If f: R rightarrow R is defined as f(x)=x^2-2 | vedclass

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(C) Given $f(x) = x^2 - 2x - 3$. For one-one: Let $f(x_1) = f(x_2)$. $x_1^2 - 2x_1 - 3 = x_2^2 - 2x_2 - 3$ $x_1^2 - x_2^2 - 2(x_1 - x_2) = 0$ $(x_1 -

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neither one-one nor onto

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(C) Given $f(x) = x^2 - 2x - 3$. For one-one: Let $f(x_1) = f(x_2)$. $x_1^2 - 2x_1 - 3 = x_2^2 - 2x_2 - 3$ $x_1^2 - x_2^2 - 2(x_1 - x_2) = 0$ $(x_1 - x_2)(x_1 + x_2) - 2(x_1 - x_2) = 0$ $(x_1 - x_2)(x_1 + x_2 - 2) = 0$ This implies $x_1 = x_2$ or $x_1 + x_2 = 2$. Since $x_1 + x_2 = 2$ allows distinct values (e.g.,$f(0) = -3$ and $f(2) = -3$),the function is not one-one. For onto: Let $y = x^2 - 2x - 3$. $y = (x-1)^2 - 4$ $(x-1)^2 = y + 4$ Since $(x-1)^2 \geq 0$,we must have $y + 4 \geq 0$,which means $y \geq -4$. The range of $f$ is $[-4, \infty)$,which is not equal to the codomain $R$. Therefore,the function is not onto. Thus,$f$ is neither one-one nor onto.

If $f: R \rightarrow R$ is defined as $f(x)=x^2-2x-3$,then $f$ is

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