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Question

If a, b, c are positive rational numbers such that a > b > c and the quadratic equation
(a+b2c)x2+(b+c2a)x+(c+a2b)=0 has a root in the interval (-1, 0), then


A

b + c > a

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B

c + a < 2b

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C

both roots of the given equation ae rational

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D

the equaiton ax2+2bx+c=0 has both negative real roots

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Solution

The correct options are
B

c + a < 2b


C

both roots of the given equation ae rational


D

the equaiton ax2+2bx+c=0 has both negative real roots


a>b>c ...(i)
and given equation is
(a+b2c)x2+(b+c2a)x+(c+a2b)=0 ...(ii)
Eq. (ii) has a root in the interval (-1, 0)
f (-1) f(0) < 0
(2abc)(c+a2b)<0 ...(iii)
From Eq. (i)
a>bab>0
and a>cac>02abc>0 .... (iv)
From Eqs. (iii) and (iv),
c+a2b<0
or c+a<2b. Alternate (b) is correct
Discriminant of Eq. (ii)
=(b+c2a)24(a+b2c)(c+a2b)={(ba)+(ca)}24{(ac)+(bc)}{cb+ab}={(ca)(ab)}24{(bc)(ca)}×{(ab)(bc)}If bc=X,ca=Y and ab=Z
then, discriminant=(YZ)24(XY)(ZX)=(2XYZ)2 (X+Y+Z=0)=(0+3X)2=9X2=9(bc)2
Both roots of the given equation are rational.
Alternate (c) is correct.
If roots of ax2+2bx+c=0 are α and β
Then α+β=2ba<0α+β<0 and αβ=ca>0 (a>b>c)
and discriminant of
ax2+2bx+c=0is 4b24ac>0 (b>c and b<a)
Hence, roots of ax2+2bx+c=0 are negative and real.
Alternative (d) is correct


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