Sets and Relations MCQ Questions & Answers in Calculus | Maths

For $$\left( {fog} \right)\left( x \right)$$   to exists range of $$g \subseteq $$  domain of $$f.$$
$$\eqalign{ & \therefore \,{\text{Domain of }}f \Rightarrow 3\left| x \right| - x - 2 \geqslant 0 \cr & \Rightarrow 3\left| x \right| - x \geqslant 2 \cr & {\text{When }}x \geqslant 0 \Rightarrow x \geqslant 1 \cr & {\text{When}}\,x < 0 \Rightarrow x < - \frac{1}{2} \cr & \therefore \,\sin \,x \geqslant 1{\text{ and }}\sin \,x < - \frac{1}{2}{\text{ for }}f\left\{ {g\left( x \right)} \right\}{\text{ to exists}}{\text{.}} \cr & {\text{i}}{\text{.e}}{\text{., }}\sin \,x = 1{\text{ and }} - 1 \leqslant \sin \,x < - \frac{1}{2} \cr & \therefore \,x = \left( {4m + 1} \right)\frac{\pi }{2}{\text{ and }}2n\pi + \frac{{7\pi }}{6} \leqslant x \leqslant 2n\pi + \frac{{11\pi }}{6} \cr & {\text{i}}{\text{.e}}{\text{., }}\left\{ {\left( {4m + 1} \right)\frac{\pi }{2}:m\, \in \,I} \right\}\mathop \cup \limits_{n\, \in \,I} \left[ {2n\pi + \frac{{7\pi }}{6} \leqslant x \leqslant \,2n\pi + \frac{{11\pi }}{6}} \right] \cr} $$

$$\left( {a,\,b} \right)\,r\,\left( {a,\,b} \right)$$   because $$a+b=b+a.$$
So, $$r$$ is reflexive.
$$\left( {a,\,b} \right)\,r\,\left( {c,\,d} \right) \Rightarrow a + d = b + c \Rightarrow c + b = d + a \Rightarrow \left( {c,\,d} \right)\,r\,\left( {a,\,b} \right)$$
So, $$r$$ is symmetric.
$$\left( {a,\,b} \right)\,r\,\left( {c,\,d} \right)$$    and $$\left( {c,\,d} \right)\,r\,\left( {e,\,f} \right) \Rightarrow a + d = b + c,\,\,c + f = d + e$$
Adding, $$a + d + c + f = b + c + d + e \Rightarrow a + f = b + e \Rightarrow \left( {a,\,b} \right)\,r\,\left( {e,\,f} \right)$$
$$\therefore \,r$$  is transitive.

$$n\left( {A \times A} \right) = n\left( A \right).n\left( A \right) = {3^2} = 9.$$       So, the total number of subsets of $${A \times A}$$  is $${2^9}$$ and a subset of $${A \times A}$$  is a relation over the set $$A.$$

$$\eqalign{ & {\text{Let }}R = \left\{ {x:x\, \in \,N,\,x{\text{ is a multiple of 3 and }}x \leqslant 100} \right\} \cr & {\text{and}}\,S = \left\{ {x:x\, \in \,N,\,x{\text{ is a multiple of 5 and }}x \leqslant 100} \right\} \cr & \therefore \,R = \left\{ {3,\,6,\,9,\,12,\,15,\,.....,\,99} \right\} \cr & {\text{and}}\,S = \left\{ {5,\,10,\,15,.....,95,\,100} \right\} \cr & {\text{Now, }}\left( {R \times S} \right) \cap \left( {S \times R} \right) \cr & = \left( {R \cap S} \right) \times \left( {S \cap R} \right) \cr & = \left( {15,\,30,\,45,\,60,\,75,\,90} \right) \times \left( {15,\,30,\,45,\,60,\,75,\,90} \right) \cr & \therefore {\text{ Number of elements in }}\left( {R \times S} \right) \cap \left( {S \times R} \right) = 6 \times 6 = 36 \cr} $$

$$\eqalign{ & n\left( {A' \cap B'} \right) = n\left( {A \cup B} \right)' \cr & = n\left( U \right) - n\left( {A \cup B} \right) \cr & = n\left( U \right) - \left[ {n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)} \right] \cr & = 700 - \left[ {200 + 300 - 100} \right] \cr & = 300 \cr} $$

