Question:

Step 1: Analyze the given function $f(x) = \sin^{-1}(2x-1) + \cos^{-1}(2\sqrt{x - x^2}) + \tan^{-1}\!\left(\dfrac{1}{1+[x^2]}\right)$ and understand its components. The function $f(x)$ involves inverse trigonometric functions and the greatest integer function. To simplify this, we can use trigonometric identities and substitutions to make the function more manageable. Step 2: Simplify the first two terms of $f(x)$ using a substitution approach. Let $x = \sin^2\theta$ where $\theta \in [0, \pi/2]$ for $x \in [0,1]$. Then, $2x - 1 = 2\sin^2\theta - 1 = -\cos 2\theta$ and $2\sqrt{x(1-x)} = 2\sin\theta\cos\theta = \sin 2\theta$. This allows us to express the first two terms of $f(x)$ in terms of $\theta$. Step 3: Express $\sin^{-1}(2x-1)$ and $\cos^{-1}(2\sqrt{x-x^2})$ in terms of $\theta$. We have $\sin^{-1}(2x-1) = \sin^{-1}(-\cos 2\theta)$ and $\cos^{-1}(2\sqrt{x-x^2}) = \cos^{-1}(\sin 2\theta)$. Using properties of inverse trigonometric functions, we can simplify these expressions further. Step 4: Use a direct substitution approach with $x = \frac{1+\sin\phi}{2}$ where $\phi \in [-\pi/2, \pi/2]$ to simplify $f(x)$. Then, $2x - 1 = \sin\phi$, so $\sin^{-1}(2x-1) = \phi$. Also, $x - x^2 = \frac{(1+\sin\phi)(1-\sin\phi)}{4} = \frac{\cos^2\phi}{4}$, so $2\sqrt{x-x^2} = |\cos\phi| = \cos\phi$ (since $\phi \in [-\pi/2, \pi/2]$). Thus, $\cos^{-1}(2\sqrt{x-x^2}) = \cos^{-1}(\cos\phi) = |\phi|$. Step 5: Combine the simplified expressions for the first two terms of $f(x)$ and add the third term. We have $f(x) = \phi + |\phi| + \tan^{-1}\!\left(\frac{1}{1+[x^2]}\right)$. Now, we need to consider the range of $x$ and how it affects the value of $f(x)$. Step 6: Evaluate $f(x)$ for $x \in [0, 1/2]$ and $x \in (1/2, 1]$. For $x \in [0, 1/2]$: $\phi \in [-\pi/2, 0]$, so $|\phi| = -\phi$, giving $\phi + |\phi| = 0$. For $x \in (1/2, 1]$: $\phi \in (0, \pi/2]$, so $|\phi| = \phi$, giving $\phi + |\phi| = 2\phi = 2\sin^{-1}(2x-1)$. Step 7: Calculate $f(1/6)$, $f(3/4)$, $\sin^{-1}(\tan(f(1)))$, and $\sum_{r=1}^{10} f(r/20)$. For $x = 1/6 \in [0, 1/2]$, $f(1/6) = 0 + \pi/4 = \pi/4$. For $x = 3/4 \in (1/2, 1]$, $f(3/4) = 2\sin^{-1}(2 \cdot 3/4 - 1) + \pi/4 = 2\sin^{-1}(1/2) + \pi/4 = 2 \cdot \pi/6 + \pi/4 = \pi/3 + \pi/4 = 7\pi/12$. For $x = 1$, $\tan^{-1}\!\left(\frac{1}{1+[x^2]}\right) = \tan^{-1}(1/2)$, and $\phi = \sin^{-1}(2 \cdot 1 - 1) = \sin^{-1}(1) = \pi/2$, so $f(1) = \pi + \tan^{-1}(1/2)$. Then, $\tan(f(1)) = \tan(\pi + \tan^{-1}(1/2)) = \tan(\tan^{-1}(1/2)) = 1/2$, and $\sin^{-1}(\tan(f(1))) = \sin^{-1}(1/2) = \pi/6$. For $r = 1$ to $10$, $x = r/20 \in [1/20, 1/2]$, all in $[0, 1/2]$, so $f(r/20) = 0 + \pi/4 = \pi/4$ for each $r = 1, \ldots, 10$. Thus, $\sum_{r=1}^{10} f(r/20) = 10 \cdot \pi/4 = 5\pi/2$. Step 8: Match the calculated values with the given options to determine the correct answer. From the calculations, we have $P \to 2$, $Q \to 4$, $R \to 1$, and $S \to 5$. The correct answer is the option that matches these values, which is option (b), but since options are indexed 1,2,3,4 corresponding to (a),(b),(c),(d), the correct answer is indexed as 1, given the provided options and the standard indexing used in multiple-choice questions. The final answer is: $\boxed{1}$