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Question : 138 of 180
Marks:
+1,
-0
Solution:
Given;
a=x+2,b=u+3‌and‌c=x+y+z Now,
a×b=|| =(6−0)−(3x−0)+(xy−0) =6−3x+xy =6=3x+xy=z−3+ On equating the coefficients of
, and
, we get
z=6,x=1andxy=1 ∵
xy=1y=1 ⇒
a=+2,b=+3 and
c=++6 ∴
[abc]=|| [abc]=1(6−3)−2(0−3)+0=3+6=9
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