Sets and Relations MCQ Questions & Answers in Calculus | Maths

If $$R$$ be the relation, $$x\,R\,y \Leftrightarrow x - y$$   is divisible by $$m.$$
$$x\,R\,x$$   because $$x-x$$   is divisible by $$m.$$  So, $$R$$ is reflexive.
$$x\,R\,y \Rightarrow y\,R\,x.$$    So, $$R$$ is symmetric.
$$x\,R\,y$$   and $$y\,R\,z \Rightarrow x - y = {k_1}m,\,\,y - z = {k_2}m$$
$$\therefore x - z = \left( {{k_1} + {k_2}} \right)m.$$     So, $$R$$ is transitive.
As $$R$$ is reflexive, symmetric and transitive, it is an equivalence relation.

$$\eqalign{ & {\text{Let }}A = \left\{ {4n + 2:n\, \in N} \right\}\,{\text{and }}B = \left\{ {3n:n\, \in \,N} \right\} \cr & \Rightarrow \,A = \left\{ {6,\,10,\,14,\,18,\,22,\,26,\,30,\,34,\,38,\,42,.....} \right\} \cr & {\text{and }}B = \left\{ {3,\,6,\,9,\,12,\,15,\,18,\,21,\,24,\,27,\,30,.....} \right\} \cr & \therefore \,A \cap B = \left\{ {6,\,18,\,30,\,42,.....} \right\} \cr & = 6 + 12n - 12 \cr & = 12n - 6 \cr & {\text{Hence, }}A \cap B = \left\{ {12n - 6:n{\text{ is a natural number}}} \right\} \cr} $$

$$P\left( \phi \right)$$  is the power set of the set $$\phi .$$
$$\therefore $$  Cardinality $$ = P\left\{ {P\left[ {P\left( \phi \right)} \right]} \right\} = 4$$

$$\eqalign{ & f\left( x \right) + 2f\left( {\frac{1}{x}} \right) = 3x\,\,\,\,\,.....\left( 1 \right) \cr & f\left( {\frac{1}{x}} \right) + 2f\left( x \right) = \frac{3}{x}\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
Adding (1) and (2)
$$ \Rightarrow \,\,f\left( x \right) + f\left( {\frac{1}{x}} \right) = x + \frac{1}{x}$$
Substracting (1) from (2)
$$ \Rightarrow \,\,f\left( x \right) - f\left( {\frac{1}{x}} \right) = \frac{3}{x} - 3x$$
On adding the above equations
$$\eqalign{ & \Rightarrow \,\,f\left( x \right) = \frac{2}{x} - x \cr & f\left( x \right) = f\left( { - x} \right) \cr & \Rightarrow \,\,\frac{2}{x} - x = \frac{{ - 2}}{x} + x \cr & \Rightarrow \,\,x = \frac{2}{x} \cr & {x^2} = 2\,\,\,\,{\text{or }}\,x = \sqrt 2 , - \sqrt 2 \cr} $$

$$\eqalign{ & n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \cr &= 12 + 9 - 4 \cr &= 17 \cr & {\text{Now, }}n\left( {{{\left( {A \cup B} \right)}^C}} \right) = n\left( U \right) - n\left( {A \cup B} \right) \cr &= 20 - 17 \cr &= 3 \cr} $$

$$\eqalign{ & {4^n} - 3n - 1 \cr & = {\left( {1 + 3} \right)^n} - 3n - 1 \cr & = \left[ {^n{C_0} + {\,^n}{C_1}.3 + {\,^n}{C_2}{{.3}^2} + ..... + {\,^n}{C_n}{{.3}^n}} \right] - 3n - 1 \cr & = 9\left[ {^n{C_2} + {\,^n}{C_3}.3 + ..... + {\,^n}{C_n}{{.3}^{n - 2}}} \right] \cr} $$
$$\therefore \,\,{4^n} - 3n - 1$$    is a multiple of 9 for all $$n.$$
∴ $$X$$ = {$$x : x$$  is a multiple of 9 }
Also, $$Y = \left\{ {9\left( {n - 1} \right):n \in N} \right\} = $$      {All multiples of 9}
Clearly $$X \subset Y.$$
$$\therefore X \cup Y = Y.$$

