ \pi /2 }{ \frac { \cos { x/ } 2 }{ \sin ^{ 5/2 }{ x } } } dx.\)
\(Put \sin { \frac { x }{ 2 } } =t\) so that  \(\cos { \frac { x }{ 2 } } .\frac { 1 }{ 2 } dx=dt\)
\(i.e.\ \cos { \frac { x }{ 2 } } dx=2dt.\)
When  \(x=\frac { \pi }{ 3 } t=\sin { \frac { \pi }{ 6 } } =\frac { 1 }{ 2 } .\)
When  \(x=\frac { \pi }{ 2 } , t=\sin { \frac { \pi }{ 4 } } =\frac { 1 }{ \sqrt { 2 } } .\)
\(\therefore I=\ \frac { 1 }{ 2 } \int _{ 1/2 }^{ 1/\sqrt { 2 } }{ \frac { 2dt }{ { t }^{ 5/2 } } } =\int _{ 1/2 }^{ 1/\sqrt { 2 } }{ { t }^{ 5/2 } } dt\)
\(={ \left[ \frac { { t }^{ -3/2 } }{ -3/2 } \right] }_{ 1/2 }^{ 1/\sqrt { 2 } }\)
\(=-\frac { 2 }{ 3 } { \left[ \frac { 1 }{ { t }^{ -3/2 } } \right] }_{ 1/2 }^{ 1/\sqrt { 2 } }=\frac { 3 }{ 2 } .\)

\(I=\int { \tan ^{ 8 }{ x } } (\sec ^{ 2 }{ x } )\)
\(=\int { \tan ^{ 8 }{ x } } (\sec ^{ 2 }{ x } )\sec ^{ 2 }{ x } dx\)
\(=\int { \tan ^{ 8 }{ x } } (1+\tan ^{ 2 }{ x } )\sec ^{ 2 }{ x } dx\)
\(=\int { \tan ^{ 8 }{ x } } \sec ^{ 4 }{ x } dx+ \int { \tan ^{ 10 }{ x } \sec ^{ 2 }{ x } } dx.\)
\(Put\ \tan { x } =t\) so that \(\sec ^{ 2 }{ x } dx=dt\)
\(\therefore I= \int { { t }^{ 8 } } dt+\int { { t }^{ 10 } } dt=\frac { { t }^{ 9 } }{ 9 } +\frac { { t }^{ 11 } }{ 11 } +c\)
\(I= \frac { \tan ^{ 11 }{ x } }{ 11 } +\frac { \tan ^{ 9 }{ x } }{ 9 } +C.\)

Let \(I=\int { \frac { dx }{ \sqrt { (x-a)(\beta -x) } } } \)
\(Put\ x-a={ t }^{ 2 }\) so that \(dx=2t\ dt\)
Also \(\beta -x=\beta -({ t }^{ 2 }+(\beta -a)=(\beta -a)-{ t }^{ 2 }\)
\(\therefore I= \int { \frac { 2t\quad dt }{ { t }^{ 2 }({ \beta -a-t }^{ 2 }) } } =\int { \frac { 2\quad dt }{ ({ \beta -a)-t }^{ 2 } } } \)
\(=2\int { \frac { dt }{ \sqrt { { (\sqrt { \beta -a } ) }^{ 2 }-{ t }^{ 2 } } } } =2\sin ^{ -1 }{ \frac { t }{ \sqrt { \beta -a } } } +c\)
\(=2\sin ^{ -1 }{ \sqrt { \frac { x-a }{ \beta -a } } } +C\)

