(i) Commutativity. for  \(a,b\in R,\quad (a,b)*(c,d)\)
\(=(a+c,b+d)\)
\(=(c+a,d+b)=(c,d)*(a,b)\)
\(\left[ \because \ a+c=c+a\quad in\ R;etc. \right] \)
Thus '*' is commutative.
Associativity. For \(a,b,c\in R,\)
\(\left[ (a,b)*(c,d) \right] *(e,f)=(a+c,b+d*(e,f)\)
\(=(a+c+e,b+d+f)\)
\((a*b)*\left[ (c,d)*(e,f) \right] =(a*b)*(c+e,(d+f)\)
\(=(a+c+e,b+d+f)\)
\(\therefore \quad [(a*b)*(c,d)*(e,f)=(a*b)*[(c,d)*(e,f)]\)
Thus '*' is associative.
(II) (I) Let (x,y) be the identity element.
Then  \(a,b)*(x,y)=(a,b)=(x,y)*(a,b)\)
\(\Rightarrow \) \((a+x,b+y)=(a,b)=(x+a,y+b)\)
\(\Rightarrow \)  \(x=0,y=0\)
Hence, (0,0) is the identity element.
(II) Let (x, y) be the inverse of \((a,b)\in A\)
Then (a,b)*(x, y) = (0, 0) = (x, y)*(a,b)
\(\Rightarrow \) (a+x, b+y) = (0, 0) = (x+a, y+b)
\(\Rightarrow \) a+x = x+a =0 and b+y = y+b = 0
\(\Rightarrow \) x = -a,y = -b
Hence, (-a, -b) is the inverse of (a, b)\(\in A\).

We have: \(f:N\rightarrow S\) , \(f(x)=9x^{ 2 }+6x-5\).
Now \(f(x_{ 1 })=f(x_{ 2 })\Rightarrow 9x_{ 1 }^{ 2 }+6x_{ 1 }-5=9x_{ 2 }^{ 2 }+6x_{ 2 }-5\)
\(\Rightarrow 9(x_{ 1 }-x_{ 2 })+6(x_{ 1 }-x_{ 2 })=0\)
\(\Rightarrow (x_{ 1 }-x_{ 2 })\left[ 9x_{ 1 }+9x_{ 2 }+6 \right] \)
\(\Rightarrow x_{ 1 }=x_{ 2 }\quad \left[ \because x_{ 1 },x_{ 2 }\in N \right] \)
\(\Rightarrow f\ \ is\quad one-one.\)
Let \(y\ \in \ S\) be an arbitrary number.
Now \(y=f(x)\Rightarrow y=9x^{ 2 }+6x-5\Rightarrow y=(3x+1)^{ 2 }-6\)
\(\Rightarrow \sqrt { y+6 } =3x+1\Rightarrow x=\sqrt { \frac { y+6-1 }{ 3 } } \in N\)
\(\Rightarrow \ x=f^{ -1 }(y)\)
Since f is one-one and onto \(\Rightarrow \) f is invertible.
Also \(f^{ -1 }(y)=\sqrt { \frac { y+6-1 }{ 3 } } \)
\(\Rightarrow \)   \(f^{ -1 }(x)=\sqrt { \frac { x+6-1 }{ 3 } } \)
\(\therefore \)    \(f^{ -1 }(43)=\frac { \sqrt { 49 } -1 }{ 3 } =\frac { 7-1 }{ 3 } =2\)
and   \(f^{ -1 }=\frac { \sqrt { 169 } -1 }{ 3 } =\frac { 13-1 }{ 3 } =4\).

We have: (a, b)R(c, d) if ad(b+c)(a+d).
(I) \((a,b)R(a,b)\Rightarrow ab(b+a)=ba(a+b)\)
\(\Rightarrow ab(a+b)=ab(a+b)\), which is true.
Thus R is reflexive.
(II) \((a,b)R(c,d)\Rightarrow ad(b+c)=bc(a+d)\)
\(\Rightarrow bc(a+d)=ad(b+c)\)
\(\Rightarrow cb(d+a)=da(c+b)\)
\(\left[ \because \quad bc=cb\quad and\quad a+d=d+a;\quad etc.\quad \forall \quad a,b,c,d\in N \right] \)
\(\Rightarrow \) (c,d) R (a,b).
Thus R is symmetric.
(III) Let (a,b) R (c,d) and (c,d) R (e,f).
\(\therefore \) ad(b+c) = bc(a+d) and cf(d+e) = de(c+f)
\(\Rightarrow \) \(\frac { b+c }{ bc } =\frac { a+d }{ ad } \)  and \(\frac { d+e }{ de } =\frac { c+f }{ cf } \)
\(\Rightarrow \) \(\frac { 1 }{ c } +\frac { 1 }{ b } =\frac { 1 }{ d } +\frac { 1 }{ a } \) and \(\frac { 1 }{ e } +\frac { 1 }{ d } =\frac { 1 }{ f } +\frac { 1 }{ c } \)
\(\Rightarrow \) \(\frac { 1 }{ c } -\frac { 1 }{ d } =\frac { 1 }{ a } -\frac { 1 }{ b } \) and \(\frac { 1 }{ c } +\frac { 1 }{ d } =\frac { 1 }{ e } +\frac { 1 }{ f } \)
\(\Rightarrow \) \(\frac { 1 }{ a } -\frac { 1 }{ b } =\frac { 1 }{ e } -\frac { 1 }{ f } \) \(\Rightarrow \) \(\frac { 1 }{ a } +\frac { 1 }{ f } =\frac { 1 }{ b } +\frac { 1 }{ e } \)
\(\Rightarrow \) be (a+f) = af(b+e) \(\Rightarrow \)af(b+e)=be(a+f)
\(\Rightarrow \) (a,b) R (e,f).
Thus R is transitive.
Hence, R is an equivalence relation.

