\({ A }^{ -1 }=\frac { 1 }{ 8 } \left[ \begin{matrix} 5 & -1 \\ -7 & 3 \end{matrix} \right] \)

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

\(tan^{ -1 }\left( \frac { x+1 }{ x-1 } \right) +tan^{ -1 }\left( \frac { x-1 }{ x } \right) =-tan^{ -1 }7\)
\(\Rightarrow tan^{ -1 }\left( \frac { \left( \frac { x+1 }{ x-1 } \right) +\left( \frac { x-1 }{ x } \right) }{ 1-\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) } \right) =-tan^{ -1 }7\)
if \(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) <1\)
\(\Rightarrow tan^{ -1 }\left[ \frac { x(x+1)+(x-1)^{ 2 } }{ \left( x-1 \right) x-\left( x+1 \right) \left( x-1 \right) } \right] =tan^{ -1 }7\)
\(\Rightarrow \frac { \left( x^{ 2 }+x \right) +\left( x^{ 2 }+1-2x \right) }{ \left( x^{ 2 }-x \right) -\left( x^{ 2 }-1 \right) } =tan\left[ -tan^{ -1 }7 \right] \)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \) 2x² - 8x + 8 = 0
\(\Rightarrow \) (x - 2)² = 0
\(\Rightarrow \) x = 2
Let us now verify whether x = 2 satisfies the condition (i)
\(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) =3\times \frac { 1 }{ 2 } =\frac { 3 }{ 2 } \) Which is not less than 1.
Hence this value does not satisfy the condition (i) there is no solution to the given trigonometric equation.

\(I=\int { \frac { 1 }{ { cos }^{ 4 }x+{ sin }^{ 4 }x } } dx\)
Dividing numerator and denominator by cos⁴x
\(=\int { \frac { sec^{ 4 }x }{ 1+tan^{ 4 }x } } dx\)
\(=\int { \frac { (1+tan^{ 2 }x)sec^{ 2 }x }{ 1+tan^{ 4 }x } } dx\)
Putting, tanx=t
\(\Rightarrow sec^{ 2 }xdx=dt\)
\(=\int { \frac { ({ t }^{ 2 }+1)dt }{ { t }^{ 4 }+1 } } \)
\(=\int { \frac { 1+\frac { 1 }{ { t }^{ 2 } } }{ { t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } } } dt\)     [dividing by t²]
\(=\int { \frac { dz }{ { z }^{ 2 }+(\sqrt { 2 } )^{ 2 } } } \) ,  where \(t-\frac { 1 }{ t } =z\)
\(=\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { z }{ \sqrt { z } } \right) +C\)
\(=\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { { t }^{ 2 }-1 }{ \sqrt { 2t } } \right) +C\)
\(=\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { { tan }^{ 2 }x-1 }{ \sqrt { 2 } tanx } \right) +C\)

Applying \({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 }\quad and\quad { C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA\cos ^{ 2 }{ B } +cosB & cosC-cosA\cos ^{ 2 }{ C } -cosC \\ \cos ^{ 2 }{ A } +cosA & \cos ^{ 2 }{ A } -cosA & -\cos ^{ 2 }{ A } -cosA \end{matrix} \right| =0\)
\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA & cosC-cosA \\ \cos ^{ 2 }{ A } -1 & \cos ^{ 2 }{ B } -\cos ^{ 2 }{ A } & \cos ^{ 2 }{ C } -cosA \end{matrix} \right| =0\)
\({ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 2 }\)
(cos B - cos A ) x ( cos C -cos B)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & 1 & 0 \\ \cos ^{ 2 }{ A } -1 & cosB+cosA & cosC-cosA \end{matrix} \right| =0\)
\(\therefore \quad (cosB-cosA)\times (cosC-cosA)\times (cosC-cosB)\)
[1-0]=0  (Expanding along C₃)
\(\therefore \ -cosB=cosA\quad or\quad cosC=cosAorcosC=cosB\)
\(\Rightarrow cosB=cosAorcosC=cosAorcosC=cosB\)
\(\Rightarrow \angle B=\angle Aor\angle C=\angle Aor\angle C=\angle B\)
\(\Rightarrow \triangle ABC\)  is an isosceles triangle.

