Let a be any element and e be the identity element.
The a * e = e * a = a
a * e = a ⇒ a + e -1 = a ⇒ e-1 = 0 ⇒ e = 1
∴ The identity element is 1

Let I = \(\int _{ 1 }^{ 2 }{ \frac { 3x }{ { 9x }^{ 2 }-1 } dx } \) ⇒ IA³| = \(\left| I \right| \)
Put t = 9x² - 1 ⇒ dt = 18x dx
\(\frac{d t}{6}=3 x d x\)

∴ \(\int _{ 8 }^{ 35 }{ \frac { dt }{ 6t } } \)
= \(\frac { 1 }{ 6 } { \left[ log \ t \right] }_{ 8 }^{ 35 }\)
= \(\frac { 1 }{ 6 } [log35-log8]\)
= \(\frac { 1 }{ 6 } \left[ log\left( \frac { 35 }{ 8 } \right) \right] \)

f(x) -|x+3| =  ∀ \(x\in R\).
Now, |x+3| ≥ 0  ∀ \(x\in R\).
⇒ f(x) ≥ 0  ∀ \(x\in R\).
So, the minimum value of f(x) is 0
Also, f(x) = |x + 3| does not have the maximum value.

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 1 \\ 1 & 1 \\ 0 & 1 \end{matrix} \right) \wedge \left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) =\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)

In this problem a ∗ b is in the quotient form. Since the division by 0 is undefined, the denominator b -1 must be nonzero.
It is clear that b −1 = 0 if b = 1. As 1∈Q, ∗ is not a binary operation on the whole of Q. However it can be found that by omitting 1 from Q, the output a ∗b exists in Q\{1}. Hence ∗ is a binary operation on Q\{1}.

Let f(x) = e^-|-x| = e^-|x| = f(x)
So f (x) is an even function.
Hence, \(\int ^{log 2}_{-log 2} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-x}\) dx
= 2(-e^-x)\(^{log2}_{0}\) = 2 (-e^-log2 + e⁰) = 2 \((-e ^{log \frac{1}{2}} + 1)\)
= 2\((-\frac {1}{2}+1)=1\).

Given y = a cos bx  ...(1)
Differentiating equation (1) w.r.t 'x', we get 
\(\frac{d y}{d x}=\mathrm{a}(-\sin \mathrm{b} x) \mathrm{b}=-\mathrm{ab} \sin \mathrm{b} x\)
Again differentiating, we get
\(\frac{d^2 y}{d x^2}  =-\mathrm{ab} \cos \mathrm{b} x \cdot \mathrm{b}   \)
\(\frac{d^2 y}{d x^2}  =-\mathrm{ab}^2 \cos \mathrm{b} x=-\mathrm{b}^2(\mathrm{a} \cos \mathrm{b} x)  \)
\(\frac{d^2 y}{d x^2}  =-\mathrm{b}^2 \mathrm{y}  \)
\(\frac{d^2 y}{d x^2}+\mathrm{b}^2 \mathrm{y}  =0\)
 Therefore, y = a cos bx is a solution of the differential equation \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ b }^{ 2 }y=0\)

Z = \(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 107 }+\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 107 }\)
\(\bar { z } =\left( \overline { \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } } \right) ^{ 107 }+\left( \overline { \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } } \right) ^{ 107 }\)
= \(\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 107 }+\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 107 }\) = z
Since z = \(\bar { z } \), Im(z) = 0

Given equation is x² + 3y² = a²
\(\div \) a² we get, \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ \frac { { a }^{ 2 } }{ 3 } } =1\)
Here a² and b2 = \(\frac{a^2}{3}\) ⇒ b = \(\frac{a}{\sqrt3}\)
Length of major axis is 2a and
Length of minor axis is 2b = \(\frac { 2a }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 2a\sqrt { 3 } }{ 3 } \)

Let \({ cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) =\theta \Rightarrow \frac { 3 }{ 5 } =cos\theta \)
\(\therefore sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 25-9 }{ 25 } } \)
= \(\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)

Agumented matrix
[A|B] \(\left[ \begin{matrix} 2 & 5 \\ 6 & 15 \end{matrix}|\begin{matrix} 7 \\ 13 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & 5 \\ 0 & 0 \end{matrix}|\begin{matrix} 7 \\ -8 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 and \(\rho\)([A|B]) = 3
∴ \(\rho\) (a) ≠ \(\rho\) ([AIB])
Hence the system is inconsistent.

