$$\eqalign{ & f\left( 2 \right) = f\left( {{3^{\frac{1}{4}}}} \right) \Rightarrow \,{\text{many to one function and}} \cr & f\left( x \right) \ne - \sqrt 3\,\,\,\forall \,x\, \in \,R \Rightarrow \,{\text{into function}} \cr} $$

Note : Number of proper subjects of $$A = {2^n} - 1$$
Given : $$A = \left\{ {1,\,2,\,3,\,4,\,5} \right\}.$$     Here $$n = 5$$
$$\therefore $$  number of proper subjects $$ = {2^5} - 1 = 31$$

Let $$f:R \to R$$   be a function defined by $$f\left( x \right) = \frac{{x - m}}{{x - n}}$$
$$\eqalign{ & {\text{For any }}\left( {x,\,y} \right)\, \in \,R.\,{\text{Let }}f\left( x \right) = f\left( y \right) \cr & \Rightarrow \frac{{x - m}}{{x - n}} = \frac{{y - m}}{{y - n}} \cr & \Rightarrow x = y\,\,\therefore \,f{\text{ is one - one}} \cr & {\text{Let }}\alpha \, \in \,R\,\,{\text{such that }}f\left( x \right) = \alpha \cr & \Rightarrow \alpha = \frac{{x - m}}{{x - n}} \cr & \Rightarrow \left( {x - n} \right)\alpha = x - m \cr & \Rightarrow x = \frac{{n\alpha - m}}{{\alpha - 1}}.{\text{ for }}\alpha = 1,\,x\, \notin \,R \cr & {\text{So, }}f{\text{ is not onto}}{\text{.}} \cr} $$

$$\eqalign{ & fog\left( x \right) = f\left\{ {g\left( x \right)} \right\} \cr &= f\left( {3x + 2} \right) \cr &= 2{\left( {3x + 2} \right)^2} \cr &= 18{x^2} + 24x + 8 \cr & \therefore \,\,c = 8 \cr} $$

The image of any given elements in $$A$$ can be any one of the image of $$n$$ element in $$B$$.
$$\therefore $$  The $$m$$ elements in $$A$$ can be assigned images $$n \times n..... \times n\left( {m{\text{ times}}} \right) = {n^m}{\text{ ways}}$$
$$\therefore $$  Total mapping from $$A$$ and $$B = {n^m}$$

For subsets $$A$$ and $$B$$ of $$U,$$
If $$\left( {A - B} \right) \cup \left( {B - A} \right) = A,$$
$$ \Rightarrow \,B = \phi $$

$$\eqalign{ & \left( {\text{A}} \right)\,\,\left| x \right| = 5\,\, \Rightarrow x = 5\,\,\left[ {\because \,x\, \in \,N\,} \right] \cr & \therefore \,{\text{ Given set is singleton}}{\text{.}} \cr & \left( {\text{B}} \right)\,\,\left| x \right| = 6\,\, \Rightarrow x = - 6,\,6\,\,\left[ {\because \,x\, \in \,Z\,} \right] \cr & \therefore \,{\text{ Given set is not singleton}}{\text{.}} \cr & \left( {\text{C}} \right)\,\,{x^2} + 2x + 1 = 0\,\, \cr & \Rightarrow {\left( {x + 1} \right)^2} = 0 \cr & \Rightarrow x = - 1,\, - 1 \cr & {\text{Since, }} - 1\, \notin \,N, \cr & \therefore \,\,{\text{Given set}} = \phi \cr & \left( {\text{D}} \right)\,\,{x^2} = 7\,\, \Rightarrow x = \pm \sqrt 7 \cr} $$

Let $$A \equiv $$  Set of Tamil speaking students and $$B \equiv $$  Hindi speaking students
$$\eqalign{ & n\left( A \right) = 400,\,n\left( B \right) = 300{\text{ and }}n\left( {A \cup B} \right) = 600 \cr & n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \cr & \Rightarrow n\left( {A \cap B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cup B} \right) \cr & \Rightarrow n\left( {A \cap B} \right) = 400 + 300 - 600 = 100 \cr} $$

$$S$$ = {1, 2, 3, 4}
Let $$P$$ and $$Q$$ be disjoint subsets of $$S$$
Now for any element $$a \in s,$$  following cases are possible
$$a \in P\,{\text{and }}a \notin Q,a \notin P\,{\text{and }}a \in Q,a \notin P\,{\text{and }}a \notin Q$$
⇒ For every element there are three option
∴ Total option $$ = {3^4} = 81$$
Here $$P \ne Q$$  except when $$P = Q = \phi $$
∴ 80 ordered pairs $$(P, Q)$$  are there for which $$P \ne Q.$$
Hence total number of unordered pairs of disjoint subsets
$$\eqalign{ & = \frac{{80}}{2} + 1 \cr & = 41 \cr} $$

$$\eqalign{ & n\left( A \right) = 4,n\left( B \right) = 2 \cr & \Rightarrow \,\,n\left( {A \times B} \right) = 8 \cr} $$
The number of subsets of $$A \times B$$  having atleast
3 = elements = $$^8{C_3} + {\,^8}{C_4} + ..... + {\,^8}{C_8}$$
$$\eqalign{ & = {2^8} - {\,^8}{C_0} - {\,^8}{C_1} - {\,^8}{C_2} \cr & = 256 - 1 - 8 - 28 \cr & = 219 \cr} $$