We know that the equation of the plane having intercepts a, band c on the three
co-ordinate axes is
\({{x}\over{a}}+{{y}\over{b}}+{{y}\over{c}}=1\)
Here, the co-ordinates of A, Band C are (a, 0, 0),  (0, b, 0) and (0, 0, c) respectively.
The centroid of \(\triangle \)ABC is \(\left({{a}\over{3}},{{b}\over{3}},{{c}\over{3}}\right).\)
Equating \(\left( {{a}\over{3}},{{b}\over{3}},{{c}\over{3}} \right)\) to \((\alpha,\beta,\gamma),\) we get a = \(3\alpha,\) b = \(3\beta\) and c = \(3\gamma\)
Thus, the equation of the plane is
\({{x}\over{3\alpha}}+{{y}\over{3\beta}}+{{z}\over{3\gamma}}=1\)
or \({{x}\over{\alpha}}+{{y}\over{\beta}}+{{z}\over{\gamma}}=3\)

We have, x = \(\frac { at }{ 1+{ t }^{ 2 } } ,y=\frac { { at }^{ 2 } }{ 1+{ t }^{ 2 } } \)
\(\frac { dx }{ dt } =\frac { { (1+t) }^{ 2 }a-at(2t) }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\Rightarrow\)  \(\frac { dx }{ dt } =\)\(\frac { a+a{ t }^{ 2 }-2a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
=\(\frac { a-a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
and \(\frac { dy }{ dt } =\frac { { (1+{ t }^{ 2 } })^{ 2 }2at-a{ t }^{ 2 }(2t) }{ { (1+{ t }^{ 2 } })^{ 2 } } \)
\(\Rightarrow\)\(\frac { dy }{ dt } =\frac { 2at+2a{ t }^{ 3 }-2a{ t }^{ 3 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } =\frac { 2at }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\therefore\) \(\frac { dy }{ dx } =\frac { 2at }{ { (1+{ t }^{ 2 } })^{ 2 } } \times \frac { { (1+{ t }^{ 2 } })^{ 2 } }{ a-a{ t }^{ 2 } } \)
\(=\frac { 2at }{ a-a{ t }^{ 2 } } =\frac { 2t }{ 1-{ t }^{ 2 } } \)
\({ (\frac { dy }{ dx } ) }_{ att=2 }\) = \(\frac { 2(2) }{ 1-4 } \)
= -4/3

(a) K then Q then J(fix case)
\(\frac { 4 }{ 52 } \times \frac { 4 }{ 51 } \times \frac { 4 }{ 50 } =\frac { 1 }{ 13 } \times \frac { 4 }{ 51 } \times \frac { 2 }{ 25 } \)
\(=\frac { 8 }{ 16575 } \)
(b) KQJ or QKJ or JQK or KJQ or OJK or JKQ = 3!
\(6\left( \frac { 4 }{ 52 } \times \frac { 4 }{ 51 } \times \frac { 4 }{ 50 } \right) =6\times \frac { 8 }{ 16575 } =\frac { 48 }{ 16575 } \)

\(\overset\rightarrow {BA}\)= Position vector of A- Position vector of B
\(\overset\rightarrow{BA}=(\overset\wedge i+2\overset\wedge j-3\overset\wedge k)-(-\overset\wedge i-2\overset\wedge j+\overset\wedge k)\)
\(\overset\rightarrow {BA}=2\overset\wedge i+4\overset\wedge j-\overset\wedge k \)
\(\left| \overset\rightarrow {BA}\right| =\sqrt{2^{2}+4^{2}+(-4)^{2}}\)
\(=\sqrt{4+16+16}\)
\(=\sqrt{36}=6 \)
Direction cosines of the vector \(\overset\rightarrow{BA}\) are \((\frac{2}{6},\frac{4}{6},-\frac{4}{6})\)
\(=(\frac{1}{3},\frac{2}{3},-\frac{2}{3})\)

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

\(x+\alpha =t\)
\(x=t-\alpha \)
\(dx=dt\)
\(\int { \frac { cos(t-\alpha )dt }{ cost } } =\int { \frac { costcos\alpha +sintsin\alpha }{ cost } } \)
\(\int { \frac { costcos\alpha dy }{ cost } } +\int { \frac { sintsin\alpha }{ cost } } dt\)
\(=cos\alpha \int { dt+sin\alpha \int { tantdt } } \)
\(=tcos\alpha +sin\alpha \quad log\left| sect \right| +c\)
\(=(x+\alpha )cos\alpha +sin\alpha \quad log\left| sec(x+\alpha \right| +c\)

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.

