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Question

The value of k for which the number 3 lies between the roots of the equation x2+(1−2k)x+(k2−k−2)=0 is given by

A
2<k<5
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B
k<2
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C
2<k<3
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D
k>5
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Solution

The correct option is A 2<k<5
Let f(x)=x2+(12k)x+k2k2 and a=1>0
The number 3 lies between the roots of the given equation, if f(3)<0
Now, f(3)=9+(12k)3+k2k2=107k+k2=k27k+10
Hence, f(3)<0k27k+10<0
(k2)(k5)<02<k<5.

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