Question:

Step 1: Set up coordinates. Place $O$ at origin, $A = (1, 0)$, $B = (1/2, \sqrt{3}/2)$ for an equilateral triangle with side 1. Step 2: Express $\overrightarrow{OP}$ in coordinates. $$\overrightarrow{OP} = (2-t)\overrightarrow{OA} + t\overrightarrow{OB} = (2-t)(1,0) + t(1/2, \sqrt{3}/2)$$ $$= \left(2-t + \frac{t}{2},\ \frac{t\sqrt{3}}{2}\right) = \left(2 - \frac{t}{2},\ \frac{t\sqrt{3}}{2}\right)$$ Step 3: Compute $\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA}$. $$\overrightarrow{AP} = \left(2 - \frac{t}{2} - 1,\ \frac{t\sqrt{3}}{2}\right) = \left(1 - \frac{t}{2},\ \frac{t\sqrt{3}}{2}\right)$$ Step 4: Compute $|\overrightarrow{AP}|^2$. $$|\overrightarrow{AP}|^2 = \left(1 - \frac{t}{2}\right)^2 + \left(\frac{t\sqrt{3}}{2}\right)^2 = 1 - t + \frac{t^2}{4} + \frac{3t^2}{4} = 1 - t + t^2$$ Step 5: Find the range of $|\overrightarrow{AP}|^2 = t^2 - t + 1$. Minimum at $t = 1/2$: $|\overrightarrow{AP}|^2_{\min} = 1/4 - 1/2 + 1 = 3/4$, so $|\overrightarrow{AP}|_{\min} = \dfrac{\sqrt{3}}{2}$. As $t \to \pm\infty$, $|\overrightarrow{AP}| \to \infty$. So $|\overrightarrow{AP}| \in \left[\dfrac{\sqrt{3}}{2}, \infty\right)$. Step 6: Check each option: - $\dfrac{1}{2} < \dfrac{\sqrt{3}}{2}$: NOT achievable. Option (a) is incorrect. - $\dfrac{1}{\sqrt{2}} \approx 0.707 < \dfrac{\sqrt{3}}{2} \approx 0.866$: NOT achievable. Option (b) is incorrect. - $\dfrac{\sqrt{3}}{2}$: achievable (minimum). Option (c) is correct. - $2 > \dfrac{\sqrt{3}}{2}$: achievable. Option (d) is correct. Also check $\dfrac{1}{2}$: $t^2 - t + 1 = 1/4 \Rightarrow t^2 - t + 3/4 = 0$, discriminant $= 1 - 3 < 0$, not achievable. So correct options are (c) and (d), i.e., options 3 and 4.