Let numbers be x, y, z then
x + y + z = -1
2y + 3z = 5
x + y – z = -1
\(x=-\frac { 7 }{ 2 } \), y = \(\frac { 5 }{ 2 } \), z = 0

Order of A = 2 x 3
\(A=\left[ \begin{matrix} 1 & -2 & -5 \\ 3 & 4 & 0 \end{matrix} \right] \)

\(\text { We have } \tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}\)
\(\text { Or }\tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)=\frac{\pi}{4}\)
\(\text { i.e. }\tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)=\frac{\pi}{4}\)
\(\text { Therefore } \quad \frac{5 x}{1-6 x^{2}}=\tan \frac{\pi}{4}=1\)
\(\text { or } \quad 6 x^{2}+5 x-1=0 \text { i.e., }(6 x-1)(x+1)=0\)
\(\text { which gives } \ x=\frac{1}{6} \text { or } x=-1 .\)
Since x = – 1 does not satisfy the equation, as the L.H.S. of the equation becomes negative\(x=\frac{1}{6}\) is the only solution of the given equation

As given \(\overrightarrow { a } .\overrightarrow { b } =\overrightarrow { a } .\overrightarrow { c } =0=\overrightarrow { a } \bot \)both \(\overrightarrow { b } \) and \(\overrightarrow { c } \)
⇒ \(\overrightarrow { a } \parallel \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Let \(\overrightarrow { a } =\lambda \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
then \(\left| \overrightarrow { a } \right| =\left| \lambda \right| \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| \)
⇒ \(\frac { \left| \overrightarrow { a } \right| }{ \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| } =\left| \lambda \right| \)
⇒ \(\left| \lambda \right| =\frac { 1 }{ sin\frac { \pi }{ 6 } } =2\)
λ = 士2
\(\overrightarrow { a } =\pm 2\left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Hence proved
(ii) Now \(\left[ \overrightarrow { a } +\overrightarrow { b } ,\overrightarrow { b } +\overrightarrow { c } ,\overrightarrow { c } +\overrightarrow { a } \right] \)
= \(\left[ \left( \overrightarrow { a } +\overrightarrow { b } \right) \times \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right] .\left( \overrightarrow { c } +\overrightarrow { a } \right) \)
= \(\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } +\left( \overrightarrow { b } \times \overrightarrow { c } \right) .\overrightarrow { a } \)
(As the scalar triple product = 0, if any two vectors are equal)
Hence, \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } \) = \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\overrightarrow { c } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= 2\(\overrightarrow { a } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= \(2\overrightarrow { a } \left( \pm \frac { 1 }{ 2 } \overrightarrow { a } \right) \)
= 士1
Hence proved

Total number of oranges = 25
number of good oranges = 20
number of bad oranges = 5
Probability of getting a bad orange,
\(P(B)=\frac { 5 }{ 25 } =\frac { 1 }{ 5 } \)
Now  \(p=\frac { 1 }{ 5 } \)
\(\\ q=1-p=\frac { 4 }{ 5 } \)
Let X be the random variable of "Number of bad oranges".
\(\Rightarrow X=0,1,2,3,4\)
\(P(X=0)={ n }_{ { C }_{ r } }{ \left( p \right) }^{ n-r }{ \left( q \right) }^{ r }\)
\(P(X=0)={ 4 }_{ { C }_{ 0 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 4 }=\frac { 256 }{ 625 } \)
\(P(X=1)=4\times \frac { 64 }{ 125 } \times \frac { 1 }{ 5 } =\frac { 256 }{ 625 } \)
\(P(X=2)={ 4 }_{ { C }_{ 2 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 2 }{ \left( \frac { 1 }{ 5 } \right) }^{ 2 }\)
\(=6\times \frac { 16 }{ 625 } =\frac { 96 }{ 625 } \)
\(P(X=3)={ 4 }_{ { C }_{ 3 } }\left( \frac { 4 }{ 5 } \right) { \left( \frac { 1 }{ 5 } \right) }^{ 3 }\)
\(=4\times \frac { 4 }{ 625 } =\frac { 16 }{ 625 } \)
\(P(X=4)={ 4 }_{ { C }_{ 4 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 0 }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }\)
\(=\frac { 1 }{ 625 } \)
\(\therefore\)   Probability distribution is

