Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α and β, respectively, with positive x-axis, where A and B are as in the diagram.
Draw AL and BM perpendicular to the x-axis. Then \(\left| \vec { OL } \right| =\left| \vec { OA } \right| \) cos α = cos α, \(\left| \vec { LA } \right| =\left| \vec { OA } \right| \) sin α =  sin α
So, \(\vec { OL } =\left| \vec { OL } \right| \)\(\hat { i } \) =  cos,α \(\hat { i } \), \(\overrightarrow { LA } \) = sin α (-\(\hat { j } \))
Therefore, \(\hat { a } =\overrightarrow { OA} = \overrightarrow { OL } +\overrightarrow { LA } \) = cos α \(\hat { i } \) - sin α \(\hat { j } \) ..(1)
Similarly \(\hat { b } \) = cos β \(\hat { i } \)+ sin β \(\hat { j } \) ....(2)
The angle between \(\hat { a } \) and \(\hat{b}\) is α + β and so,
\(\hat { a } .\hat { b } =\left| \hat { a } \right| \left| \hat { b } \right| \) cos (α + β) = cos (α + β) ... (3)

On the other hand, from (1) and (2)
\(\hat { a } .\hat { b } =(cos\alpha \hat { i } -sina\hat { j } )(cos\beta \hat { i } -sin\beta \hat { j } )\) = cos α cos β - sin α sin β....(4)
From (3) and (4), we get cos(α + β) = cos α cos β - sin α sin β

Given |z₁| = 1, |z₂|= 2, |z₃| = 3, |z₁ + z₂ +  z₃| = 1
|z₁|² = 1² ⇒ z₁ \(\overline { { z }_{ 1 } } \) = 1 ⇒ z₁ = \(\frac { 1 }{ { z }_{ 1 } } \)
|z₂|² = 4 ⇒ z₂ \(\overline { { z }_{ 2 } } \) = 1 ⇒ z₂ = \(\frac { 4 }{ { z }_{ 2 } } \)
|z₃|² = 9 ⇒ z₃ \(\overline { { z }_{ 3 } } \) = 1 ⇒ z₃ = \(\frac { 9 }{ { z }_{ 3 } } \)
∴ \(\left| 9,\frac { 1 }{ \overline { { z }_{ 1 } } } .\frac { 4 }{ \overline { { z }_{ 2 } } } +4.\frac { 1 }{ \overline { { z }_{ 1 } } } .\frac { 9 }{ \overline { { z }_{ 3 } } } +\frac { 4 }{ \overline { { z }_{ 2 } } } .\frac { 9 }{ \overline { z_{ 3 } } } \right| \)
\(\left| \frac { 36 }{ \overline { { z }_{ 1 } } \overline { { z }_{ 2 } } } +\frac { 36 }{ \overline { { z }_{ 1 } } \overline { { z }_{ 3 } } } +\frac { 36 }{ \overline { { z }_{ 2 } } \overline { { z }_{ 3 } } } \right| =\left| 36\left( \frac { \overline { { z }_{ 3 } } +\overline { { z }_{ 2 } } +\overline { { z }_{ 1 } } }{ \overline { { z }_{ 1 } } \overline { { z }_{ 2 } } \overline { { z }_{ 3 } } } \right) \right| \)
\(\\ \left[ \because |\overline { z_{ 1 } } +\overline { { z }_{ 2 } } +\overline { { z }_{ 3 } } |=|\overline { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } | \right] \)
=\(\frac { 36|\overline { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } | }{ |\overline { { z }_{ 1 } } ||\overline { { z }_{ 2 } } ||\overline { { z }_{ 3 } } | } =36\frac { |\overline { \overline { { z }_{ 1 } } +\overline { z_{ 2 } } +\overline { z_{ 3 } } | } }{ |\overline { { z }_{ 1 } } ||\overline { { z }_{ 2 } } ||\overline { { z }_{ 3 } } | } \)
\(\left[ \because |\overline { { z }_{ 1 } } |=|{ z }_{ 1 }|,|\overline { { z }_{ 2 } } =|{ z }_{ 21 }|,|\overline { { z }_{ 3 } } |=|\overline { { z }_{ 3 } } | \right] \)
\(=\frac{36(1)}{1(2)(3)}=\frac{\not 36}{\not 6}=6\)
∴ |9z₁ + z₂ + 4z₁z₃ + z₂z₃| = 6

