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

Let A be the point (5, 4, - 6)
Image A' be the point (5, 4, - 6)
\(\therefore\) A'(5, 4, - 6)
Distance between AA'
\(=\sqrt { (5-5)^{ 2 }+(4-4)^{ 2 }+(6+6)^{ 2 } } \)
\(=\sqrt { 0+0+12^{ 2 } } \)
=123 units

\((\overset\rightarrow a+\overset\rightarrow b).(\overset\rightarrow a-\overset\rightarrow b)\)
\(\overset\rightarrow a.\overset\rightarrow a+\overset\rightarrow a.\overset\rightarrow b-\overset\rightarrow b.\overset\rightarrow a-\overset\rightarrow b.\overset\rightarrow b=\left| \overset\rightarrow a \right| ^{ 2}-\left| \overset\rightarrow b \right| ^{ 2}\)
\((\overset\rightarrow a+\overset\rightarrow b).(\overset\rightarrow a-\overset\rightarrow b)=8\left|\overset\rightarrow b \right| ^{ 2 }-\left|\overset\rightarrow b \right| ^{2 }\)
\(=64\left| \overset { \rightarrow }{b } \right| ^{2 }-\left| \overset { \rightarrow }{b } \right| ^{2 }\)
\(=63\left| \overset { \rightarrow }{ b} \right| ^{ 2 }\)
\((\overset\rightarrow a+\overset\rightarrow b).(\overset\rightarrow a-\overset\rightarrow b)=8\)
\(63\left| \overset { \rightarrow }{ b} \right| ^{ 2 }=8\)
\(\left| \overset { \rightarrow }{b } \right| =\sqrt{\frac{8}{63}}=\frac{2\sqrt7}{3\sqrt7}\)
\(\left|\overset\rightarrow a \right| =\frac{8\times 2\sqrt2}{3\sqrt{7}}=\frac{16\sqrt2}{3\sqrt{7}}\)

Total number of bolts = 50
Total number of nuts = 50
Total number of rusted bolts = 25
Total number of rusted nuts = 25
Total no. of items = 100
Total no. rusted items = 50
E = rusted item, F = Bolts
\(P(E\cup F)=P(F)+P(F)-P(E\cap F)\)
\(P(E)=\left( \frac { 25 }{ 100 } \times \frac { 25 }{ 100 } \right) \)
\(=\frac { 1 }{ 4 } \times \frac { 1 }{ 4 } =\frac { 1 }{ 16 } \)
\(P(F)=\left( \frac { 50 }{ 100 } \times \frac { 50 }{ 100 } \right) =\frac { 1 }{ 4 } \)
\(P(E\cap F)=\left( \frac { 25 }{ 100 } \times \frac { 25 }{ 100 } \right) \)
\(=\frac { 1 }{ 4 } \times \frac { 1 }{ 4 } =\frac { 1 }{ 16 } \)
\(P(E\cup F)=\frac { 1 }{ 16 } +\frac { 1 }{ 4 } -\frac { 1 }{ 16 } \)
\(=\frac { 1 }{ 4 } \)

y = a cos2x + b sin2x
\(\Rightarrow \frac { dy }{ dx } =-a\sin { 2x } \times 2+b\cos { 2x } \times 2\)
\(\Rightarrow \frac { dy }{ dx } =2\left[ -a\sin { 2x } +b\cos { 2x } \right] \)
\(\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2\left[ -a\cos { 2x } \times 2-b\sin { 2x } \times 2 \right] \)
\(\Rightarrow \frac { { x }^{ 2 }y }{ { dx }^{ 2 } } =-4\left[ a\cos { 2x } +b\sin { 2x } \right] =-4y\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +4y=0\)

