\(\mathrm{LHS}=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right) \)
\(\text { Put } x^{2}=\cos 2 \theta, \text { then } \)
\(\mathrm{LHS}=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right)\) 
\( \tan ^{-1}\left(\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right)\) 
\(\tan ^{-1}\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right)\)
[divide numerator and denominator inside the bracket by \(cos \theta]\)

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

\(I=\int _{ 0 }^{ \pi }{ \frac { x }{ { a }^{ 2 }\cos ^{ 2 }{ x } +{ b }^{ 2 }\sin ^{ 2 }{ x } } } dx\) ..(i)
Apply property
\(\Rightarrow I=\int _{ 0 }^{ \pi }{ \frac { \left( \pi -x \right) }{ { a }^{ 2 }\cos ^{ 2 }{ x } +{ b }^{ 2 }\sin ^{ 2 }{ x } } } dx\)
By adding eqn. (i) and (ii),
\(2I=\pi \int _{ 0 }^{ \pi }{ \frac { 1 }{ { a }^{ 2 }\cos ^{ 2 }{ x } +{ b }^{ 2 }\sin ^{ 2 }{ x } } } dx\)
\(\Rightarrow I=\pi \int _{ 0 }^{ \pi /2 }{ \frac { \sec ^{ 2 }{ x } }{ { a }^{ 2 }+{ b }^{ 2 }\tan ^{ 2 }{ x } } } dx\)
Dividing Nr. & Dr. by cos²x
Put \(b\tan { x } =t\Rightarrow \sec ^{ 2 }{ x } dx=\frac { 1 }{ b } dt\)
Where x = 0, t = 0;
and \(x=\frac { \pi }{ 2 } ,\)
\(t=\infty ;\)
\(\Rightarrow I=\frac { \pi }{ b } \int _{ 0 }^{ \infty }{ \frac { dt }{ { a }^{ 2 }+{ t }^{ 2 } } } \)
\(\Rightarrow =\frac { \pi }{ b } \left[ \frac { 1 }{ a } \tan ^{ -1 }{ \frac { t }{ a } } \right] _{ 0 }^{ \infty }\)
\(\Rightarrow I=\frac { \pi }{ ab } \left( \frac { \pi }{ 2 } -0 \right) \)
\(=\frac { { \pi }^{ 2 } }{ 2ab } \)

\({ A }^{ -1 }=\left[ \begin{matrix} 1/a & 0 & 0 \\ 0 & 1/b & 0 \\ 0 & 0 & 1/c \end{matrix} \right] \)
The inverse is impossible. It is possible when a, b, c are all real and\(\ne\)0, otherwise |A| becomes singular
i.e., |A| = 0
and \({ A }^{ -1 }=\frac { adj\quad A }{ |A| } \)
will become meaningless.
Value: Anybody can progress till he continues working, otherwise his progress will be halted.

Let the events be:
E₁ = Choosing 1^st coin
E₂ = Choosing 2^nd coin
E₃ = Choosing 3^rd coin
A: Getting Heads
\(P({ E }_{ 1 })=P({ E }_{ 2 })=P(E_{ 3 })=\frac { 1 }{ 3 } \)
\(P(A/{ E }_{ 1 })=\frac { 40 }{ 100 } ,P(A/{ E }_{ 2 })=\frac { 75 }{ 100 } ,\)
\(P(A/{ E }_{ 3 })=\frac { 1 }{ 2 } \)
\(P({ E }_{ 1 }/A)\)
\(=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } .\frac { 40 }{ 100 } }{ \frac { 1 }{ 3 } .\frac { 40 }{ 100 } +\frac { 1 }{ 3 } .\frac { 75 }{ 100 } +\frac { 1 }{ 3 } .\frac { 1 }{ 2 } } =\frac { 8 }{ 33 } \)

let radius, height and slant height of cone be r, h and l respectively.
\(\therefore r^2+h^2=l^2\)