tan^-1 \(\left( \frac { 1-x }{ 1+x } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }\)x gives tan^-1 1-tan^-1 x = \(\frac{1}{2}\)tan^-1x. 
Therefore, \(\frac{\pi}{4}=\frac{3}{2}tan^{-1}\)x, which in turn reduces to tan⁻¹ = \(\frac{\pi}{6}\)
Thus, x = tan\(\frac{\pi}{6}=\frac{1}{\sqrt3}\)

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\(\frac { \pi }{ 3 } \)

Given lines are \(\frac { x-3 }{ 3 } =\frac { y-3 }{ -1 } \)....(1)
and z-1 = 0 
\(\Rightarrow\) z = 1
and \(\frac { x-6 }{ 2 } =\frac { z-1 }{ 3 } \) ...(2)  
and y-2 = 0\(\Rightarrow\) y = 2
Substituting y = 2 and z = 1 in (1) we get
\(\frac { x-3 }{ 3 } =\frac { 2-3 }{ -1 } =\frac { -1 }{ -1 } =1\Rightarrow x-3=3\Rightarrow x=6\)
The point of intersection is (6, 2, 1)
Let us check whether (6, 2, 1) satisfies (1) and (2)
\((1)\rightarrow \frac { 6-6 }{ 2 } =\frac { 1-1 }{ 3 } \Rightarrow 0=0\)
\((2)\rightarrow \frac { 6-3 }{ 3 } =\frac { 1-3 }{ -1 } \Rightarrow -1=-1\)
Hence, the given two lines intersect and the point of intersection is (6, 2, 1).

The plane passes through the point.
\(\vec { a } =2\hat { i } +3\hat { j } +6\hat { k } \) and parallel to the lines \(\frac{x-1}{2}\)
\(=\frac { y+1 }{ 3 } =\frac { z-3 }{ 1 } and\frac { x+3 }{ 2 } =\frac { y-3 }{ -5 } =\frac { z+1 }{ -3 } \)
\(\Rightarrow \vec { b } =2\hat { i } +3\hat { j } +\hat { k }\ and\ \vec { c } =2\hat { i } -5\hat { j } -3\hat { k } \)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 1 \\ 2 & -5 & -3 \end{matrix} \right| \)
\(=\hat { i } (-9+5)-\hat { j } (-6-2)+\hat { k } (-10-6)\)
\(=-4\hat { i } +8\hat { j } -16\hat { k } \)
The non-parametric vector equation of the plane is
\((\vec { r } .\vec { a } ).(\vec { b } \times \vec { c } )=0,\)
\(\Rightarrow [\vec { r } (2\hat { i } +3\hat { j } +16\hat { k } ).(-4\hat { i } +8\hat { j } -16\hat { k } )]=0\)
\(\Rightarrow [\vec { r } .(-4\hat { i } +8\hat { j } -16\hat { k } )]-(-8+24-96)=0\)
\(\Rightarrow \vec { r } .(-4\hat { i } +8\hat { j } -16\hat { k } )=-80\)
\(\div -4,\) We get
\(\vec { r } .(\hat { i } -2\hat { j } +4\hat { k } )=20\)
\(Let\quad \vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\Rightarrow (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } -2\hat { j } +4\hat { k } )=20\)
\(\Rightarrow x=2y+4z=20\)
\(\Rightarrow x-2y+4z-20=0\)

