Class 12 Relations And Functions · Case Study5 Marksmixed_relations_and_functions
Class 12 Maths · Relations And Functions · Case Study #67
Bike race: counting relations and functions
Part (1) 1 Marks
How many relations are possible from \(B\) to \(G\)?
Answer: Option (A): \(2^6\)
- \(|B\times G|=3\cdot2=6\), so there are \(2^6\) subsets and hence relations.
Part (2) 1 Marks
On the set \(B\) of selected boys, define \(R=\{(x,y)\in B\times B:x\text{ and }y\text{ are students of the same sex}\}\). Then \(R\) is
Answer: Option (A): Equivalence
- All members of \(B\) are boys, so \(R=B\times B\), an equivalence relation.
Part (3) 1 Marks
How many functions can be formed from \(B\) to \(G\)?
Answer: Option (D): \(2^3\)
- Each of three inputs has two choices of image, giving \(2^3\).
Part (4) 1 Marks
Let \(R:B\to G\) be defined by \(R=\{(b_1,g_1),(b_2,g_2),(b_3,g_1)\}\). Then \(R\) is
Answer: Option (B): Surjective
- Both \(g_1\) and \(g_2\) occur as images, so the function is onto. The distinct inputs \(b_1,b_3\) share an image, so it is not injective.
Part (5) 1 Marks
How many injective functions from \(B\) to \(G\) are possible?
Answer: Option (A): 0
- Three domain elements cannot be assigned distinct images in a two-element codomain. There are zero injective functions.