Given a*b = LCM(a,b)
∴ (5*7) = LCM(5, 7) = 35

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

Given a * b = a+3b²
(8 * 3) = 8 + 3(3)²
= 8 + 27 = 35
⇒ (8 ∗ 3) = 35

\(\Rightarrow 2 *(x * 5)=10\)
\( \Rightarrow 2 *\left(\frac{5 x}{5}\right)=10 \)
\( \Rightarrow 2 * x=10 \)
\( \Rightarrow \frac{2 x}{5}=10 \)
\( \Rightarrow x=\frac{50}{2} \)
\(\mathrm{x}=25 \)

\(\frac{1}{2}>\left(\frac{1}{2}\right)^3=\left(\frac{1}{2}, \frac{1}{2}\right) \in R\)
Hence R is reflexive.

Let * be a binary on NB defined as a * b = a/b, for a, b ∈ N
We notice, if b is not  a factor a then a * b ∉ N.
Hence, division is not a binary operation on N

Given) f : R →→ R defined by f(x) = [x]
For one-one: We know by definition that for
a ≤ x i.e. for x₁,x²∈ (a, a+1) ,
x₁ ≠ x₂ ⇒ f(x₁) = f(x₂) = a
Hence, not one-one.
For onto: For y(non integer) in co-domain there does not exist x ∈ R in domain such that f(x) = y. Hence, not onto.

\(a *(b * c)=a *\left(\frac{b+c}{2}\right)=\frac{a+\frac{b+c}{2}}{2}=\frac{2 a+b+c}{4} \ldots(1)\)
and \((a * b) * c=\left(\frac{a+b}{2}\right) * c=\frac{\frac{a+b}{2}+c}{2}=\frac{a+b+2 c}{4} \ldots(2)\)
From (1) and (2), we get
a * (b* c) ≠ (a * b) * c
Hence, not associative

XRY ⇒ X∩Y = X ⇒ Y∩X = X ⇒ YRX
Hence symmetric

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

\(f \circ g(x)=f(g(x))=f\left(\frac{3+x}{2}\right)=2\left(\frac{3+x}{2}\right)-3=3+x-3=x\)
\(g \circ f(x)=g(f(x))=g(2 x-3)=\frac{3+2 x-3}{2}=\frac{2 x}{2}=x\)
As fog and gof are identify functions, so one is inverse of the other

fog(x) = f(g(x))
= f(x^1/3) = 8(x^1/3)³
= 8x

\( a * b=a+3 b^2 \\ 2 * 4=2+3(4)^2 \\ =2+3 \times 16=50 \)

$fog(x)=f[g(x)]$
$=f(x-7)$  ...................[ Since, $g(x)=x-7$ ]
$=x-7+7$ ..................[ Since, $f(x)=x+7$ ]           
$\Rightarrow$  $f$$o$$g(x)=x\Rightarrow$  $f$$o$$g(7)=7$