Let $$U$$ be the set of all consumers who were questioned, $$A$$ be the set of consumers who liked product $${P_1}$$ and $$B$$ be the set of consumers who liked product $${P_2}.$$
It is given that
$$\eqalign{ & n\left( U \right) = 2000,\,\,n{\left( A \right)} = 1720,\,\,n\left( B \right) = 1450, \cr & n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \cr & = 1720 + 1450 - n\left( {A \cap B} \right) \cr & = 3170 - n\left( {A \cap B} \right) \cr & {\text{Since,}}\,A \cup B \subseteq U\,\,\,\therefore \,\,n\left( {A \cup B} \right) \leqslant n\left( U \right) \cr & \Rightarrow \,3170 - n\left( {A \cap B} \right) \leqslant 2000 \cr & \Rightarrow \,3170 - 2000 \leqslant n\left( {A \cap B} \right) \cr & \Rightarrow n\left( {A \cap B} \right) \geqslant 1170 \cr} $$
Thus, the least value of $$n\left( {A \cap B} \right)$$   is 1170.
Hence, the least number of consumers who liked both the products is 1170.

$$\eqalign{ & n\left( A \right) = 40\% {\text{ of }}10000 = 4000, \cr & n\left( B \right) = 2000, \cr & n\left( C \right) = 1000, \cr & n\left( {A \cap B} \right) = 500, \cr & n\left( {B \cap C} \right) = 300, \cr & n\left( {C \cap A} \right) = 400, \cr & n\left( {A \cap B \cap C} \right) = 200 \cr & \therefore \,\,n\left( {A \cap \overline B \cap \overline C } \right) = n\left\{ {A \cap \left( {B \cup C} \right)'} \right\} \cr & = n\left( A \right) - n\left\{ {A \cap \left( {B \cup C} \right)} \right\} \cr & = n\left( A \right) - n\left( {A \cap B} \right) - n\left( {A \cap C} \right) + n\left( {A \cap B \cap C} \right) \cr & = 4000 - 500 - 400 + 200 \cr & = 3300 \cr} $$

$$x - y$$  is an integer.
$$\because \,\,x - x = 0$$   is an integer
⇒ $$A$$ is reflexive.
Let $$x - y$$  is an integer
⇒ $$y - x$$  is an integer
⇒ $$A$$ is symmetric
Let $$x - y, y - z$$    is also an integer
⇒ $$x - y + y - z$$    is also an integer
⇒ $$x - z$$  is an integer
⇒ $$A$$ is transitive
∴ $$A$$ is an equivalence relation.
Hence statement - 1 is true.
Also $$B$$ can be considered as
$$xBy$$  if $$\frac{x}{y} = \alpha ,$$  a rational number
$$\because \,\,\frac{x}{x} = 1$$   is a rational number
⇒ $$B$$ is reflexive
But $$\frac{x}{y} = \alpha ,$$  a rational number need not imply $$\frac{y}{x} = \frac{1}{\alpha },$$  a rational number because
$$\frac{0}{1}$$ is rational
⇒ $$\frac{1}{0}$$ is not rational
∴ $$B$$ is not an equivalence relation.

From the given condition $$n\left( {M \cup F} \right) = 15,\,\,n\left( F \right) = 8,\,\,n\left( {M \cap F} \right) = 4$$
$$\eqalign{ & {\text{So, }}n\left( {M \cup F} \right) = n\left( M \right) + n\left( F \right) - n\left( {M \cap F} \right) \cr & {\text{or, }}n\left( M \right) = n\left( {M \cup F} \right) + n\left( {M \cap F} \right) - n\left( F \right) \cr & {\text{hence }}n\left( M \right) = 15 + 4 - 8 = 11 \cr} $$

$$f'\left( x \right) = ax + 3\,\cos \,x + 4\,\sin \,x$$
$$f\left( x \right)$$  is one-one if $$f'\left( x \right) > 0$$   for all $$x$$ or $$f'\left( x \right) < 0$$   for all $$x.$$
Now, $$ - 5 \leqslant 3\,\cos \,x - 4\,\sin \,x \leqslant 5$$
$$\eqalign{ & {\text{So, }}f'\left( x \right) < 0\,\forall \,x,{\text{ if }}a < - 5{\text{ and }}f'\left( x \right) > 0\,\forall \,x,{\text{ if }}a > 5 \cr & \therefore \,a\, \in \left( { - \infty ,\, - 5} \right) \cup \left( {5,\,\infty } \right) \cr} $$

$$\eqalign{ & {\text{I}}{\text{.}}\,\,\,\mathop U\limits_{n = 2}^{10} \,{A_n}{\text{ is the set of first 10 prime numbers}} \cr & = \left\{ {2,\,3,\,5,\,7,\,11,\,13,\,17,\,19,\,23,\,29} \right\} \cr & {\text{II}}{\text{.}}\,\,n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \cr & \Rightarrow 50 = 28 + 32 - n\left( {A \cap B} \right) \cr & \Rightarrow n\left( {A \cap B} \right) = 60 - 50 = 10 \cr} $$

Let $$N = $$  National Savings Certificates
$$S = $$  Shares
Total number of persons $$ = 32$$
Number of persons who invest in National Savings Certificates $$ = 30$$
Number of persons who invest in Shares $$ = 17$$
Therefore
$$n\left( {N \cup S} \right) = 32,\,\,n\left( N \right) = 30,\,\,n\left( S \right) = 17$$
We know that,
$$\eqalign{ & n\left( {N \cup S} \right) = n\left( N \right) + n\left( S \right) - n\left( {N \cap S} \right) \cr & \Rightarrow 32 = 30 + 17 - n\left( {N \cap S} \right) \cr & \Rightarrow n\left( {N \cap S} \right) = 47 - 32 = 15 \cr} $$

Let $$X$$ = {1, 2, 3, 4, 5}
Total no. of elements = 5
Each element has 3 options. Either set $$Y$$ or set $$Z$$ or none. $$\left( {\because Y \cap Z = \phi } \right)$$
So, number of ordered pairs $$ = {3^5}$$

$$\eqalign{ & \left( {A - B} \right) \cup \left( {B - A} \right) \cup \left( {A \cap B} \right) \cr & = {\text{only }}A\, \cup \,{\text{only }}B\, \cup {\text{ both }}A{\text{ and }}B \cr & = A \cup B \cr} $$

Both the statements are true.
$$\eqalign{ & {\text{II}}{\text{. }}n\left( {S \cup T} \right) = n\left( S \right) + n\left( T \right) - n\left( {S \cap T} \right) \cr & = 720 + 450 - n\left( {S \cap T} \right) \cr & = 1170 - n\left( {S \cap T} \right) \cr & \Rightarrow 1170 - n\left( {S \cap T} \right) \leqslant n\left( U \right) \cr & \Rightarrow 1170 - n\left( {S \cap T} \right) \leqslant 1000 \cr & \Rightarrow n\left( {S \cap T} \right) \geqslant 170 \cr} $$