Let E₁: Event selecting bag with 4 red & 4 black balls
E₂: Event selecting bag with 2 red & 6 black balls
A: Event selecting 2 red balls without replacement
Then, P(E₁) = P(E₂) = \(\frac { 1 }{ 2 } \)
\(P(A/{ E }_{ 1 })=\frac { 4_{ C_{ 2 } } }{ { 8 }_{ C_{ 2 } } } =\frac { 3 }{ 14 } ,\)
\(P(A/{ E }_{ 2 })=\frac { { 2 }_{ { C }_{ 2 } } }{ { 8 }_{ { C }_{ 2 } } } =\frac { 1 }{ 28 } .\)
\(P\left( \frac { { E }_{ 1 } }{ A } \right) =\frac { P({ E }_{ 1 }).P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 }) } \)
\(=\frac { \frac { 1 }{ 2 } .\frac { 3 }{ 14 } }{ \frac { 1 }{ 2 } .\frac { 3 }{ 14 } +\frac { 1 }{ 2 } .\frac { 1 }{ 28 } } =\frac { 6 }{ 7 } \)

\(cos^{ -1 }\left( \frac { 12 }{ 13 } \right) ={ tan }^{ -1 }\frac { 5 }{ 12 } ,\)
\(and\quad { cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) ={ tan }^{ -1 }\frac { 3 }{ 4 } \)
\(L.H.S.\ ={ tan }^{ -1 }\frac { 5 }{ 12 } +{ tan }^{ -1 }\frac { 3 }{ 4 } \)
\(={ tan }^{ -1 }\left( \frac { \frac { 5 }{ 12 } +\frac { 3 }{ 4 } }{ 1-\frac { 5 }{ 12 } \times \frac { 3 }{ 4 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 20+36 }{ 48 } }{ \frac { 48-15 }{ 48 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { 56 }{ 33 } \right) =R.H.S.\)

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

=\(\frac { { x }^{ 2 }+\sqrt { 1+x^{ 2 } } +1 }{ x\left( 1+\sqrt { 1+{ x }^{ 2 } } \right) } \)

\(a_{11}=2-3=-1 ; a_{12}=2-6=-4 ; a_{13}=2-9=-7 \)
\(a_{21}=4-3=1 ; a_{22}=4-6=-2 ; a_{23}=4-9=-5 \)
\(a_{31}=6-3=3 ; a_{32}=6-6=0 ; a_{33}=6-9=-3\)
Matrix is \(\left[ \begin{matrix} -1 & -4 & -7 \\ 1 & -2 & -5 \\ 3 & 0 & -3 \end{matrix} \right] \)

Here, \({\overrightarrow{a}}_{1}=\hat{i}+2\hat{j}+3\hat{k};\)  \({\overrightarrow{b}}_{1}=\hat{i}-3\hat{j}+2\hat{k}\)
and  \({\overrightarrow{a}}_{2}=4\hat{i}+5\hat{j}+6\hat{k};\)   \({\overrightarrow{b}}_{2}=2\hat{i}+3\hat{j}+\hat{k}\)
\({\overrightarrow{a}}_{2}-{\overrightarrow{a}}_{1}=3\hat{i}+3\hat{j}+3\hat{k};\)
\({\overrightarrow{b}}_{1}-{\overrightarrow{b}}_{2}=\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix}=-9\hat{i}+3\hat{j}+9\hat{k}\)
\(({\overrightarrow{b}}_{1}\times{\overrightarrow{b}}_{2}).({\overrightarrow{a}}_{2}-{\overrightarrow{a}}_{1})\) = - 27 + 9 + 27 = 9;
\(|{\overrightarrow{b}_{1}}\times{\overrightarrow{b}_{2}}|=\sqrt{171}\)
= \(\left| {{(\overrightarrow{b}_{1}\times{\overrightarrow{b}}_{2}).({\overrightarrow{a}}_{2}-{\overrightarrow{a}}_{1})}\over{|{\overrightarrow{b}}_{1}\times{\overrightarrow{b}}_{2}|}} \right| \)
= \({{9}\over{\sqrt{171}}}={{3}\over{\sqrt{19}}}\) or \({{3\sqrt{19}}\over{19}}\)

