Correct Answer - b.
`(1)/(F)-(1)/(f_(1))+(1)/(f_(1))+(1)/(f_(m))`
`(1)/(F)=(2)/(f_(1))+(2)/(f_(m))`
or `(1)/(F)=2(mu-1)((1)/(R))+(1)/(oo)`
or `F=(R)/(2mu(-1))`
Now, `-60=(R)/(2(mu-1))`
or `120(mu-1)=R` (i)
Again, `(1)/(F)-(1)/(f_(1))+(1)/(f_(1))+(1)/(f_(m))`
or `=(2)/(f_(1))+(1)/(R//2)`
or `=2(mu-1)((1)/(R))+(2)/(R)`
or `=(2)/(R)(mu-1+1)`
or`F=(R)/(2mu)`
Now, `-20=(-R)/(2mu)`
or `40mu=R` (ii)
Dividing (i) by (ii), we get
`(120(mu-1))/(40mu)=(R)/(R)=1`
or ` 120(mu-1)=40mu`
or `120mu-40mu=120`
or `80mu=120`
or `mu=(120)/(80)=(3)/(2)=1.5`