V(volume) \(=\frac{\pi}{3}r^2h\)
\(\Rightarrow V^2=\frac{\pi^2r^4h^2}{9}\)
\(\Rightarrow h^2=\frac{9V^2}{\pi^2r^4}\)
\(A=\pi r l,z=A^2=\pi^2r^2l^2\)
\(=\pi^2r^2(r^2+h^2)\)
\(let \ \ A^2=z\)
\(\therefore\ z=\pi^2r^2[r^2+\frac{9V^2}{\pi^2r^4}]\)
\( z=\pi^2[r^4+\frac{9V^2}{\pi^2r^2}]\)
\( \frac{dz}{dr}=\pi^2[4r^3-\frac{18V^2}{\pi^2r^3}]\)
\(\therefore \ \ \ \frac{dz}{dr}=0\)
\(\Rightarrow\ r=6\sqrt{\frac{9V^2}{2\pi^2}}\)
at \( r=6\sqrt{\frac{9V^2}{2\pi^2}}\)
\(\frac{d^2z}{dr^2}=\pi^2(12r^2+\frac{54V^2}{\pi^2r^4 })>0\)
\(\therefore\) curved surface area is minimum iff
\(2\pi^2r^6=9V^2\)
i.e., \(2\pi^2r^6=\pi^2r^4h^2\)

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

\(\mathrm{LHS}=(a I+b A)^{3}=\left\{\left[\begin{array}{ll} a & 0 \\ 0 & a \end{array}\right]+\left[\begin{array}{ll} 0 & b \\ 0 & 0 \end{array}\right]\right\}^{3} \)
\(=\left[\left.\begin{array}{ll} a & b \\ 0 & a \end{array}\right|^{3}=\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\right. \)
\(=\left[\begin{array}{cc} a^{2} & 2 a b \\ 0 & a^{2} \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]=\left[\begin{array}{cc} a^{3}+0 & a^{2} b+2 a^{2} b \\ 0+0 & 0+a^{3} \end{array}\right]=\left[\begin{array}{cc} a^{3} & 3 a^{2} b \\ 0 & a^{3} \end{array}\right] \)
\(\mathrm{RHS}=a^{3} I+3 a^{2} b A=\left[\begin{array}{cc} a^{3} & 0 \\ 0 & a^{3} \end{array}\right]+\left[\begin{array}{cc} 0 & 3 a^{2} b \\ 0 & 0 \end{array}\right]=\left[\begin{array}{cc} a^{3} & 3 a^{2} b \\ 0 & a^{3} \end{array}\right] \)
\(\mathrm{LHS}=\mathrm{RHS} \)
\(\text {Hence, }(a I+b l)^{3}=a^{3} I+3 a^{2} b A \)

After dividing numerator and denominator by a^3 we have,
\( \tan ^{-1}\left(\frac{3\left(\frac{x}{a}\right)-\left(\frac{x}{a}\right)^3}{1-3\left(\frac{x}{a}\right)^2}\right) \)
\(Put \mathrm{x} / \mathrm{a}=\tan \theta\ and\ \theta=\tan ^{-1}(\mathrm{x} / \mathrm{a}) \)
\( =\tan ^{-1}\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right) \)
\(=\tan ^{-1}(\tan 3 \theta) \)
\(=3 \theta \\ =3 \tan ^{-1}(\mathrm{x} / \mathrm{a})\)

Let P_T : Event that P speaks truth
and Q_T: Event that Q speaks truth.
Given, \(P\left(P_T\right)=\frac{70}{100} \), then \(P\left(\bar{P}_T\right)=1-\frac{70}{100}=\frac{30}{100}\)
and \(P\left(Q_T\right)=\frac{80}{100} \text {, }\)then \(P\left(\bar{Q}_T\right)=1-\frac{80}{100}=\frac{20}{100}\)
P (A and B are agree to each other)
\(\begin{aligned} =P\left(P_T \cap Q_T\right)+P\left(\bar{P}_T \cap \bar{Q}_T\right) \end{aligned}\)
\(\begin{aligned} =P\left(P_T\right) \cdot P\left(Q_T\right)+P\left(\bar{P}_T\right) \cdot P\left(\bar{Q}_T\right) \end{aligned}\)
[\(\because\) events P_T and Q_T are independent events]
\(\begin{aligned} =\frac{70}{100} \times \frac{80}{100}+\frac{30}{100} \times \frac{20}{100} \end{aligned}\)
\(\begin{aligned} =\frac{5600}{10000}+\frac{600}{10000} \end{aligned}\)
\(\begin{aligned} =\frac{6200}{10000}=\frac{62}{100} \end{aligned}\)
\(\therefore\) Percentage of P(A and B are agree to each other)
\(=\frac{62}{100} \times 100=62 \%\)
No, agree does not mean that they are speaking truth.

