Let X = {0, 1, 2, 3, 4, 5}.
The operation * on X is defined as:
\(a * b=\left\{\begin{array}{ll} a+b, & \text { if } a+b<6 \\ a+b-6 & \text { if } a+b \geq 6 \end{array}\right.\)
An element e ∈ X is the identity element for the operation *, if a * e = a = e * a ∀ a ∈ X
For a ∈ X we observed that
a * 0 = a + 0 =a   [a ∈ X ⇒ a + 0 < 6]
0 * a = 0 + a = a        [a ∈ X ⇒ 0 + a < 6]
a * 0 = a = 0 * a ∀ a ∈ X
Thus, 0 is the identity element for the given operation *.
An element a ∈ X is invertible if there exists b∈ X such that a * b = 0 = b * a.
ie \(\left\{\begin{array}{ll} a+b=0=b+a & \text { if } a+b<6 \\ a+6-6=0=b+a-6 & \text { if } a+b \geq 6 \end{array}\right.\)
i.e.,
a = −b or b = 6 − a
But, X = {0, 1, 2, 3, 4, 5} and a, b ∈ X. Then, a ≠ −b.

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. 
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Here  \(f(x)=y=\frac { 1 }{ x } \Rightarrow x=\frac { 1 }{ y } .\)
\(\therefore \)   \(f^{ -1 }=\{ (y,x)\} =\left\{ \left( y,\frac { 1 }{ y } \right) \right\} \)
\(\Rightarrow \)  \(f^{ -1 }(x)=\frac { 1 }{ x } .\)
\(\therefore \)  \(fof^{ -1 }(x)=f\left( \frac { 1 }{ x } \right) =\frac { 1 }{ 1/x } =x.\)
Also     \(f^{ -1 }\quad of\quad (x)=f^{ -1 }\left( \frac { 1 }{ x } \right) =\frac { 1 }{ 1/x } =x.\)
Thus  \(fof^{ -1 }(x)=f^{ -1 }\quad of\quad (x)\)
Hence, \(fof^{ -1 }=f^{ -1 }\quad of.\)

Given a * b = min {a, b]
Operation table for * is 

\(f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \) \(9x_{ 1 }^{ 2 }+6x_{ 1 }-5=9x_{ 2 }^{ 2 }+6x_{ 2 }-5\)
\(\Rightarrow \) \(9(x_{ 1 }^{ 2 }-x_{ 2 }^{ 2 })+6(x_{ 1 }-x_{ 2 })=0\)
\(\Rightarrow \) \((x_{ 1 }-x_{ 2 })\quad (9(x_{ 1 }+x_{ 2 })+6=0\)
\(\Rightarrow \) \(x_{ 1 }-x_{ 2 }=0\Rightarrow x_{ 1 }=x_{ 2 }\)
\(\Rightarrow \) \(f\) is one-one \(\Rightarrow \) \(f\) is invertible
Let \(y=f(x)=9x^{ 2 }+6x-5\)
\(\Rightarrow \) \(​​​​9x^{ 2 }+6x-(5+y)=0\)
\(\Rightarrow \) \(x=\frac { -6\pm \sqrt { 36+36(5+y) } }{ 18 } \)
\(=\frac { -6+6\sqrt { 6+y } }{ 18 } =\frac { -1\pm \sqrt { y+6 } }{ 3 } \)
Since \(x\ge 0,\)
\(\therefore \) \(x=\frac { -1-\sqrt { y+6 } }{ 3 } \) is not possible.
\(\therefore \) \(x=\frac { -1+\sqrt { y+6 } }{ 3 } \).
Hence \(f^{ -1 }(y)=\frac { \sqrt { y+6 } -1 }{ 3 } \)