\(\because \vec { r } =\vec { { a }_{ 1 } } +\lambda \vec { { b }_{ 1 } } \)
\(\vec { r } =\vec { { a }_{ 2 } } +\mu \vec { { b }_{ 2 } } \)
\(\therefore \vec { { a }_{ 1 } } =\left( \hat { i } -\hat { j } \right) , \vec { { a }_{ 2 } } =\left( 2\hat { i } +2\hat { j } -\hat { k } \right) \)
\(\therefore\vec { { a }_{ 2 } } -\vec { { a }_{ 1 } } =\left( \hat { i } +3\hat { j } -\hat { k } \right) \)
\(\vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 2 & 1 \\ 5 & -3 & 2 \end{matrix} \right| \)
\(\hat { i } \left( 4+3 \right) -\hat { j } \left( 4-5 \right) +\hat { k } \left( -6-10 \right) \left( 7\hat { i } +\hat { j } -16\hat { k } \right) \)
\(d=\left| \frac { \left( 7\hat { i } +\hat { j } -16\hat { k } \right) .\left( \hat { i } +3\hat { j } -\hat { k } \right) }{ \sqrt { { 7 }^{ 2 }+{ 1 }^{ 2 }+{ 16 }^{ 2 } } } \right| \)
\(\Rightarrow d=\left| \frac { 7+3+16 }{ \sqrt { \left( 49 \right) +\left( 1 \right) +\left( 256 \right) } } \right| =\frac { 26 }{ \sqrt { 306 } } \)
\(\therefore d=\frac { 26 }{ 3\sqrt { 34 } } units\)
The two ships will not meet any accidental collision.
Value: Carelessness leads to mishappening.

Let the manufacturer produce x packages of nuts and y packages of bolts. Therefore, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.

The profit on a package of nuts is Rs. 17.50 and on a package of bolts is Rs. 7. Therefore, the constraints are
x + 3y ≤ 12
3x + y ≤ 12
Total profit, Z = 17.5x + 7y
The mathematical formulation of the given problem is
Maximise Z = 17.5x + 7y … (1)
subject to the constraints,
x + 3y ≤ 12 … (2)
3x + y ≤ 12 … (3)
x, y ≥ 0 … (4)
The feasible region determined by the system of constraints is as follows.

The corner points are A (4, 0), B (3, 3), and C (0, 4).
The values of Z at these corner points are as follows.

The maximum value of Z is Rs. 73.50 at (3, 3).
Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit of Rs. 73.50.

Let 'x' and 'y' the number of packets of food P and Q respectively.
We have:  \(x\ge 0\)  ...(1)
\(y\ge 0\)  ....(2)
\(12x+3y\ge 240\quad i.e.4x+y\ge 80\) .....(3)
\(4x+20y\ge 460\quad i.e.x+5y\ge 115\) .....(4)
\(6x+4y\le 300\quad i.e.3x+2y\le 150\) .....(5)
The mathematical problem is as below:
Minimize Z=6x+3y subject to (1)-(5)
Draw the lines
x = 0, 4x + y = 80, x + 5y = 80, x + 5y = 115 and 3x + 2y = 150.
The lines 3x + 2y = 150 and 4x + y = 80 meet at L (2, 72)
The lines 4x + y = 80 and x + 3y = 115 meet at M (15, 20)
The lines 3x + 2y = 150 and x + 5y = 115 meet at N(40,15).

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

Hence, minimum amount of vitamin A = 150 units.

The given ellipse is \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Since (1) is symmetrical about both axes,
Therefore, area of the ellipse = 4 (Shaped area).

\(=4(area\ OAB)\)
\(But\ area\ OAB=\overset { 4 }{ \underset { 0 }{ \int { } } } ydx\quad [Taking\ vertical\ strips]\)
\(=\overset { 4 }{ \underset { 0 }{ \int { } } } \frac { 3 }{ 4 } \sqrt { 16-{ x }^{ 2 } } dx\)
\([\because \frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow \frac { { y }^{ 2 } }{ 9 } =1-\frac { { x }^{ 2 } }{ 16 } \Rightarrow y=\frac { 3 }{ 4 } \sqrt { 16-{ x }^{ 2 } } (\because y>0)]\)
\(=\frac { 3 }{ 4 } \left[ \frac { x\sqrt { 16-{ x }^{ 2 } } }{ 2 } +\frac { 16 }{ 2 } { sin }^{ -1 }\frac { x }{ 4 } \right] _{ 0 }^{ 4 }\)
\(=\frac { 3 }{ 4 } [[2(0)+8{ sin }^{ -1 }(1)]-[0-0]]\)
\(=\frac { 3 }{ 4 } \left[ 8\frac { \pi }{ 2 } \right] =3\pi \)
\(\therefore From\ (2),area\ of\ the\ ellipse =4(3\pi )=12\pi sq.units.\)

