The circuit shown in the figure is a :  

The transfer function of the circuit is given by:

\({V_o(s)\over V_i(s)}={Z_2\over Z_1+Z_2}\)

\(Z_1=R+{1\over sC}={1+sRC\over sC}\)

\(Y_2={1\over R}+{sC}={1+sRC\over R}\)

\(Z_2={1\over Y_2}={R\over 1+sRC}\)

\(Z_1+Z_2={1+sRC\over sC}+{R\over 1+sRC}\)

\(Z_1+Z_2={(1+sRC)^2+sRC\over sC(1+sRC)}\)

\({V_o(s)\over V_i(s)}={R\over 1+sRC}\times {sC(1+sRC)\over (1+sRC)^2+sRC}\)

\({V_o(s)\over V_i(s)}={sRC\over (1+sRC)^2+sRC}\)

\(|T(0)|=0\)

\(|T(∞)|=0\)

The gain of the transfer function is low for both ω = 0 and ∞.

Hence, it is a band-pass filter.

The location of the pole is:

1 + sRC = 0

s = - RC

The cut-off frequency is given by:

\(\omega_o={-1\over Pole \space location}\)

\(\omega_o={-1\over -RC}\)

\(\omega_o={1\over RC}\) rad/sec