f : R to R is defined as f(x) = begin{cases} | vedclass

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(A) For $f(x)$ to be one-one,the function must be strictly monotonic or the ranges of the two pieces must be disjoint.Case $1$: For $x \leq 0$,$f(x) =

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$( - \infty ,0)$

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(A) For $f(x)$ to be one-one,the function must be strictly monotonic or the ranges of the two pieces must be disjoint.Case $1$: For $x \leq 0$,$f(x) = x^2 + 2mx - 1$. The vertex of this parabola is at $x = -m$. For the function to be one-one on $(-\infty, 0]$,the vertex must be at $x \geq 0$,which implies $-m \geq 0$,so $m \leq 0$.Case $2$: For $x > 0$,$f(x) = mx - 1$. For this to be one-one,$m$ must be $\geq 0$. However,if $m > 0$,the range of $f(x)$ for $x > 0$ is $(-1, \infty)$. The range of $f(x)$ for $x \leq 0$ is $[-1-m^2, \infty)$. These ranges overlap,so the function would not be one-one.Case $3$: If $m  0$,$f(x) = mx - 1$ is a decreasing function with range $(-\infty, -1)$. For $x \leq 0$,$f(x) = x^2 + 2mx - 1$ is decreasing on $(-\infty, -m]$. Since $-m > 0$,the function is decreasing on $(-\infty, 0]$. The range for $x \leq 0$ is $[-1-m^2, \infty)$. Since the ranges $(-1, -\infty)$ and $[-1-m^2, \infty)$ do not overlap for $m Thus,$m \in (-\infty, 0)$.

$f : R \to R$ is defined as $f(x) = \begin{cases} x^2 + 2mx - 1, & x \leq 0 \\ mx - 1, & x > 0 \end{cases}$. If $f(x)$ is one-one,then the set of values of $m$ is:

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