\(\left| \begin{matrix} 1 & 3 & 1 \\ 2 & -1 & -1 \\ 0 & \lambda & 3 \end{matrix} \right| =0\)
\(\Rightarrow \lambda=7\)

\(\because \quad \int { { e }^{ x }\left[ f(x\_ +f'(x) \right] } dx\)
\(={ e }^{ x }(f(x)+C\)
\(\because \quad \int _{ 0 }^{ \frac { \pi }{ 2 } }{ e^{ x } } \left[ sinx+cosx \right] dx\)
\(=\left[ { e }^{ x }sinx \right] ^{ \frac { \pi }{ 2 } }_{ 0 }\)
\(={ e }^{ \frac { \pi }{ 2 } }sin\left( \frac { \pi }{ 2 } \right) -{ e }^{ 0 }sin(0)\)
\(={ e }^{ \frac { \pi }{ 2 } }(1)-{ e }^{ 0 }(0)\)
\(={ e }^{ \frac { \pi }{ 2 } }\)

Given, y = tan^-1\(\sqrt { \frac { sinx }{ 1+cosx } } \)
y = tan^-1\(\sqrt { \frac { 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } }{ 2{ cos }^{ 2 }\frac { x }{ 2 } } } \)
y = tan^-1\(\left( \sqrt { tan\frac { x }{ 2 } } \right) \)
y = tan^-1

Given, \(y=x^2-ax+7\)
\(\Rightarrow \frac{dy}{dx}=2x-a\)
\(m_1=2x-a\)
Line  \(2x-y+9=0\)
\(\Rightarrow 2-\frac{dy}{dx}=0\)
\(\Rightarrow \frac{dy}{dx}=2=m_2\)
for parallel, \(m_1=m_2\)
\(\therefore 2x-a=2\)
\(\Rightarrow 2(-1)-a=2\)
\(\Rightarrow -2-2=a\)
\(\Rightarrow a=-4\)

\({ sin }^{ -1 }x=A\ and\ { sin }^{ -1 }y=B\)
\(\Rightarrow\) x = sin A and y = sin B
\(\therefore cos\ A=\sqrt { 1-{ x }^{ 2 } } ,cos\ B=\sqrt { 1-{ y }^{ 2 } } \)
we have sin (A+B) = sin A cos B + cos A sin B
\(\Rightarrow sin(A+B)=x\sqrt { 1-{ y }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } y\)
\(\Rightarrow sin(A+B)=x\sqrt { 1-{ y }^{ 2 } } +y\sqrt { 1-{ x }^{ 2 } } \)
\(\Rightarrow A+B={ sin }^{ -1 }\left( x\sqrt { 1-{ y }^{ 2 } } +y\sqrt { 1-{ x }^{ 2 } } \right) \)
\(\therefore { sin }^{ -1 }x+sin^{ -1 }y={ sin }^{ -1 }\left( x\sqrt { 1-{ y }^{ 2 } } +y\sqrt { 1-{ x }^{ 2 } } \right) \)

Equivalence relations could be the following:
{(1, 1), (2, 2), (3, 3), (1,2), (2, 1)} and 
{(1, I), (2, 2), (3, 3), (1, 2), (1, 3), (2,1), (2, 3), (3, 1), (3,2)} 
So, only two equivalence relations.

We have, \(A=\left| \begin{matrix} 2 & 3 & -1 \\ 4 & 1 & 0 \\ 3 & 3 & 2 \end{matrix} \right| \)
\(M_{ 12 }\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =8-0=8\)
\(M_{ 21 }\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =6+3=9\)
\(C_{ 21 }=-\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =-(6+3)=-9\)
\(C_{ 12 }=-\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =-(8-0)=-8\)
\( M_{ 12 }\times M_{ 21 }+C_{ 21 }\times C_{ 12 }\)
\( 8\times (9)+(-9)(-8)=72+72\)
\(=144\)

\(AA'= \left[ \begin{matrix} 1 & 2 & 3 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right] =\left[ 1+4+9 \right] =\left[ 14 \right]\)

