a*b = \(\frac { ab }{ 5 } \). For all a, b  \(\in R-[0].\)
(i) 2*(x*5) = 6
\( \Rightarrow 2 *\left(\frac{5 x}{5}\right)=6 \)
\( \Rightarrow 2 * x=6 \)
\( \Rightarrow \frac{2 x}{5}=6 \)
\( \Rightarrow 2x=30\)
\( \mathrm{x}=15 \)
(ii) 3*(x*3) = 9
\( \Rightarrow 3 *\left(\frac{3 x}{5}\right)=9\)
\( \Rightarrow \left(\frac{3\frac{3 x}{5}}{5}\right)=9\)
⇒ 9x = 25 ×× 9
∴  x = 25

(i) Given a*b = 3a + b²
\(\Rightarrow 4*3=3(4)+(3)^{ 2 }\)
= 12 + 9
4*3 = 21
(ii) Given a*b = 3a + b²
\(\Rightarrow 5*2=3(5)+2^{ 2 }\)
\(\Rightarrow 5*2=15\)
\(\therefore 5*2=19\)

(i) At a,b \(\in Q\) 
a*b = a²+ 2b
and b*a = b²+ 2a
\(\Rightarrow b*a\neq b*a\)
* is not commutative on Q.

e ∈ R is the identity element for *if a*e = e*a = a ∀ a ∈ R ⇒ e = 2

Given a * b = \(a+4b^{ 2 }\) is a binary operation compute (-5)*(2*0)
(-5)*(2*0) = -5*[2 + 4(0)²]
= 5 * 2
= 5 + 4(2)²
= -5 +16
⇒(-5)* (2*0) = 11   

we have f = {(a, 3) (b, 2) (e, 1)}, f is a one-one function.
\(f^{ -1 },f^{ 0 }=[(3,a),(2,b)(1,c)]\)
(ii) f = {(a, 2), (b, 1)(e, 1)}. f is not one-one, f is not onto because 3 \(\epsilon\) B as it has no Pre-image.

Equivalence relations could be the following:
{(1, 1), (2, 2), (3, 3), (1,2), (2, 1)} and 
{(1, I), (2, 2), (3, 3), (1, 2), (1, 3), (2,1), (2, 3), (3, 1), (3,2)} 
So, only two equivalence relations.

Symmetric Relation : A relation R on a set A is called symmetric relation if aRb implies bRa, for every a,b \(\in a\) i.,e if (a,b)  \(\in R\) \(\Rightarrow \) (b,a)  \(\in R\)For every a,b \(\in A\)
Example 
A = (1,2,3)
A x A =(1,2) (2,1) (1,1) (2,2) (3,3) (1,3) (2,3)  (3,1) (3,2 )  \(\in R\) 
since (a,b)  \(\in R\)  (b,a)  \(\in R\) for every a, b  \(\in A\) 
Relation is said to be symmetric 

Given R=[(a,a² ) : a is a prime number less than 5}
⇒ R =[(2,8),(3,27)]
Range=[8,27]

(i) 2 * 3 = 2³ = 8
(ii) 3 * 2 = 3² = 9.

No, as \(f(-2)=(-2)^2=4\) and \(f(2)=(2)^2=4\)
i.e \(x_1 \neq x_2 \Rightarrow f\left(x_1\right)=f\left(x_2\right)\).
Not one-one

Let \(y=\frac{3 x-4}{5} \Rightarrow 5 y=3 x-4\)
\(\Rightarrow x=f^{-1}(y)=\frac{5 y+4}{3} \Rightarrow f^{-1}(x)=\frac{5 x+4}{3}\)

22 * 4 = HCF (22,4) = 2

It is given that * P(X) x P(X) -7 P(X) is defined as
\(A*B=A\cap B\forall A,B\in \ P(X)\)
we know that
\(A\cap X=A=X\cap A\forall \in \quad P(X)\)
\(\Rightarrow A*X=A=X*A\forall A\in P(X)\)
Thus, X is the identity element for the given binary operation * .
Now, an element A  P(X) is invertible if there exists B P(X) such that, A*B = X = B*A.
(AS X is the identity element)
\(A\cap B=A=B\cap A\)
Thus, X is the only invertible element in P(X) with
respect to the given operation * .
Hence, the given result is proved. 
Values:
Respect for elders,
Concern for the aged,
Lending a helping hand.

