Step 1: Understanding the Concept.
This circuit is a resistor in series with an inductor-capacitor branch that returns to ground, and the output is measured across that L-C branch. Recognizing the transfer function shape tells us the filter type directly, without plugging in frequency limits one at a time.
Step 2: Key Formula or Approach.
Using the Laplace variable $s = j\omega$, the impedance of the series L-C branch is $Z_{LC}(s) = sL + \frac{1}{sC} = \frac{s^2LC + 1}{sC}$. The transfer function of the divider is
\[ H(s) = \frac{V_{out}(s)}{V_{in}(s)} = \frac{Z_{LC}(s)}{R + Z_{LC}(s)} = \frac{s^2LC + 1}{s^2LC + sRC + 1} \]
Step 3: Detailed Explanation.
Look at the numerator: it is $s^2LC + 1$, which becomes exactly zero when $s^2 = -\frac{1}{LC}$, that is at $s = j\omega_0$ with $\omega_0 = \frac{1}{\sqrt{LC}}$. A zero of the transfer function sitting on the imaginary axis means the gain is exactly zero at that one real frequency $\omega_0$, this is the signature of a notch (band stop) filter.
Now check the two ends of the frequency axis. Substituting $s \to 0$ in $H(s)$ gives $H(0) = \frac{1}{1} = 1$, so DC passes fully. Substituting $s \to \infty$, the $s^2LC$ term dominates both top and bottom, so $H(s) \to \frac{s^2LC}{s^2LC} = 1$, so very high frequency also passes fully.
So the gain is 1 at both ends of the spectrum and drops to 0 only in a band around $\omega_0$, which is the definition of a band stop (band reject) filter, something a low pass filter (gain should keep falling with frequency) or a high pass filter (gain should stay low near DC) cannot produce.
Step 4: Final Answer.
The transfer function has unity gain at $\omega=0$ and $\omega=\infty$ with a zero in between at $\omega_0 = 1/\sqrt{LC}$, so the circuit rejects a narrow band around resonance and passes everything else. This is a band stop filter, option (D).
\[ \boxed{H(s) = \dfrac{s^2LC+1}{s^2LC+sRC+1} \Rightarrow \text{band stop filter}} \]