Let \(I= \int _{ 0 }^{ 1 }{ \log { \frac { 1-x }{ x } } dx } \)
\(=\int _{ 0 }^{ 1 }{ \log { \left( \frac { 1-(1-x) }{ 1-x } \right) } } dx\)
\(=\int _{ 0 }^{ 1 }{ \log { \left( \frac { x }{ 1-x } \right) } } dx\)
\(=-\int _{ 0 }^{ 1 }{ \log { \left( \frac { 1-x }{ x } \right) } } dx=-1\)
\(\Rightarrow 2I=0\Rightarrow I=0.\)

\(\overset { \rightarrow }{ x } .\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) +\overset { \rightarrow }{ y } .\left( \overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ z } .\left( \overset { \rightarrow }{ c } +\overset { \rightarrow }{ a } \right) \)
\(\\ =\frac { \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } .\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) +\frac { \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } .\left( \overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) +\frac { \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } .\left( \overset { \rightarrow }{ c } +\overset { \rightarrow }{ a } \right) \) 
\(=\frac { 1 }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } \left[ (\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } ).\overset { \rightarrow }{ a } +(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } ).\overset { \rightarrow }{ b } +(\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } ).\overset { \rightarrow }{ b } +(\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } ).\overset { \rightarrow }{ c } +(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } ).\overset { \rightarrow }{ c } +(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } ).\overset { \rightarrow }{ a } \right] \) 
\(=\frac { 1 }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } \left[ \left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \right] +\left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \overset { \rightarrow }{ b } \right] +\left[ \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \right] +\left[ \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \overset { \rightarrow }{ c } \right] +\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ a } \right] \right] \) 
\(=\frac { 1 }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } \left[ \left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \right] +0+\left[ \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \right] +0+\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +0 \right] \) 
\(=\frac { 3\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } =3.\)

\(\int { { \left( \frac { { x }^{ 3 }+{ 3x }+4 }{ { x }^{ 2 } } \right) } } dx\)
\(=\int { { \left( \frac { { x }^{ 3 } }{ \sqrt { x } } +3\frac { x }{ \sqrt { x } } +\frac { 4 }{ \sqrt { x } } \right) } } dx\)
\(=\int { { x }^{ \frac { 5 }{ 2 } }dx+3 } \int { { x }^{ \frac { 1 }{ 2 } }dx+4 } \int { { x }^{ \frac { 1 }{ 2 } }dx } \)
\(=\frac { { x }^{ \frac { 5 }{ 2 } +1 } }{ \frac { 5 }{ 2 } +1 } +3\frac { { x }^{ \frac { 1 }{ 2 } +1 } }{ \frac { 1 }{ 2 } +1 } +4\frac { { x }^{ -\frac { 1 }{ 2 } +1 } }{ -\frac { 1 }{ 2 } +1 } +C\)
\(=\frac { 2 }{ 7 } { x }^{ \frac { 7 }{ 2 } }+{ 2x }^{ \frac { 3 }{ 2 } }+8{ x }^{ \frac { 1 }{ 2 } }+C.\)

\((i)\lim _{ x\rightarrow 0 }{ f\left( x \right) = } \lim _{ x\rightarrow 0 }{ (5x-3) } =0-3=-3\)
\(f(0)=-3\)
\(Thus\ \lim _{ x\rightarrow 0 }{ f\left( x \right) } =f(0)\)
f is continuous at x = 0
\((ii)\lim _{ x\rightarrow -3 }{ f\left( x \right) } =\lim _{ x\rightarrow -3 }{ (5x-3) }\)
\( =\lim _{ h\rightarrow 0 }{ \left[ 5(-3+h)-3 \right] } \)
\(=\lim _{ h\rightarrow 0 }{ (-18+5h) } \)
\(=-18+5(0)=-18\)
\(Thus\ \lim _{ x\rightarrow -3 }{ f\left( x \right) =f(-3) } \)
f is continuous at x = -3
\((iii)\lim _{ x\rightarrow 5 }{ f\left( x \right) = } \lim _{ x\rightarrow 5 }{ (5x-3) } \)
\(=\lim _{ h\rightarrow 0 }{ \left[ 5(5+h)-3 \right] } \)
\(=\lim _{ h\rightarrow 0 }{ (22+5h)=22+0=22 } \)
\(f(5)=5(5)-3=25-3=22\)
\(Thus\ \lim _{ x\rightarrow 5 }{ f\left( x \right) } =f(5)\)
f is continuous at x = 5

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24.
The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and  (6, 4)
Hence, the possible orders of a matrix having 24 elements are:
1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4
(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.
Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.

