If $f : A \to B$, $f(x) = \sin^{-1}\left(\dfrac{[x]}{\{x\}}\right)$ and $g : C \to D$, $g(x) = \cos^{-1}\left(\dfrac{[x]}{\{x\}}\right)$, then which of the following is always <b>correct</b>?
[<b>Note:</b> $[\cdot]$ and $\{\cdot\}$ denotes greatest integer and fractional part function respectively.]
$A = C$
$f(x)$ and $g(x)$ both are injective
$B$ and $D$ both are singleton sets
Number of integral solution of the equation $f(x) + g(x) = \dfrac{\pi}{2}$ is zero.
Step-by-Step Solution
Step 1: Determine the domain of $f(x)$ and $g(x)$. For both functions, we need $\dfrac{[x]}{\{x\}}$ to lie in $[-1, 1]$. Also, $\{x\} \neq 0$, so $x$ must not be an integer.
Step 2: Analyze when $\dfrac{[x]}{\{x\}} \in [-1,1]$. Since $\{x\} \in (0,1)$ for non-integers, we need $[x] \in (-\{x\}, \{x\})$. Since $[x]$ is an integer and $\{x\} \in (0,1)$, the only integer in $(-1, 1)$ is $0$. So $[x] = 0$, meaning $x \in (0,1)$.
Step 3: Conclude domains. Both $f$ and $g$ have domain $(0,1)$, so $A = C = (0,1)$. Option 1 is correct in this sense, but we check further.
Step 4: Evaluate the range. For $x \in (0,1)$, $[x] = 0$, so $\dfrac{[x]}{\{x\}} = 0$ for all $x \in (0,1)$. Thus $f(x) = \sin^{-1}(0) = 0$ and $g(x) = \cos^{-1}(0) = \dfrac{\pi}{2}$ for all $x \in (0,1)$.
Step 5: Check option 3. Since $f(x) = 0$ for all $x \in (0,1)$, $B = \{0\}$. Since $g(x) = \dfrac{\pi}{2}$ for all $x \in (0,1)$, $D = \left\{\dfrac{\pi}{2}\right\}$. Both $B$ and $D$ are singleton sets. Option 3 is correct.
Step 6: Check option 2. Since $f$ and $g$ are constant functions, they are not injective. Option 2 is incorrect.
Step 7: Check option 4. $f(x) + g(x) = 0 + \dfrac{\pi}{2} = \dfrac{\pi}{2}$ for all $x \in (0,1)$. The equation holds for all $x \in (0,1)$, but the question asks for integral solutions. Since $x \in (0,1)$ contains no integers, the number of integral solutions is zero. Option 4 is correct.
Correct Answer: 3, 4