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Question

The set of values of k for which roots of the quadratic equation −x2−2(k−1)x−(k+5)=0 are less than \(1)\ is

A
R1
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B
[4,)
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C
[2,3+272]
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D
R
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Solution

The correct option is B [4,)
Let f(x)=x22(k1)x(k+5)
On comparing with standard quadratic equation y=ax2+bx+c we get
a=1,b=2(k1),c=(k+5).

We can see that a<0, so the given expression will be a downward opening parabola as shown below

Required conditions are
(i) D0
D=b24ac0
D=(2(k1))24.(1).((k+5)0
4k212k160
k23k40
(k4)(k+1)0
k(,1][4,)

(ii)f(1)<0
(1)22(k1)(1)(k+5)<0
3k+4>0
k(43,)(2)

(iii) x coordinate of the vertex will be less than 1
b2a<1
(2(k1))2(1)<1
k>0
k(0,)(3)

Final set of solution will be the intersection of (1),(2),(3).
k((,1][4,))(43,)(0,)

k[4,)

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