1.Let A=[−1,1]A = [-1, 1]A=[−1,1] and f:A→Af : A \to Af:A→A be defined as f(x)=x∣x∣f(x) = x|x|f(x)=x∣x∣ for all x∈Ax \in Ax∈A, then f(x)f(x)f(x) isa.many one and into functionb.one-many and into functionc.many-one and into functiond.one-one and onto functionLogin to continueOnly logged in users canattempt or see the solution.
