Find the pairs of equal sets, if any, give reasons:

$A = \{ 0\} ,$

$B = \{ x:x\, > \,15$ and $x\, < \,5\} $

$C = \{ x:x - 5 = 0\} ,$

$D = \left\{ {x:{x^2} = 25} \right\}$

$E = \{ \,x:x$ is an integral positive root of the equation ${x^2} - 2x - 15 = 0\,\} $

Since $0 \in A$ and $0$ does not belong to any of the sets $B, C, D$ and $E,$ it follows that, $A \neq B, A \neq C, A \neq D, A \neq E.$

Since $B =\phi$ but none of the other sets are empty. Therefore $B \neq C , B \neq D$ and $B \neq E$. Also $C =\{5\}$ but $-5 \in D$, hence $C \neq D$.

Since $E =\{5\}, C = E .$ Further, $D =\{-5,5\}$ and $E =\{5\},$ we find that, $D \neq E$

Thus, the only pair of equal sets is $C$ and $E .$

Match each of the set on the left in the roster form with the same set on the right described in set-builder form:

$(i)$ $\{1,2,3,6\}$ | $(a)$ $\{ x:x$ is a prime number and a divisor $6\} $ |

$(ii)$ $\{2,3\}$ | $(b)$ $\{ x:x$ is an odd natural number less than $10\} $ |

$(iii)$ $\{ M , A , T , H , E , I , C , S \}$ | $(c)$ $\{ x:x$ is natural number and divisor of $6\} $ |

$(iv)$ $\{1,3,5,7,9\}$ | $(d)$ $\{ x:x$ a letter of the work $\mathrm{MATHEMATICS}\} $ |

The set $A = \{ x:x \in R,\,{x^2} = 16$ and $2x = 6\} $ equals

Which of the following are examples of the null set

Set of odd natural numbers divisible by $2$

If $A$ and $B$ are any two non empty sets and $A$ is proper subset of $B$. If $n(A) = 4$, then minimum possible value of $n(A \Delta B)$ is (where $\Delta$ denotes symmetric difference of set $A$ and set $B$)

Set $A$ has $m$ elements and Set $B$ has $n$ elements. If the total number of subsets of $A$ is $112$ more than the total number of subsets of $B$, then the value of $m \times n$ is

- [JEE MAIN 2020]