In the word “CATARACT” , the distinct letters = {C, A, T, R} = {A, C, R, T} In the word “TRACT”, the distinct letters = {T, R, A, C} = {A, C, R, T} Excluding the repetition of letters, both the sets...
Set A = {1, 2} Set B = {1, 2} Set C = {3, 1} Set D = {1, 3} Set E = {1, 2} Set F = {1, 3} Equal Sets: (i) A, B, E (II) C, D, F
Equivalent sets: (i) Set A, Set E, Set H - Because of two identical elements (ii) Set B, Set D, Set G - Because of four identical elements (iii) Set C, Set F - Because of three identical elements...
Answers: (i) A = {2, 3} B = x2 + 5x + 6 = 0 x2 + 3x + 2x + 6 = 0 x(x+3) + 2(x+3) = 0 (x+3) (x+2) = 0 x = -2 and -3 x = {–2, –3} A and B are not equal. (ii) A = Every letter in WOLF A = {W, O, L, F}...
If the number of elements is same but the elements are different in a set, then it is said to be an equivalent sets. (i) A = {1, 2, 3} The number of elements = 3 (ii) B = {t, p, q, r, s} The number...
(i) A - x is the letters in the word reap A ={R, E, A, P} = {A, E, P, R} (ii) B - x is the letters in the word paper B = {P, A, E, R} = {A, E, P, R} (iii) C - x is the letters in the word rope C =...
Answers: (i) This set is an infinite set. Reason - Infinite set of concentric circles can be drawn in a plane. (ii) This set is a finite set. Reason - Only 26 letters in English Alphabets are...
When all the elements of two sets are similar, then those two sets are considered to be the same. (i) A = {1, 2, 3} (ii) B ={x ∈ R: x2–2x+1=0} x2–2x+1 = 0 (x–1)2 = 0 ∴ x = 1. B = {1} (iii) C= {1, 2,...
Answers: (i) This set is an infinite set. Reason - The integers less than 5 can be infinity. (ii) This set is an infinite set. Reason - In between two real numbers, the real numbers are...
Answers: (i) This set is an infinite set. Reason - Natural numbers greater than 5 can go till infinity. (ii) This set is a finite set. Reason - The natural numbers start from 1 and there are 199...
Answers: (i) It is an empty set. Reason - There isn't any natural number whose square is 2. (ii) It is an empty set. Reason - There isn't any natural number which is less than 8 and greater than 12....
Answers: (i) It is not an empty set Reason - All the numbers ending with 0. Except 0 is divisible by 5 and is even natural number. (ii) It is not an empty set. Reason - Two is the only even prime...
![(6,12]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-85d82cf64f306d0d0558ee7c723d02de_l3.png "Rendered by QuickLaTeX.com")
(ii) 
Solution: (i) $(6,12]=\{x, x \in \mathrm{R}, 6<x \leq 12\}$ (ii) $[-23,5)=\{x . x \in \mathrm{R},-23 \leq x<5\}$
(ii) ![[6,12]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-42de7806061f263ec283ca0ad64ee75b_l3.png "Rendered by QuickLaTeX.com")
Solution: (i) $(-3,0)=\{x, x \in \mathrm{R},-3<x<0\}$ (ii) $[6,12]=\{x, x \in \mathrm{R}, 6 \leq x \leq 12\}$
(ii) 
Solution: (i) $\{x . x \in \mathrm{R}, 0 \leq x<7\}=[0,7)$ (ii) $\{x: x \in \mathrm{R}, 3 \leq x \leq 4\}=[3,4]$
(ii) 
Solution: (i) $\{x: x \in \mathrm{R},-4<x \leq 6\}=(-4,6]$ (ii) $\{x, x \in \mathrm{R},-12<x<-10\}=(-12,-10)$
Solution: If with m elements A is a set $n(\mathrm{~A})=m$ then $n[\mathrm{P}(\mathrm{A})]=2^{m}$ If $A=\Phi$ we obtain $n(A)=0$ $n[\mathrm{P}(\mathrm{A})]=2^{0}=1$ As a result, P (A) has one...
Solution: (i) Subsets of {1, 2, 3} are Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and {1, 2, 3}. (ii) Only subset of Φ is Φ.
Solution: (i) Subsets of {a} are Φ and {a}. (ii) Subsets of {a, b} are Φ, {a}, {b}, and {a, b}.
Solution: Provided that A= {1, 2, {3, 4}, 5} (i) {Φ} ⊂ A is incorrect Φ∈ {Φ}; where, Φ ∈ A.
Solution: Provided that A= {1, 2, {3, 4}, 5} (i) Φ ∈ A is incorrect Φ is not an element of A. (ii) Φ ⊂ A is correct Φ is a subset of every set.
Solution: Provided that A= {1, 2, {3, 4}, 5} (i) {1, 2, 5} ∈ A is incorrect {1, 2, 5} is not an element of A. (ii) {1, 2, 3} ⊂ A is incorrect 3 ∈ {1, 2, 3}; in which, 3 ∉ A.
Solution: Provided that A= {1, 2, {3, 4}, 5} (i) 1⊂ A is incorrect We know that an element of a set can never be a subset of itself. (ii) {1, 2, 5} ⊂ A is correct Here each element of {1, 2, 5} is...
Solution: Provided that A= {1, 2, {3, 4}, 5} (i) {{3, 4}} ⊂ A is correct {3, 4} ∈ {{3, 4}} and {3, 4} ∈ A. (ii) 1∈A is correct 1 is an element of A.
Solution: Provided that A= {1, 2, {3, 4}, 5} (i) {3, 4} ⊂ A is incorrect As here 3 ∈ {3, 4}; in which, 3∉A. (ii) {3, 4} ∈A is correct {3, 4} is an element of A.
is an even natural number less than 6} ⊂ { is a natural number which divides 36}Solution: (i) The statement is false. Here elements of {a, b, c} are a, b, c. As a result, {a} ⊂ {a, b, c} (ii) The statement is true. {$x: x$ is an even natural number less than 6} = {2, 4} {$x: x$...
Solution: (i) The statement is false. 2 ∈ {1, 2, 3} in which, 2∉ {1, 3, 5} (ii) The statement is true. Here each element of {a} is also an element of {a, b, c}.
Solution: (i) The statement is false. Each element of {a, b} is an element of {b, c, a}. (ii) The statement is true. It is known that a, e are two vowels of the English alphabet.
Solution: (i) {x: x is an even natural number} ⊂ {x: x is an integer}
is a triangle in a plane}…{ is a rectangle in the plane} (ii) { is an equilateral triangle in a plane}… { is a triangle in the same plane}Solution: (i) {$x: x$ is a triangle in a plane} ⊄ {$x: x$ is a rectangle in the plane} (ii) {$x: x$ is an equilateral triangle in a plane} ⊂ {$x: x$ is a triangle in the same plane}
is a student of Class XI of your school} … { student of your school} (ii) { is a circle in the plane} … { is a circle in the same plane with radius 1 unit}Solution: (i) {$x: x$ is a student of Class XI of your school} ⊂ {$x: x$ student of your school} (ii) {$x: x$ is a circle in the plane} ⊄ {$x: x$ is a circle in the same plane with radius 1...
Solution: (i) {2, 3, 4} ⊂ {1, 2, 3, 4, 5} (ii) {a, b, c} ⊄ {b, c, d}