Sets and Relations MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & \frac{{12}}{{\left( {3 + \sqrt 5 } \right) + 2\sqrt 2 }} \cr & = \frac{{12}}{{\left( {3 + \sqrt 5 } \right) + 2\sqrt 2 }} \times \frac{{\left( {3 + \sqrt 5 } \right) - 2\sqrt 2 }}{{\left( {3 + \sqrt 5 } \right) - 2\sqrt 2 }} \cr & = \frac{{12\left[ {3 + \sqrt 5 - 2\sqrt 2 } \right]}}{{{{\left( {3 + \sqrt 5 } \right)}^2} - {{\left( {2\sqrt 2 } \right)}^2}}} \cr & = \frac{{12\left[ {3 + \sqrt 5 - 2\sqrt 2 } \right]}}{{9 + 5 + 6\sqrt 5 - 8}} \cr & = \frac{{12\left[ {3 + \sqrt 5 - 2\sqrt 2 } \right]}}{{6\left( {\sqrt 5 + 1} \right)}} \times \frac{{\sqrt 5 - 1}}{{\sqrt 5 - 1}} \cr & = \frac{{2\left[ {3\sqrt 5 + 5 - 2\sqrt {10} - 3 - \sqrt 5 + 2\sqrt 2 } \right]}}{{5 - 1}} \cr & = \frac{{2 + 2\sqrt 5 + 2\sqrt 2 - 2\sqrt {10} }}{2} \cr & = 1 + \sqrt 2 + \sqrt 5 - \sqrt {10} \cr} $$

$$\eqalign{ & {x^2} + {y^2} = 9 \cr & \Rightarrow {y^2} = 9 - {x^2} \cr & \Rightarrow y = \pm \sqrt {9 - {x^2}} \cr & x = 0\,\, \Rightarrow y = \pm \sqrt {9 - 0} = \pm 3\, \in \,Z \cr & x = \pm 1\,\, \Rightarrow y = \pm \sqrt {9 - 1} = \pm \sqrt 8 \, \notin \,Z \cr & x = \pm 2\,\, \Rightarrow y = \pm \sqrt {9 - 4} = \pm \sqrt 5 \, \notin \,Z \cr & x = \pm 3\,\, \Rightarrow y = \pm \sqrt {9 - 9} = 0\, \in \,Z \cr & x = \pm 4\,\, \Rightarrow y = \pm \sqrt {9 - 16} = \pm \sqrt { - 7} \, \notin \,Z{\text{ }} \cr & {\text{and so on}}.... \cr & \therefore \,R = \left\{ {\left( {0,\,3} \right),\,\left( {0,\, - 3} \right),\,\left( {3,\,0} \right),\,\left( { - 3,\,0} \right)} \right\} \cr & {\text{Domain of }}R = \left\{ {x:\left( {x,\,y} \right)\, \in \,R} \right\} = \left\{ {0,\,3,\, - 3} \right\} \cr & {\text{Range of }}R = \left\{ {y:\left( {x,\,y} \right)\, \in \,R} \right\} = \left\{ {3,\, - 3,\,0} \right\} \cr} $$

Given $$A = \left\{ {x:x \leqslant 9,\,x\, \in \,N} \right\} = \left\{ {1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9} \right\}$$
Total possible multiple of $$3$$ are $$3,\,6,\,9,\,12,\,15,\,18,\,21,\,24,\,27$$
But $$3$$ and $$27$$  are not possible because $$3$$ and $$27$$  can not be express as such that $$a + b + c$$   is multiple of $$3.$$
$$\eqalign{ & 6 \to 1 + 2 + 3 \cr & 9 \to 2 + 3 + 4,\,5 + 3 + 1,\,6 + 2 + 1, \cr & 12 \to 9 + 2 + 1,8 + 3 + 1,\,7 + 1 + 4,\,7 + 2 + 3,\,6 + 4 + 2,\,6 + 5 + 1,\,5 + 4 + 3 \cr & 15 \to 9 + 4 + 2,\,9 + 5 + 1,\,8 + 6 + 1,\,8 + 5 + 2,\,8 + 4 + 3,\,7 + 6 + 2,\,7 + 5 + 3,\,6 + 5 + 4 \cr & 18 \to 9 + 8 + 1,\,9 + 7 + 2,\,9 + 6 + 3,\,9 + 4 + 5,\,8 + 7 + 3,\,8 + 6 + 4,\,7 + 6 + 5 \cr & 21 \to 9 + 8 + 4,\,9 + 7 + 5,\,8 + 7 + 6 \cr & 24 \to 9 + 8 + 7 \cr} $$
Hence, total largest possible subsets are $$30.$$

$$\eqalign{ & {\text{We have fog}}\left( x \right) = f\left( {g\left( x \right)} \right) = \sin \left( {{{\log }_e}\left| x \right|} \right) \cr & {\log _e}\left| x \right|\,{\text{has range }}R,{\text{ for which}}\,\sin \left( {{{\log }_e}\left| x \right|} \right) \in \left[ { - 1,\,1} \right] \cr & {\text{Therefore, }}{R_1} = \left\{ {u: - 1 \leqslant u \leqslant 1} \right\} \cr & {\text{Also, gof}}\left( x \right) = g\left( {f\left( x \right)} \right) = {\log _e}\left| {\sin \,x} \right| \cr & \because \,\,0 \leqslant \left| {\sin \,x} \right| \leqslant 1{\text{ }} \cr & {\text{or }} - \infty < {\log _e}\left| {\sin \,x} \right| \leqslant 0 \cr & {\text{or }}{R_2} = \left\{ {v: - \infty < v \leqslant 0} \right\} \cr} $$

