|=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.