Q.Write the set $D = \{1, 4, 9, 16, 25, 36\}$ in set-builder form.

Showing the 12 most recent of 19 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Roster form of the set $\{x : x \text{ is an integer}, x^2 \le 4\}$ is

   (a) $\{1, 2\}$

   (b) $\{0, 1, 2\}$

   (c) $\{-2, -1, 0, 1, 2\}$

   (d) $\{-2, 2\}$

   ›Reveal solutionSolution

   Solve $x^2 \le 4$ for integers: $-2 \le x \le 2$, giving the roster form $\{-2,-1,0,1,2\}$.

   The set is defined by the rule $x^2 \le 4$ with $x$ an integer. Solving the inequality: $x^2 \le 4 \implies -2 \le x \le 2$. The integers in this range are $-2, -1, 0, 1, 2$ — five elements in total. Listing them explicitly (the roster form) gives $\{-2, -1, 0, 1, 2\}$.

   ✓Final answer

   (c) $\{-2, -1, 0, 1, 2\}$.

2. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $A = \{x : x \in N \text{ and } (x-2)(x-3)=0\}$ then $n(A) =$

   (a) 2

   (b) 3

   (c) 1

   (d) 4

   ›Reveal solutionSolution

   $A = \{x \in N : (x-2)(x-3) = 0\} = \{2, 3\}$, so $n(A) = 2$.

   The condition $(x-2)(x-3) = 0$ holds when $x = 2$ or $x = 3$ (by the zero-product rule). Both values are natural numbers, so $A = \{2, 3\}$.

   The cardinality $n(A)$ is the number of elements in $A$, which is $2$.

   ✓Final answer

   The correct option is (a) 2.

3. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $X = \{x : x \in N \text{ and } x \text{ is a prime number}\}$ then $n(X) =$

   (a) 10

   (b) 100

   (c) 1000

   (d) none of these

   ›Reveal solutionSolution

   $X$ is the set of all prime numbers in $N$, which is an infinite set, so $n(X)$ cannot be any of the finite values listed.

   $X = \{x \in N : x \text{ is a prime number}\} = \{2, 3, 5, 7, 11, \ldots\}$.

   A classical result (proved by Euclid) states there are infinitely many primes — the list never terminates. So $X$ is an infinite set and $n(X)$ is not a finite number like 10, 100, or 1000.

   ✓Final answer

   The correct option is (d) none of these.

4. CBSE 2025Set ANNUAL1 markMCQ
   Q.$A = \{x : x^2 = 4\} \Rightarrow A =$

   (a) $\{4, -4\}$

   (b) $\{2, -2\}$

   (c) $\{2, 4\}$

   (d) $\{-2, 4\}$

   ›Reveal solutionSolution

   $A = \{x : x^2 = 4\} = \{2, -2\}$, the two real solutions of $x^2 = 4$.

   Solving $x^2 = 4$: taking the square root of both sides gives $x = \pm 2$, i.e., $x = 2$ or $x = -2$. Both satisfy the original equation, so $A = \{2, -2\}$.

   ✓Final answer

   The correct option is (b) $\{2, -2\}$.

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.$A = \{x : x^2 - 5x + 6 = 0\} \Rightarrow A =$

   (a) $\{2, 3\}$

   (b) $\{2, 3, 6\}$

   (c) $\{-2, -3\}$

   (d) $\{3, -2\}$

   ›Reveal solutionSolution

   $A = \{x : x^2 - 5x + 6 = 0\} = \{2, 3\}$.

   Factorise $x^2 - 5x + 6 = (x-2)(x-3) = 0$, giving $x = 2$ or $x = 3$. So $A = \{2, 3\}$.

   ✓Final answer

   The correct option is (a) $\{2, 3\}$.

6. CBSE 2025Set ANNUAL1 markMCQ
   Q.$X = \{x : x^2 - 2x - 3 = 0\} \Rightarrow X =$

   (a) $\{3, 1\}$

   (b) $\{-3, 1\}$

   (c) $\{3, -1\}$

   (d) $\{-3, -1\}$

   ›Reveal solutionSolution

   $X = \{x : x^2 - 2x - 3 = 0\} = \{3, -1\}$.

   Factorise $x^2 - 2x - 3 = (x-3)(x+1) = 0$, giving $x = 3$ or $x = -1$. So $X = \{3, -1\}$.

   ✓Final answer

   The correct option is (c) $\{3, -1\}$.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.$X = \{x : 9x^2 - 6x + 1 = 0\} \Rightarrow X =$

   (a) $\{1/3\}$

   (b) $\{1/3, -1/3\}$

   (c) $\{3, 1\}$

   (d) $\{1/3, 1/3\}$

   ›Reveal solutionSolution

   $X = \{x : 9x^2 - 6x + 1 = 0\} = \{1/3\}$, since the quadratic is a perfect square with a repeated root.

