Question

Let $f(x)=\sin ^{-1}(2 x-1)+\cos ^{-1}\left(2 \sqrt{x-x^2}\right)+\tan ^{-1}\left(\frac{1}{1+\left[x^2\right]}\right)$ where $[k]$ denotes greatest integer less than or equal to $k$.

Column I		Column II
P	$f \left(\frac{1}{6}\right)$ is equal to	1	$\frac{\pi}{6}$
Q	$f \left(\frac{3}{4}\right)$ is equal to	2	$\frac{\pi}{4}$
R	$\sin ^{-1}(\tan ( f (1)))$ is equal to	3	$\frac{\pi}{3}$
S	$ \displaystyle\sum_{ r =1}^{10} f \left(\frac{ r }{20}\right)$ is equal to	4	$\frac{7 \pi}{12}$
		5	$\frac{5 \pi}{2}$

• A. $P \rightarrow 2$; Q $\rightarrow 4$; R $\rightarrow 1 ; S \rightarrow 3$
• B. $P \rightarrow 2 ; Q \rightarrow 4 ; R \rightarrow 1 ; S \rightarrow 5$
• C. $P \rightarrow 5 ; Q \rightarrow 4 ; R \rightarrow 1$; $S \rightarrow 3$
• D. $P \rightarrow 5 ; Q \rightarrow 4 ; R \rightarrow 1 ; S \rightarrow 5$

Answer 1

Answer: (B)

$ ( x )=\sin ^{-1}(2 x -1)+\cos ^{-1}\left(2 \sqrt{ x - x ^2}\right)+\tan ^{-1}\left(\frac{1}{1+\left[ x ^2\right]}\right)$
Domain of $f(x)=[0,1]$
$f(x)=\begin{cases}0+\frac{\pi}{4}=\frac{\pi}{4}, & x \in\left[0, \frac{1}{2}\right] \\2 \sin ^{-1}(2 x-1)+\frac{\pi}{4}, & x \in\left(\frac{1}{2}, 1\right) \\\pi+\tan ^{-1}\left(\frac{1}{2}\right), & x=1\end{cases}$
Now, verify the options.