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Question

If both the roots of x2+2(k+1)x+9k5=0 are less than 0 & k<100, then the number of integral values of k is

A
95.0
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B
95
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C
95.00
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Solution

Let f(x)=x2+2(k+1)x+9k5 and roots of f(x)=0 are α,β such that α,β<0

Conditions :
(i) D04(k+1)24(9k5)0k27k+60
(k1)(k6)0k(,1][6,) (1)

(ii) α+β<02(k+1)<0k>1 (2)

(iii) αβ>09k5>0k>59 (3)

From (1),(2) and (3), we get:


k(59,1][6,)
Since k is an integer less than 100,
k=1,6,7,,98,99
95 possible values of k

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