Let f: R rightarrow R be defined as f(x)=x-1 | vedclass

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(D) Given $f(x) = x - 1$ and $g(x) = \frac{x^2}{x^2 - 1}$.$f(g(x)) = g(x) - 1 = \frac{x^2}{x^2 - 1} - 1 = \frac{x^2 - (x^2 - 1)}{x^2 - 1} = \frac{1}{x

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neither one-one nor onto function

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(D) Given $f(x) = x - 1$ and $g(x) = \frac{x^2}{x^2 - 1}$.$f(g(x)) = g(x) - 1 = \frac{x^2}{x^2 - 1} - 1 = \frac{x^2 - (x^2 - 1)}{x^2 - 1} = \frac{1}{x^2 - 1}$.Let $h(x) = f(g(x)) = \frac{1}{x^2 - 1}$.Since $h(x) = h(-x)$,the function is an even function,which implies it is many-one.To find the range,let $y = \frac{1}{x^2 - 1}$.Then $x^2 - 1 = \frac{1}{y} \Rightarrow x^2 = \frac{1}{y} + 1 = \frac{1 + y}{y}$.For $x$ to be real,$x^2 \geq 0$,so $\frac{1 + y}{y} \geq 0$.This inequality holds for $y \in (-\infty, -1] \cup (0, \infty)$.Since the range is not equal to the co-domain $(R)$,the function is into.Thus,$f \circ g$ is neither one-one nor onto.

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

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