Let the coin be lossed n times.
P(getting a head) = \(\frac { 1 }{ 2 } \)( = p)
P(not getting no head) = \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)(=q)
P(at lest one head) = 1-P(0)
= \(1-^{ n }{ C }_{ 0 }{ q }^{ n }{ p }^{ 0 }=1-\left( 1 \right) { \left( \frac { 1 }{ 2 } \right) }^{ n }\left( 1 \right) \)
= \(1-{ \left( \frac { 1 }{ 2 } \right) }^{ n }\)
By the question, \(1-{ \left( \frac { 1 }{ 2 } \right) }^{ n }>\frac { 90 }{ 100 } \)
\(\Rightarrow 1-{ \left( \frac { 1 }{ 2 } \right) }^{ n }>0.9\Rightarrow { \left( \frac { 1 }{ 2 } \right) }^{ n }<1-0.9\)
\(\Rightarrow 1-{ \left( \frac { 1 }{ 2 } \right) }^{ n }<0.1\Rightarrow n\ge 4\)

2 log |y| = - cos 2x + 1 is the required solution.

Let 'x' and 'y' be the number of gold rings and chains respectively.
We have:
\(x\ge 0\)  ...(1) 
\(y\ge 0\)...(2)
\(x+y\le 24\)...(3) 
\(x+\frac { y }{ 2 } \le 16\) ...(4)
The objective function, or the profit, Z is:
Z = 300x + 190y ..(5)
We have to maximise Z subject to (1)-(4).
For solution set, we draw the lines:
x = 0, y = 0, x + y = 24, 2x + y = 32.
The lines x + y = 24 and 2x + y = 32 meet at E (8,16).

The shaded portion represents the feasible region, which is bounded.
Applying Corner Point Method, we have:

Hence, the maximum profit is Rs. 5,440 and it is obtained when 8 gold rings and 16 chains are manufactured.

As line AB is perpendicular to the plane therefore, direction ratios of the line are 2, 3,4 and line passes through the point A( 1, -2, 3)
\(\therefore \) equation of the perpendicular is
\(\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{4}=\lambda \text { (say) }\)
General point on the line is
\(B(2 \lambda+1,-3 \lambda-2,4 \lambda+3)\)
If the point lies on the plane, then 
\(2(2\lambda +1)-3(-3\lambda -2)+4(4\lambda +3)+9=0\Rightarrow 29\lambda =-29\Rightarrow \lambda =-1\)
Substituting in (i), we get foot of the perpendicular as B(-1, 1, -1).

\(\frac { 9 }{ 8 } \) sq units.

Let point be (x, y), then \(\frac{d y}{d t}=-\frac{d x}{d t}\)
Differentiating both sides of the ellipse
16x² + 9y² = 400 w.r.t. t, we get
\(16 \cdot 2 x \cdot \frac{d x}{d t}+18 y \frac{d y}{d t}=0 \)
\(\Rightarrow \frac{d y}{d t}=-\frac{16 x}{9 y} \cdot \frac{d x}{d t} \Rightarrow 16 x=9 y \quad \text { \{using (i) } \)
Substituting in curve, we get
\(16 x^{2}+\frac{256}{9} x^{2}=400 \)
\(\Rightarrow x^{2}=9 \rightarrow x=\pm 3, \text { then } y=\pm \frac{16}{3}\)
Hence, point are \((3,{16\over 3}),(-3,{-16\over 3})\)

Equation of plane: 2x + y + 3z = 26
Normal to the plane = \(2\hat{i}+\hat{j}+3\hat{k}\)
\(\therefore \) < 2, 1, 3 > are direction ratios of the normal to the plane.

