x = 2, y = 2

The binary operation * on N is defined as a * b = L.C.M. of a and b.
(i) 5 * 7 = L.C.M. of 5 and 7 = 35
20 * 16 = L.C.M of 20 and 16 = 80
(ii) It is known that:
L.C.M of a and b = L.C.M of b and a &mn For E; a, b ∈ N.
∴ a * b = b * a
Thus, the operation * is commutative.
(iii) For a, b, c ∈ N, we have:
(a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c
a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c
∴ (a * b) * c = a * (b * c)
Thus, the operation * is associative.
(iv) It is known that:
L.C.M. of a and 1 = a = L.C.M. 1 and a &mnForE; a ∈ N
⇒ a * 1 = a = 1 * a &mnForE; a ∈ N
Thus, 1 is the identity of * in N.
(v) An element a in N is invertible with respect to the operation * if there exists an element b in N, such that a * b = e = b * a.
Here, e = 1
This means that:
L.C.M of a and b = 1 = L.C.M of b and a
This case is possible only when a and b are equal to 1.
Thus, 1 is the only invertible element of N with respect to the operation *.

We have y=\(f\left( x \right) =\sqrt { x-2 } \) ...(1)
\(f'(x)=\frac { 1 }{ 2\sqrt { x-2 } } \)
Now \(f(a)=(2)=\sqrt { 2-2 } =0\)
\(f(b)=f(3)=\sqrt { 3-2 } =\sqrt { 1 } =1\)
By Lagrange's Theorem, we have:
\(f'(x)=\frac { f(b)-f(a) }{ b-a } \)
\(=\frac { 1 }{ 2\sqrt { x-2 } } =\frac { 1-0 }{ 3-2 } \Rightarrow \frac { 1 }{ 4(x-2) } =1\)
\(1=4x-8\Rightarrow 4x=9\)
\(x=\frac { 9 }{ 4 } \epsilon \left( 2,3 \right) \)
Putting in (1), \(y=\sqrt { \frac { 9 }{ 4 } -2 } =\sqrt { \frac { 1 }{ 4 } } =\frac { 1 }{ 2 } \)
Hence, \(\left( \frac { 9 }{ 4 } ,\frac { 1 }{ 2 } \right) \)is the required point.

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, \(fof^{ -1 }=f^{ -1 }\quad of.\)

\(\Delta=\left| \begin{matrix} -a({ b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 }) & { 2b }^{ 3 } & { 2c }^{ 3 } \\ { 2 }a^{ 3 } & -b({ c }^{ 2 }+{ a }^{ 2 }-{ b }^{ 2 }) & { 2c }^{ 3 } \\ { 2a }^{ 3 } & { 2b }^{ 3 } & -c({ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }) \end{matrix} \right|\)
Operating R₁-->R₁-R₂& R₂-->R₂-R₃, we get:
\(\Delta=\begin{vmatrix}-a(b^2+c^2-a^2)&b(a^2+b^2+c^2)&0\\0&-b(c^2+a^2+b^2)&c(a^2+b^2+c^2)\\2a^3&2b^3&-c(a^2+b^2-c^2) \end{vmatrix}\)
\(=(a^2+b^2+c^2)\begin{vmatrix} -a&b&0\\0&-b&c\\2a^3&2b^3&-c(a^2+b^2-c^2)\end{vmatrix}\)
\(=abc(a^2+b^2+c^2)^2\begin{vmatrix}-1&1&0\\0&-1&1\\2a^2&2b^2&-a^2-b^2+c^2 \end{vmatrix}\)
\(=abc(a^2+b^2+c^2)^2\begin{vmatrix}-1&1&0\\0&-1&1\\2a^2&2b^2&-a^2-b^2+c^2 \end{vmatrix}\) [Operating C₂--> C₂+C₁]
\(abc(a^2+b^2+c^2)^2(-1)[a^2+b^2-c^2-2a^2-2b^2]\) [Expanding by R₁]
\(=abc(a^2+b^2+c^2)^2(-1)(-a^2-b^2-c^2)\)
\(=abc(a^2+b^2+c^2)^3,\)
which is true

Since a, b, c are in A.P.,
b-a = c-b = d, common difference....(1)
Now \(\Delta=\begin{vmatrix} x+1&x+2&x+a\\x+2&x+3&x+b\\x+3&x+4&x+c\end{vmatrix}\)
= \(\begin{vmatrix}x+1&x+2&x+a\\1&1&b-a\\1&1&c-b \end{vmatrix}\)
= \(\begin{vmatrix}x+1&x+2&x+a\\1&1&d\\1&1&d \end{vmatrix}\)
= 0 Which is true