\(\therefore Mean=\sum { XP(X) } \)
\(=\frac { 500 }{ 625 } =\frac { 20 }{ 25 } =\frac { 4 }{ 5 } \)
\(Var.\quad (X)=E({ X }^{ 2 })-[E(X){ ] }^{ 2 }\)
\(=\frac { 800 }{ 625 } -\frac { 16 }{ 25 } \)
\(Var.(X)=\frac { 32 }{ 25 } -\frac { 16 }{ 25 } =\frac { 16 }{ 25 } \)

\(I=\int { \frac { { x }^{ 2 }+x+1 }{ (x+2)({ x }^{ 2 }+1) } } dx\)
\(=\int { \left[ \frac { A }{ (x+2) } +\frac { Bx+C }{ ({ x }^{ 2 }+1) } \right] dx } \)
\(\Rightarrow A{ x }^{ 2 }+A+B{ x }^{ 2 }+2Bx+Cx+2C\)
\(=x+{ x }^{ 2 }+1\)
\(A+B=1\)
\(2B+C=1\)
\(A+2C=1\)
\(A=3/5,\)
\(B=2/5,\)
and ​​​​​​\(C=1/5\)
\(\therefore I=\int { \left[ \frac { 3 }{ 5(x+2) } +\frac { \frac { 2 }{ 5 } x+\frac { 1 }{ 5 } }{ ({ x }^{ 2 }+1 } \right] dx } \)
\(=\frac { 1 }{ 5 } \int { \left[ \frac { 3 }{ (x+2) } +\frac { 2x+1 }{ ({ x }^{ 2 }+1) } \right] } dx\)
\(=\frac { 1 }{ 5 } \left[ \int { \frac { 3 }{ (x+2) } } dx+\int { \frac { 2x+1 }{ ({ x }^{ 2 }+1) } dx+\int { \frac { 1 }{ ({ x }^{ 2 }+1) } dx } } \right] \)
\({ I }_{ 1 }=\frac { 1 }{ 5 } \int { \frac { 3 }{ (x+2) } } dx\)
\(=\frac { 3 }{ 5 } log(x+2)+{ C }_{ 1 }\)
\({ I }_{ 2 }=\frac { 1 }{ 5 } \int { \frac { 2x }{ ({ x }^{ 2 }+1) } } dx\)
Let  \({ x }^{ 2 }+1=t,\Rightarrow 2xdx=dt\)
\(=\frac { 1 }{ 5 } \int { \frac { dt }{ t } =\frac { 1 }{ 5 } logt } \)
\(=\frac { 1 }{ 5 } log(1+{ x }^{ 2 })+{ C }_{ 2 }\)
\({ I }_{ 3 }=\frac { 1 }{ 5 } \int { \frac { 1 }{ { x }^{ 2 }+1 } dx } \)
\(=\frac { 1 }{ 5 } tan^{ -1 }x+{ C }_{ 3 }\)
\(\therefore I=\frac { 1 }{ 5 } \left[ 3log(x+2)+log(1+{ x }^{ 2 })+tan^{ -1 }x \right] +C\)

Given \(A=\left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{matrix} \right] \) 
|A| = 1(1+3)-2(-1-1)+1(3-1)
= 4 + 4 + 2 = 10
Co-factor Matrix is :
\(=\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{matrix} \right] \)
\(\therefore\) adj A = transpose of above matrix
\(=\left[ \begin{matrix} 4 & -5 & 2 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{matrix} \right] \quad \)
\(\therefore { A }^{ -1 }=\frac { adj\quad A }{ |A| } \)
\(=\frac { 1 }{ 10 } \left[ \begin{matrix} 4 & -5 & 2 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{matrix} \right] \quad \)
\(\Rightarrow { A }^{ -1 }=\left[ \begin{matrix} 2/5 & -1/2 & 1/10 \\ 1/5 & 0 & -1/5 \\ 1/5 & 1/2 & 3/10 \end{matrix} \right] \)
Given set of equations are:
x + 2y + z = 4
-x + y + z = 0
x - 3y + z = 4
\(\Rightarrow \quad \left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 0 \\ 4 \end{matrix} \right] \)
\(\Rightarrow \quad A.X=B\)
Multiplying both sides by A^-1 , we get
\({ A }^{ -1 }AX={ A }^{ -1 }B\)
\(\Rightarrow X={ A }^{ -1 }B\)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2/5 & -1/2 & 1/10 \\ 1/5 & 0 & -1/5 \\ 1/5 & 1/2 & 3/10 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 0 \\ 4 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} \frac { 8 }{ 5 } +\frac { 2 }{ 5 } \\ \frac { 4 }{ 5 } -\frac { 4 }{ 5 } \\ \frac { 4 }{ 5 } +\frac { 6 }{ 5 } \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 0 \\ 2 \end{matrix} \right] \)
\(\therefore \ x=2,y=0,z=2\)

