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Question : 34 of 160
Marks:
+1,
-0
Solution:
Let
AB=+2−3
‌AC=2+−3‌AD=3−+p‌‌ Volume ‌=‌[ABACAD]‌±12=||‌±12=1(p−3)−2(2p+9)−3(−2−3)‌±12=p−3−4p−18+15‌±12=−3p−6‌⇒‌‌p=−6‌ or ‌p=2‌⇒p=2‌ and ‌−6‌ satisfies equation ‌‌x2+4x−12=0
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