Let f: ℝ → ℝ be defined as f(x) = x - 1 and g: R → \ 1, - 1\ → ℝ be...

$fog\left(x\right) = f\left(g\left(x\right)\right)$= f\left(\dfrac{{x}^{2}}{{x}^{2} + 1}\right)$</p><p>$= \dfrac{{x}^{2}}{{x}^{2} + 1} - 1 = \dfrac{{x}^{2} - {x}^{2} - 1}{{x}^{2} + 1} = \dfrac{- 1}{{x}^{2} + 1}$</p><p>We know that,$0 \le {x}^{2} < \infty , \forall x \in R$</p><p>$⇒ 1 \le {x}^{2} + 1 < \infty , \forall x \in R ⇒ 1 \ge \dfrac{1}{{x}^{2} + 1} > 0, \forall x \in R ⇒ - 1 \le \dfrac{- 1}{{x}^{2} + 1} < 0, \forall x \in R$</p><p>So, range of$fog$\left(x\right)$ is \left[- 1, 0\right) ⊂ R$.</p><p> Hence, the function$fog\left(x\right)$is into function and$fog$\left(- x\right) = f$\left(g\left(- x\right)\right) = \dfrac{- 1}{{\left(- x\right)}^{2} + 1} = \dfrac{- 1}{{x}^{2} + 1} = f\left(g\left(x\right)\right)$</p><p>$∴ fog\left(x\right)$is an even function. So, it is a many one function.</p><p>Hence,$fog$\left(x\right)$ is neither one-one nor onto function.