Let length be x m and breadth be y m.
༜ Length of fence L = x + 2y
Let given area \(=a\Rightarrow xy=a\ or\ y=\frac{a}{x}\)
\(\Rightarrow L=x+\frac{2a}{a}\)
\(\Rightarrow \frac{dL}{dx}=1-\frac{2a}{x^2}\)
For maxima or minima,
\(\frac{dL}{dx}=0\Rightarrow x^2=2a, \ \therefore x=\sqrt{2a}\)
\(\frac{d^2L}{dx^2}=\frac{2a}{x^3}>2\)
For minimum length,
\(L=\sqrt{2a}+\frac{2a}{\sqrt{2a}}\)
\(=2\sqrt{2a}\)
\(x=\sqrt{2a},y=\frac{a}{\sqrt{2a}}=\frac{\sqrt{2a}}{2}=\frac{1}{2}x\)
\(\therefore x=2y\)

Let (a, b), (c, d) E Q x Q. Then b + ad may not be mequal to d + cd. We find that (1, 2) * (2, 3) = (2, 5), (2,3) * (1,2) = (2,7) '*(2, 5) Hence, * is not commutative.
Let, (a, b), (c, d), (e,f> E Q x Q, {(a, b) * (c, d) * (e,f) 
= (ace, b + ad + acf)
= (a, b) * {(c, d) * (e, f)
Hence * is associative. 1
(x, y) Q x Q is the identity element for * if 2
(x, y)  Q x Q is the inverse of (a, b) E Q x Q if (c, d) * (a, b) = (a, b) * (c, d) = (1,0),
i.e., (ac, b + ad) = (ca, d + cb) = (1,0)
\(\Rightarrow c=\frac { 1 }{ a } ,d=\frac { -b }{ a } \)
The inverse of (a,b) \(\in Q\) XQ \(a\neq 0\quad \left( \frac { 1 }{ a } ,\frac { -b }{ a } \right) \)

Let, \(\sin ^{ -1 }{ x } =\sin { A } \quad \Rightarrow \quad \sin { A } =x\)
\(\sin ^{ -1 }{ y } =\sin { B } \quad \Rightarrow \quad \sin { B } =y\)
\(\sin ^{ -1 }{ z } =\sin { C } \quad \Rightarrow \quad \sin { C } =z\)
\(\Rightarrow \quad A+B+C=\pi \quad \Rightarrow \quad 2A+2B+2C=2\pi \)
\(\therefore \quad sin2A+sin2B+sin2C=4sinAsinBsinC\)
[Using trigonometric property]
\(\Rightarrow \quad 2sinAcosA+2sinBcosB+2sinCcosC=4sinAsinBsinC\)
\(\Rightarrow \quad 2sinA\sqrt { 1-\sin ^{ 2 }{ A } } +2sinB\sqrt { 1-\sin ^{ 2 }{ B } } +2sinC\sqrt { 1-\sin ^{ 2 }{ C } } =4sinAsinBsinC\)
\(\Rightarrow \quad 2x\sqrt { 1-{ x }^{ 2 } } +2y\sqrt { 1-{ y }^{ 2 } } +2z\sqrt { 1-{ z }^{ 2 } } =4xyz\)
\(\Rightarrow \quad x\sqrt { 1-{ x }^{ 2 } } +y\sqrt { 1-{ y }^{ 2 } } +2z\sqrt { 1-{ z }^{ 2 } } =2xyz\)
Hence proved.