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

\(cosC.sinD=\frac { 1 }{ 2 } \left[ sin(C+D)-sin(C-D) \right] \)
\(=\frac { 1 }{ 2 } \int { \left[ sin(3x+2x)-sin(3x-2x) \right] } dx\)
\(=\frac { 1 }{ 2 } \int { (sin5x-sinx) } dx\)
\(=\frac { 1 }{ 2 } \left( -\frac { cos5x }{ 5 } \right) -\frac { 1 }{ 2 } \left[ -cosx \right] \)
\(=\frac { 1 }{ 2 } \left[ cosx-\frac { 1 }{ 5 } cos5x \right] +C\)

If A is symmetric matrix then
\(A={ A }^{ \prime }\)
\(\Rightarrow \left[ \begin{matrix} 0 & b & -2 \\ 3 & 1 & 3 \\ 2a & 3 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 3 & 2a \\ b & 1 & 3 \\ -2 & 3 & -1 \end{matrix} \right] \)
\(\therefore\)  By equality of matrices, 
b = 3 and a = - 1

Given A = Set of all human beings in a town at a particular time and R = {(x, y); x is a wife of y; x, y \(\in \) A} 
(i) Since x is a wife of x, is not true (x, x) \(\notin \)  R
So, R is not reflexive. 
(ii) x is a wife of y, but y not wife of (y, x) \(\notin \) : R. So, R is not symmetric. 
(iii) x is a wife of y, and y is wife of z But this situation does not exist. So, R is not transitive.

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|

Given \({ X }_{ m\times 3 }{ Y }_{ p\times 4 }={ Z }_{ 2\times b }\)
If XY is defined,
then \(3=p\Rightarrow p=3\)and 
\({ Z }_{ m\times 4 }={ Z }_{ 2\times b }\Rightarrow \) m = 2 and b = 4.
\(\therefore \ p=3,m=2\quad and\quad b=4.\)

Let the farmer buy x kg of fertilizer F₁ and y kg of fertilizer F₂. T`herefore,
x ≥ 0 and y ≥ 0
The given information can be complied in a table as follows.

F₁ consists of 10% nitrogen and F₂ consists of 5% nitrogen. However, the farmer requires at least 14 kg of nitrogen.
∴ 10% of x + 5% of y ≥ 14
x10 + y 20 ≥ 14
2x + y ≥ 280
F₁ consists of 6% phosphoric acid and F₂ consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.
∴ 6% of x + 10% of y ≥ 14
6x/100 + 10y/100 > = 14
3x +56y > = 700
Total cost of fertilizers, Z = 6x + 5y
The mathematical formulation of the given problem is
Minimize Z = 6x + 5y … (1)
subject to the constraints,
2x + y ≥ 280 … (2)
3x + 5y ≥ 700 … (3)
x, y ≥ 0 … (4)
The feasible region determined by the system of constraints is as follows.

It can be seen that the feasible region is unbounded.
The corner points are A(700/3, 0), B(100, 80) and C(0, 280)
The values of Z at these points are as follows.

As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.
For this, we draw a graph of the inequality, 6x + 5y < 1000, and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with
6x + 5y < 1000
Therefore, 100 kg of fertiliser F₁ and 80 kg of fertilizer F₂ should be used to minimize the cost. The minimum cost is Rs. 1000.

