If someone is applying let us say +2 volts to the input of this rectifier, how would it behave ? Am l correct to assume that D1 and D2 are always turned off ?

 

There is an error in the circuit. D1 needs to be removed.
If the op amp output's low level can go to nearly 0V, it will work as a precision full wave rectifier.

 

No, D1 is fine. See the link - http://sound.westhost.com/appnotes/an001.htm

 

No, D1 is fine. See the link - http://sound.westhost.com/appnotes/an001.htm

Sorry, I didn't even consider a half wave rectifier. I was thinking of a full wave rectifier, with a single positive supply, and an op amp whose output can (nearly) reach the negative rail.
As I said, with those conditions, it will function as a full wave rectifier if D1 is removed.

 

So how does it work ? For instance if a 1V is applied at the waveform

 

First, let me say that you will not have a chance of understanding this if you do not understand basic op amp feedback theory.
When the input swings above 0V, the op amp output goes low until D1 conducts, keeping node A at 0V (virtual GND). D2 is reverse biased, so Vo will be at the same voltage as node A, i.e., 0V.
When the input swings below 0V, the op amp output goes high until D2 conducts, again keeping node A at virtual GND. D1 is now reverse biased. All of the input current flows from the output node to the input, and since node A is 0V, Vo=-Vi.
You therefore have a half wave rectifier.