Let (S, *) be an algebraic structure and a ∈ S.  Assume that the inverse of a exists in S. It is to be proved that the inverse of a is unique. The existence of inverse in S ensures the existence of the identity element e in S. 
Let a ∈ S. It is to be proved that the inverse a (if exists) is unique.
Suppose that a has two inverses, say a₁, a₂
Treating a₁ as an inverse of a gives \(a * a_{1}=a_{1} * a=e\)  ........(1)
Next treating a₂ as the inverse of a gives \(a * a_{2}=a_{2} * a=e\)  ........(2)
\(a_{1}=a_{1} * e=a_{1} *\left(a * a_{2}\right)=\left(a_{1} * a\right) * a_{2}=e * a_{2}=a_{2}(\text { by }(1) \text { and }(2))\)
So, a₁= a₂. Hence the inverse of a is unique which completes the proof.

Let (S, *) be an algebraic structure. Assume that the identity element of S exists in S .
It is to be proved that the identity element is unique. Suppose that e₁ and e₂ be any two identity elements of S .
First treat e₁ as the identity and e₂ as an arbitrary element of S.
Then by the existence of identity property \(e_{2} * e_{1}=e_{1} * e_{2}=e_{2}\)  ..........(1)
Interchanging the role of e₁ and e₂  \(e_{2}, e_{1} * e_{2}=e_{2} * e_{1}=e_{1}\)  ..........(2)
From (1) and (2), e₁ = e₂. Hence the identity element is unique which completes the proof.

Given u (x, y) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ \sqrt { x+y } } \)
\(u({ \lambda }x,{ \lambda }y)=\frac { { \lambda }^{ 2 }{ x }^{ 2 }+{ { \lambda } }^{ 2 }{ y }^{ 2 } }{ \sqrt { { \lambda }x+{ \lambda }y } } \)
= \(\frac { { { \lambda } }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) }{ \sqrt { { \lambda } } (\sqrt { x+y } ) } \)
= \({ { \lambda } }^{ 2-\frac { 1 }{ 2 } }u(x,y)\)
= \({ { \lambda } }^{ \frac { 3 }{ 2 } }u(x,y)\)
∴ u (x, y) is a homogeneous function of degree \(\frac32\)
∴ By Euler's theorem,
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } \) = n.u ≍ \(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 3 }{ 2 } u\)
Hence, proved.

(i) Given that five fair coins are tossed once. Since the coins are fair coins the probability of getting an head in a single coin is
 \(p=\frac { 1 }{ 2 } \) and \(q=1-p=\frac { 1 }{ 2 } \) 
Let X denote the number of heads that appear in five coins. X is binomial random variable that takes on the values 0, 1, 2,  3, 4 and 5 and \(p=\frac { 1 }{ 2 } \) That is \(X\sim B\left( 5,\cfrac { 1 }{ 2 } \right) \) 
Therefore the binomial distribution is
\(f(x)=\left( \begin{matrix} n \\ x \end{matrix} \right) p*\left( 1-p \right) ^{ n-x }\), x = 0, 1, 2,..,n
becomes
\(f(x)=\left( \begin{matrix} 5 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 2 } \right) ^{ x }\left( \cfrac { 1 }{ 2 } \right) ^{ n-x }\), x = 0, 1, 2,..,5
That is 
\(f(x)=\left( \begin{matrix} 5 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 2 } \right) ^{ n }\), x = 0, 1, 2...,n
(ii) A fair die is rolled ten times and X denotes the number of times 4 appeared. X is binomial
random variable that takes on the values 0, 1, 2, 3,...10 , with n = 10 and \(p=\cfrac { 1 }{ 6 } \). That is \(X\sim B\left( 10,\cfrac { 1 }{ 6 } \right) \) 
Probability of getting a four in a die is \(p=\frac { 1 }{ 2 } \) and \(q=1-p=\frac { 5 }{ 6 } \)
Therefore the binomial distribution is 
\(f(x)=\left( \begin{matrix} 10 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 6 } \right) ^{ x }\left( \cfrac { 5 }{ 6 } \right) ^{ 10-x }\) x = 0, 1, 2,...,10