The parametric form of vector equation of a line passing through a point \(\left( \overset { \rightarrow }{ a } \right) \) and parallel to \(\overset { \rightarrow }{ b } \) is
\({ \overset { \rightarrow }{ r } }=\overset { \rightarrow }{ a } +t\overset { \rightarrow }{ b } \), t ∈ R
⇒ Here \(\overset { \rightarrow }{ a } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ b } =2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \)
\(\therefore { \overset { \rightarrow }{ r } }=\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) +t\left( 2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \right) \), t ∈ R which is the required equation of a line.

Sum of the roots = sin ∝ + cos ∝ = \(\frac{-b}{a}\)
Product of the roots = sin ∝ cos ∝ = \(\frac{c}{a}\)
Now 1 = cos²∝ + sin² ∝
= (sin ∝ +cos ∝)² - 2 sin ∝ cos ∝
\(1=\frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } \Rightarrow 1=\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \)
⇒ a² = b² - 2ac ⇒ a² + 2ac = b²
Adding c² both sides, a² +2ac+c² = b²+c²
⇒ (a+c)² = b² + c²

Equation of the circle is x² + y² - x + 2y - 3 = 0
Here 2g = -1 ⇒ g = \(\frac { -1 }{ 2 } \)
2f = 2 ⇒ f = 1 and c = -3
Centre is (-g, -f) = \(\left( \frac { 1 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
= \(\sqrt { \frac { 1 }{ 4 } +1+3 } \)
= \(\sqrt { \frac { 1 }{ 4 } +4 } =\sqrt { \frac { 1+16 }{ 2 } } \)
r = \(\sqrt { \frac { 17 }{ 2 } } \) units.

We know that sin^-1: [-1, 1] \(\rightarrow \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)is given by
sin⁻¹x = y if and only if x = sin y for −1\(\le x\le \) and -\(\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \). Thus
\({ sin }^{ -1 }\left( sin\left( \frac { 5\pi }{ 6 } \right) \right) \)= \({ sin }^{ -1 }\left( sin\left( \frac { \pi }{ 6 } \right) \right) \) = \({ sin }^{ -1 }\left( sin\frac { \pi }{ 6 } \right) \), since \(\frac{\pi}{6}\)\(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)

\({ i }^{ 59 }+\frac { 1 }{ { i }^{ 59 } } \)
i^{4 \(\times\) 14 + 3} + i^{-(4 \(\times\) 14 + 3)}
= (i⁴)¹⁴.i³ + (i⁴)^-14.i^-3
= 1.i³+1.i^-3  [∵ i⁴ = 1]
= -i + i  [∴ i³ = -i and i^-3= i]
= 0

Let \(\vec { a } =-6\hat { i } +14\hat { j } +10\hat { k } \), \(\vec { b } =14\hat { i } -10\hat { j } -6\hat { k } \) and \(\vec { c } =2\hat { i } +4\hat { j } -2\hat { k } \)
Volume of the parallelepiped having \(\vec { a } ,\vec { b } \) and \(\vec { c } \) as its co-terminus edges is \(\vec { a } .(\vec { b } \times \vec { c } )\).
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{matrix} \right| \)
= \(-6\left| \begin{matrix} -10 & -6 \\ 4 & -2 \end{matrix} \right| -14\left| \begin{matrix} 14 & -6 \\ 2 & -2 \end{matrix} \right| +10\left| \begin{matrix} 14 & -10 \\ 2 & 4 \end{matrix} \right| \)
= -6(20 + 24) - 14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760 = 720.
∴ Volume of the required parallelepiped = 720 cubic units.

Let sec^-1\(\left( -\frac { 2\sqrt { 3 } }{ 3 } \right) =\theta \).
Then, sec\(\theta\) = \(-\frac{2}{\sqrt3}\) where \(\theta\in[0,\pi]\)\{\(\frac{\pi}{2}\)}.
Thus, cos \(\theta =-\frac{\sqrt{3}}{2}\).
Now, \(cos\frac { 5\pi }{ 6 } =cos\left( \pi -\frac { \pi }{ 6 } \right) =-cos\left( \frac { \pi }{ 6 } \right) =-\frac { \sqrt { 3 } }{ 2 } .\) 
Hence, Sec-¹ \(\left( -\frac { 2\sqrt { 3 } }{ 3 } \right) =\frac { 5\pi }{ 6 } \)

