Prove that the function f is surjective, where 𝑓: 𝑁 → 𝑁 such that
f(n)={(n + 1)/2, if n is odd n/2, if n is even
Is the function injective? Justify your answer.
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
CBSE Class 12 Sample Paper for 2023 Boards
Question 2 Important
Question 3
Question 4 Important
Question 5
Question 6 Important
Question 7
Question 8 Important
Question 9
Question 10 Important
Question 11 Important
Question 12
Question 13 Important
Question 14 Important
Question 15
Question 16
Question 17 Important
Question 18 Important
Question 19 [Assertion Reasoning] Important
Question 20 [Assertion Reasoning] Important
Question 21 (Choice 1) Important
Question 21 (Choice 2) You are here
Question 22
Question 23 (Choice 1)
Question 23 (Choice 2) Important
Question 24 Important
Question 25
Question 26 Important
Question 27 (Choice 1) Important
Question 27 (Choice 2)
Question 28 (Choice 1)
Question 28 (Choice 2) Important
Question 29 (Choice 1) Important
Question 29 (Choice 2) Important
Question 30
Question 31 Important
Question 32 Important
Question 33 (Choice 1)
Question 33 (Choice 2) Important
Question 34 (Choice 1) Important
Question 34 (Choice 2) Important
Question 35
Question 36 (i) [Case Based] Important
Question 36 (ii)
Question 36 (iii) (Choice 1) Important
Question 36 (iii) (Choice 2)
Question 37 (i) [Case Based] Important
Question 37 (ii)
Question 37 (iii) (Choice 1)
Question 37 (iii) (Choice 2) Important
Question 38 (i) [Case Based] Important
Question 38 (ii)
CBSE Class 12 Sample Paper for 2023 Boards
Last updated at May 29, 2023 by Teachoo
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Question 21 (Choice 2) Prove that the function f is surjective, where 𝑓: 𝑁 → 𝑁 such that 𝑓(𝑛)={█((𝑛 + 1)/2, 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑@&𝑛/2, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛)┤ Is the function injective? Justify your answer. Check onto (surjective) f (n) = {█((𝑛 + 1)/2 ", if n is odd" @𝑛/2 ", if n is even" )┤ for all n ∈ N Let f(x) = y , such that y ∈ N When n is odd y = (𝑛 + 1)/2 2y = n + 1 2y – 1 = n n = 2y – 1 Hence, for y is a natural number , n = 2y – 1 is also a natural number When n is even y = 𝑛/2 2y = n n = 2y Hence for y is a natural number , n = 2y is also a natural number Thus, for every y ∈ N, there exists x ∈ N such that f(n) = y Hence, f is onto (surjective) Check one-one (injective) f (n) = {█((𝑛 + 1)/2 ", if n is odd" @𝑛/2 ", if n is even" )┤ for all n ∈ N. f(1) = (1 + 1)/2 = 2/2 = 1 f(2) = 2/2 = 1 Since, f(1) = f(2) but 1 ≠ 2 Both f(1) & f(2) have same image 1 ∴ f is not one-one (not injective)