Math Problem Statement

P2(b)Question: P2(b): Let X={-1,1,2} and f(x)=x^2. If M_1={-1,-1,1,1,2,2}, M_2={1,1,-2,-2,-2}, defined by f(M_i )=N_i ;i=1,2. Then, find f(M_1∪M_2), f(M_1)∪f(M_2), f(M_1∩M_2), f(M_1⊎M_2) and f(M_1⊖M_2).

Solution

Let us carefully compute each part of the problem, step by step.


Step 1: Definitions and Interpretations

We are given:

  • Set XX: X={1,1,2}X = \{-1, 1, 2\}

  • Function f(x)=x2f(x) = x^2.

  • Sets:

    • M1={1,1,1,1,2,2}M_1 = \{-1, -1, 1, 1, 2, 2\}
    • M2={1,1,2,2,2}M_2 = \{1, 1, -2, -2, -2\}
  • M1M2M_1 \cup M_2: Union of the two sets.

  • M1M2M_1 \cap M_2: Intersection of the two sets.

  • M1M2M_1 \oplus M_2: Symmetric difference of the two sets (M1M2M1M2M_1 \cup M_2 - M_1 \cap M_2).

  • f(M)f(M): Applying ff to each element of MM, resulting in a set of the images under ff.


Step 2: Applying the Function f(x)=x2f(x) = x^2

  1. Compute f(M1)f(M_1):

    • M1={1,1,1,1,2,2}M_1 = \{-1, -1, 1, 1, 2, 2\}
    • Apply f(x)=x2f(x) = x^2: f(1)=1,f(1)=1,f(2)=4f(-1) = 1, f(1) = 1, f(2) = 4.
    • f(M1)={1,4}f(M_1) = \{1, 4\}.
  2. Compute f(M2)f(M_2):

    • M2={1,1,2,2,2}M_2 = \{1, 1, -2, -2, -2\}
    • Apply f(x)=x2f(x) = x^2: f(1)=1,f(2)=4f(1) = 1, f(-2) = 4.
    • f(M2)={1,4}f(M_2) = \{1, 4\}.

Step 3: Operations on Sets

Now compute the required expressions:

  1. f(M1M2)f(M_1 \cup M_2):

    • M1M2={1,1,2,2}M_1 \cup M_2 = \{-1, 1, 2, -2\}
    • Apply f(x)f(x): f(1)=1,f(1)=1,f(2)=4,f(2)=4f(-1) = 1, f(1) = 1, f(2) = 4, f(-2) = 4.
    • f(M1M2)={1,4}f(M_1 \cup M_2) = \{1, 4\}.
  2. f(M1)f(M2)f(M_1) \cup f(M_2):

    • f(M1)={1,4},f(M2)={1,4}f(M_1) = \{1, 4\}, f(M_2) = \{1, 4\}.
    • Union: f(M1)f(M2)={1,4}f(M_1) \cup f(M_2) = \{1, 4\}.
  3. f(M1M2)f(M_1 \cap M_2):

    • M1M2={1}M_1 \cap M_2 = \{1\}
    • Apply f(x)f(x): f(1)=1f(1) = 1.
    • f(M1M2)={1}f(M_1 \cap M_2) = \{1\}.
  4. f(M1M2)f(M_1 \oplus M_2):

    • M1M2=(M1M2)(M1M2)M_1 \oplus M_2 = (M_1 \cup M_2) - (M_1 \cap M_2).
    • M1M2={1,2,2}M_1 \oplus M_2 = \{-1, -2, 2\}.
    • Apply f(x)f(x): f(1)=1,f(2)=4,f(2)=4f(-1) = 1, f(-2) = 4, f(2) = 4.
    • f(M1M2)={1,4}f(M_1 \oplus M_2) = \{1, 4\}.
  5. f(M1M2)f(M_1 \ominus M_2):

    • This is also the symmetric difference result.
    • f(M1M2)=f(M1M2)={1,4}f(M_1 \ominus M_2) = f(M_1 \oplus M_2) = \{1, 4\}.

Final Results

  • f(M1M2)={1,4}f(M_1 \cup M_2) = \{1, 4\}
  • f(M1)f(M2)={1,4}f(M_1) \cup f(M_2) = \{1, 4\}
  • f(M1M2)={1}f(M_1 \cap M_2) = \{1\}
  • f(M1M2)={1,4}f(M_1 \oplus M_2) = \{1, 4\}
  • f(M1M2)={1,4}f(M_1 \ominus M_2) = \{1, 4\}

Would you like a detailed explanation of any step?
Here are 5 related questions to expand on the concepts:

  1. How does the function f(x)f(x) affect the structure of sets in general?
  2. What is the difference between the union and symmetric difference of two sets?
  3. Why does f(M1M2)f(M_1 \cap M_2) result in a smaller set than f(M1)f(M2)f(M_1) \cup f(M_2)?
  4. How can these operations be visualized using Venn diagrams?
  5. What happens when f(x)f(x) is not one-to-one or onto?

Tip: Always simplify the sets first before applying a function like f(x)f(x) to avoid redundant computations.

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Math Problem Analysis

Mathematical Concepts

Set theory
Functions
Union and intersection of sets
Symmetric difference of sets
Image of sets under a function

Formulas

f(x) = x^2
M_1 ∪ M_2
M_1 ∩ M_2
M_1 ⊕ M_2 = (M_1 ∪ M_2) − (M_1 ∩ M_2)

Theorems

Properties of set operations
Function mapping and images

Suitable Grade Level

Grades 9-12