Equation of the line through P(2, 3, 4) and perpendicular to the give plane is
\({{x-2}\over{2}}={{y-3}\over{1}}={{z-4}\over{3}}=\lambda\)
\(\therefore\) Co-ordinates of point N are N \((2\lambda+2,\lambda+3,3\lambda+4)\)
Since N lies on the plane,
\(\Rightarrow\)  \(2(2\lambda+2)+(\lambda+3)+3(3\lambda+4)=26\)
\(\Rightarrow\)  \(4\lambda+4+\lambda+3+9\lambda+12=26\)
\(\Rightarrow\)  \(14\lambda=26-19\)
\(\Rightarrow\)  \(14\lambda=7\)
\(\Rightarrow\)  \(\lambda={{1}\over{2}}\)
\(\therefore\) Co-ordinates of the foot of the perpendicular i.e., N are
\(\left( 2\left({{1}\over{2}} \right)+2,{{1}\over{2}}+3,3\left({{1}\over{2}} \right)+4 \right)\) i.e., \(\left( 3,{{7}\over{2}},{{11}\over{2}} \right)\)
\(\therefore\) The length of perpendicular form P to given plane is
\(|NP|=\sqrt{{2-3}^{2}+{\left(3-{{7}\over{2}} \right)}^{2}+\left(4-{{11}\over{2}}\right)^{2}}\)
\(=\sqrt{1+{{1}\over{4}}+{{9}\over{4}}}\)
\(=\sqrt{{{14}\over{1}}}\)
\(=\sqrt{{7}\over{2}}\) units
Now, N is the mid-point of PP', where P'\((\alpha,\beta,\gamma)\) is the image of point P.
\(\therefore\)   \(\left(3,{{7}\over{2}},{{11}\over{2}} \right)=\left({{2+\alpha}\over{2}} ,{{3+\beta}\over{2}},{{4+\gamma}\over{2}}\right)\)
\(\Rightarrow \) \(3={{2+\alpha}\over{2}};{{7}\over{4}}={{3+\beta}\over{2}};{{11}\over{2}}={{4+\gamma}\over{2}}\)
\(\Rightarrow\)  \(6=2+\alpha;7=3+\beta;11=4+\gamma\)
\(\Rightarrow\)  \(\alpha=4;\beta=4;\gamma=7\)
\(\therefore\) Image of point P is P'(4, 4, 7).

Let x and y units of packet of mixes are purchased from 5 and T respectively. If Z is the total cost, then
Z = 10x + 4y ...(i)
is objective function which we have to minimize
Here constraints are:
4x + y \(\ge\)80..(ii)
2x + y \(\ge\)60.(iii) 
Also, x \(\ge\) 0...(iv)
y \(\ge\)0
On plotting the graph of above constraints or inequalities (ii), (iii), (iv) and (v),
we get shaded region having corner point A, P, B as feasible region. For co-ordinate of P

Point of intersection of
2x + y = 60...(vi)
and 4x + y = 80....(vii)
From (vi) - (vii),
2x + y - 4x - y = 60 - 80
\(\Rightarrow\) -2x = -20
\(\Rightarrow\) x = 10
\(\Rightarrow\) y = 40
\(\because\) Co-ordinate of P = (10, 40)
Now the value of Z is evaluated at corner point the following table:

Since feasible region is unbounded. Therefore we have to draw the graph of the inequality.
10x + 4y < 260 ...(viii)
Since the graph of inequality (viii) does not have any point common.
So the minimum value of Z is 260 at (10, 40). i.e., minimum cost of each bottle is ~ 260 if the company purchases 10 packets of mixes from 5 and 40 packets of mixes from supplier T.