The given equation is:
\(\left( x+y \right) \frac { dy }{ dx } =2{ e }^{ -y }-1\)
\(\Rightarrow\)  \(\frac { dy }{ 2{ e }^{ -y }-1 } =\frac { dx }{ x+1 } \)
Variables Separable 
Integrating, \(\int { \frac { dy }{ 2{ e }^{ -y }-1 } } =\int { \frac { dx }{ x+1 } +C } \)
\(\Rightarrow\) \(\int { \frac { dy }{ 2{ e }^{ -y }-1 } =log|x+1|+C } \)  .(1)
Now  \(I=\int { \frac { dy }{ 2{ e }^{ -y }-1 } } =\int { \frac { { e }^{ y } }{ 2-{ e }^{ y } } } dy\)
Put \({ e }^{ y }=t\) so that \({ e }^{ y }\quad dy=dt.\)
\(\therefore\) \(I=\int { \frac { dt }{ 2-t } =- } log|2-t|=-log|2-{ e }^{ y }|.\)
From(1),  \(-log2-{ e }^{ y }|=log|x+1|+C\)...(2)
When \(x=0,y=0,\)                   
\(\therefore\)  \(log|2-1|=log|0+1|+C\)
\(\Rightarrow\) \(log|1|=log|1|+C\)
\(\Rightarrow\) \(0=0+C\Rightarrow C=0.\)
Putting in (2), \(-log|2-{ e }^{ y }|=log|x+1|\) 
\(\Rightarrow\)  \(log|2-{ e }^{ y }|=log|\frac { 1 }{ x+1 } |\)
\(\Rightarrow\) \(2-{ e }^{ y }=\frac { 1 }{ x+1 } \)
\(\Rightarrow\)  \({ e }^{ y }=2-\frac { 1 }{ x+1 } =\frac { 2x+1 }{ x+1 } \)
\(\Rightarrow\) \(y=log|\frac { 2x+1 }{ x+1 } |,x\neq -1.\)
Which is the requored solution.

\(\frac { 5 }{ 2 } sq.units\)

\(We\quad know\quad that\quad A={ I }_{ 2 }A\)
\(i.e.\ \begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A\)
\(\Rightarrow \begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}A \quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ -3 & 4 \end{bmatrix}A \quad [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }-{ 3R }_{ 1 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 4 & -5 \\ -3 & 4 \end{bmatrix}A.\quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }]\)
\(Hence,\ { A }^{ -1 }=\begin{bmatrix} 4 & -5 \\ -3 & 4 \end{bmatrix}.\)

The given line is \(x=\sqrt { 3 } y\) and the given circle is x² + y² = 4
From (1) \(y=\frac { x }{ \sqrt { 3 } } \)
Putting in (2), \({ x }^{ 2 }+\frac { { x }^{ 2 } }{ 3 } =4\Rightarrow \frac { { 4x }^{ 2 } }{ 3 } =4\Rightarrow { x }^{ 2 }=3\Rightarrow x=\pm \sqrt { 3 } \)
When \(x=\pm \sqrt { 3 } ,y=\pm 1\)
Thus P is \((\sqrt { 3 } ,1)\)
Now Reqd, area = ar (OPL) + ar (PLA)
\(\overset { \sqrt { 3 } }{ \underset { 0 }{ \int { } } } \frac { x }{ \sqrt { 3 } } dx+\overset { 2 }{ \underset { \sqrt { 3 } }{ \int { } } } \sqrt { 4-{ x }^{ 2 } } dx\ [Taking\ vertical\ strips]\)
\(=\frac { 1 }{ \sqrt { 3 } } \overset { \sqrt { 3 } }{ \underset { 0 }{ \int { } } } xdx+\overset { 2 }{ \underset { \sqrt { 3 } }{ \int { } } } \sqrt { { 2 }^{ 2 }-{ x }^{ 2 } } dx\)
\(=\frac { 1 }{ \sqrt { 3 } } \left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ \sqrt { 3 } }+\left[ \frac { x\sqrt { 4-{ x }^{ 2 } } }{ 2 } +\frac { 4 }{ 2 } { sin }^{ -1 }\frac { x }{ 2 } \right] _{ \sqrt { 3 } }^{ 2 }\)
\(=\frac { 1 }{ 2\sqrt { 3 } } [3-0]+\left[ \left\{ { 0+2sin }^{ -1 }(1) \right\} -\left\{ \frac { \sqrt { 3 } (1) }{ 2 } +2{ sin }^{ -1 }\frac { \sqrt { 3 } }{ 2 } \right\} \right] \)
\(=\frac { \sqrt { 3 } }{ 2 } +2\left( \frac { \pi }{ 2 } \right) -\frac { \sqrt { 3 } }{ 2 } -2\left( \frac { \pi }{ 3 } \right) \)
\(=\pi -\frac { 2\pi }{ 3 } =\frac { \pi }{ 3 } sq.units\)

