Sets and Relations MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & n\left( T \right) = 50 \cr & n\left( D \right) = 30 \cr & n\left( H \right) = 40 \cr & n\left( T \right) = n\left( D \right) + n\left( H \right) - n\left( {DnH} \right) \cr & 50 = 30 + 40 - n\left( {D \cap H} \right) \cr & n\left( {D \cap H} \right) = 70 - 50 = 20 \cr} $$
Number of people having diabetes and high blood pressure $$ = 20$$

Since, $$\left( {X \cup Y} \right)$$  has 60 elements, $$X$$ has 38 elements and $$Y$$ has 42 elements.
We know that
$$\eqalign{ & \left( {X \cup Y} \right) = X + Y - X \cap Y \cr & {\text{or,}}\,\,60 = 38 + 42 - \left( {X \cap Y} \right) \cr & {\text{or,}}\,\,\left( {X \cap Y} \right) = 80 - 60 = 20 \cr} $$

$$\because \,\,\left( {1,1} \right) \notin R$$
⇒ $$R$$ is not reflexive (2, 3) $$ \notin R$$   but (3, 2) $$ \notin R$$
∴ $$R$$ is not symmetric

$$\eqalign{ & {\text{Let }}f\left( x \right) = ax + b.....(1) \cr & \Rightarrow f\left( {ax + b + x} \right) = x + ax + b \cr & \Rightarrow f\left( {\left( {a + 1} \right)x + b} \right) = \left( {a + 1} \right)x + b \cr & {\text{Replace }}\left( {a + 1} \right)x + b\,{\text{by }}y{\text{, we have}} \cr & \Rightarrow f\left( y \right) = \left( {a + 1} \right)\left( {\frac{{y - b}}{{a + 1}}} \right) + b \cr & {\text{or }}f\left( x \right) = \left( {a + 1} \right)\left( {\frac{{x - b}}{{a + 1}}} \right) + b \cr} $$
$$\therefore $$  required number of linear functions is 2.

$$\eqalign{ & f\left( x \right) = \frac{{\sin \left( {\left[ x \right]\pi } \right)}}{{{x^2} + x + 1}} \cr & {\text{Let }}\left[ x \right] = n\, \in \,{\text{integer}} \cr & \therefore \,\sin \left[ x \right]\pi = 0{\text{ or }}f\left( x \right) = 0 \cr & {\text{Hence, }}f\left( x \right){\text{ is constant function}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{Let, }}y = {5^{x\left( {x - 4} \right)}} \cr & \Rightarrow x\left( {x - 4} \right) = {\log _5}y \cr & \Rightarrow {x^2} - 4x - {\log _5}y = 0 \cr & \Rightarrow x = \frac{{4 \pm \sqrt {16 + 4\,{{\log }_5}y} }}{2} \cr & \Rightarrow x = \left( {2 \pm \sqrt {4 + {{\log }_5}y} } \right) \cr & {\text{But }}x \geqslant 4,{\text{ so }}x = \left( {2 + \sqrt {4 + {{\log }_5}y} } \right) \cr & \therefore \,{f^{ - 1}}\left( x \right) = 2 + \sqrt {4 + {{\log }_5}x} \cr} $$

We know, for two sets $$A$$ and $$B$$
$$\eqalign{ & A - B = A - \left( {A \cap B} \right) \cr & \therefore \,n\left( {A - B} \right) = n\left( A \right) - n\left( {A \cap B} \right) \cr & {\text{Given, }}n\left( A \right) = 115,\,n\left( B \right) = 326{\text{ and }}n\left( {A - B} \right) = 47 \cr & \Rightarrow \,47 = 115 - n\left( {A \cap B} \right) \cr & \Rightarrow \,n\left( {A \cap B} \right) = 68 \cr & {\text{Consider }}n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \cr & = 115 + 326 - 68 = 373 \cr} $$

In the given Venn diagram, shaded area between set $$P$$ and $$Q$$ is $$\left( {P \cap Q} \right) - R$$   and shaded area between set $$P$$ and $$R$$ is $$\left( {P \cap R} \right) - Q.$$   So, both the shaded area is union of these two area and is represented by $$\left( {\left( {P \cap Q} \right) - R} \right) \cup \left( {\left( {P \cap R} \right) - Q} \right).$$

$$\begin{array}{l}(\text{i})A\subseteq A\text{ ie, }ARA,\forall A\in P\left(S\right)\\ \therefore R\text{ is reflexive}\text{.}\\ (\text{ii})A\subseteq B\cancel{\Rightarrow }B\subseteq A\\ \therefore ARB\cancel{\Rightarrow }BRA\\ \text{So, }R\text{ is not symmetric}\text{.}\\ (\text{iii})ARB\text{ and }BRA\\ \Rightarrow A\subseteq B\text{ and }B\subseteq A\\ \Rightarrow A=B\\ \text{Thuse, }R\text{ is anti-symmetric}\text{.}\\ \text{(iv) }ARB\text{ and }BRC\\ \Rightarrow A\subseteq B\text{ and }B\subseteq C\\ \Rightarrow A\subseteq C\\ \Rightarrow ARC\\ \therefore R\text{ is transitive relation}\text{.}\end{array}$$
Thus, $$R$$ is partially ordered relation but not an equivalence relation.

$$R$$ is reflexive, symmetric and transitive. So, the most appropriate option is (D)