$$a$$ is not an element of $$\left\{ {\left\{ a \right\},\,b} \right\}$$
$$\therefore \,a\, \notin \,\left\{ {\left\{ a \right\},\,b} \right\}$$
$$\left\{ {b,\,c} \right\}$$  is the element of $$\left\{ {a,\,\left\{ {b,\,c} \right\}} \right\}$$
$$\eqalign{ & \therefore \,\left\{ {b,\,c} \right\}\, \in \,\left\{ {a,\,\left\{ {b,\,c} \right\}} \right\} \cr & b\, \in \,\left\{ {a,\,b} \right\}{\text{ but }}b \notin \,\left\{ {a,\,\left\{ {b,\,c} \right\}} \right\} \cr & \therefore \,\left\{ {a,\,b} \right\}\, \not\subset \,\left\{ {a,\,\left\{ {b,\,c} \right\}} \right\} \cr} $$

$$\eqalign{ & {\text{Let }}f\left( x \right) \ne 2{\text{ be true }} \cr & {\text{and }}f\left( y \right) = 2,\,f\left( z \right) \ne 1{\text{ are false}} \cr & \Rightarrow f\left( x \right) \ne 2,\,f\left( y \right) \ne 2,\,f\left( z \right) = 1 \cr & \Rightarrow f\left( x \right) = 3,\,f\left( y \right) = 3,\,f\left( z \right) = 1 \cr} $$
but the function is many one, similarly to other cases.

$$\eqalign{ & {\text{When }}x = 1,\,y = 7\, \in \,N,\,{\text{so }}\left( {1,\,7} \right) \in \,R \cr & {\text{When }}x = 2,\,y = 2 + 3 = 5\, \in \,N,\,{\text{so }}\left( {2,\,5} \right) \in \,R \cr & {\text{Again for }}x = 3,\,y = 3 + 2 = 5\, \in \,N,\,{\text{so }}\left( {3,\,5} \right) \in \,R \cr & {\text{Similarly for }}x = 4,\,y = 4 + \frac{6}{4}\, \notin \,N \cr & {\text{and for }}x = 5,\,y = 5 + \frac{6}{5}\, \notin \,N \cr & {\text{Thus, }}R = \left\{ {\left( {1,\,7} \right),\,\left( {2,\,5} \right),\,\left( {3,\,5} \right)} \right\} \cr & \therefore {\text{ Domain of }}R = \left\{ {1,\,2,\,3} \right\}{\text{ and Range of }}R = \left\{ {7,\,5} \right\} \cr} $$

$$\eqalign{ & g\left( x \right) = {x^2} + x - 2{\text{ and }}\frac{1}{2}\left( {gof} \right)\left( x \right) = 2{x^2} - 5x + 2 \cr & \Rightarrow g\left( {f\left( x \right)} \right) = 4{x^2} - 10x + 4 \cr & \Rightarrow {\left( {f\left( x \right)} \right)^2} + f\left( x \right) - 2 = 4{x^2} - 10x + 4 \cr & \Rightarrow {\left( {f\left( x \right)} \right)^2} + f\left( x \right) - \left( {4{x^2} - 10x + 6} \right) = 0 \cr & \Rightarrow f\left( x \right) = \frac{{ - 1 \pm \sqrt {16{x^2} - 40x + 25} }}{2} \cr & \Rightarrow f\left( x \right) = \frac{{ - 1 \pm \left( {4x - 5} \right)}}{2} \cr & \Rightarrow f\left( x \right) = \frac{{4x - 6}}{2}{\text{ }}\,{\text{or}}\,{\text{ }}\frac{{ - 4x + 4}}{2} \cr & \Rightarrow f\left( x \right) = 2x - 3{\text{ or }} - 2x + 2 \cr & {\text{Hence, }}f\left( x \right) = 2x - 3 \cr} $$