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Q.
Let $g(x)=x^2 -4 x -5$, then
Relations and Functions - Part 2
Solution:
Let $g(x_1)=g(x_2)$
$\Rightarrow x^{2}_{1}-4x_{1}-5=x^{2}_{2}-4x_{2}-5$
$\Rightarrow x^{2}_{1}-x^{2}_{2}=4\left(x_{1}-x_{2}\right)$
$\Rightarrow \left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}-4\right)=0$
Either $x_{1} = x_{2}$ or $x_{1}+x_{2} = 4$
Either $x_{1} = x_{2}$ or $x_{1} = 4 - x_{2}$
$\therefore $ There are two values of $x_1$, for which $g(x_1) = g(x_2)$.
$\therefore g(x)$ is not one-one $\forall\, x \in R$