Let godown A supply x and y quintals of grain to the shops D and E respectively. Then, (100 − x − y) will be supplied to shop F.
The requirement at shop D is 60 quintals since x quintals are transported from godown A. Therefore, the remaining (60 −x) quintals will be transported from godown B.
Similarly, (50 − y) quintals and 40 − (100 − x − y) = (x + y − 60) quintals will be transported from godown B to shop E and F respectively.
The given problem can be represented diagrammatically as follows.
\(60-x\ge 0\Leftrightarrow x\le 60\) ..(1)
\(50-y\ge 0\Leftrightarrow y\le 50\) .(2)
\(100-(x+y)\ge 0\Leftrightarrow x+y\le 100\) ..(3)
\(x+y-60\ge 0\Leftrightarrow x+y\ge 60\) ..(4)
and \(x,y\ge 0\)  ...(5)
The given problem can be formulated as
Minimize z = 2.5x + 1.5y + 410 … (1)
subject to the constraints,
The feasible region determined by the system of constraints is as follows.

The corner points are A (60, 0), B (60, 40), C (50, 50), and D (10, 50).
The values of z at these corner points are as follows.

The minimum value of z is 510 at (10, 50).
Thus, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.
The minimum cost is Rs. 510.

The given equation is \({ x }^{ 2 }dy+\left( 2xy+{ y }^{ 2 } \right) dx=0\)
\(\Rightarrow\) \(\frac { dy }{ dx } =-\frac { 2xy+{ y }^{ 2 } }{ { x }^{ 2 } } \)
| Homogeneous 
Put \(y=vx\) so that  \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore \) (1) becomes :\(v+x\frac { dv }{ dx } =-\frac { 2v{ x }^{ 2 }+{ v }^{ 2 }{ x }^{ 2 } }{ { x }^{ 2 } } \)
\(\Rightarrow\)  \(v+x\frac { dv }{ dx } =-2v-{ v }^{ 2 }\)
\(\Rightarrow\)  \(x\frac { dv }{ dx } =-3v-{ v }^{ 2 }\)
\(\Rightarrow\) \(\frac { dv }{ 3v-{ v }^{ 2 } } =-\frac { dx }{ x } \)
| Variables Separable 
Integrating,  \(\int { \frac { dv }{ v\left( v+3 \right) } } =-\int { \frac { dx }{ x } } +\) constant
\(\Rightarrow\) \(\frac { 1 }{ 3 } \int { \left( \frac { 1 }{ v } -\frac { 1 }{ v+3 } \right) } dv=-\int { \frac { dx }{ x } + } \) constant
[Partial Fractions]
\(\Rightarrow\) \(\frac { 1 }{ 3 } \quad log|v|-\frac { 1 }{ 3 } log|v+3|=-log\quad |x|+log\quad c'|\)
\(\Rightarrow\) \(\frac { 1 }{ 3 } log|\frac { 1 }{ v+3 } |=log\quad |\frac { c' }{ x } |\Rightarrow \frac { v }{ v+3 } ={ \left( \frac { c' }{ x } \right) }^{ 3 }\)
\(\Rightarrow\) \(\frac { \frac { y }{ x } }{ \frac { y }{ x+3 } } =\frac { { c' }^{ 3 } }{ { x }^{ 3 } } \Rightarrow \frac { y }{ y+3x } =\frac { { c' }^{ 3 } }{ { x }^{ 3 } } \)
\(\Rightarrow\)  \( { x }^{ 3 }y=c\left( 3x+y \right) \)..(2)
When \(x=1,y=1\) then (1)(1) \(=c\left( 3+1 \right) \Rightarrow c=\frac { 1 }{ 4. } \)
Putting in (2),
\({ x }^{ 3 }y=\frac { 1 }{ 4. } \left( 3x+y \right) ,\) 
​​​​​​​which is the required solution