\(Let\ y={ sin }^{ 2 }x\quad and\quad u={ \quad e }^{ cosx }\)
\( \frac { dy }{ dx } ={ \quad e }^{ cosx }.(-sinx)=-sinx{ \quad e }^{ cosx }\)
\(\frac { dy }{ du } =\frac { { dy }/{ dx } }{ { du }/{ dx } } =\frac { 2sinxcosx }{ -sinx{ \quad e }^{ cosx } } =-\frac { 2cosx }{ { \quad e }^{ cosx } } \)

The given system of equation is:
2x - y = -2
3x + 4y = 3
these can be written as A X = B
= X = A^-1B
where  \(A=\begin{bmatrix} 2&-1\\3&4\end{bmatrix}, X=\begin{bmatrix} x\\y\end{bmatrix}\ and\ B=\begin{bmatrix}-2\\3 \end{bmatrix}\)
\(|A|=\begin{bmatrix} 2&-1\\3&4\end{bmatrix}=8+3=11\neq0\Rightarrow A^{-1}\)exists.
Now \(adj\ A=\begin{bmatrix} 4&-3\\1&2\end{bmatrix}=\begin{bmatrix}4&1\\-3&2 \end{bmatrix}\)
\(A^{-1}={1\over |A|}(adj\ A)={1\over11}\begin{bmatrix}4&1\\3&2 \end{bmatrix}\)
\(X={1\over11}\begin{bmatrix}4&1\\-3&2 \end{bmatrix}\begin{bmatrix} -2\\3\end{bmatrix}\)
\(={1\over11}\begin{bmatrix} -8+.3\\6+6\end{bmatrix}={1\over11}\begin{bmatrix} -5\\12\end{bmatrix}\)
\(\Rightarrow \begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} -5/11\\12/11\end{bmatrix}\)
\(x={-5\over11},y={12\over11}\)|

(a) It is easy to see that f is one-one and onto, so that f is invertible with the inverse f ^–1 of f given by f ^–1 = {(1, 1), (2, 2), (3, 3)} = f.
(b) Since f (2) = f (3) = 1, f is not one-one, so that f is not invertible.
(c) It is easy to see that f is one-one and onto, so that f is invertible with f ^–1 = {(3, 1), (2, 3), (1, 2)}.

We have 
\(\tan ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right]=\tan ^{-1}\left[\frac{\frac{a \cos x-b \sin x}{b \cos x}}{\frac{b \cos x+a \sin x}{b \cos x}}\right]=\tan ^{-1}\left[\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b} \tan x}\right]\)
\(=\tan ^{-1} \frac{a}{b}-\tan ^{-1}(\tan x)=\tan ^{-1} \frac{a}{b}-x\)

Equation of line PR is \(\frac { x-7 }{ 2 } =\frac { y-14 }{ 4 } =\frac { z-5 }{ -1 } =\lambda \)
General point on line is R \((2\lambda +7,4\lambda +14,-\lambda +5)\)
If this point lies in plane 2x + 4y - z =  2,
then \(2(2\lambda +7)+4(4\lambda +14)-(-\lambda +5)=2\Rightarrow 21\lambda =-63\Rightarrow \lambda =-3\)
Substituting in (i), Foot of perpendicular is R (1,2,8).
\(PR=\sqrt { { (7-1) }^{ 2 }+{ (14-2) }^{ 2 }+{ (5-8) }^{ 2 } } =\sqrt { 36+144+9 } =\sqrt { 189 } \) units
From (i) and (iii), we get 
\(\left( \frac { 7+\alpha }{ 2 } ,\frac { 14+\beta }{ 2 } ,\frac { 5\gamma }{ 2 } \right) =(1,2,8)\)
\(\Rightarrow \frac { 7+\alpha }{ 2 } =1,\frac { 14+\beta }{ 2 } =2,\frac { 5+\gamma }{ 2 } =8\Rightarrow \alpha =-5,\beta =-10,\gamma =11.\)
Image of point P (7,14,5) in plane 2x + 4y - z= 2 is Q (-5, -10, 11).

