h(x) =x3-x i.e eg h(1)=h(-1)=0
g(x)=x2sinx i.e whenever sinx become 0
f(x)=lnx+x lets assume many to one
then lnx1+x1=lnx2+x2 =>lnx1-lnx2=x2-x1
=>ln(x1/x2)=x2-x1
=>x1/x2=e(x2-x1) [Eq 1]
As x1!=x2 so assume x1>x2 LHS of Eq1 become >1 but RHS <1
X2<X1 not possible LHS<1 but RHS>1,So f(x) is not many to one.