If f: [-6, 6] -> RR is defined by f(x)=x...

If `f: [-6, 6] -> RR` is defined by `f(x)=x^2-3` for `x in RR` then `(fofof)(-1)+(fofof)(0)+(fofof)(1)=`

A

`f (4 sqrt2)`

B

`f (3 sqrt2)`

C

`(f (2 sqrt2)`

D

`f (sqrt2)`

Verified by Experts

The correct Answer is:

A

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