We have z = (2+3i)(1−i) = (2+3)+(3−2)i = 5+i
\(\Rightarrow\) \({ z }^{ -1 }=\frac { 1 }{ z } =\frac { 1 }{ 5+i } \)
Multiplying the numerator and denominator by the conjugate of the denominator, we get
\({ z }^{ -1 }=\frac { \left( 5-i \right) }{ \left( 5+i \right) \left( 5-i \right) } =\frac { 5-i }{ { 5 }^{ 2 }+{ I }^{ 2 } } =\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)
\(\Rightarrow\)\({ z }^{ -1 }=\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)

Let A be a non-singular matrix of order 2m+1, where m = 0, 1, 2,... Then, we get |A| ≠ 0 and, by property (ii), we have |adj A| = |A|^(2m+1) − 1 = |A|^2m.
Since |A|^2m is always positive, we get that |adj A| is positive.

Let X be the random variable of number of tails when three coins tossed.
S = {HHH, HHT, THH, HTH, HTT, THT, TTH,TTT}
n(S) = 8
Let X denote the number of tarits occured.
X^-1 (0) {HHH} = 1
X^-1 (1) = {HHT, THT, HTH} = 3
X^-1 (2) =, {HTT, THT, TTH} = 3
X^-1 (3) = {TTT} = 1
ஃ X takes the values 0, 1, 2, 3.

Let p(x) = x⁹ -5x² - 14x⁷ = 0
p(x) has only one sign change
Also p(-x) = (-x)⁹ - 5 (-x)⁸ - 14(-x)⁷ = 0
⇒ p(-x) = -x⁹ -5x⁸ + 14x⁷ = 0
p(-x) has only one sign change
∴ p(-x) has at most one positive and one negative root.

Let A = \(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 0 & 5 \end{matrix} \right] \). Applying elementary row operations, we get
A \(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -3 & -2 \\ 0 & -6 & -4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -3 & -2 \\ 0 & 0 & 0 \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has two non-zero rows. So, ρ(A) = 2.

Since the open box has square area, let the length, breadth and height of the box be I, l and b cm respectively.
ஃ Surface area = l² + 41b = 108
\(\Rightarrow l+4b=\frac { 108 }{ l } \) 
\(\Rightarrow 4b=\frac { 108 }{ l } -1\) 
\(\Rightarrow b=\frac { 108 }{ 4l } -\frac { l }{ 4 } \) 
\(\Rightarrow b=\frac { 27 }{ l } -\frac { l }{ 4 } \) 
Let f(1) = Volume of the box = 1 \(\times\) 1\(\times\) b = l² b
=  \({ l }^{ 2 }\left( \frac { 27 }{ l } -\frac { l }{ 4 } \right) \)
\(f(1)=27l-\frac { { l }^{ 3 } }{ 4 } \) 
\(f'(l)=27-\frac { { 3l }^{ 2 } }{ 4 } \) 
f'(1) = 0
\(\Rightarrow 27-\frac { 3{ l }^{ 2 } }{ 4 } =0\) 
\(\Rightarrow \frac { 3l^{ 2 } }{ 4 } =27\)
\(\Rightarrow l^{2}=\not 27 \times \frac{4}{\not 3}=36\) 
\(\Rightarrow 1=\pm 6\) 
\(\Rightarrow 1=6\) 
ஃThe critical number is 6
\(f''(l)=\frac { 6l }{ 4 } =\frac { 3l }{ 2 } \) 
\(\therefore f''(6)=-\frac { 3(6) }{ 2 } <0\)
ஃ f"(1) is maximum when l = 6
When l = 6,\(b=\frac { 27 }{ 6 } -\frac { 6 }{ 4 } \) 
= \(\frac { 9 }{ 2 } -\frac { 3 }{ 2 } =\frac { 6 }{ 2 } =3cm\)
Hence the dimensions of the required box are 6 cm, 6 cm and 3 cm respectively.

2x + 3y − z = 0, x − y − 2z = 0, 3x + y + 3z = 0
Reducing the augmented matrix to row - echelon form we get
[A|0]=\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & -1 & -2 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 2 & 3 & -1 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 4 & 9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 4 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 0 & \frac { 33 }{ 5 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|0] = 3
So, \(\rho \)(A) = \(\rho \)(A|0]) = 3 = Number of unknowns Hence, the system is consistent with unique solutions.
Thus, the system has trivial solution only.
x = 0, y = 0, z = 0