(A) is the correct answer.
Reason: Since A and B are symmetric matrices,
\(\therefore \ A\prime =A\quad and\quad B\prime =B ...(1)\)
\(Now\ (AB-BA)\prime =(AB)\prime -(BA)\prime \)
\(=B\prime A\prime -A\prime B\prime\)
\(=BA-AB\)
\(=-(AB-BA)\)
\(\Rightarrow (AB-BA)\) is skew-symmetric.

Let y=tan(2x+3)=tan t, where t=2x+3
\(\frac { dy }{ dt } ={ sec }^{ 2 }\ t\ and\ \)
\(\frac { dt }{ dx } =2(1)+0=2.\)
\(By\quad chain\quad rule,\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } \)
\(={ sec }^{ 2 }\ t.2\)
\(=2 { sec }^{ 2 }(2x+3)\)

\(\left| \begin{matrix} x & { x }^{ 2 } & yz \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right| \)
= \(\left| \begin{matrix} x-y & { x }^{ 2 }-y^2 & yz-zx \\ y-z & { y }^{ 2 }-z^2 & zx-xy \\ z & { z }^{ 2 } & xy \end{matrix} \right| \)
[Operating R\(\rightarrow\)R₁-R₂ & R\(\rightarrow\)R₂-R₃] =
\(\left| \begin{matrix} x-y & (x-y)(x+y) & z(y-x) \\ y-z &(y-z)(y+z)& x(z-y)\\ z & { z }^{ 2 } & xy \end{matrix} \right| \)
= \((x-y)(y-z)\left| \begin{matrix}0 &x-z&-z+x \\ 1 & y+z &-x \\ z & { z }^{ 2 } & xy \end{matrix} \right| \)
[Taking (z-x)common from R₁]
\((x-y)(y-z)(z-x)\left| \begin{matrix} 0 &-1 & -1 \\ 1 & y+z & -x \\ z & z^2 & xy \end{matrix} \right| \)
\((x-y)(y-z)(z-x)\left| \begin{matrix} 0 & 0 & -1 \\ 1 & x+y+z & -x \\ z & z^2-xy & xy \end{matrix} \right| \)
\((x-y)(y-z)(z-x)(-1)\left| \begin{matrix} 1 & x+y+z \\z &z^2-xy\end{matrix} \right| \)
= (x-y) (y-z) (z-x) (-1) [z²-xy-zx-yz-z²]
= (x-y) (y-z) (z-x) (1) (-xy-yz-zx)
= (x-y) (y-z) (z-x) (xy+yz+zx)

Let sin^-1 x = \(\theta\) then  sin^-1 x = \(\theta\). we have
sin^-1 \((2x\sqrt { 1-{ x }^{ 2 } } )\) = sin^-1\(2\sin { \theta } \sqrt { 1-{ sin }^{ 2 } } \)
= sin^–1 (2sinθ cosθ) = sin^–1 (sin2θ) = 2θ = 2 sin^–1 x

\(={112\over3}\)