\(A=\left[ \begin{matrix} -2 & 1 \\ 3 & 4 \end{matrix} \right] \)
then \(adj\quad A={ \left[ \begin{matrix} 4 & -3 \\ -1 & -2 \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }\)
Taking \(A(adj\quad A)=\left[ \begin{matrix} -2 & 1 \\ 3 & 4 \end{matrix} \right] { \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }\)
\(=\left[ \begin{matrix} -11 & 0 \\ 0 & -11 \end{matrix} \right] =-11\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
Taking \((adj\ A)A={ \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }{ \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }\)
\(=\left[ \begin{matrix} -11 & 0 \\ 0 & -11 \end{matrix} \right] =-11\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
Getting \(\left| A \right| =\left[ \begin{matrix} -2 & 1 \\ 3 & 4 \end{matrix} \right] =-11\)
\(\therefore \left| A \right| =-11\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
A.(adj A) = (adj A).A = |A| Hence proved.

Given equation \(\left| \begin{matrix} p & q & p\alpha +q \\ q & r & q\alpha +r \\ p\alpha +q & q\alpha +r & 0 \end{matrix} \right| =0\)
Given equation \(\Rightarrow \frac { 1 }{ pq } \left| \begin{matrix} pq & { q }^{ 2 } & pq\alpha +{ q }^{ 2 } \\ pq & pr & pq\alpha +pr \\ p\alpha +q & q\alpha +r & 0 \end{matrix} \right| =0\)
\(({ R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 })\)
\(\Rightarrow \frac { 1 }{ pq } \left| \begin{matrix} 0 & { q }^{ 2 }-pr & { q }^{ 2 }-pr \\ pq & pr & pq\alpha +pr \\ p\alpha +q & q\alpha +r & 0 \end{matrix} \right| =0\)
\(\Rightarrow \frac { { q }^{ 2 }-pr }{ pq } \left| \begin{matrix} 0 & 1 & 1 \\ pq & pr & pq\alpha +pr \\ p\alpha +q & q\alpha +r & 0 \end{matrix} \right| =0\)
\(\Rightarrow \frac { { q }^{ 2 }-pr }{ pq } p\left| \begin{matrix} 0 & 1 & 1 \\ q & r & q\alpha +r \\ p\alpha +q & q\alpha +r & 0 \end{matrix} \right| =0\)
\(({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 3 })\)
\(\Rightarrow \frac { { q }^{ 2 }-pr }{ pq } p\left| \begin{matrix} 0 & 1 & 1 \\ q & -q\alpha & q\alpha +r \\ p\alpha +q & q\alpha +r & 0 \end{matrix} \right| =0\)
\(\Rightarrow \frac { { q }^{ 2 }-pr }{ pq } p({ q }^{ 2 }\alpha +rq+pq{ \alpha }^{ 2 }+{ q }^{ 2 }\alpha )=0\)
\(\Rightarrow({ q }^{ 2 }-pr)(2q\alpha +r+p{ \alpha }^{ 2 }+{ q }^{ 2 }\alpha )=0\)
\(\Rightarrow { q }^{ 2 }-pr=0\)
(i.e., p,q,r are in GP) or \(2q\alpha +r+p{ \alpha }^{ 2 }=0\)
(i.e.,\(\alpha\) is a root of the equation 2qx + r + px² = 0)

Let \(\frac { { x }^{ 2 }+x+1 }{ (x+1)^{ 2 }(x+2) } =\frac { A }{ x+1 } +\frac { B }{ (x+1)^{ 2 } } +\frac { C }{ x+2 } \)
Solving for A, B, C, we get
A = -2, B = 1, C = 3
\(\therefore \frac { { x }^{ 2 }+x+1 }{ (x+1)^{ 2 }(x+2) } dx\)
\(=-2\int { \frac { 1 }{ x+1 } dx+\int { \frac { 1 }{ (x+1)^{ 2 } } } dx+ } 3\int { \frac { 1 }{ x+2 } } dx\)
\(=-2log\left| x+1 \right| -\frac { 1 }{ x+1 } +3log\left| x+2 \right| +C\)

