Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 1.2 , 4 - Chapter 1 Class 12 Relation and Functions
Last updated at June 5, 2023 by Teachoo
Ex 1.2
Ex 1.2, 1
Ex 1.2, 2 (i) Important
Ex 1.2, 2 (ii) Important
Ex 1.2, 2 (iii)
Ex 1.2, 2 (iv)
Ex 1.2, 2 (v) Important
Ex 1.2 , 3
Ex 1.2 , 4 You are here
Ex 1.2, 5 Important
Ex 1.2 , 6 Important
Ex 1.2, 7 (i)
Ex 1.2, 7 (ii)
Ex 1.2 , 8 Important
Ex 1.2 , 9
Ex 1.2 , 10 Important
Ex 1.2 , 11 (MCQ) Important
Ex 1.2, 12 (MCQ)
Transcript
Ex 1.2 , 4 Show that the Modulus Function f: R → R given by f(x) =|𝑥| , is neither one-one nor onto, where |𝑥| is x, if x is positive or 0 and |𝑥| is − x, if x is negative. f(x) =|𝑥| = {█( 𝑥 , 𝑥≥0 @−𝑥 , 𝑥<0)┤ Check one-one Example f (1) = |1| = 1 f (– 1) = |1| = 1 Since, different elements 1, –1, have the same image 1 , ∴ f is not one-one. Check onto f: R → R f(x) = |𝑥| Let f(x) = y such that y ∈ R y = |𝑥| Hence value of y is defined only if y is positive, But y is a real number Hence, if y is negative, there is not corresponding element of x Hence, f is not onto