We have \(x\in [-1,\quad 1]\)
Let  x = sin \(\theta\)             \(\therefore \quad \theta \in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
\(\Rightarrow \quad \quad \theta ={ sin }^{ -1 }x\)
\(\Rightarrow \quad \quad -\frac { \pi }{ 2 } \le \theta \le \frac { \pi }{ 2 } \)
\(\Rightarrow \quad \quad \frac { \pi }{ 2 } \ge -\theta \ge -\frac { \pi }{ 2 } \)
\(\Rightarrow \quad \frac { \pi }{ 2 } +\frac { \pi }{ 2 } \ge \frac { \pi }{ 2 } -\theta \ge -\frac { \pi }{ 2 } +\frac { \pi }{ 2 } \)
\(\Rightarrow \quad\pi \ge \frac { \pi }{ 2 } -\theta \ge 0\)
\(\Rightarrow \quad cos\left( \frac { \pi }{ 2 } -\theta \right) =sin\quad \theta =x\)
\(\Rightarrow \quad \frac { \pi }{ 2 } -\theta ={ cos }^{ -1 }x\)
\(\Rightarrow { \quad cos }^{ -1 }x=\frac { \pi }{ 2 } -{ sin }^{ -1 }x\)
\(\Rightarrow { \quad sin }^{ -1 }x+{ cos }^{ -1 }x=\frac { \pi }{ 2 } if\quad x\in [-1,\quad 1]\)

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, \(A=\left| \begin{matrix} 6 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right| \)
|A| = 6(4-0)-0(0-0)+1(0-0)
|2A| = \(\left| \begin{matrix} 12 & 0 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 8 \end{matrix} \right| \)
= 12(16-0)-0(0-0)+2(0-0)
= 192
|2A| = 8 x 24 = 8|A|

\([ac+bd]+[{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }]=[ac+bd+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }] \)

Let 'x' units of food A and 'y' units of food B be used.
Then LLP problem is:
Minimize:  Z = 5x + 4y
Subject to: \(200x+100y\ge 4000\) i.e. \(2x+y\ge 40\) ,
\(x+2y\ge 50,\)
\(40x+40y\ge 4000\) i.e. \(x+y\ge 35\)
and  \(x\ge 0,y\ge 0.\)

The feasible region (shaded) is unbounded.
Let us evaluate Z at the corner points:
C(50, 0), G(20, 15),
[Point of intersection of x + 2y = 50, x + y = 35]H(5, 30)
[Point of intersection of 2x + y = 40, x + y = 35] and B(0, 40).
Applying Corner Point Method, we have:

Hence, least is Rs. 145 when 5 units of food A and 30 units of food B are used.

