Sets and Relations MCQ Questions & Answers in Calculus | Maths

We have,
$$\eqalign{ & R = \left\{ {\left( {x,\,y} \right):y = x - 3,\,x = 11{\text{ or }}12{\text{ or }}13,\,y = 8{\text{ or }}10{\text{ or }}12} \right\} \cr & \,\,\,\,\,\,\, = \left\{ {\left( {11,\,8} \right),\,\left( {13,\,10} \right)} \right\} \cr & \therefore \,{R^{ - 1}}\left\{ {\left( {y,\,x} \right):\left( {x,\,y} \right)\, \in \,R} \right\} = \left\{ {\left( {8,\,11} \right),\,\left( {10,\,13} \right)} \right\} \cr} $$

$$\eqalign{ & A \times B = \left\{ {\left( {1,\,5} \right),\left( {1,\,7} \right),\left( {1,\,9} \right),\left( {2,\,5} \right),\left( {2,\,7} \right),\left( {2,\,9} \right),\left( {3,\,5} \right),\left( {3,\,7} \right),\left( {3,\,9} \right),\left( {4,\,5} \right),\left( {4,\,7} \right),\left( {4,\,9} \right)} \right\} \cr & B \times A = \left\{ {\left( {5,\,1} \right),\left( {5,\,2} \right),\left( {5,\,3} \right),\left( {5,\,4} \right),\left( {7,\,1} \right),\left( {7,\,2} \right),\left( {7,\,3} \right),\left( {7,\,4} \right),\left( {9,\,1} \right),\left( {9,\,2} \right),\left( {9,\,3} \right),\left( {9,\,4} \right)} \right\} \cr & A \times B \ne B \times A{\text{ but }}n\left( {A \times B} \right) = n\left( {B \times A} \right) = 12 \cr} $$

Since, $$A$$ and $$B$$ are subsets of set $$X$$ therefore
$$\eqalign{ & A \subseteq X{\text{ and }}B \subseteq X \cr & {\text{Consider}}\,\left\{ {A \cap \left( {X - B} \right)} \right\} \cup B \cr & = \left( {A \cap B'} \right) \cup B \cr & = A \cup B\,\,\,\,\,\,\,\,\,\left( {\because \,\,B' \cap B = B} \right) \cr} $$

$$\eqalign{ & {\text{Given, }} \cr & f\left( x \right) = ax + b,\,g\left( x \right) = cx + d{\text{ and }}f\left\{ {g\left( x \right)} \right\} = g\left\{ {f\left( x \right)} \right\} \cr & \Rightarrow f\left( {cx + d} \right) = g\left( {ax + b} \right) \cr & \Rightarrow a\left( {cx + d} \right) + b = c\left( {ax + b} \right) + d \cr & \Rightarrow acx + ad + b = cax + bc + d \cr & \Rightarrow ad + b = bc + d \cr & \Rightarrow f\left( d \right) = g\left( b \right) \cr} $$

$$\eqalign{ & \left( {A - B} \right) \cup \left( {B - A} \right) \cup \left( {A \cap B} \right) \cr & A - B = A \cap B' \cr & B - A = B \cap A' \cr} $$
$$\left( {A - B} \right) \cup \left( {B - A} \right) = $$     the elements that are only in $$A$$ and $$B$$, which are not common in $$A$$ and $$B$$.
$$\left( {A \cap B} \right) = $$   elements common in $$A$$ and $$B$$.
$$\left( {A - B} \right) \cup \left( {B - A} \right) \cup \left( {A \cap B} \right) = $$       elements in $$A$$ and $$B$$ and common in both $$ = A \cup B$$
Hence, option A is correct.

$$\eqalign{ & {\text{We have}} \cr & A = \left\{ {0,\,1,\,2} \right\} \cr & B = \left\{ {2,\,3} \right\} \cr & C = \left\{ {3,\,4} \right\} \cr & \left( {A \cup B} \right) = \left\{ {0,\,1,\,2,\,3} \right\} \cr & \left( {A \cup B} \right) \times C = \left\{ {\left( {0,\,3} \right),\,\left( {0,\,4} \right),\,\left( {1,\,3} \right);\left( {1,\,4} \right);\left( {2,\,3} \right),\,\left( {2,\,4} \right),\,\left( {3,\,3} \right);\left( {3,\,4} \right)} \right\} \cr & \therefore \,n\left[ {\left( {A \cup B} \right) \times C} \right] = 8 \cr} $$