\(\mathrm{LHL}=\lim _{x=0} f(0-h)=\lim _{h \rightarrow 0} \frac{e^{-1 / h}-1}{e^{-1 / h}+1} \)
\(=\lim _{h \rightarrow 0} \frac{\frac{1}{e^{1 / h}}-1}{\frac{1}{e^{1 / h}}+1}=\frac{0-1}{0+1}=-1\)
\(\mathrm{RHL} =\lim _{h=0} f(0+h)=\lim _{h \rightarrow 0} \frac{e^{1 / h}-1}{e^{1 / h}+1}=\lim _{h \rightarrow 0} \frac{1-\frac{1}{e^{1 / h}}}{1+\frac{1}{e^{1 / h}}} \)
\(=\frac{1-0}{1+0}=1\)
\(\begin{matrix} LHL \\ x=0 \end{matrix}\neq \begin{matrix} RHL \\ x=0 \end{matrix} \)
\(\therefore \) Discontinuous.

Let  \((3x+1)=A\frac { d }{ dx } (4-3x-2x^{ 2 })+B\)
\(\Rightarrow 3x+1=A(-3-4x)+B\)
When \(x=-\frac { 3 }{ 4 } \)
\(3\left( -\frac { 3 }{ 4 } \right) +1=A\left[ -3-4\left( -\frac { 3 }{ 4 } \right) \right] +B\)
\(\Rightarrow -\frac { 9 }{ 4 } +1=B\)
\(\Rightarrow B=-\frac { 5 }{ 4 } \)
When x = 0,
3(0) + 1 = A(-3-0) + B
= -3A + B
\(\Rightarrow 1+\frac { 5 }{ 4 } =-3A\)  \(\left( \because B=-\frac { 5 }{ 4 } \right) \)
\(\Rightarrow \frac { 9 }{ 4 } \times \left( -\frac { 1 }{ 3 } \right) =A\)
\(\Rightarrow A=-\frac { 3 }{ 4 } \)
\(\therefore 3x+1=-\frac { 3 }{ 4 } (-3-4x)-\frac { 5 }{ 4 } \)
\(\therefore I=\int { \left[ -\frac { 3 }{ 4 } (-3-4x)-\frac { 5 }{ 4 } \right] \sqrt { 4-3x-2x^{ 2 }dx } } \)
\(=\int { -\frac { 3 }{ 4 } (-3-4x)\sqrt { 4-3x-2x^{ 2 }dx } } -\int { \frac { 5 }{ 4 } } \sqrt { 4-3x-2x^{ 2 }dx } \)
\(=-\frac { 3 }{ 4 } \frac { (4-3x-2x^{ 2 })^{ 3/2 } }{ \frac { 3 }{ 2 } } -\frac { 5 }{ 4 } \int { \sqrt { 2\left( 2-\frac { 3x }{ 2 } -{ x }^{ 2 } \right) } } dx\)
\(=-\frac { 1 }{ 2 } (4-3x-2x^{ 2 })-\frac { 5\sqrt { 2 } }{ 4 } \int { \sqrt { \left( \frac { \sqrt { 41 } }{ 4 } \right) ^{ 2 }-\left( x+\frac { 3 }{ 4 } \right) ^{ 2 } } } dx\)
\(=\frac { 1 }{ 2 } (4-3x-2x^{ 2 })-\frac { 5\sqrt { 2 } }{ 4 } \left[ \frac { 4x+3 }{ 8 } \sqrt { \left( \frac { \sqrt { 41 } }{ 4 } \right) ^{ 2 }-\left( x+\frac { 3 }{ 4 } \right) ^{ 2 } } +\frac { \frac { 41 }{ 16 } }{ 2 } { sin }^{ -1 }\frac { x+\frac { 3 }{ 4 } }{ \frac { \sqrt { 41 } }{ 16 } } \right] +C\)
\(\therefore I=-\frac { 1 }{ 2 } (4-3x-2x^{ 2 })^{ 3/2 }\)
\(=-\frac { 5\sqrt { 2 } }{ 4 } \left[ \frac { 4x+3 }{ 8 } \sqrt { \left( \frac { \sqrt { 41 } }{ 4 } \right) ^{ 2 }-\left( x+\frac { 3 }{ 4 } \right) ^{ 2 } } +\frac { 41 }{ 32 } sin^{ -1 }\frac { 4x+3 }{ \sqrt { 41 } } \right] \)

The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O (0, 0), A (2, 0), B (0, 3), and C\(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)
The values of Z at these corner points are as follows.

Therefore, the maximum value of Z is \(\frac { 235 }{ 19 } \) at \(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \).