We have: \(y={ X }^{ x }\)
Taking logs \(logy\quad y=x\quad log\quad x\)
\(loy\quad y=x\quad log\quad x\)
\(\frac { 1 }{ y } \frac { dy }{ dx } =x.\frac { 1 }{ x } +log\quad x.1\)
\(\frac { dy }{ dx } =y(1+log\quad x)...(1)\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =y.\frac { d }{ dx } (1+log\quad x)+\frac { dy }{ dx } (1+log\quad x)\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =y.\left( 0+\frac { 1 }{ x } \right) +\frac { dy }{ dx } .\frac { \frac { dy }{ dx } }{ y } \)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { y }{ x } +\frac { 1 }{ y } { \left( \frac { dy }{ dx } \right) }^{ 2 }\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { 1 }{ y } { \left( \frac { dy }{ dx } \right) }^{ 2 }-\frac { y }{ x } =0\)

(i) When \(a,b\in Z^{ + }\) , then a-b may not belong to \(Z^{ + }\).
\(\left[ For\quad Ex.\quad 5-7=-2\notin Z^{ + } \right] \)
Hence "*' is not a binary operation on \(Z^{ + }\).
(ii) When \(a,b\in Z^{ + }\) , then \(a,b\in Z^{ + }\)
Hence, "*' is a binary operation on \(Z^{ + }\).
(iii) When \(a,b\in R\) , then \(ab^{ 2 }\) \(\in R\) .
Hence, '*' is a binary operation on R.
(iv) When \(a,b\in Z^{ + }\), \(\left| a-b \right| \in Z^{ + }\).
 \(\left[ \because \quad \left| a-b \right| \ge 0 \right] \)
Hence, '*' is a binary operation on \(Z^{ + }\).
(v) When \(a,b\in Z^{ + }\) , then \(a,\in Z^{ + }\).
Hence, '*'  is a binary operation on \(Z^{ + }\) .

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

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

We have: \(A\prime =\begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}.\quad and\ B=\ \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}\quad \)
Then  \(A=\begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}.\)
Now  \(A+2B=\begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}+2\begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}\)
\(=\begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}+\begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}\)
\( =\begin{bmatrix} -2-2 & 1+0 \\ 3+2 & 2+4 \end{bmatrix}=\begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}\)
Hence \((A+2B)\prime =\begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}.\)

We get \(\begin{vmatrix}1&a&bc\\0&b-a&c(a-b)\\0&c-a&b(a-c) \end{vmatrix}\)
Taking factors (b – a) and (c – a) common from R₂ and R₃, respectively, we get
\((b-a)(c-a)\begin{vmatrix} 1&a&bc\\0&1&-c\\0&1&-b\end{vmatrix}\)
= (b – a) (c – a) [(– b + c)] (Expanding along first column)
= (a – b) (b – c) (c – a)

Let \({ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) =y\) where \(y\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
\(\Rightarrow siny=-\frac { \pi }{ 2 } \)
\(\Rightarrow siny=-sin\frac { \pi }{ 6 } =sin\left( -\frac { \pi }{ 6 } \right) \)
\(\Rightarrow y=-\frac { \pi }{ 6 } \)
Hence, the required principal value = \(-\frac { \pi }{ 6 } \)

\(\Rightarrow\left(1+x^{2}\right) y=\underline{x}, \sqrt{x^{2}+4}+2 \log \left|x+\sqrt{x^{2}+4}\right|\) is the required solution.