Let the events be as felow:
 E₁ : He knows the answer
 E₂ : He guesses
 E₃ : He copies
And A : He answerd correctly.
Then \(P({ E }_{ 1 })=\frac { 1 }{ 2 } ,P({ E }_{ 2 })=\frac { 1 }{ 4 } ,P({ E }_{ 3 })=\frac { 1 }{ 4 } \)
Also \(P(A/{ E }_{ 1 })=1,P(A/{ E }_{ 2 })=\frac { 1 }{ 4 } ,P(A/{ E }_{ 3 })=\frac { 3 }{ 4 } \)
By Bayes' theorem
\(({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 2 } \times 1 }{ 1\times \frac { 1 }{ 2 } +\frac { 1 }{ 4 } \times \frac { 1 }{ 4 } +\frac { 1 }{ 4 } \times \frac { 3 }{ 4 } } \)
= \(\frac { \frac { 1 }{ 2 } }{ \frac { 1 }{ 2 } +\frac { 1 }{ 16 } +\frac { 3 }{ 16 } } =\frac { 1 }{ 2 } \times \frac { 16 }{ 8+1+3 } \)
= \(\frac { 8 }{ 12 } =\frac { 2 }{ 3 } \)
(I) If Arjun copies the answer, then there is violation of honesty in his character.
(II) If Arjun gusses the answer, it may give him negative marking for wrong answer
(III) Arjun should accept the question the way as it is and leave tha question unanswered because cheating spoils his character in the long run.

The given differential equation can be expressed as
\( \frac{d y}{d x} =\frac{x+2 y}{x-y} \) ... (1)
Let \(\mathrm{~F}(x, y) =\frac{x+2 y}{x-y} \)
Now \(\mathrm{~F}(\lambda x, \lambda y) =\frac{\lambda(x+2 y)}{\lambda(x-y)}=\lambda^0 \cdot f(x, y)\) 
Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation
Alternatively,
 \(\frac{d y}{d x}=\left(\frac{1+\frac{2 y}{x}}{1-\frac{y}{x}}\right)=g\left(\frac{y}{x}\right)\)  ... (2)
 R.H.S. of differential equation (2) is of the form \(g\left(\frac{y}{x}\right)\) and so it is a homogeneous function of degree zero. Therefore, equation (1) is a homogeneous differential equation. To solve it we make the substitution
 y = vx ... (3)
Differentiating equation (3) with respect to, x we get
\(\frac{d y}{d x}=v+x \frac{d v}{d x}\) 
Substituting the value of y and \(\frac{d y}{d x}\) in equation (1) we get
 \(v+x \frac{d v}{d x}=\frac{1+2 v}{1-v}\) 
or \(x \frac{d v}{d x}=\frac{1+2 v}{1-v}-v\) 
or \(x \frac{d v}{d x}=\frac{v^2+v+1}{1-v}\)
\(\frac{v-1}{v^2+v+1} d v=\frac{-d x}{x}\) 
Integrating both sides of equation (5), we get
\(\int \frac{v-1}{v^2+v+1} d v=-\int \frac{d x}{x}\)
\(\frac{1}{2} \int \frac{2 v+1-3}{v^2+v+1} d v=-\log |x|+\mathrm{C}_1\)
\( or \ \frac{1}{2} \int \frac{2 v+1}{v^2+v+1} d v-\frac{3}{2} \int \frac{1}{v^2+v+1} d v=-\log |x|+\mathrm{C}_1 \\ or\ \frac{1}{2} \log \left|v^2+v+1\right|-\frac{3}{2} \int \frac{1}{\left(v+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} d v=-\log |x|+\mathrm{C}_1 \)
\( \frac{1}{2} \log \left|v^2+v+1\right|-\frac{3}{2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 v+1}{\sqrt{3}}\right)=-\log |x|+\mathrm{C}_1 \\ \frac{1}{2} \log \left|v^2+v+1\right|+\frac{1}{2} \log x^2=\sqrt{3} \tan ^{-1}\left(\frac{2 v+1}{\sqrt{3}}\right)+\mathrm{C}_1 \)
Replacing v by \(\frac{y}{x}\) we get
\(\frac{1}{2} \log \left|\frac{y^2}{x^2}+\frac{y}{x}+1\right|+\frac{1}{2} \log x^2=\sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+\mathrm{C}_1\)
or \(\frac{1}{2} \log \left|\left(\frac{y^2}{x^2}+\frac{y}{x}+1\right) x^2\right|=\sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+\mathrm{C}_1\)
or \(\log \left|\left(y^2+x y+x^2\right)\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+2 \mathrm{C}_1\)
or \(\log \left|\left(x^2+x y+y^2\right)\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3} x}\right)+\mathrm{C}\)
which is the general solution of the differential equation (1)

