IIT JEE Main Maths -Unit 1- Power set- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 1- Power set – Study Notes – New syllabus
IIT JEE Main Maths -Unit 1- Power set – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Power Set
Power Set
The power set of a set \( A \) is the set of all possible subsets of \( A \), including the empty set \( \varnothing \) and the set \( A \) itself.
Notation: The power set of \( A \) is denoted by \( P(A) \) or \( 2^A \).
If \( A = \{a_1, a_2, a_3, \dots, a_n\} \), then
\( P(A) = \{ B \mid B \subseteq A \} \)
Number of elements in Power Set:
If \( n(A) = k \), then \( n(P(A)) = 2^k \).
This is because each element of \( A \) can either be included or not included in a subset, giving \( 2 \) choices per element.
Example
Let \( A = \{1, 2, 3\} \). Find \( P(A) \) and the total number of subsets.
▶️ Answer / Explanation
Step 1: List all possible subsets of \( A \).
\( P(A) = \{\varnothing, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\} \)
Step 2: Count the number of subsets.
\( n(A) = 3 \Rightarrow n(P(A)) = 2^3 = 8 \)
Step 3: Verify inclusion:
All subsets, from none to all elements, are included in \( P(A) \).
Conclusion: \( P(A) \) contains \( 8 \) subsets in total.
Example
Find the power set of \( A = \{1, 2\} \).
▶️ Answer / Explanation
Step 1: List all subsets of \( A \).
- \( \varnothing \)
- \( \{1\} \)
- \( \{2\} \)
- \( \{1,2\} \)
Step 2: Write the power set.
\( P(A) = \{\varnothing, \{1\}, \{2\}, \{1,2\}\} \)
Step 3: Count elements: \( n(A) = 2 \Rightarrow n(P(A)) = 2^2 = 4 \).
Example
If \( B = \{x, y, z\} \), find \( P(B) \) and the number of subsets that contain element \( x \).
▶️ Answer / Explanation
Step 1: Write all subsets of \( B \).
\( P(B) = \{\varnothing, \{x\}, \{y\}, \{z\}, \{x,y\}, \{x,z\}, \{y,z\}, \{x,y,z\}\} \)
Step 2: Count subsets containing \( x \).
- \( \{x\}, \{x,y\}, \{x,z\}, \{x,y,z\} \)
Number of such subsets: 4.
Step 3: Verify total subsets: \( 2^3 = 8 \).
Example
Let \( C = \{1, 2, 3, 4\} \). Find the number of subsets of \( P(C) \) that contain exactly 8 elements.
▶️ Answer / Explanation
Step 1: The number of elements in \( P(C) \) is:
\( n(P(C)) = 2^{n(C)} = 2^4 = 16 \).
Step 2: A subset of \( P(C) \) with exactly 8 elements means we choose 8 subsets out of 16.
The number of such possible subsets is given by the combination formula:
\( \text{Number of subsets} = \dbinom{16}{8} = 12870 \)
Conclusion: There are 12,870 subsets of \( P(C) \) that contain exactly 8 elements.