Let (a, b), (c, d) E Q x Q. Then b + ad may not be mequal to d + cd. We find that (1, 2) * (2, 3) = (2, 5), (2,3) * (1,2) = (2,7) '*(2, 5) Hence, * is not commutative.
Let, (a, b), (c, d), (e,f> E Q x Q, {(a, b) * (c, d) * (e,f) 
= (ace, b + ad + acf)
= (a, b) * {(c, d) * (e, f)
Hence * is associative. 1
(x, y) Q x Q is the identity element for * if 2
(x, y)  Q x Q is the inverse of (a, b) E Q x Q if (c, d) * (a, b) = (a, b) * (c, d) = (1,0),
i.e., (ac, b + ad) = (ca, d + cb) = (1,0)
\(\Rightarrow c=\frac { 1 }{ a } ,d=\frac { -b }{ a } \)
The inverse of (a,b) \(\in Q\) XQ \(a\neq 0\quad \left( \frac { 1 }{ a } ,\frac { -b }{ a } \right) \)

Let a, b \(\in A\), a " b = a + b + ab
Commutatively : for all a, b \(\in A\)
.a * b = a + b + ab
= b + a + ba
= b * a
Associatively: Let a, b, c \(\in A\)
(a * b) " c = (a + b + ab) " c
= (a + b + ab) + c + (a + b + ab)c
(a * b) " c = a + b + c + ab + bc + ac + abc
a * (b * c) = a * (b + c + bc)
= a + b + c + ab + bc + ac + abc
Clearly
(a * b) * c = a * (b "c)  \(\forall \) a, b, c  \(\in A\)}
 * is associative.
Identity: Let e \(\in A\)}such that
a * c = a
c + a + ea = a
c = 0
Identity element of A is e = O.
Inverse: Let b \(\in A\)such that
a*b = b*a = e
\(\Rightarrow \) a + b + ab = 0 and b + a + ba = 0
\(\Rightarrow \)  a = -b-ab
\(\Rightarrow \)  a = -b(l + a)
\(\Rightarrow \) b = \(\frac { -a }{ 1+a } \) \([\because \quad a\in A\therefore a\neq -1]\)
Invertible element of A is \(\frac { -a }{ 1+a } \) for all \(a\in A\)

Assume, \(f(x)=x^{ 2 }+4x+7\) 
= \((x+2)^{ 2 }+3\)
\(\Rightarrow y-3=(x+2)^{ 2 }\)
\(\Rightarrow x+2=\pm \sqrt { y-3 } \)
\(\Rightarrow x=\sqrt { y-3 } -2\quad y>3\)
\(g:s\rightarrow N\)
\(g(y)=\sqrt { y-3 } -2\)
\(gof(x)=g[f(x)]=g(x^{ 2 }+4x+7)\)
\(=g[(x+2)^{ 2 }+3]\)
\(=\sqrt { (x+3)^{ 2 }+3-3-2 } \)
\(=x+2-2=x\)
\(fog(y)=f[g(y)]\)
\( =f\sqrt { y-3 } -2\)
\(=(\sqrt { y-3 } -2+2)^{ 3 }=y\)
\( \Rightarrow gof=I_{ N }\)
\(gog=I_{ s }\)
\(\Rightarrow f^{ -1 }=g=\sqrt { x-3 } -2\)
Yes, interest & knowledge have bijective relation Value Interest leads to knowledge

Let x,y \(\in W\)
If x and y both are even f(x) = f(y)
x + 1 = y + 1
x = y
If x and y both are odd, f(x) = f(y)
x - 1 = y - 1
x = y
If s is odd and y is even i.e., \(x\neq y\) (x-1) is even (y+1) is odd
\(\Rightarrow x\neq y\Rightarrow f(x)\neq f(y)\)
Similarly for X is even and y is odd f is one-one
Range of f = [f(0) ,f(1) ,f(2) .....}
= [1, 0, 3, 2....] = W = co-domain
f is onto
Hence f is invertible
\(f^{ 2 }:W\rightarrow W.f^{ -1 }(x)=\begin{cases} x-1,xis\quad odd \\ x+1,xis\quad even \end{cases}\)

We have: \(f(x)=\frac { (4x+3) }{ 6x-4) } ,x\neq \frac { 2 }{ 3 } .\)
(a) \(fof(x)=f(f(x))=f\frac { (4x+3) }{ 6x-4) } \)
\(=\frac { 4\left( \frac { (4x+3) }{ 6x-4) } \right) +3 }{ 6\left( \frac { (4x+3) }{ 6x-4) } \right) -4 } \)
\(=\frac { 16x+12+x8x-12 }{ 24x+18-24x+16 } =\frac { 34x }{ 34 } =x.\)
(b) Let \(y=\frac { (4x+3) }{ 6x-4) } \)
\(\Rightarrow \) \(6xy-4y=4x+3\Rightarrow (6y-4)x=4y+3\)
\(\Rightarrow \)  \(x=\frac { (4y+3) }{ 6y-4) } \)
\(\Rightarrow \)  \(g(y)=f^{ -1 }(y)=\frac { (4y+3) }{ 6y-4) } \)
\(\therefore \)   \(f^{ -1 }(x)=\frac { (4x+3) }{ 6x-4) } =f(x)\)
Hence, \(f^{ -1 }=f\) .