We have: \(\frac { dy }{ dx } +\frac { { y }^{ 2 }+y+1 }{ { x }^{ 2 }+x+1 } =0\)
\(\Rightarrow \frac { dy }{ { y }^{ 2 }+y+1 } +\frac { dx }{ { x }^{ 2 }+x+1 } =0\)  |Variables Separable
Integrating, \(\int { \frac { dy }{ { y }^{ 2 }+y+1 } } +\int { \frac { dx }{ { x }^{ 2 }+x+1 } = } \) Constant
\(\Rightarrow \int { \frac { dy }{ \left( { y }^{ 2 }+y+\frac { 1 }{ 4 } \right) +\frac { 3 }{ 4 } } } +\int { \frac { dx }{ \left( { x }^{ 2 }+x+\frac { 1 }{ 4 } \right) +\frac { 3 }{ 4 } } } =\) Constant
\(\Rightarrow \int { \frac { dy }{ { \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }+{ \left( y+\frac { 1 }{ 2 } \right) }^{ 2 } } } \)\(+\int { \frac { dx }{ { \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }+{ \left( x+\frac { 1 }{ 2 } \right) }^{ 2 } } } =\) Constant
\(\Rightarrow \frac { 1 }{ \frac { \sqrt { 3 } }{ 2 } } { tan }^{ -1 }\frac { y+\frac { 1 }{ 2 } }{ \frac { \sqrt { 3 } }{ 2 } } +\frac { 1 }{ \frac { \sqrt { 3 } }{ 2 } } { tan }^{ -1 }\frac { x+\frac { 1 }{ 2 } }{ \frac { \sqrt { 3 } }{ 2 } } =\frac { 1 }{ \frac { \sqrt { 3 } }{ 2 } } { tan }^{ -1 }A\)
\(\Rightarrow { tan }^{ -1 }\frac { 2y+1 }{ \sqrt { 3 } } +{ tan }^{ -1 }\frac { \left( 2x+1 \right) }{ \sqrt { 3 } } ={ tan }^{ -1 }A\)
\({ tan }^{ -1 }\frac { \frac { 2y+1 }{ \sqrt { 3 } } +\frac { 2x+1 }{ \sqrt { 3 } } }{ 1-\left( \frac { 2y+1 }{ \sqrt { 3 } } \right) \left( \frac { 2x+1 }{ \sqrt { 3 } } \right) } { tan }^{ -1 }A'\)
\(\Rightarrow \frac { \left( 2y+1 \right) +\left( 2x+1 \right) }{ \left[ 3-\left( 2x+2y+4xy+1 \right) \right] } =\sqrt { 3 } A'\)
\(=2\left( x+y+1 \right) =A\left( 2-2x-2y-4xy \right) ,\)
where  \(\sqrt { 3 } A'=A\)
\(\Rightarrow x+y+1=A\left( 1-x-y-2xy \right) ,\)
which is true.

\(y = \frac {{x}^{3}}{4}-\frac {2}{x}\)

(I + A)²-3A = I
Alternative Mwthod:
Given A² = A,
(I+A)²-3A = I² + 2IA + A²-3A
= I + 2A + A - 3A = I

Given, \(\left| a\right| =\left| b \right| \)
and a.b \(=\frac{1}{2}\)
Let \(\theta \) be thae angle between \(\overrightarrow a\) and \(\overrightarrow b\)
then, \(\cos \theta=\frac{a.b}{\left| a\right| \left| b \right| }\)
\(\cos 60^{ \circ }=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right| .\left| \overrightarrow { a} \right| }\)
\(\Rightarrow \frac{1}{2}=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right|^{ 2} }\)
\(\Rightarrow \left| \overrightarrow { a } \right| ^{2 }=\frac{\frac{1}{2}}{\frac{1}{2}}=1\)
\(\Rightarrow \left| \overrightarrow { a} \right| =1 \)
\(\Rightarrow\left| \overrightarrow { a} \right| =\left| \overrightarrow { b} \right| =1\)

We have, y = \(log\left( tanx\frac { x }{ 2 } \right) \)
\(\frac { dy }{ dx } =\frac { 1 }{ \frac { tanx }{ 2 } } \times { sec }^{ 2 }\frac { x }{ 2 } \times \frac { 1 }{ 2 } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { \frac { cosx }{ 2 } }{ \frac { sinx }{ 2 } } \times \frac { 1 }{ { cos }^{ 2 }\frac { x }{ 2 } } \times \frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } } =\frac { 1 }{ sinx } \)
\(\Rightarrow \frac { dy }{ dx } =cosecx\)

Given equation
x = ay + b, Z = cy + d
\(\frac { x-b }{ a } =y,\frac { x-d }{ c } =y\)
\(\Rightarrow \frac { x-b }{ z } =\frac { y-0 }{ 1 } =\frac { z-d }{ c } \)
Direction ratios are (a, 1, c) and a point on the given line is (b, 0, d).