$$\eqalign{ & f\left( x \right) = f\left( { - x} \right).{\text{ So, }}f{\text{ is many - one}}{\text{.}} \cr & {\text{Also, }}f\left( x \right) = 1 - \frac{5}{{{x^2} + 1}} > 1 - 5 = - 4 \cr & {\text{So, }}f{\text{ is into}}{\text{.}} \cr} $$

$$\eqalign{ & n\left( {M \cup P} \right) = 20,\,\,n(M) = 12,\,\,n\left( {M \cap P} \right) = 4 \cr & n\left( {M \cup P} \right) = n(M) + n(P) - n\left( {M \cap P} \right) \cr & \Rightarrow 20 = 12 + n\left( P \right) - 4 \cr & \therefore n\left( P \right) = 12 \cr} $$
So, the required number $$ = n\left( P \right) - n\left( {M \cap P} \right) = 12 - 4 = 8$$

$$\eqalign{ & {\text{Let }}A = \left\{ {1,\,2} \right\} \cr & B = \left\{ {3,\,4,\,0} \right\} \cr & C = \left\{ {5,\,6,\,0} \right\} \cr & D = \left\{ {7,\,8} \right\} \cr & {\text{Such that }}\left( {A \cap B} \right) = \left( {C \cap D} \right) = \phi \cr & \Rightarrow \left( {A \cup C} \right) = \left\{ {1,\,2,\,5,\,6,\,0} \right\} \cr & \Rightarrow \left( {B \cup D} \right) = \left\{ {3,\,4,\,7,\,8,\,0} \right\} \cr & \Rightarrow \left( {A \cup C} \right) \cap \left( {B \cup D} \right) = \left\{ 0 \right\} \cr & {\text{So }}\left( {A \cup C} \right)\,{\text{and }}\left( {B \cup D} \right)\,{\text{are not always disjoint}} \cr & \Rightarrow \left( {A \cap C} \right) = \phi {\text{ and }}\left( {B \cap D} \right) = \phi \cr & {\text{So }}\left( {A \cap C} \right){\text{ and }}\left( {B \cap D} \right){\text{ are always disjoint}}{\text{.}} \cr} $$

Total number of students $$ = 500$$
Let $$H$$ be the set showing number of students who can speak Hindi $$ = 475$$  and $$B$$ be the set showing number of students who can speak Bengali $$ = 200$$
So, $$n\left( H \right) = 475$$   and $$n\left( B \right) = 200$$   and given that $$n\left( {B \cup H} \right) = 500$$
we have
$$\eqalign{ & n\left( {B \cup H} \right) = n\left( B \right) + n\left( H \right) - n\left( {B \cap H} \right) \cr & \Rightarrow \,500 = 200 + 475 - n\left( {B \cap H} \right) \cr & {\text{So, }}n\left( {B \cap H} \right) = 175 \cr} $$
Hence, persons who speak Hindi only $$ = n\left( H \right) - n\left( {B \cap H} \right) = 475 - 175 = 300$$

Let there be three non empty, non overlapping sets; inside a universal set $$U.$$ This creates $$8$$ regions marked as: $$a,\,b,\,c,\,d,\,e,\,f,\,g,\,h$$

$$\eqalign{ & {\bf{Statement}}\,{\bf{1:}}\,\,A - \left( {B - C} \right) = \left( {A - B} \right) \cup C \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{. represent region }}'a',\,{\text{R}}{\text{.H}}{\text{.S}}{\text{. represents }}'a,\,d,\,g' \cr & {\text{Hence, this is not correct}}{\text{.}} \cr & {\bf{Statement}}\,{\bf{2:}}\,\,A - \left( {B \cup C} \right) = \left( {A - B} \right) - C \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{. represent region }}'a',\,{\text{R}}{\text{.H}}{\text{.S}}{\text{. also represents }}'a' \cr & {\text{Hence, only statement 2 is correct}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{Let }}x < 0 \cr & \therefore \,\left( {gof} \right)\left( x \right) = g\left( {f\left( x \right)} \right) \cr & = g\left( {{x^3} + 1} \right) \cr & = {\left[ {\left( {{x^3} + 1} \right) - 1} \right]^{\frac{1}{3}}}\,\,\left( {\because x < 0 \Rightarrow {x^3} + 1 < 1} \right) \cr & = {\left( {{x^3}} \right)^{\frac{1}{3}}} \cr & = x \cr & {\text{Let }}x \geqslant 0 \cr & \therefore \,\left( {gof} \right)\left( x \right) = g\left( {f\left( x \right)} \right) \cr & = g\left( {{x^2} + 1} \right) \cr & = {\left( {\left( {{x^2} + 1} \right) - 1} \right)^{\frac{1}{2}}}\,\,\left( {\because x \geqslant 0 \Rightarrow {x^2} + 1 \geqslant 1} \right) \cr & = {\left( {{x^2}} \right)^{\frac{1}{2}}} \cr & = \left| x \right| \cr & = x\,\,\,\,\,\,\left( {\because x \geqslant 0} \right) \cr & \therefore \,\left( {gof} \right)\left( x \right) = x,\,\forall \,x\, \in \,R \cr} $$