Given f(x) = \(\frac { \sqrt { 4-{ x }^{ 2 } } }{ \sqrt { { x }^{ 2 }-9 } } \)
When x = 2, f(x) = 0
When x = -2, f(x) = 0
For all the other values, we get negative value in the square root which is not possible.
\(\therefore\) Domain = {2, -2}

Given functions are f(x) = x³ and g(x) = \(x^{ ^{ \frac { 1 }{ 3 } } }\)
Now, f o g(x) = f(g(x)
= \(f\left( { x }^{ \frac { 1 }{ 3 } } \right) \)
= \(\left( { x }^{ 3 } \right) ^{ \frac { 1 }{ 3 } }=x\)

Since f o g(x) = x is symmetric about the line y = x, g(x) is the inverse of f(x)
∴ g(x) = f^-1(x).
where f o g(x) = g o f(x) = x so that
(i) f o g is bijective.
(ii) both f(x) and g(x) also bijective.
(iii) f and g are symmetrical about y = x

Given f(x) = 3x - 4
Let y = 3x - 4 ⇒ y + 4 = 3x
\(⇒ x={y+4\over 3}\)
Let g(y) = \(y+4\over 3\)
Now gof(n) = g(f(n)) = g(3\(\times\) -4) = \({3x-4+4\over 3}={3x\over 3}=x\)
and fog(y) = f(g(y)) = \(f\left(y+4\over 4\right)=3\left(y+4\over 3\right)-4=y+4-4=y\)
Thus, gof(x) = I_x and fog (y) = I_y
This implies that f and g are bijections and inverses to each other
Hence f is bijection and \(f^{-1} (x)={y+4\over 3}\)
Replacing y by x, we get f^-1 (x) = \(\frac { x+4 }{ 3 } \)

Hence, the graph of y = f^-1(x) is the reflection of the graph of f in y = x

We know that n(A\(\cup\) B) = n(A-B)+n(B-A)+n(A\(\cap\)B) if A and B are not disjoint.
\(\Rightarrow\) n(A-B)+n(B-A) = n(A\(\cup\)B) - n(A\(\cap\)B)
\(\Rightarrow\) n(A\(\Delta \)B) = 10-3
\(\Rightarrow\) \(\therefore\) n(A\(\Delta \)B) = 7
\(\therefore\) n[P(A\(\Delta \)B)] = 2⁷= 128

The relation R defined on the set of all positive integers by "mRn" if m divides n".
Given relation is "mRn if m divides n".
Reflexivity :  mRm since m divides m for all positive integers m.
\(\therefore\)  R is reflexive.
Symmetricity: mRn \(\Rightarrow\)  nRm.
m divides n \(\Rightarrow\)  n divides m but 'n' does not divide 'm'
\(\therefore\) R is not symmetric
Transitive : mRn and nRp \(\Rightarrow\)  mRp.
m divides n and n divides p \(\Rightarrow\) m divides p.
\(\therefore\) R is transitive.
\(\therefore\) R is reflexive, and transitive.