$$\eqalign{ & {\text{We have}} \cr & n\left( {A \cup B \cup C} \right) = n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {C \cap A} \right) + n\left( {A \cap B \cap C} \right) \cr & = 10 + 15 + 20 - 8 - 9 - n\left( {C \cap A} \right) + n\left( {A \cap B \cap C} \right) \cr & = 28 - \left\{ {n\left( {C \cap A} \right) - n\left( {A \cap B \cap C} \right)} \right\}.....({\text{i}}) \cr & {\text{Since }}\,n\left( {C \cap A} \right) \geqslant n\left( {A \cap B \cap C} \right) \cr & {\text{We have }}n\left( {C \cap A} \right) - n\left( {A \cap B \cap C} \right) \geqslant 0.....({\text{ii}}) \cr & {\text{From (i) and (ii) : }}n\left( {A \cup B \cup C} \right) \leqslant 28.....({\text{iii}}) \cr & {\text{Now, }}n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \cr & = 10 + 15 - 8 = 17 \cr & {\text{and }}n\left( {B \cup C} \right) = n\left( B \right) + n\left( C \right) - n\left( {B \cap C} \right) \cr & = 15 + 20 - 9 = 26 \cr & {\text{Since, }}n\left( {A \cup B \cup C} \right) \geqslant n\left( {A \cup C} \right)\,{\text{and }}n\left( {A \cup B \cup C} \right) \geqslant n\left( {B \cup C} \right){\text{,}} \cr & {\text{we have }}n\left( {A \cup B \cup C} \right) \geqslant 17{\text{ and }}n\left( {A \cup B \cup C} \right) \geqslant 26 \cr & {\text{Hence }}n\left( {A \cup B \cup C} \right) \geqslant 26.....({\text{iv}}) \cr & {\text{From (iii) and (iv) we obtain}} \cr & {\text{26}} \leqslant n\left( {A \cup B \cup C} \right) \leqslant 28 \cr & {\text{Also }}n\left( {A \cup B \cup C} \right){\text{ is a positive integer}} \cr & \therefore \,n\left( {A \cup B \cup C} \right) = 26{\text{ or }}27{\text{ or }}28 \cr} $$

$$\eqalign{ & \rho \,{\text{is reflexive, since }}\left| {a - a} \right| = 0 < \frac{1}{2}{\text{ for all }}a\, \in \,R \cr & \rho \,{\text{is symmetric, since }} \Rightarrow \left| {b - a} \right| < \frac{1}{2} \cr & \rho \,{\text{is not transitive}}{\text{.}} \cr & {\text{For, if we take three numbers }}\frac{3}{4},\frac{1}{3},\frac{1}{8} \cr & {\text{Then,}}\,\,\left| {\frac{3}{4} - \frac{1}{3}} \right| = \frac{5}{{12}} < \frac{1}{2}{\text{ and }}\left| {\frac{1}{3} - \frac{1}{8}} \right| = \frac{5}{{24}} < \frac{1}{2} \cr & {\text{But, }}\left| {\frac{3}{4} - \frac{1}{8}} \right| = \frac{5}{8} > \frac{1}{2} \cr & {\text{Thus, }}\frac{3}{4}\rho \frac{1}{3}{\text{ and }}\frac{1}{3}\rho \frac{1}{8}{\text{ but }}\frac{3}{4}\left( { \sim \rho } \right)\frac{1}{8} \cr} $$

$$\eqalign{ & N\Delta {N_p} = \left( {N - {N_p}} \right) \cup \left( {{N_p} - N} \right) \cr & = \left\{ {1,\,2,.....} \right\} \cup \left\{ {.....,\, - 2,\, - 1} \right\} \cr & = I - \left\{ 0 \right\} \cr} $$

$$\eqalign{ & {\text{I}}{\text{.}}\,A = {\text{Set of first elements }} = \left\{ {a,\,b,\,c} \right\} \cr & \,\,\,\,\,B = {\text{Set of second elements }} = \left\{ {1,\,2} \right\} \cr & {\text{II}}{\text{.}}\,B \cap \phi = \phi \,\,\,\therefore \,A \times \phi = \phi \cr} $$