The given equation is
\(\left( 1+{ x }^{ 2 } \right) \frac { dy }{ dx } ={ e }^{ m\_ tan-1_{ x } }-y,\)
\(\Rightarrow \)  \(\frac { dy }{ dx } +\frac { y }{ 1+{ x }^{ 2 } } =\frac { { e }^{ m\_ tan-1_{ x } } }{ 1+{ x }^{ 2 } } \)..(1)
| Linear Equation           
Comparing with \(\frac { dy }{ dx } +Py=Q,\) we have:
\('P'=\frac { 1 }{ 1+{ x }^{ 2 } } \) and \('Q'=\frac { { e }^{ m\_ tan-1_{ x } } }{ 1+{ x }^{ 2 } } .\)    
\(\therefore\)  \(I.F.={ e }^{ \int { P\quad dx } }={ e }^{ \int { \frac { 1 }{ 1+{ x }^{ 2 } } dx } }={ e }^{ m\_ tan-1_{ x } }.\)
Multiplying (1) by \({ e }^{ tan-1_{ x } }\) we get:
\({ e }^{ tan-1_{ x } }.\frac { dx }{ dy } +\frac { y }{ 1+{ x }^{ 2 } } { e }^{ tan-1_{ x } }={ e }^{ tan-1_{ x } }.\frac { { e }^{ m\_ tan-1_{ x } } }{ 1+{ x }^{ 2 } } \)
\(\Rightarrow \)  \(\frac { d }{ dx } \left( y.{ e }^{ tan-1_{ x } } \right) =\frac { { e }^{ \left( m+1 \right) tan-1_{ x } } }{ 1+{ x }^{ 2 } } .\)                    
Integrating,   \(y.{ e }^{ \left( m+1 \right) tan-1_{ x } }=\int { \frac { { e }^{ \left( m+1 \right) tan-1_{ x } } }{ 1+{ x }^{ 2 } } } dx+c\)....(2)
Now \(I=\int { \frac { { e }^{ \left( m+1 \right) tan-1_{ x } } }{ 1+{ x }^{ 2 } } } dx.\)                              
Put  \({ tan }^{ -1 }x=t\)  so that \(\frac { 1 }{ 1+{ x }^{ 2 } } dx=dt\)
\(\therefore\)   \(I=\int { { e }^{ \left( m+1 \right) t }dt= } \frac { { e }^{ \left( m+1 \right) t } }{ m+1 } =\frac { { e }^{ \left( m+1 \right) tan-1_{ x } } }{ m+1 } \)
\(\therefore\)     From(2),\(y.{ e }^{ { tan-1 }_{ x } }=\frac { { e }^{ \left( m+1 \right) tan-1_{ x } } }{ m+1 } +c\)
\(\Rightarrow \)    \(y=\frac { { e }^{ \left( m+1 \right) tan-1_{ x } } }{ m+1 } +c{ e }^{ { tan-1 }_{ x } }\)...(3)
When  \(y=1,x=0,\therefore 1=\frac { 1 }{ m+1 } +c\)
\(\Rightarrow \)  \(c=1-\frac { 1 }{ m+1 } =\frac { m }{ m+1 } .\)
Putting in(3),  \(y=\frac { { e }^{ { tan-1 }_{ x } } }{ m+1 } +\frac { m }{ m+1 } { e }^{ { tan-1 }_{ x } },\)
which is the required solution

We have: \({ x }^{ y }={ e }^{ x-y }\)
Taking logs., \(ylogx=(x-y)\)
\(y(1+log\quad x)=x\)
\(y=\frac { x }{ 1+log\quad x } \)
\(\frac { dy }{ dx } =\frac { \left( 1+log\quad x \right) .1-x\left( 0+\frac { 1 }{ x } \right) }{ { \left( 1+logx \right) }^{ 2 } } \)
\(=\frac { 1+log\quad x-1 }{ { \left( 1+logx \right) }^{ 2 } } =\frac { log\quad x }{ { \left( 1+logx \right) }^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { log\quad x }{ { \left( log\quad e+log\quad x \right) }^{ 2 } } \)

y² = 4x is right - handed parabola.

Therefore, Required area, ABCD = \(\overset { 4 }{ \underset { 1 }{ \int { } } } ydx\) [Taking vertical strips]
\(\overset { 4 }{ \underset { 1 }{ \int { } } } 2\sqrt { x } dx\\ \)
[\({ y }^{ 2 }=4x\Rightarrow y=\pm 2\sqrt { x } .\) But region ABCD lies in 1^st quadrant, Therefore y is +ve]
\(=2\overset { 4 }{ \underset { 1 }{ \int { } } } { x }^{ 1/2 }dx=2\left[ \frac { { x }^{ 3/2 } }{ 3/2 } \right] _{ 1 }^{ 4 }=\frac { 4 }{ 3 } [{ 4 }^{ 3/2 }-1]\)
\(=\frac { 4 }{ 3 } [8-1]=\frac { 28 }{ 3 } =9\frac { 1 }{ 3 } sq.units.\)