\(={(2x+2)^{3/2}\over 3}-{(2x+1)^{3/2}\over3}+c\)

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

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.

\(\int { \frac { \sqrt { x^{ 2 }+1 } \left\{ log({ x }^{ 2 }+1)-2logx \right\} }{ x^{ 4 } } } dx\)
\(=\sqrt { 1+\frac { 1 }{ { x }^{ 2 } } } \left( log\left( 1+\frac { 1 }{ x^{ 2 } } \right) \right) \frac { 1 }{ x^{ 3 } } dx\)
Let, \(1+\frac { 1 }{ { x }^{ 2 } } ={ t }^{ 2 }\)
\(\Rightarrow \frac { -2 }{ { x }^{ 3 } } dx=2tdt\)
\(\Rightarrow \frac { 1 }{ { x }^{ 3 } } dx=-tdt\)
\(=-\int { t(2logt)t\quad dt+ } -2\int { logt.{ t }^{ 2 }dt } \)
\(=-2logt.\frac { { t }^{ 3 } }{ 3 } +\int { 2\frac { 1 }{ t } .\frac { t^{ 3 } }{ 3 } dt } \)
\(=-\frac { 2 }{ 3 } logt.{ t }^{ 3 }+\frac { 2 }{ 9 } { t }^{ 3 }+C\)
\(=\frac { 2 }{ 3 } \left( 1+\frac { 1 }{ { x }^{ 2 } } \right) ^{ 3 }\left( -log\left( 1+\frac { 1 }{ { x }^{ 2 } } \right) +\frac { 1 }{ 3 } \right) +C\)

LHS = \(\frac { 1 }{ abc } \left| \begin{matrix} { (a+b) }^{ 2 } & c & c \\ a & { (b+c) }^{ 2 } & a \\ b & b & { (c+a) }^{ 2 } \end{matrix} \right| \)
\({ C }_{ 1 }\rightarrow { C }_{ 1 }-{ C }_{ 3 },{ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 3 }\)
\(=\frac { 1 }{ abc } \left| \begin{matrix} (a+b+c)(a+b-c) & 0 & { c }^{ 2 } \\ 0 & (b+c+a)(b+c-a) & { a }^{ 2 } \\ (b+c+a)(b-c-a) & (b+c+a)(b-c-a) & { (c+a) }^{ 2 } \end{matrix} \right| \)
\(=\frac { { (a+b+c) }^{ 2 } }{ abc } \left| \begin{matrix} (a+b-c) & 0 & { c }^{ 2 } \\ 0 & (b+c-a) & { a }^{ 2 } \\ (-2a) & (-2c) & { 2ca } \end{matrix} \right| \)
\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }-{ R }_{ 2 }\)
\(=\frac { { (a+b+c) }^{ 2 } }{ abc } \left| \begin{matrix} (ac+bc-{ c }^{ 2 }) & 0 & { c }^{ 2 } \\ 0 & (b+c-a) & { a }^{ 2 } \\ (-2a) & (-2c) & { 2ca } \end{matrix} \right| \)
\(({ C }_{ 1 }\rightarrow { C }_{ 1 }+{ C }_{ 3 },{ C }_{ 2 }\rightarrow { C }_{ 2 }+{ C }_{ 3 })\)
\(=\frac { { (a+b+c) }^{ 2 } }{ abcca } \left| \begin{matrix} (ac+bc) & { c }^{ 2 } & { c }^{ 2 } \\ { a }^{ 2 } & (ba+ca) & { a }^{ 2 } \\ 0 & 0 & { 2ca } \end{matrix} \right| \)
\(=\frac { { (a+b+c) }^{ 2 }2{ c }^{ 2 }{ a }^{ 2 } }{ abcca } \left| \begin{matrix} (a+b) & c & c \\ a & (b+c) & a \\ 0 & 0 & { 1 } \end{matrix} \right| \)
\(=\frac { { (a+b+c) }^{ 2 }2 }{ b } (ab+ac+{ b }^{ 2 }+bc-ac)\)
\(=2{ (a+b+c) }^{ 3 }\)

