f(x)=|x+1| going through options (a) f(x)2=|x2+1| {f(x)}2=(x+1)2 Which implies f(x)2≠{f(x)}2 (b) f(|x|)=||x|+1| |f(x)|=||x+1||=|x+1| Which implies f(|x|)≠|f(x)| (c) f(x+y)=|x+y+1| f(x)+f(y)=|x+1|+|y+1| Which implies f(x+y)≠f(x)+f(y) Option d is correct