   Factorise: $9x^2 - 6x + 1 = (3x - 1)^2 = 0$, so $x = 1/3$ (a repeated/double root). As a set, an element is listed only once regardless of multiplicity, so $X = \{1/3\}$.

   ✓Final answer

   The correct option is (a) $\{1/3\}$.

8. CBSE 2025Set ANNUAL1 markMCQ
   Q.$A = \{x : x^3 - 4x = 0\} \Rightarrow A =$

   (a) $\{0, 4\}$

   (b) $\{0, 2, -2\}$

   (c) $\{0, 2, 4\}$

   (d) $\{0, 4, -4\}$

   ›Reveal solutionSolution

   $A = \{x : x^3 - 4x = 0\} = \{0, 2, -2\}$.

   Factorise: $x^3 - 4x = x(x^2 - 4) = x(x-2)(x+2) = 0$. This gives $x = 0$, $x = 2$, or $x = -2$.

   So $A = \{0, 2, -2\}$.

   ✓Final answer

   The correct option is (b) $\{0, 2, -2\}$.

9. CBSE 2025Set ANNUAL1 mark
   Q.Write True or False: The collection of the best eleven batsmen of the world is an example of a set.

   ›Reveal solutionSolution

   A set requires a well-defined collection of objects; "best eleven batsmen" is subjective and varies from person to person, so it is not a set.

   For a collection to be a set, membership must be well-defined — for any given object, it must be possible to say definitely whether it belongs to the collection or not.

   "The best eleven batsmen of the world" is a matter of opinion: different critics or fans would choose different players. Since there is no universally agreed, objective criterion, this collection is not well-defined and hence is not a set.

   ✓Final answer

   False.

10. CBSE 2024Set ANNUAL1 markMCQ
    Q.The solution set of the equation $x^2 - 3x + 2 = 0$ in roster form is

    (a) $\{1, -2\}$

    (b) $\{3, 4\}$

    (c) $\{1, 2\}$

    (d) None of these

    ›Reveal solutionSolution

    Factor the quadratic to get the two roots, then list them as a set — roster form.

    We are given $x^2 - 3x + 2 = 0$. Factor by splitting the middle term: we need two numbers that multiply to $2$ and add to $-3$, which are $-1$ and $-2$.

    $x^2 - x - 2x + 2 = 0$

    $x(x-1) - 2(x-1) = 0$

    $(x-1)(x-2) = 0$

    So $x = 1$ or $x = 2$. The solution set written in roster (listing) form is $\{1, 2\}$.

    ✓Final answer

    (c) $\{1, 2\}$.

11. CBSE 2024Set ANNUAL1 mark
    Q.Write the set of the letters of the word "TRIGONOMETRY".

    ›Reveal solutionSolution

    The set of distinct letters in "TRIGONOMETRY" is $\{T, R, I, G, O, N, M, E, Y\}$, a set of 9 elements.

    Writing out the letters of TRIGONOMETRY: T, R, I, G, O, N, O, M, E, T, R, Y. A set cannot contain repeated elements, so each letter is listed only once even though T, R and O each occur twice. Collecting the distinct letters in the order they first appear gives $\{T, R, I, G, O, N, M, E, Y\}$.

    ✓Final answer

    $\{T, R, I, G, O, N, M, E, Y\}$.

12. CBSE 2023Set ANNUAL1 markMCQ
    Q.Set $B=\{x : x \text{ is an integer}, -\frac{1}{2}<x<\frac{9}{2}\}$. The list of all elements of $B$ will be:

    (a) $\{0,1,2,3,4\}$

    (b) $\{1,2,3,4\}$

    (c) $\{-1,0,1,2,3\}$

    (d) $\{-1,0,1,2,3,4\}$

    ›Reveal solutionSolution

    $B=\{0,1,2,3,4\}$, the integers strictly between $-\frac12$ and $\frac92$.

    We are given $B=\{x : x \text{ is an integer},\ -\dfrac{1}{2}<x<\dfrac{9}{2}\}$.

    Since $-\dfrac12=-0.5$ and $\dfrac92=4.5$, we list every integer strictly greater than $-0.5$ and strictly less than $4.5$. The smallest integer greater than $-0.5$ is $0$ (not $-1$, since $-1$ is less than $-0.5$), and the largest integer less than $4.5$ is $4$. So the integers satisfying the condition are $0,1,2,3,4$.

    $B=\{0,1,2,3,4\}$

    ✓Final answer

    The correct option is (a) $\{0,1,2,3,4\}$.