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,
\(A=2 \int_{0}^{2} y d x=2 \int_{0}^{2} \sqrt{4-x^{2}} d x=2 \int_{0}^{2} \sqrt{2^{2}-x^{2}} d x\) 
\(=2\left[\frac{x}{2} \sqrt{2^{2}-x^{2}}+\frac{2^{2}}{2} \cdot \sin ^{-1} \frac{x}{2}\right]_{0}^{2} \)
\(=2\left[\frac{2}{2} \cdot 0+2 \cdot \frac{\pi}{2}-\frac{0}{2} \cdot 2-2 \sin ^{-1}(0)\right] \)
\(=2\left[2 \cdot \frac{\pi}{2}+0\right]=2 \pi \text { sq units } \)

\(I =\int _{ 0 }^{ 1 }{ { (\tan ^{ -1 }{ x) } }^{ 2 } } .xdx\)
\(={ \left[ { (\tan ^{ -1 }{ x) } }^{ 2 }.\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ 2 } (\tan ^{ -1 }{ x) } \frac { 1 }{ 1+{ x }^{ 2 } } .\frac { { x }^{ 2 } }{ 2 } dx\)
[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, \(I=\frac { { \pi }^{ 2 }-4\pi }{ 16 } +\log { \sqrt { 2 } } \)

\(Put\ { b }^{ 2 }+{ c }^{ 2 }{ x }^{ 2 }=t\) so that \(2{ c }^{ 2 }dx=dt\)
\(i.e.\ xdx=\frac { dt }{ 2{ c }^{ 2 } } .\)
\(\therefore I=\int { \frac { 3a }{ t } . } \frac { dt }{ 2{ c }^{ 2 } } =\frac { 3a }{ 2{ c }^{ 2 } } \int { \frac { dt }{ t } } \)
\(=\frac { 3a }{ 2{ c }^{ 2 } } \log { \left| t \right| } +k\)
\( =\frac { 3a }{ 2{ c }^{ 2 } } \log { \left| { b }^{ 2 }+{ c }^{ 2 }{ x }^{ 2 } \right| } +k.\)

\(I= \int { \frac { { x }^{ 3 }+x }{ { x }^{ 4 }-9 } } dx.\)
\(=\frac { { x }^{ 3 } }{ { x }^{ 4 }-9 } dx+\int { \frac { xdx }{ { x }^{ 4 }-9 } } ..(1)\)
Now, \({ I }_{ 1 }=\frac { { x }^{ 3 } }{ { x }^{ 4 }-9 } dx.\)
\(Put\ { x }^{ 4 }=t\)so that  \(4{ x }^{ 3 }dx=dt\ i.e. { x }^{ 3 }dx=\frac { 1 }{ 4 } dt.\)
\(\therefore { I }_{ 1 }=\int { \frac { \frac { 1 }{ 4 } dt }{ t-9 } } =\frac { 1 }{ 4 } \int { \frac { dt }{ t-9 } } \)
\(=\frac { 1 }{ 4 } \log { \left| t-9 \right| } +{ c }_{ 1 }=\frac { 1 }{ 4 } \log { \left| { x }^{ 4 }-9 \right| } +{ c }_{ 1 }.\)
And \({ I }_{ 2 }=\int { \frac { x }{ { x }^{ 4 }-9 } } dx.\)
\(Put\ { x }^{ 2 }=u\)so that \(2xdx=du\ i.e. xdx=\frac { 1 }{ 2 } du.\)
\(\therefore { I }_{ 2 }=\int { \frac { \frac { 1 }{ 2 } du }{ { u }^{ 2 }-9 } } =\frac { 1 }{ 2 } \int { \frac { du }{ { u }^{ 2 }-3^{ 2 } } } \)
\(=\frac { 1 }{ 2 } .\frac { 1 }{ 2(3) } \log { \left| \frac { u-3 }{ u+3 } \right| } +{ c }_{ 2 }= \frac { 1 }{ 12 } \log { \left| \frac { { x }^{ 2 }-3 }{ { x }^{ 2 }+3 } \right| } +{ c }_{ 2 }.\)
\(\therefore \ From(1), I=\ \frac { 1 }{ 4 } \log { \left| { x }^{ 2 }-9 \right| } +\frac { 1 }{ 12 } \log { \left| \frac { { x }^{ 2 }-3 }{ { x }^{ 2 }+3 } \right| } +{ c }.\)
Where  \({ c }_{ 1 }+{ c }_{ 2 }=c\)

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

The system of constraints is:
\(x+y\le 4\)..(1)
\(x\ge 0,y\ge 0\) ..(2)
It is observed that the feasible region OAB is bounded.