We know that, \(P={ I }_{ 2 }P\)
\(\Rightarrow \ \begin{bmatrix} 10 & -2 \\ -5 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}P\)
\(\Rightarrow \begin{bmatrix} 1 & { -1 }/{ 5 } \\ -5 & 1 \end{bmatrix}=\begin{bmatrix} { 1 }/{ 10 } & 0 \\ 0 & 1 \end{bmatrix}P.\ [\text{Applying} { R }_{ 1 }\rightarrow \left( \frac { 1 }{ 10 } \right) { R }_{ 1 }]\)
\(\begin{bmatrix} 1 & { -1 }/{ 5 } \\ -5 & 1 \end{bmatrix}=\begin{bmatrix} { 1 }/{ 10 } & 0 \\ { 1 }/{ 2 } & 1 \end{bmatrix}P. [\text {Applying} { R }_{ 2 }\rightarrow { R }_{ 2 }+5{ R }_{ 1 }]\)
Since second row of LHS matrix has all zeros, 
\(\therefore\ { P }\) does not exist.

Let \(A=\left[\begin{matrix}-1&5\\ -3&2\end{matrix}\right]\)
\(|A|=\left[\begin{matrix}-1&5\\ -3&2\end{matrix}\right]=-2+15=13\neq0\)
A is non-singular  = A^-1 exists.
Now \(A_{11}=(-1)^{1+1}M_{11}(+1)(1 )=2\)
 \(A_{12}=(-1)^{1+2}M_{12}\\ =(-1)(-3)=3;\)
\(A_{21}=(-1)^{2+1}M_{21}=(-1)(5)=-5\)
\(A_{22}=(-1)^{2+2}M_{22}\\ =(+1)(-1)=-1\)
\(\therefore\ adj\ A=\begin{bmatrix}2&3\\-5&-1 \end{bmatrix}=\begin{bmatrix}2&-5\\3&-1 \end{bmatrix}\)
\(\therefore\ A^{-1}={1\over|A|}(adj\ A)\)
\(={1\over13}\begin{bmatrix} 2&-5\\3&-1\end{bmatrix}\)
=\(\left[\begin{matrix}{2\over13}&{-5\over13}\\ {3\over13}&-{1\over13}\end{matrix}\right]\)

\(\text {Let } \sin ^{-1} \frac{12}{13}=x, \quad \cos ^{-1} \frac{4}{5}=y, \tan ^{-1} \frac{63}{16}=z\)
\( \sin x=\frac{12}{13}, \quad \cos y=\frac{4}{5}, \quad \tan z=\frac{63}{16}\)
\(\cos x=\frac{5}{13}, \sin y=\frac{3}{5}, \tan x=\frac{12}{5} \text { and } \tan y=\frac{3}{4}\)
\(\text {We have }\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5} \times \frac{3}{4}}=-\frac{63}{16}\)
Hence tan(x + y) = − tan z
i.e., tan (x + y) = tan (–z) or tan (x + y) = tan (p – z)
Therefore x + y = – z or x + y = p – z
Since x, y and z are positive, x + y \(\ne\) – z (Why?)
\(\text {Hence }x+y+z=\pi \text { or } \sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi\)