\( \frac { d }{ dx } \left( x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } -\log { \left| x+1 \right| } +C \right) \)
\(=1-\frac { 1 }{ 2 } (2x)+\frac { 1 }{ 3 } (3{ x }^{ 2 })-\frac { 1 }{ x+1 } +0\)
\(=1-x+{ x }^{ 2 }-\frac { 1 }{ 1+x } =\frac { 1-x+{ x }^{ 2 }(1+x)-1 }{ 1+x } \)
\(=\frac { 1+{ x }^{ 3 }-1 }{ 1+x } =\frac { { x }^{ 3 } }{ x+1 } .\)
\(\left( x-\frac { 1 }{ 2 } { x }^{ 2 }+\frac { 1 }{ 3 } { x }^{ 3 }-\log { \left| x+1 \right| } +C \right) \)
\(=\int { \frac { { x }^{ 3 } }{ x+1 } } dx.\)

The feasible region is unbounded.
∴ Minimum value of Z may or may not exist.
Applying Corner Point Method, we have:

We graph 3x + 2y < 13.
It is observed that the open half plane determined by 3x + 2y < 13 and R do not have a common point.
Hence, Minimum value of Z = 13.

Sum of vectors \(\overrightarrow { b } =2\hat { i } +4\hat { j } -5\hat { k } \) and \(\overrightarrow { c } =\lambda \hat { i } +2\hat { j } +3\hat { k } \) 
\(=(2+\lambda )\hat { i } +6\hat { j } -2\hat { k } \)  
\(\therefore\)  Unit vector along the sum \(=\frac { (2+\lambda )\hat { i } +6\hat { j } -2\hat { k } }{ \sqrt { { (2+\lambda ) }^{ 2 }+36+4 } } \)
by the question \(\left( \hat { i } +\hat { j } +\hat { k } \right) =\left( \frac { (2+\lambda )\hat { i } +6\hat { j } -2\hat { k } }{ \sqrt { { (2+\lambda ) }^{ 2 } } +40 } \right) =1\)
\(\Rightarrow (1){ (2+\lambda ) }+(1)(6)+(1)(-2)=\sqrt { { (2+\lambda ) }^{ 2 }+40 } \)
\(\Rightarrow 2+\lambda +6-2\quad =\sqrt { { (2+\lambda ) }^{ 2 }+40 } \)
\(\Rightarrow 6+2= \sqrt { { (2+\lambda ) }^{ 2 }+40 } \)
Squaring \({ (\lambda +6) }^{ 2 }={ { (2+\lambda ) } }^{ 2 }+40\quad \)
\(\Rightarrow { \lambda }^{ 2 }+12+36=4+{ \lambda }^{ 2 }+4\lambda +40\)
\(\Rightarrow 12\lambda +36=4\lambda +44\Rightarrow 8\lambda =8\) 
Hence  \(\lambda =1\)
(ii) Unit vector along \(\overrightarrow { b } +\overrightarrow { c } =\frac { (2+1)\hat { i } +2\hat { j } -6\hat { k } }{ \sqrt { { (2+1) }^{ 2 }+36+4 } } \) 
\(=\frac { 1 }{ 7 } (3\hat { i } +6\hat { j } -2\hat { k } ).\) 

Here, we have \(l=\cos \frac{\pi}{4}, m=\cos \frac{\pi}{2} \text { and } n=\cos \theta\)
\(\Rightarrow l=\frac{1}{\sqrt{2}}, m=0 \text { and } n=\cos \theta\)
We know that, l² + m² + n² = 1
\(\begin{aligned} \Rightarrow \quad\left(\frac{1}{\sqrt{2}}\right)^2+(0)^2+n^2=1 \Rightarrow \frac{1}{2}+n^2=1 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad n^2=1-\frac{1}{2}=\frac{1}{2} \Rightarrow n= \pm \frac{1}{\sqrt{2}} \Rightarrow n=\frac{1}{\sqrt{2}} \end{aligned}\)
                              [\(\because\) \(\theta\) is an acute angle with Z-axis]
\(\therefore \quad \cos \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}\)
Thus, the DC's of a line are \(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\)
\(\begin{aligned} \therefore \text { Vector } \vec{a} & =|\vec{a}|((\hat{i}+m \hat{j}+n \hat{k}) \end{aligned}\)
\(\begin{aligned} = & 5 \sqrt{2}\left(\frac{1}{\sqrt{2}} \hat{i}+(0) \hat{j}+\frac{1}{\sqrt{2}} \hat{k}\right)[\because|\vec{a}|=5 \sqrt{2}, \text { given }] \end{aligned}\)
\(\begin{aligned} 5 \hat{i}+5 \hat{k} \end{aligned}\)