Let ABCD be a rectangle inscribed in a given circle with centre at 0 and radius a.
Let AB = 2x and BC = 2y

Then, OA² = OM² + AM² 
\(\Rightarrow a^2=y^2+x^2\)
\(\Rightarrow y=\sqrt{a^2-x^2}\)
Let A be the area of the rectangle.
\(\therefore A=4xy=4x\sqrt{x^2-x^2}\)
\(\Rightarrow \ \ \frac{dA}{dx}=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}\)
For maximum or minimum value of A,
\(\frac{dA}{dx}=0\)
\(=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}=0\Rightarrow\ \ x=\frac{a}{\sqrt{2}}\)
Now, \(\frac{d^2A}{dx^2}=4\frac{d}{dx}\{(a^2-2x^2)(a^2-x^2)^{-1/2}\}\)
\(\Rightarrow \frac{d^2A}{dx^2}=4[-4x(a^2-x^2)^{-1/2}+(a^2-2x^2)\times(-1/2)(a^2-x^2)^(-3/2)(-2x)]\)
\(=[\frac{-4x}{\sqrt{a^2-x^2}}+\frac{x(a^2-2x^2)}{(a^2-x^2)^{3/2}}]\)
\(\therefore\ (\frac{d^2A}{dx^2})_{x=\frac{a}{\sqrt2}}=-16<0\)
Thus A is maximum when \(x=\frac{a}{\sqrt2}\)
putting \(x=\frac{a}{\sqrt2}\) in (i) \(y=\frac{a}{\sqrt2}\)
Therefore \(x=y=\frac{a}{\sqrt2}\)
Hence area is maximum when x = y ⇒ 2x = 2y
i.e., the rectangle is a square.

E₁ = Event that 1 occur;
E₂ = Event that 1 does not occur;
A = Event that the man reports that 1 occur.
\(P({ E }_{ 1 })=\frac { 1 }{ 6 } ;P({ E }_{ 2 })=\frac { 5 }{ 6 } ;\)
\(P(A/{ E }_{ 1 })=\frac { 3 }{ 5 } ;P(A/{ E }_{ 2 })=\frac { 2 }{ 5 } ;\)
\(P({ E }_{ 1 }/A)=\frac { \frac { 1 }{ 6 } .\frac { 3 }{ 5 } }{ \frac { 1 }{ 6 } .\frac { 3 }{ 5 } +\frac { 5 }{ 6 } .\frac { 2 }{ 5 } } =\frac { 3 }{ 13 } \)

