1. **Problem statement:** We have a piecewise function defined as $$f(x) = \begin{cases} x^2 + 2mx - 1 & \text{for } x \leq 0 \\ mx - 1 & \text{for } x > 0 \end{cases}$$ We want to find the values of $m$ such that $f$ is one-one (injective). 2. **Recall:** A function is one-one if it never takes the same value twice, i.e., it is strictly monotonic (strictly increasing or strictly decreasing). 3. **Analyze each piece:** - For $x \leq 0$, the function is quadratic: $f_1(x) = x^2 + 2mx - 1$. - For $x > 0$, the function is linear: $f_2(x) = mx - 1$. 4. **Monotonicity of $f_1$ on $(-\infty,0]$:** Calculate derivative: $$f_1'(x) = 2x + 2m$$ For $f_1$ to be one-one on $(-\infty,0]$, it must be strictly monotonic there. Since $x \leq 0$, the minimum of $f_1'(x)$ on this interval is at $x=0$: $$f_1'(0) = 2m$$ - If $f_1'(x)$ is always nonnegative or always nonpositive on $(-\infty,0]$, then $f_1$ is monotonic there. Check sign of $f_1'(x)$ for $x \leq 0$: - $f_1'(x) = 2x + 2m \leq 2m$ since $x \leq 0$. If $2m \leq 0$, i.e. $m \leq 0$, then $f_1'(x) \leq 0$ for all $x \leq 0$, so $f_1$ is decreasing on $(-\infty,0]$. If $m > 0$, then at $x=0$, $f_1'(0) = 2m > 0$, but for very negative $x$, $f_1'(x)$ can be negative (since $2x$ is very negative), so $f_1'(x)$ changes sign and $f_1$ is not monotonic on $(-\infty,0]$. Therefore, for $f_1$ to be monotonic on $(-\infty,0]$, we need $m \leq 0$. 5. **Monotonicity of $f_2$ on $(0,\infty)$:** $f_2(x) = mx - 1$ is linear with slope $m$. For $f_2$ to be one-one, $m \neq 0$. 6. **Check continuity and one-one across $x=0$:** At $x=0$: $$f_1(0) = 0^2 + 2m \cdot 0 - 1 = -1$$ $$f_2(0^+) = m \cdot 0 - 1 = -1$$ So $f$ is continuous at $0$. 7. **Check if the function is one-one across the boundary:** Since $f_1$ is decreasing on $(-\infty,0]$ for $m \leq 0$, and $f_2$ is linear with slope $m$ on $(0,\infty)$, for $f$ to be one-one, the slope $m$ must not contradict the monotonicity. - If $m < 0$, then $f_1$ is decreasing on $(-\infty,0]$ and $f_2$ is decreasing on $(0,\infty)$, so $f$ is strictly decreasing on entire $\mathbb{R}$. - If $m = 0$, then $f_2(x) = -1$ constant on $(0,\infty)$, not one-one. - If $m > 0$, $f_1$ is not monotonic on $(-\infty,0]$, so $f$ is not one-one. 8. **Conclusion:** The function $f$ is one-one if and only if $m < 0$. **Answer:** $m \in (-\infty,0)$ which corresponds to option (A).