Sets and Relations MCQ Questions & Answers in Calculus | Maths

Following venn diagram shows the relation $$A - \left( {B - C} \right)$$

In the above venn diagram, horizontal lines shows $$\left( {A - B} \right)$$  and vertical lines shows $$\left( {A \cap C} \right)$$
$$\therefore \,\left( {A - B} \right) \cup \left( {A \cap C} \right) = A - \left( {B - C} \right)$$

Given that $$F\left( n \right) = $$   set of all divisors of $$n$$ except $$1$$
$$\eqalign{ & \therefore \,F\left( {20} \right) = \left\{ {2,\,4,\,5,\,10,\,20} \right\}{\text{ and }}F\left( {16} \right) = \left\{ {2,\,4,\,8,\,16} \right\} \cr & \therefore \,F\left( {20} \right) \cap F\left( {16} \right) = \left\{ {2,\,4,\,5,\,10,\,20} \right\} \cap \left\{ {2,\,4,\,8} \right\} = \left\{ {2,\,4} \right\} \cr & {\text{Also, }}\left\{ {F\left( {20} \right) \cap F\left( {16} \right)} \right\} \subseteq F\left( y \right) \cr & {\text{So, least value of }}y = 2 \cr} $$

Number of elements given $$ = 6$$
Number of elements taken at a time $$ = 2{\text{ i}}{\text{.e}}{\text{.,}}\left( {p\,\& \,q} \right)$$
$$ \Rightarrow $$  Cardinality of the set $$\left( s \right) = {6^2} = 36$$
(because numbers are repeated).

The function $$f\left( x \right) = {x^2} + 2,\,x\, \in \,\left( { - \infty ,\,\infty } \right)$$       is not injective as $$f\left( 1 \right) = f\left( { - 1} \right){\text{ but }}1 \ne - 1.$$
The function $$f\left( x \right) = \left( {x - 4} \right)\left( {x - 5} \right),\,x\, \in \left( { - \infty ,\,\infty } \right)$$        is not one-one as $$f\left( 4 \right) = f\left( 5 \right),{\text{ but}}\,4 \ne 5.$$
The function, $$f\left( x \right) = \frac{{4{x^2} + 3x - 5}}{{4 + 3x - 5{x^2}}},\,x\, \in \left( { - \infty ,\,\infty } \right)$$        is also not injective as $$f\left( 1 \right) = f\left( { - 1} \right),{\text{ but }}1 \ne - 1$$
For the function, $$f\left( x \right) = \left| {x + 2} \right|,\,x\, \in \left[ { - 2,\,\infty } \right)$$
$$\eqalign{ & {\text{Let }}f\left( x \right) = f\left( y \right),\,x,\,y\, \in \left[ { - 2,\,\infty } \right) \cr & \Rightarrow \left| {x + 2} \right| = \left| {y + 2} \right| \cr & \Rightarrow x + 2 = y + 2 \cr & \Rightarrow x = y \cr} $$
So, $$f$$ is an injective.

$$\eqalign{ & {2^m} = {2^n} + 56 \cr & \Rightarrow \,{2^m} - {2^n} = 64 - 8 = {2^6} - {2^3} \cr} $$

If any of the inequations hold, it must hold for any real numbers $${x_1},{x_2},.....,{x_n}\,$$   and any $$n \in N.$$
∴ let $${x_1} = 1,{x_2} = 2,{x_3} = 3;n = 3$$      then we can check none of the inequalities (A), (B) or (C) are satisfied.

$$\eqalign{ & \left( {ho\left( {gof} \right)} \right)\left( x \right) = h\left\{ {g\left\{ {f\left( x \right)} \right\}} \right\} \cr & = h\left\{ {\tan \,{x^2}} \right\} \cr & = \log \left\{ {\tan \,{x^2}} \right\} \cr & \therefore {\text{ At }}x = \sqrt {\frac{\pi }{4}} \cr & \Rightarrow \left( {ho\left( {gof} \right)} \right)\left( x \right) = \log \,\tan \,\frac{\pi }{4} \cr & \Rightarrow \left( {ho\left( {gof} \right)} \right)\left( x \right) = \log \,1 = 0 \cr} $$

Case - I : $$x \in \left[ {0,9} \right]$$
$$\eqalign{ & 2\left( {3 - \sqrt x } \right) + x - 6\sqrt x + 6 = 0 \cr & \Rightarrow \,\,x - 8\sqrt x + 12 = 0 \cr & \Rightarrow \,\,\sqrt x = 4,2 \cr & \Rightarrow \,\,x = 16,4 \cr & {\text{Since}}\,{\text{ }}x \in \left[ {0,9} \right] \cr & \therefore \,\,x = 4 \cr} $$
Case - II : $$x \in \left[ {9,\infty } \right]$$
$$\eqalign{ & 2\left( {\sqrt x - 3} \right) + x - 6\sqrt x + 6 = 0 \cr & \Rightarrow \,\,x - 4\sqrt x = 0 \cr & \Rightarrow \,\,x = 16,0 \cr & {\text{Since }}x \in \left[ {9,\infty } \right] \cr & \therefore \,\,x = 16 \cr & {\text{Hence, }}x = 4\,\& 16 \cr} $$

$$A \cup P\left( A \right) = P\left( A \right)$$     is correct.
Since $$A$$ is a subset of its power set.

$$\eqalign{ & {\text{We have, }}R = \left\{ {\left( {a,\,b} \right):\frac{n}{{\left( {a - b} \right)}}:a,\,b\, \in \,Z} \right\} \cr & {\text{Let }}z\, \in \,Z.\,\therefore \,a - a = 0 = n \times 0\,\,{\text{i}}{\text{.e}}{\text{.,}}\,\,\frac{n}{{\left( {a - a} \right)}} \cr & \therefore \,\left( {a,\,a} \right)\, \in \,R{\text{ i}}{\text{.e}}{\text{., }}R{\text{ is reflexive}}{\text{.}} \cr & {\text{Let }}\left( {a,\,b} \right)\, \in \,R.\therefore \,\frac{n}{{\left( {a - b} \right)}} \cr & \Rightarrow \frac{n}{{ - \left( {a - b} \right)}} \Rightarrow \frac{n}{{\left( {b - a} \right)}} \Rightarrow \left( {b,\,a} \right)\, \in \,R \cr & \therefore \,R{\text{ is symmetric}}{\text{.}} \cr & {\text{Let }}\left( {a,\,b} \right),\left( {b,\,c} \right)\, \in \,R. \cr & \therefore \,\frac{n}{{\left( {a - b} \right)}},\,\frac{n}{{\left( {b - c} \right)}} \cr & \Rightarrow \frac{n}{{\left( {a - b} \right) + \left( {b - c} \right)}} \Rightarrow \frac{n}{{a - c}} \Rightarrow \left( {a,\,c} \right)\, \in \,R \cr & \therefore \,R{\text{ is transitive}}{\text{.}}\,\,\,\,\,\,\therefore \,R{\text{ is an equivalence relation}}{\text{.}} \cr} $$