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If, in a ABC, (a+b+c)(b+ca)=λbc, then

a
λ<0
b
λ>4
c
λ>0
d
0<λ<4

detailed solution

Correct option is D

We have,(a+b+c)(b+c−a)=λbc⇒2s(2s−a)=λbc⇒s(s−a)bc=λ4⇒cos2⁡A2=λ4                                             ⇒0<λ4<1⇒0<λ<4⇒∵cos2⁡A2≤1                           ∵cos2⁡A2≤1

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