Let 'x' hectares land be allocated to crop X and 'y' hectares to crop Y.
We have: \(x\ge 0\) .....(1)
\(y\ge 0\) .....(2)
\(x+y\le 50\) .....(3)
\(20x+10y\le 800\)
\(\therefore \) i.e.  \(2x+y\le 80\)....(4)
The mathematical formulation of the problem is as be low:
Maximize Z = 10500x + 9000y subject to the constraints (1)-(4).
For solution set, we draw the lines:
x = 0, y = 0, x + y = 50 and 2x + y = 80.
The lines x + y = 50 and 2x + y = 80 meet at E (30, 20).

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

Hence, the maximum profits is Rs. 4,95,000 by allocating 30 hectares for crop X and 20 hectares for crop Y.

Let 'x' kg of food I and 'y' kg of food II be mixed. We have the table:

Thus LLP problem is as below:
Minimize:  Z=50x+70y ....(1)
Subject to:\(2x+y\ge 8\)   ...(2)
\(x+2y\ge 10\)  ...(3)
and  \(x,y\ge 0\) ....(4)

First of all, we locate the region represented by (2)-(4).
The shaded region, as shown above, is feasible region.
Applying Corner Point Method, we have:

Hence, the minimum cost = Rs. 380 when 2 kg of food I and 4kg of food II are mixed.

We have a*b = H.C.F. of a and b.
The composition table is as below:

(i) For all a, b, c \(\in \) R
(a + b) + c = a + (b + c)
and (a x b) x c = a x (b x c)
Hence, '+' and 'x' are associative binary operations on R
(ii) For a, b, c \(\in \) R
(a - b) - c \(\neq \)a - (b - c)
and (a/b) / c \(\neq \)a/(b/c)
Hence, '-' and '/' are not binary operations on R

\(a*b=a+2ab\) and \(b*a=a+2a\)
\(\therefore a*b\neq b*a\)   \(\left[ \because a+2b\neq b+2d \right] \)
Hence, '*' is not communicative

(i) For all \(a,b\in R\),
a + b = b + a and \(a\times b=b\times a\).
Hence, '+' and '\(\times \)' are commutative binary operations.
(i) For all \(a,b\in R\),
\(a-b\neq b-a\)    \(\left[ For\quad ex.4-5\neq 5-4 \right] \)
and \(a\div b\neq b\div a\)    \(\left[ For\quad ex.\frac { 4 }{ 5 } \neq \frac { 5 }{ 4 } \right] \)
Hence, ' - ' and ' \(\div \)' are not commutative binary operations.

\(\vee :R\times R\rightarrow R\) given by \((a,b)\rightarrow \)min. {a,b},
which is a unique element \(viz\). max. of \((a,b)\rightarrow \). 
Hence, \(\vee \) is a binary operation.
(ii) \(\wedge :R\times R\rightarrow R\) given by \((a,b)\rightarrow \)min. {a,b},
which is a unique element \(viz\).max. of \(a,b\rightarrow R\).
Hence, \(\wedge \) is a binary operation.

suppose  \((gof)(x_{ 1 })=(gof)(x_{ 2 })\)
\(\Rightarrow \)  \(g(f(x_{ 1 }))=g(f(x_{ 2 }))\)
\(\Rightarrow \)  \(f(x_{ 1 })=f(x_{ 2 })\)   [\(\because \) g is one-one]
​​​\(\Rightarrow \)  \(x_{ 1 }=x_{ 2 }\)   [​​​​​​\(\because \)f is one-one]
Hence, gof is one-one.

f is not one-one, as f(1) = f(2) = 1. But f is onto, as given any y \(\in \) N, y \(\ne\) 1, we can choose x as y + 1 such that f (y + 1) = y + 1 – 1 = y. Also for 1 \(\in \) N, we have f(1) = 1.