Let A (x₁ ,y₁ ,z₁ ), B(x₂, y₂, z₂) and C(x₃, y₃, z₃)be the vertices of
\(\Delta \)ABC. If D, E are mid-points of [AB] and [AC] respectively, then their co-ordinates are
\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } ,\frac { { z }_{ 1 }+{ z }_{ 2 } }{ 2 } \right) \)
\(and\ \left( \frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } ,\frac { { z }_{ 1 }+{ z }_{ 3 } }{ 2 } \right) \) respectively
Now direction-ratios of BC are:
\(<{ x }_{ 3 }-{ x }_{ 2 },{ y }_{ 3 }-{ y }_{ 2 },{ z }_{ 3 }-{ z }_{ 2 }>\)

and direction-ratios of DE are:
\(<\frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } -\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } -\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } ,\frac { { z }_{ 1 }+{ z }_{ 3 } }{ 2 } -\frac { { z }_{ 1 }+{ z }_{ 2 } }{ 2 } >\)
\(i.e.<\frac { { x }_{ 3 }-{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 3 }-{ y }_{ 2 } }{ 2 } ,\frac { { z }_{ 3 }-{ z }_{ 2 } }{ 2 } >\)
\(i.e.<{ x }_{ 3 }-{ x }_{ 2 },{ y }_{ 3 }-{ y }_{ 2 },{ z }_{ 3 }-{ z }_{ 2 }>\)
Thus DE are||BC
Now
\(|DE|=\sqrt { { \left( \frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } -\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \right) }^{ 2 }+{ \left( \frac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } -\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) }^{ 2 }+{ \left( \frac { { z }_{ 1 }+{ z }_{ 3 } }{ 2 } -\frac { { z }_{ 1 }+{ z }_{ 2 } }{ 2 } \right) }^{ 2 } }\)
\(=\frac { 1 }{ 2 } \sqrt { { \left( { x }_{ 3 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-{ y }_{ 2 } \right) }^{ 2 }+{ \left( { z }_{ 3 }-{ z }_{ 2 } \right) }^{ 2 } } \)
\(=\frac { 1 }{ 2 } |BC|\)
Hence, DE||BC and |DE| = 1/2BC

Let 'x' and 'y' be the number of pieces of Model A and Model B respectively.
We have: \(x\ge 0\) ....(1)
\(y\ge 0\) .....(2)
\(9x+12y\le 180\quad i.e.3x+4y\le 60\) .....(3)
\(x+3y\le 30\)....(4)
The mathematical formulation of the problem is as be low:
Maximize Z = 8000x + 12000y subject to the constraints (1)-(4).
For solution set, we draw the lines:
x = 0, y = 0, 3x + 4y = 60 and 3y = 30.
The lines 3x + 4y = 60 and x + 3y = 30 meet at E (12, 6).

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

Hence, the maximum profit is Rs. 1,68,000 when 12 pieces of Model A and 6 pieces of Model B are manufactured per week. 

\(We\quad know\quad that\quad A={ I }_{ 2 }A\)
\(i.e.\ \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A\)
\( \Rightarrow \begin{bmatrix} -2 & 1 \\ 6 & -3 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}A\quad [Applying\quad { R }_{ 1 }\leftrightarrow { R }_{ 2 }]\)
\( \Rightarrow \begin{bmatrix} -2 & 1 \\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 1 & 3 \end{bmatrix}A\quad [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }+3{ R }_{ 1 }]\)
Since second row of LHS matrix has all zero,
\(\therefore \ { A }^{ -1 }\) does not exist.

(i) As usual, area of the field - 6 sq. units.
(ii) No.of trees = 4 \(\times\)26 = 104.
(iii) Plants provide us oxygen and play major role in rains, so plantation is necessary for all human beings.

Let 'P' be the principal at any time t.
By the question.\(\frac { dP }{ dt } =\frac { 5 }{ 100 } P\)            
\(\Rightarrow \)\(\frac { dP }{ P } =\frac { 1 }{ 20 } dt\) | Variables Separable
Integrating, \(\int { \frac { dP }{ P } =\frac { 1 }{ 20 } \int { 1.dt+c } } \)
\(\Rightarrow \) \({ log }_{ e }P=\frac { 1 }{ 20 } t+c\)....(1)
\(\left[\therefore P>0\right]\)
When \(t=0,P=1000,{ log }_{ e }1000=0+c\Rightarrow c={ log }_{ e }1000.\)
Putting in (1), \({ log }_{ e }P=\frac { 1 }{ 20 } t+{ log }_{ e }1000\) ..(2)
When \(P=2000,{ log }_{ e }2000=\frac { 1 }{ 20 } t+{ log }_{ e }1000\)
\(\Rightarrow \)\(\frac { 1 }{ 20 } t={ log }_{ e }\frac { 2000 }{ 1000 } ={ log }_{ e }\quad 2\Rightarrow t=20{ log }_{ e }2.\)=
Hence, Rs 1000 double in 20 \({ log }_{ e }\) 2 years.

