The correct option is D sgn(x)x3
f(x)=sin(sin−1x)=x ∀x∈[−1,1], which is one-one and onto.
f(x)=2πsin(sin−1x)=2πx
The range of the function for x∈[−1,1] is [−2π,2π], which is a subset of [−1,1].
Hence, the function is one-one but not onto and, hence, is not bijective.
f(x)=sgn (x)ln(ex)=(sgn (x))x={x, x>0 −x, x<0 0, x=0
This function has range [0,1] which is a subset of [−1,1].
Hence, the function is into. Also, the function is many-one.
f(x)=x3sgn (x)=⎧⎨⎩x3, x>0 −x3, x<0 0, x=0
which is many-one and into.