\(2{ tan }^{ -1 }(cosx)={ tan }^{ -1 }(2cosec\ x)\)
Given \({ 2tan }^{ -1 }(cosx)\quad \left[ \because { tan }^{ -1 }+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \right] \)
Let us find 2 tan(cos x)
= \({ 2tan }^{ -1 }(cosx)\ \left[ \because { tan }^{ -1 }+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \right] \)
= \({ tan }^{ -1 }\left( \frac { cosx+cosx }{ 1-{ cos }^{ 2 }x } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2cosx }{ { sin }^{ 2 }x } \right) \)
\(2{ tan }^{ -1 }(cosx)={ tan }^{ -1 }\left( 2cosecx \right) =0\)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2cosx }{ { sin }^{ 2 }x } \right) -{ tan }^{ -1 }(2cosecx)=0\)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2cosx }{ { sin }^{ 2 }x } \right) -{ tan }^{ -1 }\left( \frac { 2 }{ sinx } \right) =0\)
\(\Rightarrow { tan }^{ -1 }\left( \frac { \frac { 2cosx }{ { sin }^{ 2 }x } -\frac { 2B }{ sinx } }{ 1+\frac { 2cosx }{ { sin }^{ 2 }x } .\frac { 2 }{ sinx } } \right) =0\)
\(\Rightarrow 2sin\ x\ cosx-2{ sin }^{ 2 }x=0\)
\(\Rightarrow 2sin\ x\left( cosx-sinx \right) =0\)
\(\Rightarrow sinx=0\ or\ cosx=sinx\)
\(\Rightarrow sinx=0\ or\ tanx-1\)
\(\Rightarrow sinx=0\ or\ tanx=tan\cfrac { \pi }{ 4 } \)
The solutions for sin x = 0 is \(x=n\pi ,\theta \varepsilon zd\)  and the solution for
\(tan\ x=tan\frac { \pi }{ 4 } \ x=n\pi +\frac { \pi }{ 4 } ,n\varepsilon Z\)
\(\left[ \therefore tan\ x=tan\ \alpha \Rightarrow tan\ x=2n\pi +\alpha ,n\varepsilon Z \right] \)
Hence, the solutions are \(x=n\pi \) or 
\(x=n\pi +\frac { \pi }{ 4 } ,n\varepsilon z.\)

Given slope curve
\(\Rightarrow \frac { dy }{ dx } =\frac { y-1 }{ { x }^{ 2 }+x } \) ..... (1)
\(\Rightarrow \frac { dy }{ y-1 } =\frac { dx }{ { x }^{ 2 }+x } \)
\( \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^2+x}=\frac{d x}{x^2+2\left(\frac{x}{2}\right)+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
\(ie) \frac{d y}{y-1}=\frac{d x}{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
Integrating on both sides, we get
 \( \int \frac{d y}{\log (y-1)}=\int \frac{d x}{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \\  \text { ie) } \log (y-1)=\left(\frac{1}{2\left(\frac{1}{2}\right)}\right) \log \left[\frac{\left(x+\frac{1}{2}\right)-\left(\frac{1}{2}\right)}{\left(x+\frac{1}{2}\right)+\left(\frac{1}{2}\right)}\right]+\log C \\  \log (y-1)=\log \left(\frac{x}{x+1}\right)+\log C\\   ie)  (y-1)=\frac{C x}{x+1}  \) 
\(\Rightarrow log(y-1)=log\left( \frac { cx }{ x+1 } \right) \)
\(\Rightarrow y-1=\frac { cx }{ x+1 } \)
Since the curve passes through (1, 0) we get,
\(0-1=\frac { c }{ 2 } \Rightarrow c=-2\)
\(\Rightarrow y-1=\frac { -2x }{ x+1 } \)
\(\Rightarrow y=1-\frac { -2x }{ x+1 } \)
\(\Rightarrow y=\frac { x+1-2x }{ x+1 } =\frac { 1-x }{ x+1 } \)
\(\therefore y=\frac { 1-x }{ 1+x } \)