$$\eqalign{ & f\left( x \right) = 5\,{\log _5}x \Rightarrow {f^{ - 1}}\left( x \right) = {5^{\frac{x}{5}}} \cr & {f^{ - 1}}\left( {\alpha - \beta } \right) = {5^{\frac{{\alpha - \beta }}{5}}} = \frac{{{5^{\frac{\alpha }{5}}}}}{{{5^{\frac{\beta }{5}}}}} = \frac{{{f^{ - 1}}\left( \alpha \right)}}{{{f^{ - 1}}\left( \beta \right)}} \cr} $$

$$\eqalign{ & X \cap {\left( {X \cup Y} \right)^c} \cr & = X \cap \left( {{X^c} \cap {Y^c}} \right) \cr & = \left( {X \cap {X^c}} \right) \cap {Y^c} \cr & = \phi \cap {Y^c} \cr & = \phi \cr} $$

The total number of bijective functions from a set $$A$$ containing 106 elements to itself is $$\left( {106} \right)!$$

$$\eqalign{ & {\text{Given set is}} \cr & \left\{ {\left( {a,\,b} \right):2{a^2} + 3{b^2} = 35,\,a,\,b\, \in \,Z} \right\}, \cr & {\text{We can see that,}} \cr & 2{\left( { \pm 2} \right)^2} + 3{\left( { \pm 3} \right)^2} = 35{\text{ and }}2{\left( { \pm 4} \right)^2} + 3{\left( { \pm 1} \right)^2} = 35 \cr & \therefore \,\left( {2,\,3} \right),\,\left( {2,\, - 3} \right),\,\left( { - 2,\, - 3} \right),\,\left( { - 2,\,3} \right),\,\left( {4,\,1} \right),\,\left( {4,\, - 1} \right),\,\left( { - 4,\, - 1} \right),\,\left( { - 4,\,1} \right) \cr & {\text{are 8 elements of the set}}{\text{.}} \cr & \therefore \,\,n = 8 \cr} $$

$$\eqalign{ & \left( {\bf{A}} \right){\text{Let }}\left( {a,\,b} \right),\,\left( {b,\,c} \right)\, \in \,R \cup S \cr & {\text{It is possible that }}\left( {a,\,b} \right)\, \in R - S{\text{ and }}\left( {b,\,c} \right)\, \in S - R \cr & {\text{In such case, we cannot say that }}\left( {a,\,c} \right)\, \in R\,\,{\text{or }}\left( {a,\,c} \right)\, \in S \cr & \therefore \,\left( {a,\,c} \right){\text{ may not be in }}R \cup S \cr & \therefore \,R \cup S{\text{ in not transitive}}{\text{.}} \cr & \left( {\bf{B}} \right){\text{Let }}\left( {a,\,b} \right),\,\left( {b,\,c} \right)\, \in \,R \cap S \cr & \therefore \,\left( {a,\,b} \right),\,\left( {b,\,c} \right)\, \in R{\text{ and }}\left( {a,\,b} \right),\,\left( {b,\,c} \right)\, \in S \cr & \therefore \,\left( {a,\,c} \right)\, \in R\,\,{\text{and }}\left( {a,\,c} \right)\, \in S \cr & \therefore \,\left( {a,\,c} \right){\text{ }} \in \,{\text{ }}R \cap S \cr & \therefore \,R \cap S{\text{ in transitive}}{\text{.}} \cr & \left( {\bf{C}} \right){\text{Let }}\left( {a,\,b} \right)\, \in \,R \cup S \cr & \therefore \,\left( {a,\,b} \right)\, \in R{\text{ or }}\left( {a,\,b} \right)\, \in S \cr & {\text{Now, }}\left( {a,\,b} \right)\, \in R \Rightarrow \left( {b,\,a} \right)\, \in R\,\,\,\left( {\because \,R{\text{ is symmetric}}} \right) \cr & \left( {a,\,b} \right)\, \in \,S \Rightarrow \left( {b,\,a} \right)\, \in \,S\,\,\,\left( {\because \,S{\text{ is symmetric}}} \right) \cr & \therefore \,\left( {b,\,a} \right)\, \in \,R \cup S \cr & \therefore \,R \cup S{\text{ is symmetric}}{\text{.}} \cr & \left( {\bf{D}} \right){\text{ Let }}a\, \in \,A \cr & \therefore \,\left( {a,\,a} \right)\, \in \,R\,{\text{and}}\left( {a,\,a} \right)\, \in S \cr & \therefore \,\left( {a,\,a} \right)\, \in \,R \cap S \cr & \therefore \,R \cap S{\text{ in reflexive}}{\text{.}} \cr} $$