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NCERT Exemplar · Q1

Q.Write the following sets in the roster form:

(i) A={x:x∈R, 2x+11=15}A = \{x : x \in \mathbb{R},\ 2x + 11 = 15\}
(ii) B={x∣x2=x, x∈R}B = \{x \mid x^2 = x,\ x \in \mathbb{R}\}
(iii) C={x∣x is a positive factor of a prime number p}C = \{x \mid x \text{ is a positive factor of a prime number } p\}
Rajasthan RbseShort· 2mImportance★★★★★est
57% · 75/132 Questions
✓ Free question

To write sets in roster form, we identify all elements that satisfy the given conditions. For set A, we solve a linear equation; for set B, a quadratic equation; and for set C, we list the factors of a prime number. The sets are A={2}\boxed{A = \{2\}}, B={0,1}\boxed{B = \{0, 1\}}, and C={1,p}\boxed{C = \{1, p\}}.

When a set is given in set-builder notation, like S={x:condition(x)}S = \{x : \text{condition}(x)\}, it describes the properties that elements xx must possess to be part of the set. To write such a set in roster form, we need to find all specific values of xx that satisfy these conditions and then list them explicitly within curly braces. This process involves solving equations or understanding definitions related to the conditions.

Let's break down each set.

(i) A={x:x∈R, 2x+11=15}A = \{x : x \in \mathbb{R},\ 2x + 11 = 15\}

  1. Understand the condition: The set AA consists of all real numbers xx such that 2x+11=152x + 11 = 15. The core task is to solve this linear equation for xx.

  2. Solve the equation:

    We have the equation:

    2x+11=152x + 11 = 15

    Subtract 1111 from both sides:

    2x=15−112x = 15 - 11

    2x=42x = 4

    Divide by 22:

    x=42x = \frac{4}{2}

    x=2x = 2

  3. Check the domain: The condition states x∈Rx \in \mathbb{R} ( xx is a real number). Our solution x=2x=2 is indeed a real number.

  4. Write in roster form: Since x=2x=2 is the only value that satisfies the condition, the set AA contains only this element.

    A={2}A = \{2\}

(ii) B={x∣x2=x, x∈R}B = \{x \mid x^2 = x,\ x \in \mathbb{R}\}

  1. Understand the condition: The set BB consists of all real numbers xx such that x2=xx^2 = x. We need to solve this quadratic equation.

  2. Solve the equation:

    The equation is:

    x2=xx^2 = x

    To solve a quadratic equation, it's best to bring all terms to one side and set the expression to zero:

    x2−x=0x^2 - x = 0

    Now, factor out the common term xx:

    x(x−1)=0x(x - 1) = 0

    For the product of two terms to be zero, at least one of the terms must be zero. So, we have two possibilities:

    x=0orx−1=0x = 0 \quad \text{or} \quad x - 1 = 0

    Solving the second part:

    x−1=0  ⟹  x=1x - 1 = 0 \implies x = 1

    So, the solutions are x=0x=0 and x=1x=1.

    Watch out

    A common mistake here is to divide both sides of x2=xx^2 = x by xx, leading to x=1x=1. This loses the solution x=0x=0. Always move all terms to one side and factor when solving equations involving variables in denominators or when a variable could be zero.

  3. Check the domain: The condition states x∈Rx \in \mathbb{R}. Both 00 and 11 are real numbers.

  4. Write in roster form: The set BB contains the elements 00 and 11.

    B={0,1}B = \{0, 1\}

(iii) C={x∣x is a positive factor of a prime number p}C = \{x \mid x \text{ is a positive factor of a prime number } p\}

  1. Understand the condition: The set CC consists of all xx such that xx is a positive factor of a prime number pp. This means we need to understand what a prime number is and what its factors are.

  2. Define a prime number:

    Important

    A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself.

    Examples: 2,3,5,7,11,…2, 3, 5, 7, 11, \dots

  3. Identify factors of a prime number:

    Let pp be any prime number. By its definition, the only positive numbers that divide pp evenly are 11 and pp.

    For example:

    • If p=2p=2, its positive factors are 1,21, 2.
    • If p=3p=3, its positive factors are 1,31, 3.
    • If p=7p=7, its positive factors are 1,71, 7.

    In general, for any prime number pp, its positive factors are always 11 and pp.

  4. Write in roster form: The elements of set CC are these positive factors.

    C={1,p}C = \{1, p\}

✓Final answer

The sets in roster form are A={2}\boxed{A = \{2\}}, B={0,1}\boxed{B = \{0, 1\}}, and C={1,p}\boxed{C = \{1, p\}}.

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