\(\tan ^{ -1 }{ (x-1) } +\tan ^{ -1 }{ (x+1) } =\tan ^{ -1 }{ 3x } -\tan ^{ -1 }{ x } \)
\(\Rightarrow \tan ^{ -1 }{ \left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) } =\tan ^{ -1 }{ \left( \frac { 2x }{ 2-3{ x }^{ 2 } } \right) } \)
\(\Rightarrow \frac { 2x }{ 2-{ x }^{ 2 } } =\frac { 2x }{ 2-3{ x }^{ 2 } } \)
\(\Rightarrow 2x(1+3{ x }^{ 2 }-2+{ x }^{ 2 })=0\)
\(\Rightarrow x=0,\frac { 1 }{ 2 } ,-\frac { 1 }{ 2 } \)
Alternative method:
We have, 
\(\tan ^{ -1 }{ (x-1) } +\tan ^{ -1 }{ x } +\tan ^{ -1 }{ (x+1) } =\tan ^{ -1 }{ 3x } \)
\(\tan ^{ -1 }{ (x-1) } +\tan ^{ -1 }{ (x+1) } +\tan ^{ -1 }{ x } =\tan ^{ -1 }{ 3x } \)
Taking LHS
\(\tan ^{ -1 }{ \left( \frac { (x-1)+(x+1) }{ 1-(x-1)(x+1) } \right) } +\tan ^{ -1 }{ x } \)
\(= \tan ^{ -1 }{ \left( \frac { 2x }{ 1-{ x }^{ 2 }+1 } \right) } +\tan ^{ -1 }{ x } \)
\(=\tan ^{ -1 }{ \left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) +\tan ^{ -1 }{ x } } \)
\(=\tan ^{ -1 }{ \left[ \frac { \frac { 2x }{ 2-{ x }^{ 2 } } +x }{ 1-\frac { 2x }{ 2-{ x }^{ 2 } } \times x } \right] } \)
\(= \tan ^{ -1 }{ \left( \frac { 2x+2x-{ x }^{ 3 } }{ 2-{ x }^{ 2 }-2{ x }^{ 2 } } \right) } \)
\(\therefore \tan ^{ -1 }{ \left( \frac { 4x-{ x }^{ 3 } }{ 2-3{ x }^{ 2 } } \right) } =\tan ^{ -1 }{ 3x } \)
\(\Rightarrow \frac { 4x-{ x }^{ 3 } }{ 2-3{ x }^{ 2 } } =3x\)
\(\Rightarrow 6x-9{ x }^{ 3 }=4x-{ x }^{ 3 }\)
\(\Rightarrow 8{ x }^{ 2 }-2x=0\)
\(\Rightarrow x=0\quad or\quad 4{ x }^{ 2 }-1=0\)
\(\Rightarrow x=0\quad or\quad 4{ x }^{ 2 }=1\)
\(\Rightarrow x=0\quad or\quad x=\pm \frac { 1 }{ 2 } \)
\(\Rightarrow x=0,\quad x=\frac { 1 }{ 2 } \quad x=-\frac { 1 }{ 2 } \)
\(\Rightarrow x=0\quad or\quad { x }^{ 2 }=\frac { 1 }{ 4 } \)

Continuous in [0, 2\(\pi\)] as difference of two continuous functions is continuous
f'(x) = cos x - 2cos 2x, exists in (0, 2\(\pi\)), hence f is derivable;
\( f(0)=0, f(2 \pi)=0, \text { conditions satisfy } \)
\(\therefore \ f^{\prime}(c)=0 \Rightarrow \cos c-2 \cos 2 c=0 \)
\(\Rightarrow \cos c-2\left(2 \cos ^{2} c-1\right)=0 \)
\(\Rightarrow 4 \cos ^{2} c-\cos c-2=0 \Rightarrow \cos c=\frac{1 \pm \sqrt{1+32}}{8} \)
\(c={ cos }^{ -1 }\left\{ \frac { 1\pm \sqrt { 33 } }{ 8 } \right\} \)

As distance = \(\left| \frac { 5-\frac { 2 }{ 3 } }{ \sqrt { 4+1+4 } } \right| =\frac { 13 }{ 9 } \) units

T₁ and T₃ are similar as their sides are proportional.