Since \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } =0\)
\(\therefore\) \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } =-\overset { \rightarrow }{ c } \)
\(\Rightarrow { \overset { \rightarrow }{ a } }^{ 2 }+{ \overset { \rightarrow }{ b } }^{ 2 }+2\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } ={ \overset { \rightarrow }{ c } }^{ 2 }\)
\(\Rightarrow { \overset { \rightarrow }{ |a| } }^{ 2 }+{ \overset { \rightarrow }{ |b| } }^{ 2 }+2\overset { \rightarrow }{ |a| } \overset { \rightarrow }{ |b| } cos\theta ={ \overset { \rightarrow }{ |c| } }^{ 2 }\)
Where '\(\theta\)' is the angle between \(\overset { \rightarrow }{ a } \) and \(\overset { \rightarrow }{ b } \).
\(\Rightarrow \) (3)2 + (5)2 + 2(3)(5) cos \(\theta\)= (7)2
\(\Rightarrow \) 9 + 25 + 30 + cos \(\theta\)= 49
\(\Rightarrow \) 30 cos \(\theta\) = 49-34  \(\Rightarrow \) cos \(\theta\) \(1\over2 \)
\(\Rightarrow \) \(\theta\) = 60\(°\)
hence, the angle between  \(\overset { \rightarrow }{ a } \) and \(\overset { \rightarrow }{ b } \) is 60\(°\)

Let \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +\overset { \wedge }{ k } \ and\ \overset { \rightarrow }{ b } =-\overset { \wedge }{ i } +3\overset { \wedge }{ j } +4\overset { \wedge }{ k } \) 
Then    \(\overset { \rightarrow }{ a } *\overset { \rightarrow }{ b } =\left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ 1 & 2 & 1 \\ -1 & 3 & 4 \end{matrix} \right| \) 
\(\therefore |\overset { \rightarrow }{ a } *\overset { \rightarrow }{ b } |=\sqrt { { (5) }^{ 2 }+{ (-5) }^{ 2 }+{ (5) }^{ 2 } } =\sqrt { { 3(5) }^{ 2 } } \) 
\(\pm 10\sqrt { 3 } \left( \frac { 5\overset { \wedge }{ i } -5\overset { \wedge }{ j } +5\overset { \wedge }{ k } }{ 5\sqrt { 3 } } \right) i.e\pm 10(\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } )\)
\(\therefore\) The unit vector perpendicular to the plane \(\overset { \rightarrow }{ a } \)  and  \(\overset { \rightarrow }{ b } \) is given by:
Hence, the vector of magnitude  10\(\sqrt { 3 }\)  that are perpendicular to the plane of  \(\overset { \rightarrow }{ a } \) and \(\overset { \rightarrow }{ b } \) are