Let T represent the temperature of the boiling water and Tm represents the temperature of the kitchen.
By Newton's law of cooling
\(\Rightarrow \int { \frac { dT }{ T-{ T }_{ m } } =K\int { dt } } \)
\(\Rightarrow log(T-{ T }_{ m })=Kt+logC\)
\(\Rightarrow log(T-{ T }_{ m })-logC=Kt\)
\(\Rightarrow log\left( \frac { T-{ T }_{ m } }{ C } \right) =Kt\)
\(\Rightarrow T-{ T }_{ m }={ Ce }^{ Kt } ...(1)\)
when t=0,T=100
\(\therefore 100-{ T }_{ m }={ Ce }^{ 0 }\)
\(\Rightarrow C=100-{ T }_{ m }\)
\(\Rightarrow becomes,\ T-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ Kt }\)
Also when t = 5, T = 80
\(\therefore 80-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ 5K }\)
\(\Rightarrow { e }^{ 5K }=\frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } ..(2)\)
When t = 10, T = 65
(2) \(\Rightarrow\) 65 - T = (100-T_m)e^10K
= (100-T_m)(e^5K)²
\(=(100-{ T }_{ m }){ \left( \frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } \right) }^{ 2 }\)
[using(2)]
\(\Rightarrow 65-{ T }_{ m }=\frac { { (80-{ T }_{ m } })^{ 2 } }{ 100-{ T }_{ m } } \)
\(\Rightarrow\) 6500-65T_m-100T_m+T_m² = 6400+T_m²-160T_m
\(\Rightarrow\) 6500-6400 = 165Tm-160Tm
\(\Rightarrow\) 100 = 5Tm
\(\\ \Rightarrow { T }_{ m }=\frac { 100 }{ 5 } ={ 20 }^{ o }C\)
Hence the temperature of the kitchen is 20^oC

W(x, y, z) = xy + yz + zx, x =u -v, y = uv, z = u + v; y = uv; z = u
\(\frac { \partial W }{ \partial x } \) = y + z; \(\frac { \partial W }{ \partial y } \) = x + z
∴ \(\frac { \partial W }{ \partial x } \) = uv + u +  v;
\(\frac { \partial W }{ \partial y} \) = u - v + u + v;
\(\frac { \partial W }{ \partial z} \) = uv +  u - v
\(\frac { dx }{ du } =1;\frac { dy }{ du } =v;\frac { dz }{ du } =1\)
\(\frac { dx }{ dv } =1;\frac { dy }{ dv } =v;\frac { dz }{ dv} =1\)
By chain rule
\(\frac { \partial W }{ \partial u } =\frac { \partial w }{ \partial x } .\frac { dx }{ du } +\frac { \partial w }{ \partial y } .\frac { dy }{ du } +\frac { \partial w }{ \partial z } .\frac { dz }{ du } \)
= (uv +u +v) (1) +2u (v) + (uv +u - v)(1)

\(\frac { \partial W }{ \partial u } \) = 4uv + 2u = 12u (2v + 1)
\({ \left( \frac { \partial W }{ \partial u } \right) }_{ \left( \frac { 1 }{ 2 } ,1 \right) }\) = 2 x \(\frac12\) (2+ 1) = 1(2+ 1) = 3
= (uv + u + v) (-1) + (2u) (u) + (uv + u - v)(1)

= 2u² - 2v = 2 (u² - v)
∴  \({ \left( \frac { \partial W }{ \partial v } \right) }_{ \left( \frac { 1 }{ 2 } ,1 \right) }\) = \(2\left( \frac { 1 }{ 4 } -1 \right) =2\left( -\frac { 3 }{ 4 } \right) =-\frac { 3 }{ 2 } \)

It is known that Z₅ = {[0], [1], [2], [3], [4]}. The table corresponding to addition modulo 5 is as follows: We take reminders {0,1,2,3,4} to represent the classes {[0], [1], [2], [3], [4]}.

