The correct option is C 1.75,1.875,1.8125
We have to find the second,third and fourth approximation of root of the
equation x3−x−4=0 in the interval (1,2) using successive
Bisection method.
Iteration 1: k=0
c0=a0+b02=1+22=1.5
Since f(c0)f(a0)=f(1.5)f(1)>0
Therefore set a1=1.5,b1=b0
Iteration 2: k=1
c1=a1+b12=1.5+22=1.75
Since f(c1)f(a1)=f(1.75)f(1.5)>0
Therefore set a2=c1,b2=b1
Iteration 3: k=2
c2=a2+b22=1.75+22=1.875
Since f(c2)f(a2)=f(1.875)f(1.75)<0
Therefore set a3=a2,b3=c2
Iteration 4: k=3
c3=a3+b32=1.75+1.8752=1.8125
Thus the second,third and fourth approximations are 1.75,1.875,1.8125 respectively.