∴ By Corner Point Method, we have:

Hence, Z_max = 12 at (0, 4).

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.

Since  \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \ and\ \overset { \rightarrow }{ c } +\overset { \rightarrow }{ a } \) are complannar.
\(\therefore\) \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } ).\left[ (\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } )\times (\overset { \rightarrow }{ c } +\overset { \rightarrow }{ a } ) \right] =0\)
\(\Rightarrow (\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } ).\left[ \overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } +\overset { \rightarrow }{ c } +\overset { \rightarrow }{ a } \right] =0\) 
\(\Rightarrow (\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } ).(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } +\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ a } +\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } )=0 \left[ \because \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ 0 } \right] \) 
\(\Rightarrow \overset { \rightarrow }{ a } .(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } )+\overset { \rightarrow }{ a } .(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ a } )+\overset { \rightarrow }{ a } .(\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } )+\overset { \rightarrow }{ b } .(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } )+\overset { \rightarrow }{ b } .(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ a } )+\overset { \rightarrow }{ b } .(\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } )=0\) 
\(\Rightarrow 2\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +\overset { \rightarrow }{ 0 } +\overset { \rightarrow }{ 0 } +\overset { \rightarrow }{ 0 } +\overset { \rightarrow }{ 0 } +\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =0\)
\(\Rightarrow 2\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =0\Rightarrow \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =0\) 
\(\Rightarrow \) \(\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } \ and\ \overset { \rightarrow }{ c } \) are coplanar.

LHS\(=\left[ \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ c } +\overset { \rightarrow }{ a } \right] \)
\(=\left[ \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) *\left( \overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) \right] .\left( \overset { \rightarrow }{ c } +\overset { \rightarrow }{ a } \right) \) 
\(=\left[ \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } +\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } +\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ b } \right] .\left( \overset { \rightarrow }{ c } +\overset { \rightarrow }{ a } \right) \) 
\(=\left[ \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } +\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } +\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right] .\left( \overset { \rightarrow }{ c } +\overset { \rightarrow }{ a } \right) \)  
\(=\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } \right) .\overset { \rightarrow }{ c } +\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } \right) .\overset { \rightarrow }{ a } +\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } \right) .\overset { \rightarrow }{ c } +\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } \right) .\overset { \rightarrow }{ a } +\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) .\overset { \rightarrow }{ c } +\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) .\overset { \rightarrow }{ a } \\ \)
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ a } \right] +\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ c } \overset { \rightarrow }{ c } \right] +\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \right] +\left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \overset { \rightarrow }{ c } \right] +\left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \right] \) 
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +\overset { \rightarrow }{ 0 } +\overset { \rightarrow }{ 0 } +\overset { \rightarrow }{ 0 } +\overset { \rightarrow }{ 0 } +\left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \right] \) 
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =2\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =RHS\\ \) 

Let
\(\overset { \rightarrow }{ a } =2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } ,\quad \overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } -\overset { \wedge }{ k } \) 
\(\overset { \rightarrow }{ c } =3\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) 
\(\therefore\) Volume of the paralleopiped \(=[\overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } ]\)
\(=\left| \begin{matrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{matrix} \right| \)
\(=2(4-1)+3(2+3)+4(-1-6)\)  [Expanding by R₁ ]
\(=6+15-28=-7=7\)    [rejecting -ve sign]

Here \(\overrightarrow { r } *\hat { i } \) \(=(x\hat { i } +y\hat { j } +z\hat { k } )*\hat { i } \)
\(=x(i*\hat { j } )+y(\hat { j } *i)+z(\hat { k } *i)\)
\(=x\overrightarrow { (0) } +y(-\hat { k } )+z(\hat { j } )\) 
\(=-y\hat { k } +z\hat { j } \)  ...(1)
and   \(\overrightarrow { r } *\hat { j } =(x\hat { i } +y\hat { j } +z\hat { k } )*\hat { i } \) 
\(=x(i*\hat { j } )+y(\hat { j } *\hat { j } )+z(\hat { k } *\hat { j } )\)
\(=x(\hat { k } )+y\overrightarrow { (0) } +z(-\hat { i } )\)
\(=-y\hat { k } -z\hat { i } \) ...(2)
\(\therefore (\overrightarrow { r } *\hat { i } ).(\overrightarrow { r } *\hat { j } )+xy\)
\(=(-y\hat { k } +z\hat { j } ).(x\hat { k } +z\hat { j } )+xy\)
\(=-yx(\hat { k } .\hat { k } )+yz(\hat { k } .\hat { i } )+zx(\hat { j } .\hat { k } )-{ z }^{ 2 }(\hat { j } .\hat { i } )+xy\) 
\(=-yx(1)+yz(0)+zx(0)-{ z }^{ 2 }(0)+xy\)
\(=-xy+xy=0\) 

