Answer C.

Let's prove that f is increasing on [-1,1] hence there are no a and b (a!=b) such that f(a)=f(b)

On [-1,0], x<0 and ln(|x|+1)>0 

For any a<b<0

|a|>|b|>0

ln(|a|+1)>ln(|b|+1)>0

-a>-b>0

Multiply the inequalities with positive terms.

-aln(|a|+1)>-bln(|b|+1)>0

Therefore

aln(|a|+1)<bln(|b|+1)<0

and f is increasing on [-1,0]

On (0,1]

x is a positive increasing function.

|x| is an increasing function therefore 1+|x| is an increasing function and 1+|x|>1

Hence ln(1+|x|) is a positive increasing function. The product of 2 positive increasing functions is an increasing function.

f is an increasing function on (0,1]

In conclusion, f is increasing on [-1,1] therefore there are no `a!=b` such that f(a)=f(b).