Sets and Relations MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Since }}f:\left( {4,\,6} \right) \to \left( {6,\,8} \right) \Rightarrow f\left( x \right) = x + 2 \cr & \therefore \,{f^{ - 1}}\left( x \right) = x - 2 \cr} $$

$$\eqalign{ & n\left( C \right) = 20,\,\,n\left( B \right) = 50,\,\,n\left( {C \cap B} \right) = 10 \cr & {\text{Now, }}n\left( {C \cup B} \right) = n\left( C \right) + n\left( B \right) - n\left( {C \cap B} \right) \cr & = 20 + 50 - 10 \cr & = 60 \cr} $$
Hence, required number of persons $$ = 60\% .$$

$$\eqalign{ & {\text{We have, }}R = \left\{ {\left( {a,\,b} \right):1 + ab > 0,\,ab\, \in \,R} \right\} \cr & {\text{Let }}a\, \in \,R,\,\,\therefore \,{a^2} \geqslant 0{\text{ or }}1 + {a^2} > 0{\text{ or }}\left( {a,\,a} \right)\, \in \,R \cr & \therefore \,R{\text{ is reflexive}}{\text{.}} \cr & {\text{Let }}\left( {a,\,b} \right)\, \in \,R \Rightarrow 1 + ab > 0 \Rightarrow 1 + ba > 0 \cr & \Rightarrow \left( {b,\,a} \right)\, \in \,R \cr & \therefore \,R{\text{ is symmetric}}{\text{.}} \cr & \left( {2,\,\frac{1}{3}} \right)\, \in \,R{\text{ because }}1 + 2\left( {\frac{1}{3}} \right) = \frac{5}{3} > 0 \cr & \left( {\frac{1}{3},\, - 1} \right)\, \in \,R{\text{ because }}1 + \frac{1}{3}\left( { - 1} \right) = \frac{2}{3} > 0 \cr & {\text{Now, }}\left( {2,\, - 1} \right)\, \in \,R{\text{ if }}1 + 2\left( { - 1} \right) = - 1 < 0{\text{ which is not true}}{\text{.}} \cr & \therefore \,R{\text{ is not transitive}}{\text{.}} \cr} $$

\[\begin{array}{l} {\rm{Since,\,\, }}f\left( x \right) = \left\{ \begin{array}{l} 2x + x,\,\,\,\,\,x \ge 0\\ 2x - x,\,\,\,\,\,x < 0 \end{array} \right. = \left\{ \begin{array}{l} 3x,\,\,\,\,\,x \ge 0\\ \,\,x,\,\,\,\,\,x < 0 \end{array} \right.\\ {\rm{and \,\,}}g\left( x \right) = \frac{1}{3}\left\{ \begin{array}{l} 2x - x,\,\,\,\,\,x \ge 0\\ 2x + x,\,\,\,\,\,x < 0 \end{array} \right. = \left\{ \begin{array}{l} \frac{x}{3},\,\,\,\,\,x \ge 0\\ x,\,\,\,\,\,\,x < 0 \end{array} \right.\\ \therefore \,f\left( {g\left( x \right)} \right) = \left\{ \begin{array}{l} 3\left( {\frac{x}{3}} \right),\,\,\,\,\,x \ge 0\\ \,\,\,\,x,\,\,\,\,\,\,\,\,\,\,x < 0 \end{array} \right.\\ \therefore \,f\left( {g\left( x \right)} \right) = x,\,\forall \,x\, \in \,R\\ \therefore \,h\left( x \right) = x\\ \Rightarrow {\sin ^{ - 1}}\left( {h\left( {h\left( {h.....\left( {h\left( x \right).....} \right)} \right)} \right)} \right) = {\sin ^{ - 1}}x \end{array}\]
Thus, domain of $${\sin ^{ - 1}}\left( {h\left( {h\left( {h\left( {h.....h\left( x \right).....} \right)} \right)} \right)} \right){\text{ is }}\left[ { - 1,\,1} \right]$$