n = 5, X B{n, p)
P(X = 1) = 0.4096
P(X = 2) 0.2048
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, .., n
ஃnC₁,p¹q₄ 0.4096
⁵C₁, p²q⁴ = 0.4096
⁵C² p²q³ = 0.2048
5pq⁴ = 0.4096  .....(1)
10 p² q³ = 0.2048  ....(2) 
Dividing (2) by (1) we get 
\(\cfrac { 5{ pq }^{ 4 } }{ 10{ p }^{ 2 }{ q }^{ 3 } } =2\) 
q = 4p
q = 4(1- q)
q = 4 - 4q
5q = 4
q = 4/5
\(p=1-q= p=\frac { 1 }{ 5 } \) 
\(Mean=np=5\times \frac { 1 }{ 5 } =1\)
\( Variance =n p q=\not 5 \times \frac{1}{\not 5} \times \frac{4}{5}=\frac{4}{5}\) 
Distribution
(i) \(P(X=x)= ^5C_{ x }\left( \frac { 1 }{ 5 } \right) ^{ x }\left( \frac { 4 }{ 5 } \right) ^{ 5-x }\) , x = 0,1,2..n

Given E(X + 3) = 10
E(aX + b) = aE(X) + b
⇒ E(X) + 3 = 10
E(X) + 3 = 10
⇒ E(X) = 7
⇒μ = 7 ...(1)
E(X + 3)² = 116
E(X² + 6x + 9) 116
E(X²) + 6E(X) + 9 = 116 [ஃ E(9) = 9]
E(X²) + 6(7) + 9 = 116
E(X²) + 116 - 42 - 9 116 - 51
E(X²) = 65 ...(2)
Var(X) = E(X²) - [E(X)²]
65 - 7² = 65 - 49 = 16
ஃμ = 7 and σ² = 16.

By (14), Linear approximation is given by
L (x, y, z) =  U(x₀, y₀, z₀) + \({ \frac { { \partial }U }{ { \partial x } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (x-x₀)+\({ \frac { { \partial }U }{ { \partial y } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (y-y₀)+\({ \frac { { \partial }U }{ { \partial z } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (z-z₀)
Now U_x = 2x -y, Uy = -xand U_z = 3cos z.
Here (x₀, y₀, z₀) = (2,−1,0 )
hence U_x (2, −1,0) = 5, U_y (2, −1,0) = −2 and U_z (2,-1,0) = 3.
Thus L(x, y, z) = 6 + 5(x − 2) − 2( y +1) + 3(z − 0) = 5x − 2y + 3z − 6 is the required linear approximation for U at (2,−1,0).

log₁₀e = 0.4343 to find log₁₀g 1003
f(1000) = log₁₀1000 = log₁₀10³ = 3log₁₀10³ = 3 log₁₀10
= 3(1) = 3
f'(x) = \(\frac1x\). log10^e
f'(1000) = \(\frac{1}{1000}\)(0.4343)
∴ L(x) = f(x₀) f'(x₀) (x - x₀)
= 3 + \(\frac{1}{1000}\) (0.4343) (3)
= 3 + \(\frac{1.3029}{1000}\)
= 3 + 0.0013029
log₁₀1003 = 3.0013029

We know that volume of a sphere is given by V = \(\frac43\) π r³. where r > 0 is the radius. So the differential dV = 4 ㅠr² dr and hence
Δ ≈ dv = 4π(10)² (9.9-10) cm³
= 4π10² (-0.1) cm³
= −40π cm³ 
Note that we have used dr = (9.9 −10) cm, because radius decreases from 10 to 9.9. Again the negative sign in the answer indicates that the volume of the sphere decreases about 40π cm³.

Let x be the number
Let y = f(x) = \(x^\frac{1}{n}\)
Then log y = \(\frac1n\) log x
Taking differential on both sides we get,
\(\frac { 1 }{ y } dy=\frac { 1 }{ n } \times \frac { 1 }{ x } dx\)
i.e. \(\frac { \Delta y }{ y } \simeq \frac { dy }{ y } =\frac { 1 }{ n } .\frac { dx }{ x } \)
\(\therefore \frac { \Delta y }{ y } \times 100\simeq \frac { 1 }{ n } \left( \frac { dx }{ x } \times 100 \right) \)
\(\simeq \frac { 1 }{ n } \) times the percentage error in the number. Hence, percentage error in the nth root of a number is approximately \(\frac1n\) times the percentage error in the number

