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If c<1and the system of equations x+y1=0,2xyc=0 andbx+3byc=0 is consistent, then the possible real values of bare

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a
b∈−3,34
b
b∈−32,4
c
b∈−34,3
d
b∈−32,34

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detailed solution

Correct option is C

Solving (1) and (2), x=c+13, y=2−c3,substitute in equation(3).(3) ⇒ −b(c+1)3+3b2−c3−c=0-bc-b+6b-3bc-3c=0-4bc-3c=-5b-c(4b+3)=-5bc=5b4b+3 Given, c<1⇒5b4b+3-1<0∣⇒(b-3)(4b+3)<0⇒b∈(-3/4,3)


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