Opposte roots

let 2^x =z
now roots are of opposite sign. therefore if x1 and x2 be the roots of the eqn. then x1>0 and x2<0

hence z1 is greater than 1 and z2 is less than 1 (since z=2^x)
therefore the eqn is - z² +2(a-1)z + a=0;
now product of roots >0 for z
hence a>0

now the number "1" lies between the roots of this eqn, hence 1.f(1)<0
or 1+2(a-1)+a<0
or a<1/3

hence reqd soln is aE(0,1/3) [d]

Thank you very much!