Given absolute error = 2%
⇒ \(\frac { dl }{ l } =2 \% =\frac { 2 }{ 100 } =0.02\)
Given T = 2ㅠ\(\sqrt { \frac { 1 }{ g } } \)
Taking logarithm on both sides,
log T = log 2ㅠ + \(\frac12\) log l - \(\frac12\) log g
Taking differential on both sides we get,
\(\frac{1}{T}dT=0+\frac{1}{2}.\frac{1}{l}.dl\)
\(\frac { \Delta T }{ T } =\frac { 1 }{ 2 } (.02)\)
\(\frac { \Delta T }{ T } =0\)
ஃ Percentage error = \(\frac { \Delta T }{ T } \times100=.01\times100=1 \%\)

Given \(f(x)=\begin{cases} \begin{matrix} kxe^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & for\quad x\le 0 \end{matrix} \end{cases}\)
Since the given function is a probability density function 
\(\int _{ -\infty }^{ \infty }{ f(x)dx } \) = 1 
\(\Rightarrow k\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx=1 } \)
\(\Rightarrow k \frac { 1! }{ \left( 2 \right) ^{ 2 } } =1\) 
[\(\int _{ 0 }^{ \infty }{ { x }^{ n }e^{ -ax } } =\frac { n! }{ { a }^{ +1 } } \), Here a = 2, n = 1]
\(\Rightarrow \frac { k }{ 4 } =1\\ \Rightarrow k=4\)

Volume of sphere = \(\frac43\)πr²
Given r = 10 cm 
\(\frac{dr}{dt}\) = - 0.2
V = \(\frac43\)πr³
Change in Volume
= \(\frac{4}{\not 3} \pi . \not 3 r^{2} \frac{d r}{d t}\)
= 4π(10)² (-0.2)
= 400 π (-0.2) = -80 πcm³
∴ Volume decreases by 80 π cm³
Surface area of sphere = 4πr²
Change 10 surrace area = 4 π2r\(\frac{dr}{dt}\)
= 8π(10) (-0.2)
= -\(\frac{80π\times2}{10}\) = -16 π cm²
∴ Surface area decreases by 16 π cm²

Let f(x) = \(f(x)={ x }^{ \frac { 2 }{ 3 } },{ x }_{ 0 }=125,\triangle x=-2\)
∴ (123)^\(\frac23\) = f(125) +1'(125) (-2) ... (1)
\(f(125)={ (125) }^{ \frac { 2 }{ 3 } }={ { (5 }^{ 3 }) }^{ \frac { 2 }{ 3 } }\) = 5² = 25
\({ f }^{ ' }(x)={ \frac { 2 }{ 3 } x }^{ \frac { 2 }{ 3 } -1 }={ \frac { 2 }{ 3 } x }^{ \frac { 1 }{ 3 } }=\frac { 2 }{ { 3x }^{ \frac { 1 }{ 3 } } } \)
\({ f }^{ ' }(125)=\frac { 2 }{ { 3(125)x }^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ { 3{ (5 }^{ 3 } })^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ 3(5) } =\frac { 2 }{ 15 } \)
∴ \({ (123) }^{ \frac { 2 }{ 3 } }=25+\frac { 2 }{ 15 } (-2)\)
\(=25-\frac { 4 }{ 15 } =25-0.27\)
\({ (123) }^{ \frac { 2 }{ 3 } }=24.73\)

Let X be the random variable denotes the Winning amount.
X (Both are black balls) = Rs. 2 (-10) = Rs. -20
X (one red and oneblack ball) = Rs.15-Rs. 10 = Rs. 5
X (both are red ball) = Rs. 2 (15) = Rs. 30
= {-20, 5, 30}
The sample space consists of ¹⁴C₂ = 91
X = -20, Both are black balls= ⁸C₁ = 28
X = 5, One black, one redball = ⁸C₁ x ⁶C₁ = 8 x 6 = 48
X = 30, Both are white balls = ⁶C₁ = 15

Given a*b = a + b + ab -7, ∀ a,b ∈R
If a ∈R, b∈R then ab ∈ R
(a*b) = a +b+ ab - 7 ∈R
For example, let 1, 2 ∈ R
(1*2) = 1+2+(1)(2)-7
= 2 ∈ R
* a binary operation on R
\(3 *\left(\frac{-7}{15}\right)=3-\frac{7}{15}+\gamma^1\left(\frac{-7}{15_5}\right)-7\)
[Here a = 3, b = \(\frac{-7}{15}\)]
\(=3-\frac { 7 }{ 15 } -\frac { 21 }{ 15 } -7\)
\(\therefore 3*\left( \frac { -7 }{ 15 } \right) =\frac { -88 }{ 15 }\)

The entries in the columns corresponding to p → q and ㄱp ν q are identical and hence they are equivalent.

