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Question : 128 of 180
Marks:
+1,
-0
Solution:
Let
a=x+y+zNow,
aâ‹…=(x+y+z)â‹…=xaâ‹…(+)=(x+y+z)â‹…(+)=x+yand
aâ‹…(++)=(x+y+z)â‹…(++)=x+y+zNow, according to the question,
‌a⋅=a(+)=a⋅(++)=1⇒x=x+y=x+y+z=1⇒x=1,y=0,z=0∴‌‌‌a=
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