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Let \mathrm{f: \mathbb{R} \rightarrow \mathbb{R}} be defined as \mathrm{f(x)=x-1} and \mathrm{g: \mathbb{R}-\{1,-1\} \rightarrow \mathbb{R}} be defined as \mathrm{g(x)=\frac{x^{2}}{x^{2}-1}} Then the function f \circ g is

Option: 1

one-one but not onto


Option: 2

onto but not one-one


Option: 3

both one-one and onto


Option: 4

neither one-one nor onto


Answers (1)

best_answer

\begin{aligned} &f(x)=x-1 \\ &g(x)=\frac{x^{2}}{x^{2}-1} \\ &f \circ g=f(g(x))=\frac{x^{2}}{x^{2}-1}-1=\frac{1}{x^{2}-1} \end{aligned}
since f \circ g(x)  is even function hance many one function
and also Range \in(-\infty,-1] \cup(0, \infty) which is nat equal ts co-domain.

hence into function.

Posted by

shivangi.shekhar

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