We know that the slope of the tangent to the curve is \(\frac { dy }{ dx } .\)
By the question, \(\frac { dy }{ dx } =x+y\)
\(\Rightarrow\) \(\frac { dy }{ dx } -y=x\) .(1)
Here  \('P'=-1\) and \('Q'=x.\)
\(\therefore\) I.F. \(={ e }^{ \int { Pdx } }={ e }^{ \int { -1.dx } }={ e }^{ -x }\)
Multiplying (1) by \({ e }^{ -x },\) we get:
\({ e }^{ -x }.\frac { dy }{ dx } -y.{ e }^{ -x }=x.{ e }^{ -x }.\)
\(\Rightarrow\) \(\frac { d }{ dx } \left( y.{ e }^{ -x } \right) =x{ e }^{ -x }.\)
Integrating,
\(y.{ e }^{ -x }=\int { x.{ e }^{ -x } } dx+C\)
\(=x.\frac { { e }^{ -x } }{ -1 } -\int { \left( 1 \right) } \frac { { e }^{ -x } }{ -1 } dx+C\)
[Integrating by Parts]
\(=-x{ e }^{ -x }+\int { { e }^{ -x } } dx+C\)
\(=-x{ e }^{ -x }+\frac { { e }^{ -x } }{ -1 } +C\)
\(\Rightarrow\) \(y=-x-1+C\quad { e }^{ x }\)
Since the curve passes through of the origin(0,0),
\(\therefore\) \(0=-0-1+C\quad { e }^{ 0 }\Rightarrow C=1.\)
Putting in (2),\(y=-x-1+{ e }^{ x }\)
\(\Rightarrow x+y+1={ e }^{ x },\)
Which is the requested equation of the curve.

We have : \(f(x)\)  = tan x - 4x.
\(f'(x)=sec^{ 2 }x-4\)
When  \((\frac { -\pi }{ 3 } )\)
Thus for \( (\frac { -\pi }{ 3 } )\)  
Hence 'f' is strictly decreasing on \(\left( -\frac { \pi }{ 3 } ,\frac { \pi }{ 3 } \right) \)

Let 'r' be the radius of spherical ball of salt.
By the question.
\(\frac { dV }{ dt } =-ks\)
\(\Rightarrow \frac { d }{ dt } \left( \frac { 4 }{ 3 } \pi r^{ 3 } \right) =-k(4\pi r^{ 2 })\)
\(\Rightarrow \frac { 4 }{ 3 } \pi (3r^{ 2 })\frac { dr }{ dt } =-k(4\pi r^{ 2 })\)
\(\Rightarrow 4\pi r^{ 2 }\frac { dr }{ dt } =-k(4\pi r^{ 2 })\)
\(\Rightarrow \frac { dt }{ dt } =-k(-ve\quad constant)\)
Hence, the radius is decreasing at a constant rate.

We have : x = at²  ..(1)
and  y = 2at ...(2)
From (2) \(t=\frac { y }{ 2a } \)
Putting in (1), \(x=a\left( \frac { { y }^{ 2 } }{ { 4a }^{ 2 } } \right) \Rightarrow { y }^{ 2 }=4ax\)
When t = 1, then from (1), x = a.
When t = 2, then from (2), x = 4a
Therefore, Reqd. area = 2 (area ABCD)
\(=2\overset { 4a }{ \underset { a }{ \int { } } } ydx=2\overset { 4a }{ \underset { a }{ \int { } } } 2\sqrt { ax } dx\)
\(=4\sqrt { a } \left[ \frac { { x }^{ 3/2 } }{ 3/2 } \right] _{ a }^{ 4a }\)
\(=\frac { 8 }{ 3 } \sqrt { a } \left[ { (4a) }^{ 3/2 }-{ (a) }^{ 3/2 } \right] \)
\(=\frac { 8 }{ 3 } { a }^{ 2 }\left[ 8-1 \right] =\frac { 56 }{ 3 } { a }^{ 2 }sq.units\)

The given parabola is y² = 2x and the given st. line is x - y = 4
Solving (1) and (2):
From (2), x = 4 + y
Putting in (1), y² = 8 + 2y \(\Rightarrow\) y² - 2y - 8 = 0 \(\Rightarrow\) (y - 4) (y + 2) = 0 \(\Rightarrow\) y = 4, - 2
When y = 4, then from (3), x = 4 + 4 = 8
When y = -2, then from (3), x = 4 - 2 = 2
Thus the line (2) cuts parabola (2) in the points A (2, -2) and B(8, 4).