\(we\ have\ :\ { sin }^{ 2 }y+cosxy=\pi \)
\(Diff\ w.r.t.x,\)
\(\Rightarrow 2sin\ y\ cos\ y\frac { dy }{ dx } -sin\ xy\left( x.\frac { dy }{ dx } +y.1 \right) =0\)
\(\Rightarrow (2\ siny\ cos\ y-x\ sinxy)\frac { dy }{ dx } =y\ sin\ xy\)
\(Hence,\ \frac { dy }{ dx } =\frac { y\ sin\ xy }{ sin\ 2y-x\ sin\ xy } \)

The given parabola is :
y² = 4ax
and the given line is y = mx
Putting the value of y from (2) in (1), we get:
\({ m }^{ 2 }{ x }^{ 2 }=4ax\Rightarrow x({ m }^{ 2 }x-4a)=0\Rightarrow 0,\frac { 4a }{ { m }^{ 2 } } \)
When x = 0, then from (2), y = 0.
When \(x=\frac { 4a }{ { m }^{ 2 } } \), then from (2), \(y=\frac { 4a }{ m } \).

Thus the line (2) meets the parabola at O (0,0) and 
\(P\left( \frac { 4a }{ { m }^{ 2 } } ,\frac { 4a }{ m } \right) \)
\(\therefore \ Reqd.area=ar(OCPL)-ar(ODPL)\)
\(=\overset { \frac { 4a }{ { m }^{ 2 } } }{ \underset { 0 }{ \int { } } } (\sqrt { 4ax } -mx)dx\)
\(=2\sqrt { a } \left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] _{ 0 }^{ \frac { 4a }{ { m }^{ 2 } } }-m\left[ \frac { { x }^{ 2 } }{ 2 } \right] ^{ \frac { 4a }{ { m }^{ 2 } } }_{ 0 }\)
\(=\frac { 4 }{ 3 } \sqrt { a } \left[ \left( \frac { 4a }{ { m }^{ 2 } } \right) ^{ \frac { 3 }{ 2 } }-0 \right] -\frac { m }{ 2 } \left[ \frac { { 16a }^{ 2 } }{ { m }^{ 4 } } -0 \right] \)
\(=\frac { 4 }{ 3 } \sqrt { a } \frac { { 8a }^{ \frac { 3 }{ 2 } } }{ { m }^{ 3 } } -\frac { { 8a }^{ 2 } }{ { m }^{ 3 } } =\frac { 32 }{ 3 } \frac { { a }^{ 2 } }{ { m }^{ 3 } } -\frac { { 8a }^{ 2 } }{ { m }^{ 3 } } \)
\(=\left( \frac { 32 }{ 3 } -8 \right) \frac { { a }^{ 2 } }{ { m }^{ 3 } } =\frac { { 8a }^{ 2 } }{ { 3m }^{ 3 } } sq.units\)

The given equations are:
2x - y = 5
x + y = 4
Here |A| = \(\left|\begin{matrix}2&-1\\1&1\end {matrix}\right|\)
= 2 + 1 = 3 \(\neq\)0
Hence the given system of equations is consistent.