\({x\over2}cos^{-1}\ x-{1\over2}\sqrt{1-x^2}+c\)

\(I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt [ 3 ]{ sinx } }{ \sqrt [ 3 ]{ cosx } +\sqrt [ 3 ]{ sinx } } } dx\)
Apply the property
\(\int _{ 0 }^{ b }{ f(x)\quad dx } =\int { f(a+b-x)dx } \)
\(I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt [ 3 ]{ sin\left( \frac { \pi }{ 2 } -x \right) dx } }{ \sqrt [ 3 ]{ cos\left( \frac { \pi }{ 2 } -x \right) } +\sqrt [ 3 ]{ sin\left( \frac { \pi }{ 2 } -x \right) } } } \)
as   \(\frac { \pi }{ 3 } +\frac { \pi }{ 6 } =\frac { 2\pi +\pi }{ 6 } =\frac { 3\pi }{ 6 } =\frac { \pi }{ 2 } \)
\(I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt [ 3 ]{ cosx } }{ \sqrt [ 3 ]{ cosx } +\sqrt [ 2 ]{ sinx } } } dx\)
by adding eqn. (i) and (ii),
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ dx } =\left[ x \right] ^{ \pi /3 }_{ \pi /6 }\)
\(=\left[ \frac { \pi }{ 3 } -\frac { \pi }{ 6 } \right] =\frac { \pi }{ 6 } \)
\(\Rightarrow 2I=\frac { \pi }{ 6 }\)
\( \Rightarrow I=\frac { \pi }{ 12 } \)

\(\cos \theta=\frac{\overrightarrow a.\overrightarrow b}{\left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right| }=\frac{12}{10\times 2}=\frac{3}{5}\)
\(\cos \theta =\frac{3}{5}\)
\(\sin \theta=\sqrt{1-\cos ^{2}\theta} =\sqrt{1-\frac{9}{25}}\)
\(\sin \theta=\frac{\left| \overrightarrow { a } \times \overrightarrow { b } \right| }{\left| \overrightarrow {a } \right| \left| \overrightarrow {b } \right| }\)
\(\Rightarrow \left| \overrightarrow { a } \times \overrightarrow { b } \right| =\left| \overrightarrow { a } \right| \left| \overrightarrow { b} \right| \sin\theta\)
\(\Rightarrow \left| \overrightarrow { a } \times \overrightarrow { b } \right| =10\times2\times\frac{4}{5}=16\)

We have, y = a^x +x^a+x^x+a^a
Let v = x^x
log v = x log x
\(\frac { 1 }{ v } \frac { dv }{ dx } =x+\frac { 1 }{ x } +logx\)
\(\frac { dv }{ dx } =v\left| 1+logx \right| \)
= x^x(1+lodx)
\(\frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+\frac { dv }{ dx } +0\)
\(\Rightarrow \frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+{ x }^{ x }(1+logx)\)

\(\frac { dy }{ dx } +y={ e }^{ -x }\)
Let \(\frac { dy }{ dx } +Py=Q\)
Where P = 1, Q = e^-x
\(y\left( I.F. \right) =\int { I.F.\times Qdx } \)
\(\Rightarrow I.F.={ e }^{ \int { Pdx } }={ e }^{ \int { Pdx } }={ e }^{ x }\)
\(\Rightarrow y{ e }^{ x }=\int { { e }^{ x }.{ e }^{ -x } } dx=\int { dx } \)
\(\Rightarrow y{ e }^{ x }=x+C\)

\(\alpha =60^{ \circ },\beta =?,\gamma =45^{ \circ }\)
Let \(\alpha \) makes with x-axis, y makes with z-axis and \(\beta \) makes with y-axis
\(\alpha =cos\quad 60^{ \circ },m=cos\quad \beta ,y=cos45^{ \circ }\)
\(\alpha ^{ 2 }+\beta ^{ 2 }+\gamma ^{ 2 }=1\)
\(\Rightarrow cos^{ 2 }60^{ \circ }+cos^{ 2 }\beta +cos^{ 2 }45^{ \circ }\)
\(\Rightarrow \frac { 1 }{ 4 } +cos^{ 2 }\beta +\frac { 1 }{ 2 } =1\)
\(\Rightarrow cos^{ 2 }\beta =1-\frac { 1 }{ 4 } -\frac { 1 }{ 2 } \)
\(=\frac { 4-1-2 }{ 4 } \)
\(=\frac { 1 }{ 4 } \)
\(\Rightarrow cos \beta =\pm \frac { 1 }{ 2 } \)
\(\beta =\frac { \pi }{ 3 } or\ \frac { 2\pi }{ 3 } \)
But angle between two lines in the interval \(\left( 0,\frac { \pi }{ 2 } \right) \)
Hence required angles is \(\frac { \pi }{ 3 } \)