For \(a\in Z,a-a=0,\) which is divisible by 5.
\(\therefore \) \((a,a)\in R\forall a\in Z\).
Thus R is reflexive.
Now let \((a,b)\in R\)  \(\Rightarrow \) a-b is divisible by 5
\(\Rightarrow \) b-a is divisible by 5 \(\Rightarrow \) \((b,a)\in R\).
Thus R is symmetric.
Again let \((a,b)\in R,\ (b,c)\in R\)
\(\Rightarrow \) a-b and b-c are divisible by 5
\(\Rightarrow \) (a-b) + (b-c) = a-c is divisible by 5
\(\Rightarrow \) \((a,c)\in R\).
Thus R is transitive.
Hence, R is an equivalence relation.

Clearly, (x, y) R (x, y), ∀ (x, y) ∈ A, since xy = yx.
This shows that R is reflexive.
Further, (x, y) R (u, v) ⇒ xv = yu ⇒ uy = vx and hence (u, v) R (x, y).
This shows that R is symmetric. Similarly, (x, y) R (u, v) and (u, v) R (a, b) ⇒ xv = yu and 
\(u b=v a \Rightarrow x v \frac{a}{u}=y u \frac{a}{u} \Rightarrow x v \frac{b}{v}=y u \frac{a}{u} \Rightarrow x b=y a\) and hence (x, y) R (a, b).
Thus, R is transitive. Thus, R is an equivalence relation.

Let 'x' and 'y' be the number of pieces of Model A and Model B respectively.
We have: \(x\ge 0\) ....(1)
\(y\ge 0\) .....(2)
\(9x+12y\le 180\quad i.e.3x+4y\le 60\) .....(3)
\(x+3y\le 30\)....(4)
The mathematical formulation of the problem is as be low:
Maximize Z = 8000x + 12000y subject to the constraints (1)-(4).
For solution set, we draw the lines:
x = 0, y = 0, 3x + 4y = 60 and 3y = 30.
The lines 3x + 4y = 60 and x + 3y = 30 meet at E (12, 6).

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

Hence, the maximum profit is Rs. 1,68,000 when 12 pieces of Model A and 6 pieces of Model B are manufactured per week. 

Let \(x_{ 1 },x_{ 2 }\in R\).
Now \(f(x_{ 1 })=f(x_{ 2 })\Rightarrow cosx_{ 1 }=cosx_{ 2 }\)
\(\Rightarrow \)  \(x_{ 1 }=(2n\pi +x_{ 2 })\)
\(\Rightarrow \) '\(f\)' is not one-one.
(ii) Since cos \(x\) lies in [-1,1],
\(\therefore \) R is not fully covered.
Hence, '\(f\)'  is not onto.

We have: \(f(x)=x^{ 2 }-3x+2\) ....(1)
\(\therefore \) \(f(f(x))=(f(x))^{ 2 }-3f(x)+2\)  [Using (1)]
\(=(x^{ 2 }-3x+2)^{ 2 }-3(x^{ 2 }-3x+2)+2\)
\(=(x^{ 4 }+9x^{ 2 }+4-6x^{ 3 }-12x+4x^{ 2 })+(-3x^{ 2 }+9x-6)+2=x^{ 4 }-6x^{ 3 }+10x^{ 2 }-3x.\)

f : R → R is given by,
\(f(x)=|x|=\left[\begin{array}{cl} x & \text { if } \quad x \geq 0 \\ -x & \text { if } \quad x<0 \end{array}\right]\)
It is seen that f(-1) = |-1| = 1, f(1) = |1| = 1
∴ f(−1) = f(1), but −1 ≠ 1.
∴ f is not one-one.
Now, consider −1 ∈ R.
It is known that f(x) = |x| is always non-negative. Thus, there does not exist any element x in domain Rsuch that f(x) = |x|  = −1.
∴ f is not onto.
Hence, the modulus function is neither one-one nor onto.

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

We have gof (x) = g(f (x)) = g(cos x) = 3 (cos x)² = 3 cos² x. Similarly, fog(x) = f (g(x)) = f (3x²) = cos (3x²). Note that 3cos² x \(\ne\) cos 3x², for x = 0. Hence, gof \(\ne\) fog.