\(h(x,y)=\frac { 6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y }{ 2020{ x }^{ 2 }+2019{ y }^{ 2 } } \)
Given \(h(x,y)=\frac { 6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y }{ 2020{ x }^{ 2 }+2019{ y }^{ 2 } } \)
\(h(\lambda x,\lambda y)=\frac { 6{ \lambda }^{ 2 }{ x }^{ 2 }{ \lambda }^{ 3 }{ y }^{ 3 }-\pi { \lambda }^{ 5 }{ y }^{ 5 }+9{ \lambda }^{ 4 }{ x }^{ 4 }\lambda y }{ 2020{ \lambda }^{ 2 }{ x }^{ 2 }+2019{ \lambda }^{ 2 }{ y }^{ 2 } } \)
\(=\frac { { \lambda }^{ 5 }(6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y) }{ { \lambda }^{ 2 }(2020{ x }^{ 2 }+2019{ y }^{ 2 }) } \)
\( =\lambda^{3} \mathrm{~h}(\mathrm{x}, \mathrm{y}) \)
Thus f is homogeneous with degree 3.

\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 2 } e^{ -\frac { x }{ 2 } } & for\quad x>0 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\)
\(\int _{ 0 }^{ \infty }{ x.f(x)dx } =\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ x.{ e }^{ \frac { -x }{ 2 } } } dx\) 
\(\left[ \int _{ 0 }^{ \infty }{ { e }^{ -ax }.{ x }^{ n }dx=\cfrac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(\frac { 1 }{ 2 } \times \frac { 1! }{ \left( \frac { 1 }{ 2 } \right) ^{ 2 } } =\frac { 1 }{ 2 } \times \frac { 1 }{ \frac { 1 }{ 4 } } \) 
= \(\frac { 1 }{ 2 } \times \frac { 4 }{ 1 } =2\) 
\(E({ X }^{ 2 })=\int _{ 0 }^{ \infty }{ { x }^{ 2 }.f(x) } dx\) 
= \(\int _{ 0 }^{ \infty }{ { x }^{ 2 }.\frac { 1 }{ 2 } { e }^{ -\frac { x }{ 2 } } } dx\) 
= \(\frac { 1 }{ 2 } \int { { x }^{ 2 }.{ e }^{ -\frac { x }{ 2 } }dx } \) 
= \(\frac { 1 }{ 2 } \times \frac { 2! }{ \left( \frac { 1 }{ 3 } \right) ^{ 3 } } =\frac { 1 }{ 2 } \times \frac { 2 }{ \frac { 1 }{ 8 } } \) 
= \(\frac { 1 }{ 2 } \times 2\times 8=8\) 
ஃVar(X)=E(X²) - [E(x)]²
= 8-2²
= 8 - 4 = 4

Volume of sphere = \(\frac43\) πr³
Given r = 5 mm
⇒ dr = (5.3 - 5) = 0.3 mm
\(\text { Approximate volume }=\frac{4}{\not 3} \pi \cdot \not 3 r^{2} d r\)
= 4π (5²) (0.3)
= 100 π (0.3)
= 30π mm³

x = 2 and dx = 0.1
Taking differentials,
df = (2x + 3) dx
Whenx = 2, dx = 0.1
df = (2(2) + 3)(0.1) = 7(0.1) = 0.7

Given y = \(​​y=\frac { { \left( 1-2x \right) }^{ 3 } }{ 3-4x } \)
Taking differentials
\(dy=\frac { (3-4x)[3{ (1-2x) }^{ 2 }(-2)]-({ 1-2x) }^{ 3 }(-4) }{ { (3-4x) }^{ 2 } } dx\)
= \(\frac { { 2(1-2x) }^{ 2 }[-3(3-4x)+2(1-2x)] }{ (3-{ 4x) }^{ 2 } } dx\)
= \(\frac { { 2(1-2x) }^{ 2 }[-9+12x+2-4x] }{ { (3-4x) }^{ 2 } } dx\)
\(dy=\frac { { 2(1-2x) }^{ 2 }[8x-7] }{ { (3-4x) }^{ 2 } } dx\)