(i) Since each box in the table is filled by exactly one element of Z₅, the output a +₅ b is unique and hence +₅ is a binary operation.
(ii) The entries are symmetrically placed with respect to the main diagonal. So +₅ has commutative property
(iii) The table cannot be used directly for the verification of the associative property. So it is to be verified as usual
For instance, (2+₅3)+₅ 4 = 0+₅ 4 = 4(mod 5)
and 2+₅(3+₅4) = 2 +₅ 2 = 4(mod5)
Hence (2+₅3)+₅4 = 2+₅(3+₅4)
Proceeding like this one can verify this for all possible triples and ultimately it can be shown that +₅ is associative
(iv) The row headed by 0 and the column headed by 0 are identical. Hence the identity element is 0.
(v) The existence of inverse is guaranteed provided the identity 0 exists in each row and each column. From Table, it is clear that this property is true in this case. The method of finding the inverse of any one of the elements of Z₅, say 2 is outlined below.
First find the position of the identity element 0 in the III row headed by 2. Move horizontally along the III row and after reaching 0, move vertically above 0 in the IV column, because 0 is in the III row and IV column. The element reached at the topmost position of IV column is 3. This element 3 is nothing but the inverse of 2, because, 2+₅ 5+ = 0 (mod5). In this way, the inverse of each and every element of Z₅ can be obtained. Note that the inverse of 0 is 0, that of 1 is 4, that of 2 is 3, that of 3 is 2, and, that of 4 is 1.

~(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.

y=log(x+1)+log(x+1)-x+5

\(\mathrm{u}=\sec ^{-1}\left(\frac{x^{3}-y^{3}}{x+y}\right)\)
u is not a homogeneous
\( f =\sec u=\frac{x^{3}-y^{3}}{x+y} \)
\(f(x, y) =\frac{x^{3}-y^{3}}{x+y} \)
\(f(t x, t y) =\frac{t^{3} x^{3}-t^{3} y^{3}}{t x+t y} \)
\( =\frac{t^{3}\left(x^{3}-y^{3}\right)}{t(x+y)} \)
\(f(t x, t y) =t^{2} f(x, y)\)
f is homogeneous of degree 2
\( x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=p f \)
\(x \frac{\partial(\sec u)}{\partial x}+y \frac{\partial(\sec u)}{\partial y}=2 \sec u\)
\( \sec u \tan \mathrm{u}\left(x \frac{\partial u}{\partial x}\right)+\sec u \tan x\left(y \frac{\partial u}{\partial x}\right)=2 \sec u \)
\(x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=\frac{2 \sec u}{\sec u \tan u} \)
\(x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=2 \cot u\)
Hence proved

\(\overline { (2-i)^{ 12 }+(2+i)^{ 12 } } =\overline { (2-i)^{ 12 } } +\overline { (2+i)^{ 12 } } \)
\(\left( 2+i\sqrt { 3 } \right) ^{ 10 }-\left( 2-i\sqrt { 3 } \right) ^{ 10 }\)
Now \(\overline { z } \) = \(\overline { (2+\sqrt { 3 } )^{ 10 }-(2-i\sqrt { 3 } )^{ 10 } } \)
\(\overline { z } \) = \(\overline { (2+i\sqrt { 3 } )^{ 10 } } -(2-i\sqrt { 3 } )^{ 10 }\)
[∵ \(\overline { { z }_{ 1 }-{ z }_{ 2 } } =\overline { { z }_{ 1 } } -\overline { { z }_{ 2 } } \)]
= \((2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 }\)
= -\(\left[ (2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 } \right] \)
∴ \(\overline { z } \) = -\(\ { z } \) ⇒ z is purely imaginary
Hence \((2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 }\)is purely imaginary

  \(\vec { a } \)= \(a\hat { i } +a\hat { j } +c\hat { k } , \vec{b}=\hat { i } +\hat { k } \), \(\vec { c } \)= \(c\hat { i } +c\hat { j } +b\hat { k } \)
Given \(\vec { a } ,\vec { b } \) and \(\vec { c } \) are co-planar
\(\vec { a } .(\vec { b } \times \vec { c } )\) = 0
⇒ \(\left| \begin{matrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{matrix} \right| \) = 0
⇒ \(a\left| \begin{matrix} 0 & 1 \\ c & b \end{matrix} \right| -a\left| \begin{matrix} 1 & 1 \\ c & b \end{matrix} \right| +c\left| \begin{matrix} 1 & 0 \\ c & c \end{matrix} \right| \) = 0
⇒ a(0-c)-a(b-c)+c(c-0) = 0
\(\Rightarrow-\not a c-a b+\not a c+c^{2}=0\)
⇒ c² = ab ⇒ c =\(\sqrt { ab } \).
Hence c is the geometric mean of a and b.