$$\eqalign{ & {\text{We have,}} \cr & f\left( x \right) = {\sin ^2}x + {\sin ^2}\left( {x + \frac{\pi }{3}} \right) + \cos \,x\,\cos \left( {x + \frac{\pi }{3}} \right) \cr & = \frac{{1 - \cos \,2x}}{2} + \frac{{1 - \cos \left( {2x + \frac{{2\pi }}{3}} \right)}}{2} + \frac{1}{2}\left\{ {2\,\cos \,x\,\cos \left( {x + \frac{\pi }{3}} \right)\,} \right\} \cr & = \frac{1}{2}\left[ {\frac{5}{2} - \left\{ {\cos \,2x + \cos \left( {2x + \frac{{2\pi }}{3}} \right)} \right\} + \cos \left( {2x + \frac{\pi }{3}} \right)} \right] \cr & = \frac{1}{2}\left[ {\frac{5}{2} - 2\,\cos \left( {2x + \frac{\pi }{3}} \right)\cos \frac{\pi }{3} + \cos \left( {2x + \frac{\pi }{3}} \right)} \right] \cr & = \frac{5}{4}{\text{ for all }}x \cr & gof\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {\frac{5}{4}} \right) = 1\,\,\,\left[ {\because \,g\left( {\frac{5}{4}} \right) = 1\left( {{\text{given}}} \right)} \right] \cr & {\text{Hence, }}gof\left( x \right) = 1,{\text{ for all }}x. \cr} $$

$$\eqalign{ & g\left( x \right) = f\left[ {f\left( x \right)} \right] \cr & = f\left[ {\left| {x - 2} \right|} \right] \cr & = f\left( {x - 2} \right){\text{ as }}x > 20 \cr & = \left| {x - 2 - 2} \right| \cr & = x - 4 \cr & \therefore \,g'\left( x \right) = 1 \cr} $$

$$\eqalign{ & A \times B = \left\{ {1,\,2} \right\} \times \left\{ {1,\,3} \right\} = \left\{ {\left( {1,\,1} \right),\,\left( {1,\,3} \right),\,\left( {2,\,1} \right),\,\left( {2,\,3} \right)} \right\} \cr & B \times A = \left\{ {1,\,3} \right\} \times \left\{ {1,\,2} \right\} = \left\{ {\left( {1,\,1} \right),\,\left( {1,\,2} \right),\,\left( {3,\,1} \right),\,\left( {3,\,2} \right)} \right\} \cr & \therefore \,\,\left( {A \times B} \right) \cup \left( {B \times A} \right) \cr & = \left\{ {\left( {1,\,1} \right),\,\left( {1,\,3} \right),\,\left( {2,\,1} \right),\,\left( {2,\,3} \right),\,\left( {1,\,2} \right),\,\left( {3,\,1} \right),\,\left( {3,\,2} \right)} \right\} \cr} $$

$$\eqalign{ & n(C) = 21,\,\,n(H) = 26,\,\,n(F) = 29,\,\,n\left( {H \cap C} \right) = 14, \cr & n(H \cap F) = 15,\,\,n(F \cap C) = 12,\,\,n(C \cap H \cap F) = 8 \cr & \therefore n(C \cup H \cup F) = n(C) + n(H) + n(F) - n(H \cap C) - n(H \cap F) - n(F \cap C) + n(C \cap H \cap F) \cr & = 21 + 26 + 29 - 14 - 15 - 12 + 8 \cr & = 43. \cr} $$

$$\eqalign{ & {\text{Given that }}f\left( x \right) = 2x + \sin \,x,\,x\, \in \,R \cr & \Rightarrow f'\left( x \right) = 2 + \cos \,x \cr & {\text{But }} - 1 \leqslant \cos \,x \leqslant 1 \cr & \Rightarrow 1 \leqslant 2 + \cos \,x \leqslant 3 \cr & \therefore \,f'\left( x \right) > 0,\,\forall \,x\, \in \,R \cr} $$
$$ \Rightarrow \,f\left( x \right)$$   is strictly increasing and hence one-one.
Also as $$x \to \infty ,\,f\left( x \right) \to \infty {\text{ and }}x \to - \infty ,\,f\left( x \right) \to - \infty $$
$$\therefore $$  Range of $$f\left( x \right) = R = $$   domain of $$f\left( x \right) \Rightarrow f\left( x \right)$$   is onto.
Thus, $$f\left( x \right)$$  is one-one and onto.

Obviously, the relation is not reflexive and transitive but it is symmetric, because $${x^2} + {y^2} = 1\,\, \Rightarrow {y^2} + {x^2} = 1$$