Question:

Step 1: Use the symmetry condition $f(-1-x) = f(-1+x)$ to determine that $f$ is symmetric about $x = -1$. So $f(x) = a(x+1)^2 + k$ for some constants $a$ and $k$. Step 2: Compute $f(1)$, $f(-1)$, and $f'(-1)$. $f(1) = a(2)^2 + k = 4a + k$. $f(-1) = a(0)^2 + k = k$. $f'(x) = 2a(x+1)$, so $f'(-1) = 0$. Step 3: Apply the given condition $(f(1)-5)^2 + (f(-1)-1)^2 = f'(-1) = 0$. This means both squares must be zero: $f(1) = 5$ and $f(-1) = 1$. Step 4: Solve for $a$ and $k$. From $f(-1) = k = 1$ and $f(1) = 4a + 1 = 5 \Rightarrow a = 1$. So $f(x) = (x+1)^2 + 1$. Step 5: Analyze the range of $f(x)$. Since $f(x) = (x+1)^2 + 1 \geq 1$ for all $x \in R$, with minimum value $1$ at $x = -1$. Step 6: Check option (a): $[\sin^{-1} f(x)]$. Since $f(x) \geq 1$, $\sin^{-1} f(x)$ is defined only when $f(x) = 1$, i.e., at $x = -1$, giving $\sin^{-1}(1) = \pi/2$. $[\pi/2] = 1$. But this is only at one point, not wherever defined in general. Actually wherever defined, $f(x) = 1$ and $[\sin^{-1}(1)] = [\pi/2] = 1$. So option (a) equals 1 wherever defined. Step 7: Check option (b): $[\text{sgn}(f(x))]$. Since $f(x) \geq 1 > 0$, $\text{sgn}(f(x)) = 1$ always, so $[\text{sgn}(f(x))] = [1] = 1$. Option (b) is correct. Step 8: Check option (c): $\left[\tan^{-1}\dfrac{1}{f(x)}\right]$. Since $f(x) \geq 1$, $\dfrac{1}{f(x)} \in (0, 1]$. $\tan^{-1}\left(\dfrac{1}{f(x)}\right) \in (0, \pi/4]$. So $\left[\tan^{-1}\dfrac{1}{f(x)}\right] = 0 \neq 1$. Option (c) is incorrect. Step 9: Check option (d): $\left[\cot^{-1}\left[\dfrac{1}{2^{f(x)}}\right]\right]$. Since $f(x) \geq 1$, $2^{f(x)} \geq 2$, so $\dfrac{1}{2^{f(x)}} \leq \dfrac{1}{2}$. $\left[\dfrac{1}{2^{f(x)}}\right] = 0$. $\cot^{-1}(0) = \pi/2$. $[\pi/2] = 1$. Option (d) is correct. Step 10: The correct options are (b) and (d), i.e., options 2 and 4.