Let E₁ , E₂ and E₃ be the events that boxes I, II and III are chosen, respectively.
Then P(E₁) = P(E₂) = P(E₃ ) = \(\frac { 1 }{ 3 } \)
Also, let A be the event that ‘the coin drawn is of gold’
Then P(A|E₁) = P(a gold coin from bag I) = \(\frac { 2 }{ 2 } \)= 1,
P(A|E₂) = P(a gold coin from bag II) = 0 and
p(A|E₃) = P(a gold coin from bag III) = \(\frac { 1 }{ 2 } \)
Now, the probability that the other coin in the box is of gold = the probability that gold coin is drawn from the box I.
= P(E₁ |A)
By Bayes' theorem, we know that
P(E₁|A) \(=\frac { P({ E }_{ 1 })P(A|{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A|{ E }_{ 1 })+P({ E }_{ 2 })P(A|{ E }_{ 2 })+P({ E }_{ 3 })P(A|{ E }_{ 3 }) } \)
= \(\frac { \frac { 1 }{ 3 } \times1}{ \frac { 1 }{ 3 } \times1 +\left( \frac { 1 }{ 3 } \right) \times 0 +\frac { 1 }{ 3 } \times \frac { 1 }{ 2 } } =\frac { 2 }{ 3 } \)

Substituting in (i) we get
(sec x + tan x)y = sec x + tan x - x + 1 is the required nsolution.

Curves are 4y² = 9x and 3x² = 16y
Eliminating y from two equations, we get
\(4\left(\frac{3 x^{2}}{16}\right)^{2}=9 x\)
\(\Rightarrow x^{4}= \frac{9 \times 16^{2}}{4 \times 9} x=64 x \Rightarrow x=4 \text { or } x=0 \)
\(\therefore \text { area } =\int_{0}^{4}\left(y_{1}-y_{2}\right) d x=\int_{0}^{4}\left(\sqrt{\frac{9 x}{4}}-\frac{3 x^{2}}{16}\right) d x \)
\(=4 \text { sq units } \)

\(\int \frac{e^{x}}{\sqrt{5-4 e^{x}-e^{2 x}}} =\int \frac{1}{\sqrt{5-4 t-t^{2}}} d t=\int \frac{1}{\sqrt{9-(t+2)^{2}}} d t \)
\(=\sin ^{-1}\left(\frac{t+2}{3}\right)+C=\sin ^{-1}\left(\frac{e^{x}+2}{3}\right)+C \)

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

Let C₁, C₂, C₃, C₄ be the events that the lost card is of heart, spades, diamond or club respectively.
Obviously P(C₁) = P(C₂) = P(C₃) = P(C₄)
\(={13\over 52}={1\over 4}\)
Let S be the event of drawing two cards of heart from the remaining 51 cards.We wish to find \(P\left(C_1\over S\right)\)
Now \(P\left(C_1\over S\right)\) is the probability of drawing two heart cards from 51 cards given that one heart card is lost
\(={^{12}C_2\over ^{51}C_2}={12\times11\over 1\times2}\times{1\times2\over 51\times50}={22\over 425}\)
\(P\left( S\over C_3\right)=P\left( S\over C_3\right)=P\left( S\over C_4\right)={^{13}C_2\over ^{51}C_2}\)
\(={13\times12\over 1\times2}\times{1\times2\over 51\times50}={26\over 425}\)
By Bayes' Theorem
\(P\left(C_1\over S\right)={P(C_1).P\left(S\over C_1\right)\over \sum P(C_1).P\left(S\over C_1\right)}\)
\(={{1\over4}\times{22\over 425}\over{1\over 4}\times{22\over 425}+{1\over 4}\times{26\over 425}+{1\over4}\times{26\over 425}}\)
\(={22\over 22+26+26+26}\)
\(={11\over 50}\)

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

Let, \(I=\int _{ 0 }^{ \pi }{ \frac { x\tan { x } }{ \sec { x }+ \tan { x } } } \)...(i)
\(\Rightarrow I=\int _{ 0 }^{ \pi }{ \frac { \left( \pi -x \right) \tan { \left( \pi -x \right) } }{ \sec { \left( \pi -x \right) }+ \tan { \left( \pi -x \right) } } } \)
\(=\int _{ 0 }^{ \pi }{ \frac { \left( \pi -x \right) \tan { x } }{ \sec { x } +\tan { x } } } dx\)
Add eqn. (i) & (ii),
\(2I=\pi \int _{ 0 }^{ \pi }{ \frac { \tan { x } }{ \sec { x } +\tan { x } } } dx\)
Multiply & divide by (secx - tanx) on RHS,
\(\Rightarrow 2I=\pi \int _{ 0 }^{ \pi }{ \tan { x } \left( \sec { x } -\tan { x } \right) } dx\)
\(=\pi \int _{ 0 }^{ \pi }{ \left( \sec { x } .\tan { x } -\sec ^{ 2 }{ x } +1 \right) } dx\)
\(\Rightarrow 2I=\left[ \pi \left( \sec { x } -\tan { x } +x \right) \right] _{ 0 }^{ \pi }\)
\(=\pi \left\{ \left( -1-0+\pi -1 \right) \right\} =\pi \left( \pi -2 \right) \)
\(\therefore \ I=\frac { { \pi }^{ 2 }-2\pi }{ 2 } \)