Take x = cos θ, then proceeding as above, we get, \({ \sin }^{ -1 }(2x\sqrt { 1-{ x }^{ 2 } } )\) = 2 cos^–1 x

Let the events Be:
E₁ : coin is two headed
E₂ : coin is biased(heads 75%)
E₃ : Coin is biased(tails 40%)
and A : coin shows up head
\(P({ E }_{ 1 })=P({ E }_{ 2 })=P({ E }_{ 3 })=\frac { 1 }{ 3 } \)
\(P(A/{ E }_{ 1 })=1,P(A/{ E }_{ 2 })=\frac { 75 }{ 100 } =\frac { 3 }{ 4 } ,\)
By Bayes' theorem
\(P({ E }_{ 1 }​/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } \times 1 }{ \frac { 1 }{ 3 } \times 1+\frac { 1 }{ 3 } \times \frac { 3 }{ 4 } +\frac { 1 }{ 3 } \times \frac { 3 }{ 5 } } \)
\(=\frac{4}{4+3+2}=\frac{4}{9}\)

\(\Rightarrow \) y - log |y+1| = log|e^x + 1| + c

1, 20

\(\Rightarrow I=\int _{ 0 }^{ 1 }{ x{ e }^{ x } } dx\)
\(\Rightarrow I=\left[ \left[ x\int { { e }^{ x }dx } -\int { \left[ \frac { d }{ dx } x\int { { e }^{ x }dx } \right] dx } \right] \right] ^{ 1 }_{ 0 }\)
\(\Rightarrow I=\left[ x{ e }^{ x }-\int { 1.{ e }^{ x }.dx } \right] ^{ 1 }_{ 0 }\)
\(\Rightarrow I=\left[ xe^{ x }-e^{ x } \right] ^{ 1 }_{ 0 }\)
\(\Rightarrow I=\left[ (1e^{ 2 }-e^{ 1 })-({ 0e }^{ 0 }-{ e }^{ 0 } \right] \)
\(\Rightarrow I=(0)-(0-1)=1\)

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

Position vector of the point (1, 2, 3)
\(\therefore \overset { \rightarrow }{ a } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \rightarrow }{ k } \)
and parallel to vector \(3\overset { \wedge }{ i } +2\overset { \wedge }{ j } -3\overset { \wedge }{ k } \)
Required vector equation of line
\(\overset { \rightarrow }{ r } =\overset { \rightarrow }{ a\quad } +\lambda \overset { \rightarrow }{ b\quad } \)
\(\Rightarrow \overset { \rightarrow }{ r } =\overset { \rightarrow }{ i } +2\overset { \rightarrow }{ j } +3\overset { \rightarrow }{ k } +l(3\overset { \rightarrow }{ i } +2\overset { \rightarrow }{ j } -3\overset { \rightarrow }{ k } )\)
Cartesian equation
\(\frac { x-1 }{ 3 } =\frac { y-2 }{ 2 } =\frac { z-2 }{ -3 } \)

P(E) = \(\frac { 7 }{ 13 } \)
P(F) = \(\frac { 9 }{ 13 } \)
and P(E\(\cap\)F) = \(\frac { 4 }{ 13 } \)
(a) \(P(\bar { E } /F)=\frac { P(\bar { E } \cap F) }{ P(F) } \)
\(=\frac { P(F)-P(E\cap F)) }{ P(F) } \)
\(=1-\frac { P(E\cap F) }{ P(F) } =1-\frac { \frac { 4 }{ 13 } }{ \frac { 9 }{ 13 } } \)
\(\Rightarrow P(\bar { E } /F)=1-\frac { 4 }{ 9 } =\frac { 5 }{ 9 } \)
(b) \(P(\bar { E } /\bar { F } )=\frac { P(\bar { E } \cap \bar { F } ) }{ P(\bar { F } ) } \)
\(=\frac { P\overline { (E\cup F) } }{ P(\bar { F } ) } \)
\(=\frac { 1-P(E\cup F) }{ 1-P(F) } \)
\(P(E\cup F)=P(E)+P(F)-P(E\cap F)\)
\(=\frac { 7 }{ 13 } +\frac { 9 }{ 13 } -\frac { 4 }{ 13 } \)
\(=\frac { 12 }{ 13 } \)
\(P(E/F)=\frac { 1-\frac { 12 }{ 13 } }{ 1-\frac { 9 }{ 13 } } \)
\(=\frac { \frac { 1 }{ 13 } }{ \frac { 4 }{ 13 } } \)
\(=\frac { 1 }{ 4 } \)

\(\overset\rightarrow a=3\overset\wedge i+7\overset\wedge j+2\overset\wedge k\)
\(\left|\overset\rightarrow a \right| =\sqrt{9+49+4}=\sqrt{62}\)
and 
\(b=7\overset\wedge i-2\overset\wedge j+3\overset\wedge k\)
\(\left|\overset\rightarrow b \right| =\sqrt{9+49+4}=\sqrt{62}\)
\(\left|\overset\rightarrow a \right| =\left|\overset\rightarrow b \right| \)
but \(\overset\rightarrow a \neq\overset\rightarrow b\)because their corresponding components are different.