\({ sin }^{ -1 }(-1)+{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ cot }^{ -1 }(2)\)
Let \({ sin }^{ -1 }\left( -1 \right) =x\)
\(\Rightarrow -1=sin\ x\)
\(\Rightarrow sin\ x=-1=-sin\frac { \pi }{ 2 } =sin\left( \frac { -\pi }{ 2 } \right) \)
\(\Rightarrow x=\frac { -\pi }{ 2 } \)
\(\Rightarrow x=\frac { -\pi }{ 2 } \)
\(\Rightarrow \frac { 1 }{ 2 } =cos\ y\Rightarrow cos\frac { \pi }{ 3 } \)
\(\Rightarrow y=\frac { \pi }{ 3 } \)
\(\therefore { sin }^{ -1 }\left( -1 \right) +{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ cot }^{ -1 }\left( 2 \right) \)
\(\frac { -\pi }{ 2 } +\frac { \pi }{ 3 } +{ cot }^{ -1 }\left( 2 \right) \)
\({ cot }^{ -1 }(2)+\frac { -3\pi +2\pi }{ 0 } ={ cot }^{ -1 }\left( 2 \right) -\frac { \pi }{ 6 } \)

Since the plane passing through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) and is normal to the vector \(\overset { \rightarrow }{ n } =2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \)
the vector equation of the plane is \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ n } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ n } \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) =\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 2 - 3 + 4 = 3
\(\therefore { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 3

Given that \(\frac{dy}{dx}+2y\) = e^-x
This is a linear differential equation
Here P = 2 ; Q = e^−x.
\(\int { pdx } =\int { 2dx } =2x\)
Thus, I.F.\(={ e }^{ \int { pdx } }={ e }^{ 2x }\)
Hence the solution of (1) is \({ ye }^{ \int { pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C } \)
That is, \({ ye }^{ 2x }=\int { { e }^{ -x }{ e }^{ 2x }dx+C } or\quad { ye }^{ 2x }={ e }^{ x }+C\quad or\quad y={ e }^{ -x }+{ Xe }^{ -2x }\) required solution

Then A∨ B =\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\vee \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\vee 1 & 1\vee 1 \\ 1\vee 0 & 1\vee 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)
\(A\wedge B=\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\wedge \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\wedge 1 & 1\wedge 1 \\ 1\wedge 0 & 1\wedge 1 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}\)

Let y = f(x) = \(\sqrt x\)
Let x_o = 25, dx = 25.2 - 25 = 0.2
y = \(\sqrt x\)
dy = \(\frac{1}{2\sqrt{x}}\) dx
dy = \(\frac{1}{2\sqrt{x}}\) (0.2) = 0.02
∴\(\sqrt{25.2}\) = f(x₀) + f'(x₀) dx
= \(\sqrt{25}\) + 0.02
= 5 + 0.02 = 5.02

|xy| = ce^x

\( \mathrm{x}^{2}+\mathrm{y}^{2} =\mathrm{r}^{2} \cos ^{2} \theta+\mathrm{r}^{2} \sin ^{2} \theta \)
\(\mathrm{x}^{2}+\mathrm{y}^{2} =\mathrm{r}^{2} \)
\(2 \mathrm{x} =2 \mathrm{r} \frac{\partial r}{\partial x} \)
\(\frac{x}{r} =\frac{\partial r}{\partial x}\)
\( \frac{r \cos \theta}{r}=\frac{\partial r}{\partial x} \)
\(\therefore \frac{\partial r}{\partial x}=\cos \theta\)

If a quadrilateral is a square then it is a rectangle.
Converse statement :
If a quadrilateral is a rectangle, then it a square.
Inverse statement :
If a quadrilateral is not a square then it is not a rectangle.
Contrapositive statement :
If a quadrilateral is not a rectangle, the it is not a square.

