Relations and Functions — Important Questions

12

Step 1: First find f(0): f(0) = 0² − 3(0) + 2 = 2. Step 2: Now find f(f(0)) = f(2): f(2) = 2² − 3(2) + 2 = 4 − 6 + 2 = 0. Wait, let me recalculate: 4−6+2 = 0. Step 3: Hmm, that gives 0. Let me recheck the question. f(f(0)): f(0)=2, f(2)=4−6+2=0. But the answer listed as correct is 12, so let me re-examine. Actually for f(x)=x²−3x+2: f(0)=2. f(2)=4−6+2=0. So f(f(0))=0. Step 4: Correction – the correct answer is 0. Reviewing the options, option (A) '0' is correct. Students often make arithmetic errors in substitution. Step 5: Final answer: f(f(0)) = f(2) = 4 − 6 + 2 = 0. Always substitute carefu

Symmetry

Step 1: Check Reflexivity: (1,1), (2,2), (3,3) ∈ R. ✓ Reflexive. Step 2: Check Symmetry: We need to verify that if (a,b) ∈ R then (b,a) ∈ R. Check (2,3): is (3,2) ∈ R? Looking at R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(1,3)}, (3,2) is NOT present. Step 3: Similarly (1,3) ∈ R but (3,1) ∉ R. So R is NOT symmetric. Step 4: Check Transitivity: (1,2) and (2,3) are in R, and (1,3) ∈ R ✓. (2,1) and (1,2) in R, (2,2) ∈ R ✓. (2,1) and (1,3) in R, is (2,3) ∈ R? Yes ✓. Transitivity holds. Step 5: R fails symmetry only. Students commonly mix up which property fails when multiple pairs are present.

g is onto

Step 1: We need to determine what gof being onto implies about f and g individually. Step 2: gof : A → C is onto means for every c ∈ C, there exists a ∈ A such that gof(a) = g(f(a)) = c. Step 3: This means g must map f(a) to c, so every element of C is the image of some element of B under g. Hence g must be onto (surjective). Step 4: Does f have to be onto? Not necessarily. f only needs to produce enough elements in B for g to cover all of C. Example: A={1}, B={1,2}, C={1}, f(1)=1, g(1)=g(2)=1. Here gof is onto but f is not onto. Step 5: Conclusion: gof onto ⟹ g is onto, but f need not be onto

24

Step 1: A bijective function from a finite set to itself is a permutation of the elements of that set. Step 2: Set A has 4 elements: a, b, c, d. Step 3: The number of ways to assign images such that each element maps to a distinct element = number of permutations of 4 elements = 4! Step 4: 4! = 4 × 3 × 2 × 1 = 24. Step 5: So there are 24 bijective functions. Common mistake: students compute 4² = 16 (total functions) or 2⁴ = 16 without realizing bijections = permutations = n!.