Writing the given differential equation as
\(\frac { dx }{ dy } =\frac { \tan ^{ -1 }{ y } -x }{ 1+{ y }^{ 2 } } \)
or \(\frac { dx }{ dy } +\frac { x }{ 1+{ y }^{ 2 } } =\frac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } } \)
The above equation is linear in 'x'.
Getting \(I.F.{ e }^{ \int { \frac { 1 }{ 1+{ y }^{ 2 } } dy } }\)
\(={ e }^{ \tan ^{ -1 }{ y } }\)
Multiplying both sides by I.F.and integrating
we get \(x{ e }^{ \tan ^{ -1 }{ y } }=\int { { e }^{ \tan ^{ -1 }{ y } } } \frac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } } dy\)
\(\Rightarrow x{ e }^{ \tan ^{ -1 }{ y } }=\int { { e }^{ t }.tdt } \)
\(\left( Assuming\quad \tan ^{ -1 }{ y } =t\Rightarrow \frac { 1 }{ 1+{ y }^{ 2 } } dy=dt \right) \)
\(\Rightarrow x{ e }^{ \tan ^{ -1 }{ y } }=t{ e }^{ t }-{ e }^{ t }+C\)
\(\Rightarrow x{ e }^{ \tan ^{ -1 }{ y } }=\left( \tan ^{ -1 }{ y } \right) \left( { e }^{ \tan ^{ -1 }{ y } } \right) -{ e }^{ \tan ^{ -1 }{ y } }+C\)
\(\Rightarrow x=\tan ^{ -1 }{ y } -1+C{ e }^{ -\tan ^{ -1 }{ y } }\)

Given equation exists, if
x² + x \(\ge\) 0 and 0 < \(\sqrt { { x }^{ 2 }+x+1 } \le 1\)
[\(\because\)x² + x + 1 is always greater than zero]
Now, x² + x \(\ge\)0 and x² + x + 1 \(\le\)1
\(\Rightarrow\) x² + x \(\ge\)0 and x² + x \(\le\)0
\(\Rightarrow\) x² + x = 0 i.e., x(x + 1) = 0
Hence, x = 0 and -1 are the solution of the given equation.

We know that A = I₃A
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow 2{ R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 2 & 4 & 4 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow 4{ R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 8 & 4 & 4 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 0 & 1 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 4 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 3 & 4 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 3 & 2 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2}\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 3 & 4 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 3 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 1 }\rightarrow { R }_{ 1 }+{ 3R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 4 & -12 & 0 \\ -1 & 3 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }+{ 4R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 4 & -12 & 0 \\ 3 & -13 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 1 }\rightarrow { R }_{ 1 }-\frac { 4 }{ 3 } { R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 16 }{ 3 } & -\frac { 8 }{ 3 } \\ 3 & -13 & 2 \\ 1 & -4 & 0 \end{matrix} \right] \)
Applying \({ R }_{ 1 }\rightarrow \frac { 1 }{ 8 } { R }_{ 1 },{ R }_{ 2 }\rightarrow \frac { 1 }{ 3 } { R }_{ 2 }\) and R₃ = - R₃
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 3 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] A\)
\({ A }^{ -1 }=\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 3 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] \)
Matrix representation of given linear equation is
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 19 \\ 5 \\ 7 \end{matrix} \right] \)
\(\Rightarrow \)  AX = B
\(\Rightarrow \)  X = A^-1B
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 1 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] \left[ \begin{matrix} 19 \\ 5 \\ 7 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0+\frac { 10 }{ 3 } -\frac { 7 }{ 3 } \\ 19-\frac { 65 }{ 3 } +\frac { 14 }{ 3 } \\ -19+20+0 \end{matrix} \right] \)
\(\therefore \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right] \)
\(\therefore \)  x = 1, y = 2, z = 1

