Match the following columns:
Column 1:
(i) Range of $sgn\{x\}$ is: (where $\{.\}$ represents fractional part function)
(ii) Domain of $\sin^{-1} x + \sin^{-1}(1-x)$ is:
(iii) Range of $\sqrt{\frac{2\tan^{-1} x}{\pi}}$ is:
(iv) Range of $\frac{2}{\pi} \sin^{-1}[x^2 + x + 1]$ is: (where $[.]$ represent greatest integer function)
Column II:
(a) $\{1\}$
(b) $[0, 1)$
(c) $0, 1]$
(d) $[0, 1]$
Step-by-Step Solution
Key Concept: Range analysis for compositions requires careful tracking of intermediate ranges and domain restrictions at each step.
(i) When $\{x\} = 0$ (i.e., $x$ is an integer), $\text{sgn}\{x\} = 0$; when $0 < \{x\} < 1$, $\text{sgn}\{x\} = 1$, so the range is $\{0, 1\}$. (ii) For the domain $-1 \leq x \leq 1$ and $-1 \leq 1 - x \leq 1$, we get $0 \leq x \leq 1$, making the domain $[0, 1]$. (iii) Since $-\frac{\pi}{2} < \tan^{-1} x < \frac{\pi}{2}$ for all $x \in \mathbb{R}$, we have $0 \leq \frac{2\tan^{-1} x}{\pi} < 1$, giving range $[0, 1)$. (iv) With $\frac{3}{4} \leq x^2 + x + 1 < \infty$ but $\sin^{-1}[x^2 + x + 1]$ requiring $-1 \leq x^2 + x + 1 \leq 1$, we get $\frac{3}{4} \leq x^2 + x + 1 < 2$.
Correct Answer: 3,4,2,3