Question:

Step 1: Define the coordinates of points $P$, $Q$, $R$, and $O$. Let $P$ be a point $(t, t^2)$ on curve $C_2: y=x^2$. A line parallel to the x-axis from $P$ is $y=t^2$. This line intersects $C_1: y^2-x=0$ (or $x=y^2$) at $Q$. Thus, $x_Q = (t^2)^2 = t^4$. So $Q=(t^4, t^2)$. A line parallel to the y-axis from $P$ is $x=t$. This line intersects $C_3: y=f(x)$ at $R$. Thus, $y_R = f(t)$. So $R=(t, f(t))$. The origin is $O=(0,0)$. Step 2: Calculate the area of region $OPQO$. The region $OPQO$ is bounded by $C_1: x=y^2$ and $C_2: y=x^2$ (which can be written as $x=\sqrt{y}$ for $x,y \ge 0$). The area is calculated by integrating with respect to $y$ from $y=0$ to the y-coordinate of $P$, which is $t^2$. For $y \in (0,1)$, $\sqrt{y} > y^2$. $$ \text{Area } OPQO = \int_0^{t^2} (\sqrt{y} - y^2) dy $$ $$ \text{Area } OPQO = \left[\frac{2}{3}y^{3/2} - \frac{y^3}{3}\right]_0^{t^2} $$ $$ \text{Area } OPQO = \frac{2}{3}(t^2)^{3/2} - \frac{(t^2)^3}{3} $$ $$ \text{Area } OPQO = \frac{2}{3}t^3 - \frac{1}{3}t^6 $$ Step 3: Calculate the area of region $OPRO$. The region $OPRO$ is bounded by $C_2: y=x^2$ and $C_3: y=f(x)$. The area is calculated by integrating with respect to $x$ from $x=0$ to the x-coordinate of $P$, which is $t$. Since $f(x)<0$ for $x \in (0, \frac{\sqrt{3}}{2})$, the curve $y=x^2$ is above $y=f(x)$. $$ \text{Area } OPRO = \int_0^t (x^2 - f(x)) dx $$ $$ \text{Area } OPRO = \left[\frac{x^3}{3}\right]_0^t - \int_0^t f(x) dx $$ $$ \text{Area } OPRO = \frac{t^3}{3} - F(t) $$ where $F(t) = \int_0^t f(x) dx$. Step 4: Apply the given condition and find $F(t)$. The condition states that the area of region $OPRO$ is twice the area of region $OPQO$. $$ \text{Area } OPRO = 2 \times \text{Area } OPQO $$ $$ \frac{t^3}{3} - F(t) = 2 \left( \frac{2}{3}t^3 - \frac{1}{3}t^6 \right) $$ $$ \frac{t^3}{3} - F(t) = \frac{4}{3}t^3 - \frac{2}{3}t^6 $$ Rearranging to solve for $F(t)$: $$ F(t) = \frac{t^3}{3} - \frac{4}{3}t^3 + \frac{2}{3}t^6 $$ $$ F(t) = -\frac{3}{3}t^3 + \frac{2}{3}t^6 $$ $$ F(t) = -t^3 + \frac{2}{3}t^6 $$ Step 5: Find $f(t)$ by differentiating $F(t)$. By the Fundamental Theorem of Calculus, $f(t) = \frac{d}{dt} F(t)$. $$ f(t) = \frac{d}{dt} \left( -t^3 + \frac{2}{3}t^6 \right) $$ $$ f(t) = -3t^2 + \frac{2}{3} \cdot 6t^5 $$ $$ f(t) = -3t^2 + 4t^5 $$ Step 6: Evaluate $\left|32f\left(\frac{1}{2}\right)\right|$. Substitute $t=\frac{1}{2}$ into the expression for $f(t)$: $$ f\left(\frac{1}{2}\right) = -3\left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right)^5 $$ $$ f\left(\frac{1}{2}\right) = -3\left(\frac{1}{4}\right) + 4\left(\frac{1}{32}\right) $$ $$ f\left(\frac{1}{2}\right) = -\frac{3}{4} + \frac{4}{32} $$ $$ f\left(\frac{1}{2}\right) = -\frac{3}{4} + \frac{1}{8} $$ To combine the fractions, find a common denominator: $$ f\left(\frac{1}{2}\right) = -\frac{6}{8} + \frac{1}{8} $$ $$ f\left(\frac{1}{2}\right) = -\frac{5}{8} $$ This value is negative, which is consistent with the condition $f(x)<0$ for $x \in \left(0,\frac{\sqrt{3}}{2}\right)$. Finally, calculate $\left|32f\left(\frac{1}{2}\right)\right|$: $$ \left|32f\left(\frac{1}{2}\right)\right| = \left|32\left(-\frac{5}{8}\right)\right| $$ $$ \left|32f\left(\frac{1}{2}\right)\right| = |-4 \cdot 5| $$ $$ \left|32f\left(\frac{1}{2}\right)\right| = |-20| $$ $$ \left|32f\left(\frac{1}{2}\right)\right| = 20 $$