Question:

Step 1: Determine the form of the derivative based on monotonicity conditions. Since $f(x)$ is a cubic polynomial that increases on $(-\infty, 0) \cup (1, \infty)$ and decreases on $(0, 1)$, the derivative $f'(x)$ must be positive outside the interval $[0, 1]$ and negative inside it. This means $f'(x)$ has roots at $x = 0$ and $x = 1$, with a positive leading coefficient. Therefore: $$f'(x) = a \cdot x(x-1) = a(x^2 - x)$$ where $a > 0$. Step 2: Find the value of the constant $a$ using the condition $f'(2) = 6$. Substitute $x = 2$ into the derivative: $$f'(2) = a(2^2 - 2) = a(4 - 2) = 2a$$ Given that $f'(2) = 6$: $$2a = 6 \implies a = 3$$ Therefore: $$f'(x) = 3x^2 - 3x$$ Step 3: Find the cubic polynomial $f(x)$ by integrating $f'(x)$. Integrate the derivative: $$f(x) = \int (3x^2 - 3x) \, dx = x^3 - \frac{3x^2}{2} + C$$ Step 4: Determine the constant $C$ using the condition $f(2) = 2$. Substitute $x = 2$ into $f(x)$: $$f(2) = 2^3 - \frac{3(2)^2}{2} + C = 8 - 6 + C = 2 + C$$ Given that $f(2) = 2$: $$2 + C = 2 \implies C = 0$$ Therefore: $$f(x) = x^3 - \frac{3x^2}{2}$$ Step 5: Calculate the required function values. $$f(0) = 0^3 - \frac{3(0)^2}{2} = 0$$ $$f(1) = 1^3 - \frac{3(1)^2}{2} = 1 - \frac{3}{2} = -\frac{1}{2}$$ $$f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^3 - \frac{3}{2}\left(\frac{3}{2}\right)^2 = \frac{27}{8} - \frac{3}{2} \cdot \frac{9}{4} = \frac{27}{8} - \frac{27}{8} = 0$$ Step 6: Evaluate the sum of inverse tangent functions. $$\tan^{-1}(f(1)) + \tan^{-1}(f(3/2)) + \tan^{-1}(f(0)) = \tan^{-1}\left(-\frac{1}{2}\right) + \tan^{-1}(0) + \tan^{-1}(0)$$ $$= \tan^{-1}\left(-\frac{1}{2}\right) + 0 + 0 = \tan^{-1}\left(-\frac{1}{2}\right)$$ Step 7: Simplify using the relationship between inverse tangent and inverse cotangent. Using the identity $\tan^{-1}(-x) = -\tan^{-1}(x)$: $$\tan^{-1}\left(-\frac{1}{2}\right) = -\tan^{-1}\left(\frac{1}{2}\right)$$ Using the complementary angle relationship $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$, we have: $$\tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2} - \cot^{-1}\left(\frac{1}{2}\right)$$ Since $\cot^{-1}\left(\frac{1}{2}\right) = \tan^{-1}(2)$: $$\tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2} - \tan^{-1}(2)$$ Therefore: $$-\tan^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{2} + \tan^{-1}(2) = \tan^{-1}(2) - \frac{\pi}{2} = -\cot^{-1}(2)$$ The final answer is $\boxed{-\cot^{-1}(2)}$, which corresponds to **Option 4**.