Let a1>a2>a3>a4>a5>a6P=a1+a2+−−−+a6q=a1a3+a3a5+a5a1+a2a4+a4a6+a2a6r=a1a3a5+a2a4a6, then the equation2x3−px2+qx−r=0 has

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Q.

Let $a_{1} > a_{2} > a_{3} > a_{4} > a_{5} > a_{6}$
$P = a_{1} + a_{2} + - - - + a_{6}$
$q = a_{1} a_{3} + a_{3} a_{5} + a_{5} a_{1} + a_{2} a_{4} + a_{4} a_{6} + a_{2} a_{6}$
$r = a_{1} a_{3} a_{5} + a_{2} a_{4} a_{6},$ then the equation
$2 x^{3} - p x^{2} + q x - r = 0$ has

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a

Two roots lies between $(a_{3}, a_{5})$

b

Two roots between $(a_{1}, a_{3})$

c

One root between $(a_{1}, a_{2})$

d

Two roots lies between $(a_{1}, a_{4})$

answer is A, C.

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Detailed Solution

$x^{3} - (a_{1} + a_{3} + a_{5}) x^{2} + - - - + x^{3} - (a_{2} + a_{4} + a_{6}) x^{2} - = 0$

$(x - a_{1}) (x - a_{3}) (x - a_{5})$ = $(x - a_{2}) (x - a_{4}) (x - a_{6}) = 0$

$f (a_{3}) > 0, f (a_{1}) < 0$

$f (a_{4}) < 0, f (a_{5}) < 0$

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