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 \(\overrightarrow { b } +\overrightarrow { c } =\frac { (2+1)\hat { i } +2\hat { j } -6\hat { k } }{ \sqrt { { (2+1) }^{ 2 }+36+4 } } \) 
\(=\frac { 1 }{ 7 } (3\hat { i } +6\hat { j } -2\hat { k } ).\) 

We have : \(\Rightarrow { \overset { \rightarrow }{ |a| } =1=\overset { \rightarrow }{ |b| } }\ and\ \overset { \rightarrow }{ |a| } +\overset { \rightarrow }{ |b| } =1\) 
sqaring  \(|\overset { \rightarrow }{ a } +{ \overset { \rightarrow }{ { b| }^{ 2 } } }=1\Rightarrow (\overset { \rightarrow }{ a } +{ \overset { \rightarrow }{ { b) }^{ 2 } } }\)
\(\Rightarrow \overset { \rightarrow }{ { b| }^{ 2 } } +\overset { \rightarrow }{ { b| }^{ 2 } } +2|\overset { \rightarrow }{ a| } |\overset { \rightarrow }{ b| } =1\)
\( \Rightarrow { a }^{ 2 }+{ b }^{ 2 }+2|\overset { \rightarrow }{ a| } |\overset { \rightarrow }{ b| } cos\theta =1\)
Where '\(\theta \)' is the angle between  \(\overset { \rightarrow }{ a } \) and \(\overset { \rightarrow }{ b } \)
\(\Rightarrow 1+1+2(1)(1)cos\theta =1\)
\(\Rightarrow cos\theta =-\frac { 1 }{ 2 } \)
Hence, \(\theta =120°\)

Given, \(|\vec{a}+\vec{b}|=13 \text { and }|\vec{a}|=5\)
Now, \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)
\(\begin{array}{cl} \Rightarrow \quad & |\vec{a}+\vec{b}|^2=|\vec{a}|^2+0+0+|\vec{b}|^2 \end{array}\)
\(\begin{array}{cl} {\left[\because \vec{x} \cdot \vec{x}=|\vec{x}|^2, \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}=0 \text { as } \vec{a} \perp \vec{b}\right]} \end{array}\)
\(\begin{array}{ll} \Rightarrow & (13)^2=(5)^2+|\vec{b}|^2 \end{array}\)
\(\begin{array}{ll} \Rightarrow & 169=25+\left.|\vec{b}|^2 \Rightarrow|69-25=| \vec{b}\right|^2 \end{array}\)
\(\begin{array}{ll} \Rightarrow & 144=|\vec{b}|^2 \Rightarrow|\vec{b}|=12 \end{array}\)
[\(\because\) length cannot be '-' ve]

The position vector of P
\(=\frac { 1.(6\overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } )+(2\overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } ) }{ 1+2 } \)
\(=\frac { 6\overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } +4\overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } }{ 3 } =\frac { 3\overset { \rightarrow }{ a } }{ 3 } =\overset { \rightarrow }{ a } \)

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 \(\left( -\frac { \pi }{ 3 } ,\frac { \pi }{ 3 } \right) \)

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 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.

Let \(x_{ 1 },x_{ 2 }\in R\).
Now \(f(x_{ 1 })=f(x_{ 2 })\Rightarrow cosx_{ 1 }=cosx_{ 2 }\)
\(\Rightarrow \)  \(x_{ 1 }=(2n\pi +x_{ 2 })\)
\(\Rightarrow \) '\(f\)' is not one-one.
(ii) Since cos \(x\) lies in [-1,1],
\(\therefore \) R is not fully covered.
Hence, '\(f\)'  is not onto.

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