Total number of cards = 2 + 6 + 8 = 16
(a) Number of red balls = 2
\(\therefore\)  Required probability = \(\frac { 2 }{ 16 } =\frac { 1 }{ 8 } \)
(b) Number of black balls = 6
\(\therefore\)  Required probability = \(\frac { 6 }{ 16 } =\frac { 3 }{ 8 } \)
(c) number of green balls = 8
\(\therefore\)  Required probability = \(\frac { 8 }{ 16 } =\frac { 1 }{ 2 } \)
(d) Number of non-red balls = 14
 \(\therefore\) Required probability = \(\frac { 14 }{ 16 } \)

\(Let\quad { sec }^{ -1 }x=\theta ,\quad then\quad x=sec\theta \quad and\quad for\quad x<-1,\)
\(\frac { \pi }{ 2 } <\theta <\pi \)
Given expression = cot^-1(-cot \(\theta \))
\(={ cot }^{ -1 }\left[ cot\left( \pi -\theta \right) \right] =\pi -{ sec }^{ -1 }x\quad as\quad 0<\pi -\theta <\frac { \pi }{ 2 } \)

We shall prove the result by mathematical induction on n.
Step 1 : When n = 1, by the definition or integral powers of a matrix, we have
\({ A }^{ 1 }=\left[ \begin{matrix} 1+2\left( 1 \right) & -4n \\ n & 1-2\left( 1 \right) \end{matrix} \right] =\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \end{matrix} \right] \)
So, the result is true for n = 1.
Step 2 : Let the result be true for n = m. Then,
\({ A }^{ m }=\left[ \begin{matrix} 1+2m & -4m \\ m & 1-2m \end{matrix} \right] \)
Now, we will show that the result is true for n = m + 1, i.e.,
\({ A }^{ m+1 }=\left[ \begin{matrix} 1+2\left( m+1 \right) & -4\left( m+1 \right) \\ \left( m+1 \right) & 1-2\left( m+1 \right) \end{matrix} \right] \)
By the definition of integral powers of a square matrix, we have
\({ A }^{ m+1 }={ A }^{ m }.A\)
\(\Rightarrow { A }^{ m+1 }=\left[ \begin{matrix} 1+2m & -4m \\ m & 1-2m \end{matrix} \right] \left[ \begin{matrix} 3 & -4 \\ 1 & -1 \end{matrix} \right] \)
[by supposition (i)]
\(\Rightarrow { A }^{ m+1 }=\left[ \begin{matrix} 3+6m-4m & -4-8m+4m \\ 3m+1-2m & -4m-4+2m \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1+2\left( m+1 \right) & -4\left( m+1 \right) \\ \left( m+1 \right) & 1-2\left( m+1 \right) \end{matrix} \right] \)
This shows that the result is true for n = m + 1, whenever it is true for n = m.
Hence, by the principle of mathematical induction, the result is true for any positive integer n.

\(k=-\frac { 3 }{ 2 }\)
Alternative Method :
Since diagonal elements in a skew symmetric matrix are zero.
We can compare diagonal elements to zero.
i.e., 2k + 3 = 0
⇒  2k = - 3
∴ \(k=-\frac { 3 }{ 2 }\)

Symmetric Relation : A relation R on a set A is called symmetric relation if aRb implies bRa, for every a,b \(\in a\) i.,e if (a,b)  \(\in R\) \(\Rightarrow \) (b,a)  \(\in R\)For every a,b \(\in A\)
Example 
A = (1,2,3)
A x A =(1,2) (2,1) (1,1) (2,2) (3,3) (1,3) (2,3)  (3,1) (3,2 )  \(\in R\) 
since (a,b)  \(\in R\)  (b,a)  \(\in R\) for every a, b  \(\in A\) 
Relation is said to be symmetric 

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