Given f(x) = \(\frac { 4x+3 }{ 6x-4 } \)
To show f is one-one
Let   \(f(x_{ 1 })=f(x_{ 2 })\)
then \(\frac { 4x_{ 1 }+3 }{ 6x_{ 1 }-4 } =\frac { 4x_{ 2 }+3 }{ 6x_{ 3 }-4 } \)
\(\Rightarrow (4x_{ 1 }+3)(6x_{ 2 }-4)=(6x_{ 1 }-4)(4x_{ 2 }+3)\)
\(\Rightarrow 24_{ x1 },x_{ 2 }-16x_{ 1 }+18x_{ 2 }-12=24x_{ 1 }x_{ 2 }+18x_{ 1 }-16x_{ 2 }-12\)
\(\Rightarrow -16x_{ 1 }+18x_{ 2 }=18x_{ 1 }-16x_{ 2 }\)
\(\Rightarrow -16x_{ 1 }+18x_{ 2 }=-18x_{ 1 }-16x_{ 2 }\)
\(\Rightarrow -34x_{ 1 }=-34{ x }_{ 2 }\)
\(\Rightarrow x_{ 1 }=x_{ 2 }\)
f is one-one
Let \(y\in B\)
y = f(x)
\(\Rightarrow y=\frac { 4x+3 }{ 6x-4 } \)
\(\Rightarrow y(6x-4)=4x+3\)
\(\Rightarrow 6xy-4y=4x+3\)
\(\Rightarrow 6xy-4x=4y+3\)
\(\Rightarrow x(6y-4)=4y+3\)
\(\Rightarrow x=\frac { 4y+3 }{ 6y-4 } \in B=R-\left( \frac { 2 }{ 3 } \right) \)
For every value of y except y = \(\left( \frac { 2 }{ 3 } \right) \)there is a per image x = \(\frac { 4y+3 }{ 6y-4 } =g(y)\)

Let investment in first type of bond be Rs. x.
\(\therefore\)  The investment in second type of bond = Rs. 35,000 - x)
\(\therefore \ \left[ \begin{matrix} x & 35,000-x \end{matrix} \right] \left[ \begin{matrix} \frac { 8 }{ 100 } \\ \frac { 10 }{ 100 } \end{matrix} \right] =\left[ 3,200 \right] \)
\(\Rightarrow \ \frac { 8 }{ 100 } x+\left( 35,000-x \right) \frac { 10 }{ 100 } =3,200\)
\(\Rightarrow \ x=Rs.15,000\)
\(\therefore\)  investment in first bond = Rs. 15,000
and investment in second bond 
= Rs. (35,000 - 15,000)
= Rs. 20,000
Value : Trust provides financial support to the person who is disable and unable to earn sufficient income to support him or herself.

Method I : Cost Price
\(=\left[ \begin{matrix} 2.00 & 1.00 & 0.50 \end{matrix} \right] \left[ \begin{matrix} 10000 \\ 2000 \\ 18000 \end{matrix} \right] \)
\(=\left[ 20000+2000+9000 \right] \\ \) 
\( =\left[ 31000 \right] \) 
= Rs. 31,000
Sale price
\(=\left[ \begin{matrix} 2.50 & 1.50 & 1.00 \end{matrix} \right] \left[ \begin{matrix} 10000 \\ 2000 \\ 18000 \end{matrix} \right] \)
\(=\left[ 25000+3000+18000 \right] \)
\(=\left[ 46000 \right] \)
= Rs. 46,000
Profit = Rs. (46,000 - 31,000) = Rs. 15,000.
Method II. Cost Price
\(=\left[ \begin{matrix} 2.00 & 1.00 & 0.50 \end{matrix} \right] \left[ \begin{matrix} 6000 \\ 2000 \\ 8000 \end{matrix} \right] \)
\(=\left[ 12000+2000+4000 \right] \)
\(=\left[ 36000 \right] \)
= Rs. 36,000
Sale Price
\(=\left[ \begin{matrix} 2.50 & 1.50 & 1.00 \end{matrix} \right] \left[ \begin{matrix} 6000 \\ 20000 \\ 8000 \end{matrix} \right] \)
\(=\left[ 15000+30000+8000 \right] =\left[ 53000 \right] \)
= Rs. 53,000
(a) Total revenue = Rs. (46,000 + 53,000)
= Rs. 99,000.
(b) Gross profit = Rs. (15,000 + 17,000)
= Rs. 32,000.

\(\Rightarrow \)y^'=\(\frac { 1 }{ 2\sqrt { 1-{ x }^{ 2 } } } \)