|=1(−4−k2)+2(−8−6k) =−4−k2−16−12k=−k2−12k−20 If ∆≠0, then it has unique solution i.e. 4−k2≠0 ⇒k≠±2 for unique solution. Also at k=2 ∆x=−22−12×2−20=−48≠0 Then, in this case it has no solution. Hence, statement (A) and statement (D) both are correct. ⇒k≠±2 for unique solution. Also at k=2 Then, in this case it has no solution. Hence, statement (A) and statement (D) both are correct.
