Question:

Step 1: Define the areas of the regions. The curves are given as $C_1: y^2-x=0 \implies x=y^2$ or $y=\sqrt{x}$ (for $y \ge 0$), $C_2: y-x^2=0 \implies y=x^2$, and $C_3: y=f(x)$ with $f(x)<0$ for $x \in \left(0,\frac{\sqrt{3}}{2}\right)$. The problem defines the areas of region $OPQO$ and region $OPRO$. Based on the integral expressions, these areas are interpreted as follows: Area $OPQO$ is the area between curve $C_1$ ($y=\sqrt{x}$) and curve $C_2$ ($y=x^2$) from $x=0$ to $x=t$. For $x \in \left(0,\frac{\sqrt{3}}{2}\right)$, $\sqrt{x} > x^2$. $$ \text{Area}(OPQO) = \int_0^t (\sqrt{x} - x^2) dx $$ Area $OPRO$ is the area between curve $C_2$ ($y=x^2$) and curve $C_3$ ($y=f(x)$) from $x=0$ to $x=t$. Since $f(x)<0$, $x^2 - f(x)$ is positive, ensuring the integral represents a positive area. $$ \text{Area}(OPRO) = \int_0^t (x^2 - f(x)) dx $$ Step 2: Calculate the integral expressions for the areas. For $\text{Area}(OPQO)$: $$ \text{Area}(OPQO) = \left[\frac{2}{3}x^{3/2} - \frac{x^3}{3}\right]_0^t = \frac{2t^{3/2}}{3} - \frac{t^3}{3} $$ For $\text{Area}(OPRO)$, let $F(t) = \int_0^t f(x) dx$. $$ \text{Area}(OPRO) = \left[\frac{x^3}{3}\right]_0^t - \int_0^t f(x) dx = \frac{t^3}{3} - F(t) $$ Step 3: Apply the given condition relating the areas. The problem states that $\text{Area}(OPRO)$ is twice $\text{Area}(OPQO)$. $$ \frac{t^3}{3} - F(t) = 2 \left(\frac{2t^{3/2}}{3} - \frac{t^3}{3}\right) $$ $$ \frac{t^3}{3} - F(t) = \frac{4t^{3/2}}{3} - \frac{2t^3}{3} $$ Rearranging the terms to solve for $F(t)$: $$ F(t) = \frac{t^3}{3} - \frac{4t^{3/2}}{3} + \frac{2t^3}{3} $$ $$ F(t) = t^3 - \frac{4}{3}t^{3/2} $$ Step 4: Determine the function $f(x)$. By the Fundamental Theorem of Calculus, $f(t) = F'(t)$. $$ f(t) = \frac{d}{dt}\left(t^3 - \frac{4}{3}t^{3/2}\right) $$ $$ f(t) = 3t^2 - \frac{4}{3} \cdot \frac{3}{2}t^{1/2} $$ $$ f(t) = 3t^2 - 2\sqrt{t} $$ Step 5: Evaluate $\left|32f\left(\frac{1}{2}\right)\right|$. Substitute $t=\frac{1}{2}$ into the expression for $f(t)$: $$ f\left(\frac{1}{2}\right) = 3\left(\frac{1}{2}\right)^2 - 2\sqrt{\frac{1}{2}} $$ $$ f\left(\frac{1}{2}\right) = 3\left(\frac{1}{4}\right) - 2\left(\frac{1}{\sqrt{2}}\right) $$ $$ f\left(\frac{1}{2}\right) = \frac{3}{4} - \frac{2\sqrt{2}}{2} $$ $$ f\left(\frac{1}{2}\right) = \frac{3}{4} - \sqrt{2} $$ Now, calculate $\left|32f\left(\frac{1}{2}\right)\right|$: $$ \left|32f\left(\frac{1}{2}\right)\right| = \left|32\left(\frac{3}{4} - \sqrt{2}\right)\right| $$ $$ = |24 - 32\sqrt{2}| $$ Since $\sqrt{2} \approx 1.414$, $32\sqrt{2} \approx 45.25$. Therefore, $24 - 32\sqrt{2}$ is a negative value. $$ |24 - 32\sqrt{2}| = -(24 - 32\sqrt{2}) = 32\sqrt{2} - 24 $$