GATE Electronics and Communications (EC) 2012 Solved Paper

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The circuit shown is a

low pass filter with $f_{3dB} = \;{1}/{(R_{1} + R_{2})C}\:rad\/ s$
high pass filter with $f_{3dB} = \;{1}/{R_{1}C}\:rad\/ s$
low pass filter with $f_{3dB} = \;{1}/{R_{1}C}\:rad\/ s$
high pass filter with $f_{3dB} = \;{1}/{(R_{1} + R_{2})C}\:rad\/ s$

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