It is enough to prove that the sides of the triangle are equal.
Let z₁ = 1, \({ z }_{ 2 }=\frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \) and \({ z }_{ 3 }=\frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \)
The length of the sides of the triangles are
\(\left| { z }_{ 1 }-{ z }_{ 2 } \right| =\left| 1-\left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\left| \cfrac { 3 }{ 2 } -\cfrac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\frac { 2\sqrt { 3 } }{ 2 } =\sqrt { 3 } \)
\(\left\lfloor { z }_{ 2 }-{ z }_{ 3 } \right\rfloor =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -\left( \frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\sqrt { \left( \sqrt { 3 } \right) ^{ 2 } } =\sqrt { 3 } \)
\(\left| { z }_{ 3 }-{ z }_{ 1 } \right| =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -1 \right| =\left| \frac { -3 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\sqrt { 3 } \)
Since the sides are equal, the given points form an equilateral triangle

Resultant of the given forces is \(\vec { F } \) = \((\hat { 3i } -\hat { 2j } +\hat { 2k } )\) + \((\hat { 2i } +\hat { j } -\hat { k } )\) = \(​​\hat { 5i } -\hat { j } +\hat { k } \)
The displacement of the particle is given by
\(\vec { d } \) =  \((\hat { 4i } -\hat { j } +\hat { \lambda k } )-(\hat { i } +3\hat { j } -\hat { k } )\) =  \((3\hat { i } -\hat { 4j } +(\lambda +1)\hat { k } )\)
As the work done by the forces is 16 units, we have
\(\vec { F } \).\(\vec { d } \) = 16
That is \((\hat { 5i } -\hat { j } +\hat { k } ).(3\hat { i } -\hat { 4j } +(\lambda +1))\hat { k } \)  = 16 ⇒ λ + 20 = 16
So, λ = - 4

\(c\ is\frac { 2\pi }{ 9 } +c\ is\frac { 4\pi }{ 9 } +c\ is\frac { 6\pi }{ 9 } +c\ is\frac { 8\pi }{ 9 } +c\ is\left( \frac { 10\pi }{ 9 } \right) +c\ is\frac { 12\pi }{ 9 } +c\ is\frac { 14\pi }{ 9 } +c\ is\frac { 16\pi }{ 9 } \)
=\(\ c\ is\left( \frac { 2\pi }{ 9 } +\frac { 4\pi }{ 9 } +\frac { 6\pi }{ 9 } +\frac { 8\pi }{ 9 } +\frac { 10\pi }{ 9 } +\frac { 12\pi }{ 9 } +\frac { 14\pi }{ 9 } +\frac { 16\pi }{ 9 } \right) \)

= \(\left[ \because 1+2+3+....+n=\frac { n(n+1) }{ 2 } \right] \)
= c is 8π = [cos(8π)+i sin 8π]
= -1 + i(0) [∴ cos8π = -1 and sin 8π = 0 = -1

Let the points be A (2, 3, 4), B (-1, 4, 5) and C (8, 1, 2)
Equation of the line joining A and B is
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
⇒ \(\frac { x-2 }{ -1-2 } =\frac { y-3 }{ 4- } =\frac { z-4 }{ 5-4 } \)
⇒ \(\frac { x-2 }{ -3 } =\frac { y-3 }{ 1 } =\frac { z-4 }{ 1 } \)
Substitute the point C (8, 1, 2) in line (1),
\(\frac { 8-2 }{ -3 } =\frac { 1-3 }{ 1 } =\frac { 2-4 }{ 1 } \)
⇒ -2 = -2 = -2
Since the point C satisfies the equation of line joining A and B, all the three points lie on the same line.
Hence the given points are collinear.

g(x) = 2sin⁻¹(2x−1)−\(\frac{\pi}{4}\)
From the definition of sin^-1x,
\(-1\le 2x-1\le 1\)
\(\Rightarrow 1+1\le 2x\le 1+1\)
\(\Rightarrow 0\le 2x\le 2\)
\(\Rightarrow 0\le x\le 1\)
\(\therefore \) Domain = [0, 1]

