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Q.

Ifa,b,carethesidesofatrianglethentherangeofa2+b2+c2ab+bc+ca  is

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a

[1,  2)

b

(,  1][2,  )

c

[2,  3]

d

(1,  3)

answer is A.

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Detailed Solution

InΔleasinA=bsinB=csinC  a=csinAsinC,  b=CsinBsinCab+bc+caa2+b2+c2=sinAsinB+sinBsinC+sinCsinAsin2A+sin2B+sin2C=2sinAsinB+2sinBsinC+2sinCsinA2(sin2A+sin2B+sin2C)=12[(sinA+sinB+sinC)2(sin2A+sin2B+sin2C)sin2A+sin2B+sin2C]=12[(sinA+sinB+sinC)2sin2A+sin2B+sin2C1]TheTriangleEquilateralthenwegetmaximumvaluesinA=sinB=sinC=32=12[931]=12(31)=1Theminimumvalueofa2+b2+c2ab+bc+ca=1

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