Indian Statistical Institute B.Math & B.Stat : Continuity and Bijections

$Problem-1.$ Let \(f: \mathbb{R} \to \mathbb{R} \) be given by \(f(x) = x(x-1)(x+1) \). Then show that $f$ is onto but not 1-1. $$$$ The injectivity part is very much obvious. See that \( f(0)=f(1)=f(-1) = 0 \). Now for the surjectivity, observe that $f$ is an odd degree polynomial. Hence it must be surjective! (Think if not, then $\exists$ a real number which has no pre-image (say $r$ ). Now consider the polynomial \( g(x) = f(x)-r \)!! what else g(x) is also an odd degree polynomial with no real roots!!! ) $$$$ $Problem-2.$ Another Problem with a kick of continuity. Let \(f: \mathbb{R} \to \mathbb{R} \) be given by \(f(x) = x^3-3x^2+6x-5 \). Then show that $f$ is both onto and 1-1. $$$$ Again note that $f$ is an odd degree polynomial so it is surjective. Since $f$ is a polynomial it is continuous on $\mathbb{R}$. Also \(f'(x)=3x^2-6x+6 = 3(x^2-2x+2) = 3\{(x-1)^2 + 1 \} > 0 \) \( \forall x \in \mathbb{R}\). This shows that $f$ is strictly increasing. Thus $f$ is continuous and strictly increasing hence must be 1-1. $$$$ $Problem-3.$ Let \(f: \mathbb{R} \to \mathbb{R} \) be given by \(f(x) = x^2 - \frac{x^2}{1+x^2} \). Then show that $f$ is neither onto nor 1-1. $$$$ Clearly $f$ is not injective since \(f(x)=f(-x) \). Also \( f(x) = \frac{x^4}{1+x^2} \) which shows that $f$ assume non-negative values. Thus the \(Range_f \subset \mathbb{R} = Co-domain_f\). Thus $f$ is neither surjective. $$$$ $Problem-4.$ Let \( \phi :[0,1] \to [0,1] \) be a continuous and 1-1 function. Let \( \phi(0) = 0, \phi(1) = 1, \phi \big(\frac{1}{2}\big) = c, \phi \big(\frac{1}{4}\big) = d.\) Then show that $ c > d $. $$$$ This follows form the property of continuous functions. On an interval I ( not necessarily compact ), if a function is continuous and 1-1 it is strictly increasing. To prove this you can use the Intermediate Value Theorem.