Let \({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) =\theta \Rightarrow cos\theta =\frac { 4 }{ 5 } \)
\(\therefore tan\theta =\frac { opp }{ adj } =\frac { 3 }{ 4 } \)
\(\Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \)
LHS = \({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ tan }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) +{ tan }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)
\(\left[ \because { tan }^{ -1 }x+tan^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \right] \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { 3 }{ 4 } +\frac { 3 }{ 5 } }{ 1-\left( \frac { 3 }{ 4 } \right) \left( \frac { 3 }{ 5 } \right) } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { 15+12 }{ 20 } }{ \frac { 20-9 }{ 20 } } \right) ={ tan }^{ -1 }\left( \frac { 27 }{ 20 } \times \frac { 20 }{ 11 } \right) \)
=  \({ tan }^{ -1 }\left( \frac { 27 }{ 11 } \right) =RHS\)
Hence proved

Given that \((1+{ x }^{ 2 })\frac { dy }{ dx } =1+{ y }^{ 2 }\) ..(1)
The given equation is written in the variables separable form 
\(\frac { dy }{ { 1+y }^{ 2 } } =\frac { dx }{ { 1+x }^{ x } } \) ...(2)
Integrating both sides of (2), we get tan⁻¹ tan⁻¹x +C.
But tan^-1 y - tan^-1 x = tan^-1 \(\left( \frac { y-x }{ 1+xy } \right) .\) ...(4)
Using (4) in (3) leads to tan^-1 \(\left( \frac { y-x }{ 1+xy } \right)\) = C, which implies \(\frac { y-x }{ 1+xy } \) = tan C = a (say).
Thus, y − x = a(1+ xy) gives the required solution

The given differential equation can be written as
\(\frac{\cos x}{\cos x} \frac{d y}{d x}+y \frac{\sin x}{\cos x}   =\frac{1}{\cos x}  \)
\(\frac{d y}{d x}+\left(\frac{\sin x}{\cos x}\right) y   =\sec x  \)
\(\frac{d y}{d x}+(\tan x) y   =\sec x\)
This is of the form \(\frac{d y}{d x}+P y=Q \)
where
\(\mathrm{P}  =\tan x  \)
\(\mathrm{Q}  =\sec x\)
Thus, the given differential equation is linear.
\(I.F=e^{\int \operatorname{Pdx}}=e^{\int \operatorname{Lin} x d x}=e^{\log (\sec x)}=\sec x\)
So, the required solution is given by 
\({[\mathrm{y} \times \mathrm{I} . \mathrm{F}] }  =\int[Q \times I F] d x+c  \)
\(\mathrm{y} \times \sec x  =\int \sec x \times \sec x d x+c  \)
\(\mathrm{y} \sec x  =\int \sec ^2 x d x+c \\ \mathrm{y} \sec x  =\tan x+\mathrm{c}  \)
\(\div \sec x, \frac{y \sec x}{\sec x}  =\frac{\tan x}{\sec x}+\frac{c}{\sec x} \)
\(\mathrm{y}  =\frac{\sin x}{\cos x} \times \frac{1}{\sec x}+\frac{c}{\sec x}  \)
\(\mathrm{y}  =\frac{\sin x}{\cos x} \times \cos x+c \cos x \)
\(=\sin x+c \cos x\)

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
\(\frac { \partial U }{ \partial z } =\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(\therefore \frac { \partial U }{ \partial x } +\frac { \partial U }{ \partial y } +\frac { \partial U }{ \partial z } =\frac { { 3x }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(=\frac { { 3({ x }^{ 2 }+y }^{ 2 }+{ z }^{ 2 }) }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)

Let (S, *) be an algebraic structure. Assume that the identity element of S exists in S .
It is to be proved that the identity element is unique. Suppose that e₁ and e₂ be any two identity elements of S .
First treat e₁ as the identity and e₂ as an arbitrary element of S.
Then by the existence of identity property \(e_{2} * e_{1}=e_{1} * e_{2}=e_{2}\)  ..........(1)
Interchanging the role of e₁ and e₂  \(e_{2}, e_{1} * e_{2}=e_{2} * e_{1}=e_{1}\)  ..........(2)
From (1) and (2), e₁ = e₂. Hence the identity element is unique which completes the proof.