\(A=\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right] \)
|A| = 2 x 0-(-3) x -2+5 x 1
= -6 + 5 = -1
a₁₁ = (-4 + 4) = 0
a₁₂ = -(-6 + 4) = 2
a₁₃ = (3 - 2) = 1
a₂₁ = -(6 - 5) = -1
a₂₂ = (-4 - 5) = -9
a₂₃ = -(2 + 3) = -5
a₃₁ = (12 - 10) = 2
a₃₂ = -(-8 - 15) = 23
a₃₃ = (4 + 9) = 13
\({ A }^{ -1 }=\frac { 1 }{ -1 } \left[ \begin{matrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{matrix} \right] \)
Given set of equations may be written as
\(AX=B,\)
where \(B=\left[ \begin{matrix} 11 \\ -5 \\ -3 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
\(\Rightarrow X={ A }^{ -1 }B\)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{matrix} \right] \left[ \begin{matrix} 11 \\ -5 \\ -3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0-5+6 \\ -22-45+69 \\ 11-25+39 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right] \)
Hence x = 1, y = 2 and x = 3.

Proving" is commutative 
Proving" is associative 
Getting identity element as (0, 0) 
Getting inverse of (a, b) as (- a, - b)

\(\lim _{ x\rightarrow 0 }{ (\frac { 1-cos4x }{ 8{ x }^{ 2 } } ) } \) = \(\lim _{ x\rightarrow 0 }{ (\frac { 2{ sin }^{ 2 }2x }{ 8{ x }^{ 2 } } ) } \)
\(=\lim _{ x\rightarrow 0 }{ { (\frac { sin2x }{ 2x } ) }^{ 2 } } =1\)
f is continuous at x=0
\(\therefore\)  \(\lim _{ x\rightarrow 0 }{ f\left( x \right)=f(0) } \)
f(x)=k
\(\therefore\) 1=k or  k=1

\(A^{2}+a I=b A \)
\(\Rightarrow\left[\begin{array}{ll} 3 & 1 \\ 7 & 5 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 7 & 5 \end{array}\right]+a\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=b\left[\begin{array}{ll} 3 & 1 \\ 7 & 5 \end{array}\right] \)
\(\Rightarrow\left[\begin{array}{cc} 9+7 & 3+5 \\ 21+35 & 7+25 \end{array}\right]+\left[\begin{array}{cc} a & 0 \\ 0 & a \end{array}\right]=\left[\begin{array}{cc} 3 b & b \\ 7 b & 5 b \end{array}\right] \)
\(\Rightarrow\left[\begin{array}{cc} 16 & 8 \\ 56 & 32 \end{array}\right]+\left[\begin{array}{cc} a & 0 \\ 0 & a \end{array}\right]=\left[\begin{array}{cc} 3 b & b \\ 7 b & 5 b \end{array}\right] \)
\(\Rightarrow\left[\begin{array}{cc} 16+a & 8 \\ 56 & 32+a \end{array}\right]=\left[\begin{array}{cc} 3 b & b \\ 7 b & 5 b \end{array}\right] \)
\(\Rightarrow 16+a=3 b, b=8,7 b=56 ; 32+a=5 b \)
\(\Rightarrow b=8 \text { and } 16+a=24 \Rightarrow a=8 \)
a = 8, b = 8 satisfy all the above equations