1. **Problem Statement:** We have a piecewise function: $$f(x) = \begin{cases} x^2 + 2mx - 1 & \text{if } x \leq 0 \\ mx - 1 & \text{if } x > 0 \end{cases}$$ We want to find the interval of $m$ such that $f$ is one-one (injective). 2. **Understanding One-One Functions:** A function is one-one if it never takes the same value twice, i.e., it is strictly monotonic (either strictly increasing or strictly decreasing). 3. **Analyze each piece:** - For $x > 0$, $f(x) = mx - 1$ is linear. It is strictly increasing if $m > 0$, strictly decreasing if $m < 0$, and constant if $m=0$. - For $x \leq 0$, $f(x) = x^2 + 2mx - 1$ is quadratic with leading coefficient $1 > 0$, so it is a parabola opening upwards. 4. **Monotonicity of the quadratic piece:** The derivative is: $$f'(x) = 2x + 2m$$ For $x \leq 0$, the derivative is: $$f'(x) = 2x + 2m \leq 2(0) + 2m = 2m$$ - If $m \leq 0$, then $f'(x) \leq 2m \leq 0$ for all $x \leq 0$, so $f$ is decreasing on $(-\infty, 0]$. - If $m > 0$, then $f'(x)$ can be positive or negative depending on $x$, so the quadratic is not monotonic on $(-\infty, 0]$. 5. **Check continuity and injectivity at $x=0$:** At $x=0$, $$f(0) = 0^2 + 2m \cdot 0 - 1 = -1$$ From the right side, $$\lim_{x \to 0^+} f(x) = m \cdot 0 - 1 = -1$$ So $f$ is continuous at $0$. 6. **Check if the function is one-one for $m \leq 0$:** - On $(-\infty, 0]$, $f$ is decreasing. - On $(0, \infty)$, $f(x) = mx - 1$ is decreasing if $m < 0$, constant if $m=0$. If $m=0$, then $f(x) = -1$ for $x > 0$, which is constant and not one-one. If $m < 0$, both pieces are decreasing, so the whole function is strictly decreasing and hence one-one. 7. **Check if the function is one-one for $m > 0$:** - Quadratic piece is not monotonic on $(-\infty, 0]$. - Linear piece is increasing. Since the quadratic is not monotonic, $f$ is not one-one. 8. **Conclusion:** The function $f$ is one-one if and only if $m < 0$. **Answer:** The interval is $(-\infty, 0)$, which corresponds to option (A).