\(\frac { dy }{ dx } ={ e }^{ x-y }+x{ e }^{ -y }\)
\(={ e }^{ -y }\left[ { e }^{ x }+x \right] \)
\(\Rightarrow \frac { dy }{ dx } =\left( { e }^{ x }+x \right) { e }^{ -y }\)
\(\Rightarrow \int { { e }^{ y }dy } =\int { \left( { e }^{ x }+x \right) { dx } } \)
\(\Rightarrow { e }^{ y }={ e }^{ x }+\frac { { x }^{ 2 } }{ 2 } +C\)

\(L.H.S.={ sin }^{ -1 }\frac { 5 }{ 13 } +{ cos }^{ -1 }\frac { 3 }{ 5 } \)
\(\because { sin }^{ -1 }x={ tan }^{ -1 }\left( \frac { x }{ \sqrt { 1-{ x }^{ 2 } } } \right) \)
\({ sin }^{ -1 }\frac { 5 }{ 13 } ={ tan }^{ -1 }\left( \frac { 5/13 }{ \sqrt { 1-25/169 } } \right) \)
\(={ tan }^{ -1 }\frac { (5/13) }{ \sqrt { \frac { 144 }{ 169 } } } ={ tan }^{ -1 }\frac { 5 }{ 12 } \)
\(and\quad { cos }^{ -1 }x={ tan }^{ -1 }\left( \frac { \sqrt { 1-{ x }^{ 2 } } }{ x } \right) \)
\(\therefore \quad { cos }^{ -1 }\frac { 3 }{ 5 } ={ tan }^{ -1 }\left( \frac { \sqrt { 1-\frac { 9 }{ 25 } } }{ \frac { 3 }{ 5 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \sqrt { \frac { 16 }{ 25 } } }{ \frac { 3 }{ 5 } } \right) ={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)
\({ tan }^{ -1 }\frac { 5 }{ 12 } +{ tan }^{ -1 }\frac { 4 }{ 3 } ={ tan }^{ -1 }\left[ \frac { \frac { 5 }{ 12 } +\frac { 4 }{ 2 } }{ 1-\frac { 5\times 4 }{ 12\times 3 } } \right] \)
\(\left[ { tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\frac { x+y }{ 1-xy } \right] \)
\(={ tan }^{ -1 }\left| \frac { \frac { 15+48 }{ 36 } }{ \frac { 36-20 }{ 36 } } \right| \)
\(={ tan }^{ -1 }\left( \frac { 63 }{ 16 } \right) =R.H.S.\)

A relation R on a non-empty set A is called a transitive relation if (a, b), (b, c) \(\in R\) then (a, c) \(\in R\) , i.e., aRb, bRc implies aRc.
Thus a relation R on a non empty set A is said to be transitive if there exist a, b, c \(\in A\) such that (a, b)(b, c) \(\in R\)  implies (a, c) . \(\in R\)
Example
Let A = (1, 2, 3, 6)
R = (3, 6) (6, 1) (3, 1)
3 R 6 and 6 R 1 \(\Rightarrow \)(3, 1)  \(\in R\)
\(\therefore\) A is transitive.

We have, \(A=\left[ \begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix} \right] \) and \(B=\left[ \begin{matrix} -1 & 4 \\ 0 & 2 \end{matrix} \right] \)
\(AB=\left[ \begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix} \right] \left[ \begin{matrix} -1 & 4 \\ 0 & 2 \end{matrix} \right] \)
\(AB=\left[ \begin{matrix} -2+0 & 8+2 \\ -3+0 & 12+8 \end{matrix} \right] \)
\(\left[ \begin{matrix} -2 & 10 \\ -3 & 20 \end{matrix} \right] \)
and \(BA=\left[ \begin{matrix} -1 & 4 \\ 0 & 2 \end{matrix} \right] \left[ \begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 10 & 15 \\ 6 & 8 \end{matrix} \right] \)
\(\therefore\)   \(AB\neq BA\)

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

As on the y-axis, x-coordinate and z-coordinate are zeroes

Total injective mappings/functions
= ⁴ P₃ = 4! = 24.