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Question

Let x1,x2 (x1x2) be the roots of the equation x2+2(m3)x+9=0. If 6<x1,x2<1, then m lies in the interval

A
(6,274]
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B
(274,9)
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C
(2,274)
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D
(6,274)
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Solution

The correct option is D (6,274)
Let f(x)=x2+2(m3)x+9,
6<x1, x2<1


Conditions:
(i) Δ>04(m3)236>0
m26m>0m(m6)>0m(,0)(6,) (1)

(ii) f(6)>012m81<0m<274 (2)

(iii) f(1)>02m+4>0m>2 (3)

(iv) 6<b2a<16<3m<12<m<9 (4)

From (1),(2),(3) and (4), we get
m(6,274)

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