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Question

Let ABC be a triangle such that ACB=π6 and let a, b and c denote the lengths of the sides opposite to A, B and C respectively. The value(s) of x for which a=x2+x+1, b=x21 and c=2x+1 is (are)

A
(2+3)
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B
1+3
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C
2+3
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D
43
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Solution

The correct option is B 1+3
Using cosine rule for C

cosC=a2+b2c22ab
32=(x2+x+1)2+(x21)2(2x+1)22(x2+x+1)(x21)=(2x4+2x33x22x+1)2(x2+x+1)(x21)=(2x2+2x1)(x21)2(x2+x+1)(x21)
3=2x2+2x1x2+x+1
(32)x2+(32)x+(3+1)=0
x=(23)±32(32)
x=(2+3) , 1+3x=1+3 as (x>0).

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