\(P(\bar { A } /\bar { B } )=\frac { P(\bar { A } \cap \bar { B } ) }{ P(\bar { B } ) } \)
\(=\frac { 1-P(A\cup B) }{ 1-P(B) } \)
\(=\frac { 1-[P(A)+P(B)-P(A\cap B)] }{ 1-P(B) } \)
\(=\frac { 7 }{ 10 } \)

\(xdy=\left( \frac { y+1 }{ y } \right) dx\)
\(\Rightarrow \frac { ydy }{ y+1 } =\frac { dx }{ x } \)
Integrating both the sides,
\(\int { \frac { y }{ y+1 } } dy=\int { \frac { dx }{ x } } \)
\(\Rightarrow \int { \frac { y+1-1 }{ y+1 } } dy=\int { \frac { dx }{ x } } \)
\(\Rightarrow \int { dy } -\int { \frac { dy }{ y+1 } } =\int { \frac { dx }{ x } } \)
\(\Rightarrow y-\log { \left( y+1 \right) } =\log { x } +C\)

\(\int { { e }^{ x } } tan^{ -1 }xdx+\int { \frac { { e }^{ x } }{ 1+{ x }^{ 2 } } } dx\)
\(={ tan }^{ -1 }x\int { { e }^{ x } } dx-\int { \left( \frac { d }{ dx } { tan }^{ -1 }x\int { { e }^{ x }dx } \right) } dx+\int { \frac { { e }^{ x } }{ 1+{ x }^{ 2 } } } dx\)
\(={ e }^{ x }{ tan }^{ -1 }x-\int { \frac { 1 }{ 1+{ x }^{ 2 } } \times } { e }^{ x }dx+\int { \frac { { e }^{ x } }{ 1+{ x }^{ 2 } } } \)
\(={ e }^{ x }{ tan }^{ -1 }x+C\)

\(f(x)=x^{ 2 }+1,g(x)=2x+1\)
\(fog=f[(g(x)\} =f(2x+1)\)
\(=4x^{ 2 }+4x+1+1\)
\(=4x^{ 2 }+4x+2\)
\(gof=g(f(x)=g(x^{ 2 }+1)\)
\(=2(x^{ 2 }+1)+1\)
\(=2x^{ 2 }+2+1\)
\(=2x^{ 2 }+3\)

\(R.H.S.=2{ sin }^{ -1 }\sqrt { \frac { 1-x }{ 2 } } \quad \quad \begin{cases} Let\quad x=cos\theta \\ \Rightarrow \theta ={ cos }^{ -1 }x \end{cases}\)
\(=2{ sin }^{ -1 }\left( \sqrt { \frac { 1-cos\theta }{ 2 } } \right) \)
\( \begin{cases} as\quad cos\theta =1=2{ sin }^{ 2 }\frac { \theta }{ 2 } \\ \Rightarrow 1-cos\theta =2sin^{ 2 }\frac { \theta }{ 2 } \end{cases}\)
\(=2{ sin }^{ -1 }\left( sin\frac { \theta }{ 2 } \right) \)
\(=2\left( \frac { \theta }{ 2 } \right) =\theta ={ cos }^{ -1 }x\)
L.H.S = R.H.S

We have, \(A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)
\({ A }^{ \prime }=\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{matrix} \right] \)
\({ A+A }^{ \prime }=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\Rightarrow \frac { 1 }{ 2 } { A+A }^{ \prime }=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] =0\)
and \({ A-A }^{ \prime }=\left[ \begin{matrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{matrix} \right] \)
\(\Rightarrow \frac { 1 }{ 2 } { A-A }^{ \prime }=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{matrix} \right] \)
\(\Rightarrow A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)

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|

As for y² = x, 2yy^' = 1⇒ y']_(1,1) = \(\frac12\)
for x² = y, y' = 2x ⇒ y']_(1,1)= 2
\(\therefore tan\theta =\left| \frac { \frac { 1 }{ 2 } -2 }{ 1+\frac { 1 }{ 2 } .2 } \right| \) \(\Rightarrow tan\theta =\left| \frac { -\frac { 3 }{ 2 } }{ 2 } \right| \)
\(\Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \) or \({ cot }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)