Question:

Step 1: Recall that for $x > 1$, at non-integer points both $[x]$ and $\{x\}$ are locally constant and smooth respectively. Potential discontinuities occur at integer values $x = 2, 3, 4, \ldots$ Step 2: Check option (a): $f(x) = [x] + \{x\}^2$. Note that $x = [x] + \{x\}$, so $f(x) = [x] + \{x\}^2$. At integer $n$: left limit as $x \to n^-$: $[x] = n-1$, $\{x\} \to 1$, so limit $= (n-1) + 1 = n$. Right limit as $x \to n^+$: $[x] = n$, $\{x\} \to 0$, so limit $= n + 0 = n$. $f(n) = n$. Continuous. ✓ Step 3: Check option (b): $f(x) = [x]^2 + \{x\} + 2[x]\{x\} = ([x] + \{x\})^2 = x^2$. This is continuous everywhere. ✓ Step 4: Check option (c): $f(x) = \left[\dfrac{\{x\}}{e^x}\right]$. For $x > 1$, $\{x\} \in [0,1)$ and $e^x > e > 2$, so $\dfrac{\{x\}}{e^x} < \dfrac{1}{e} < 1$. Also $\dfrac{\{x\}}{e^x} \geq 0$. So $\left[\dfrac{\{x\}}{e^x}\right] = 0$ for all $x > 1$. Constant function, hence continuous. ✓ Step 5: Check option (d): $f(x) = \left[\dfrac{\sin \pi x}{2}\right]\sin(\pi\{x\})$. At integer $x = n$: $\sin(\pi\{x\}) = \sin(0) = 0$, so $f(n) = 0$. Near $x = n$, $\{x\} \approx x - n$ and $\sin(\pi x)$ changes sign. The factor $\left[\dfrac{\sin \pi x}{2}\right]$ can jump at points where $\sin(\pi x)/2$ crosses integers, causing discontinuity. For example near $x = 1.5$: $\sin(1.5\pi) = -1$, $\left[\dfrac{-1}{2}\right] = [-0.5] = -1$. This function is not continuous for all $x > 1$. ✗ Step 6: The continuous functions for all $x > 1$ are options (a), (b), and (c).