The region is shown as shaded in the above figure.
Therefore, Reqd.area
\(=\overset { 4 }{ \underset { -2 }{ \int { } } } \left( 4+y\frac { { y }^{ 2 } }{ 2 } \right) dy=\left[ 4y+\frac { { y }^{ 2 } }{ 2 } -\frac { { y }^{ 3 } }{ 6 } \right] _{ -2 }^{ 4 }\)
\(=\left( 4(4)+\frac { 16 }{ 2 } -\frac { 64 }{ 6 } \right) -\left( -8+\frac { 4 }{ 2 } +\frac { 8 }{ 6 } \right) \)
\(=\left( 16+8\frac { 64 }{ 6 } \right) -\left( -8+2+\frac { 8 }{ 6 } \right) \)
\(=\left( 24-\frac { 64 }{ 6 } \right) +\left( 6-\frac { 8 }{ 6 } \right) =30-\frac { 72 }{ 6 } =30-12=18sq.units\)

The given curve is ay² = x³
The region is shown as shaded in the figure:

Therefore, Required area, ALMB
\(=\overset { 2a }{ \underset { a }{ \int { } } } xdy=\overset { 2a }{ \underset { a }{ \int { } } } { a }^{ 1/3 }{ y }^{ 2/3 }dy\)
\(={ a }^{ 1/3 }\left[ \frac { { y }^{ 5/3 } }{ 5/3 } \right] _{ a }^{ 2a }=\frac { 3 }{ 5 } { a }^{ 1/3 }\left[ { (2a) }^{ 5/3 }-{ a }^{ 5/3 } \right] \)
\(=\frac { 3 }{ 5 } { a }^{ 1/3 }{ a }^{ 5/3 }\left[ { 2.2 }^{ 2/3 }-1 \right] =\frac { 3 }{ 5 } { a }^{ 2 }({ 2.2 }^{ 2/3 }-1)sq.units\)

No.
For  \(1,2\in Z,\quad 1*2=1-2+1.2=1\)
while   \(2*1=2-1+2.1=3\)
Hence ​​​​​​​\(1*2\neq 2*1\)

(i) a*b = 1 + ba = b*a.
Thus '*' is commutative.
(a*b)*c = (1 + bc) = a + 1 + abc = 1 + a + abc.
\(\therefore \) (a*b) * c \(\neq \) a*(b*c).
Thus '*' is not associative.
Hence '*' is commutative but not associative.

Partition the set N into pairwise disjoint subsets. The equivalent classes are as given by:
\(A_{ 0 }=\{ 5,10,15,20,............\} \)
\(A_{ 1 }=\{ 1,16,11,16,21,.......\} \)
\(A_{ 2 }=\{ 2,7,12,17,22,.........\} \)
\(A_{ 3 }=\{ 3,8,13,18,23,.........\} \)
\(A_{ 4 }=\{ 4,9,14,19,24,.........\} \)
Clearly the above five sets are pairwise disjoint and \(A_{ 0 }\cup A_{ 1 }\cup A_{ 2 }\cup A_{ 3 }\cup A_{ 4 }=\overset { 4 }{ \underset { i=0 }{ \cup } } A_{ i }=N\)

In g : \(2\rightarrow 5\) and in f, \(5\rightarrow 2\); etc.
fog = {(2, 2), (3, 3)}.

(i) 'f' is a function.
[∵ each element of A in the first place in the ordered pair is related to only one element of A in the second place]
(ii) 'g' is a not function.
[∵ 1 is related to two elements of A namely 2 and 3

\(\Delta=\begin{vmatrix} x-p&p&q\\p-x&x&q\\0&q&x\end{vmatrix}\)
\(=(x-p)\begin{vmatrix} 1&p&q\\-1&x&q\\0&q&x\end{vmatrix}\)
\(=(x-p)\begin{vmatrix} 1&p&q\\0&x+p&2q\\0&q&x\end{vmatrix}\)
\((x-p)(1)\begin{vmatrix} x+p&2q\\q&x\end{vmatrix}\)
\(=(x-p)[(x^2+px)-2q^2]\)
\((x-p)(x^2+px-2q^2)\) which is true