The correct option is A −1
(1−k)x2−4kx+2=0
a=1−k,b=−4k,c=2
Since the quadratic equation has real and equal roots,
D=b2−4ac=0
(−4k)2−4×(1−k)×2=0
16k2−8(1−k)=0
16k2−8+8k=0
2k2+k−1=0
2k2+2k−k−1=0
(k+1)(2k−1)=0
k=−1,12
Since, we have to find the inetger value
Hence, k=-1.
(a) is correct