Question

If $ A = (0),{\text{B = (x:x > 15 and x < 5), c = (x:x - 5 = 0), D = (x:}}{{\text{x}}^2} = 25)\;{\text{E = (x:x is a root of equation }}{{\text{x}}^2} - 2x - 15 = 0, $ then pair of equal sets is:
I.A, B
II.B, C
III.C, D
IV.C, E

Answer 1

Hint: Set is the collection of the distinct objects or the elements which have common property. Here we will convert all the given sets in the roster form where the elements of a set are listed and separated by commas and then will check for the equal sets.

Complete step-by-step answer:
Let us all the given sets in the form of a roster.
 $ A = (0) $
 $ \Rightarrow A = \{ 0\} $ .... (I)
Similarly for all the other sets-
 $ {\text{B = (x:x > 15 and x < 5)}} $
No number can satisfy the given condition being the number greater than $ 15 $ and less than $ 5 $ simultaneously.
Therefore, $ B = \{ \} $
B is the null set and it can be written as –
 $ B = \phi $ .... (II)
Now, $ {\text{, c = (x:x - 5 = 0)}} $
Given condition is: $ x - 5 = 0 $
Make the required “x” the subject and take another term on the right hand side of the equation. Remember when the term is moved from one side to another, the sign of the term is also changed. Positive terms becomes negative and vice-versa.
 $ \Rightarrow x = 5 $
Therefore, $ C = \{ 5\} $ .... (III)
Now, $ {\text{ D = (x:}}{{\text{x}}^2} = 25)\; $
Given condition is : $ {x^2} = 25 $
Take square-root on both the sides of the equation.
 $ \Rightarrow \sqrt {{x^2}} = \sqrt {25} $
Square and square-root cancel each other on the left hand side of the equation.
 $ \Rightarrow x = \pm 5 $
So, $ D = \{ - 5, + 5\} $ ..... (IV)
Similarly, $ {\text{E = (x:x is a root of equation }}{{\text{x}}^2} - 2x - 15 = 0 $
Given condition is:
 $ {{\text{x}}^2} - 2x - 15 = 0 $
The above equation can be re-written as –
 $ \underline {{{\text{x}}^2} - 5x} + \underline {3x - 15} = 0 $
Find the common from the pair of the terms –
 $ \Rightarrow x(x - 5) + 3(x - 5) = 0 $
Take out the common multiple –
 $ \Rightarrow (x - 5)(x + 3) = 0 $
We get,
 $
  x - 5 = 0\;or\;{\text{x + 3 = 0}} \\
   \Rightarrow {\text{x = 5}}\;{\text{or x = - 3}} \\
  $
Also, given that “x” is positive.
So, $ E = \{ 5\} $ .... (V)
From the equations (I), (II), (III), (IV).
We can observe that –
Only $ C = E $ are exactly the same.
So, the correct answer is “Option IV”.

Note: Always remember the basic concepts of the algebraic factorization. Follow the given conditions properly while converting the set-builder to roster form. Set builder notation is used to describe a set by enumerating the elements or stating the properties to satisfy for its members.