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Question

If the equation x4(k1)x2+(2k)=0 has three distinct real roots, then the possible value(s) of k is/are

A
{2}
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B
{21,2}
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C
{51}
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D
{22,32}
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Solution

The correct option is A {2}
x4(k1)x2+(2k)=0 (1)
Assuming x2=t,
t2(k1)t+(2k)=0 (2)
Let roots of the equation (2) be t1,t2 (t1t2)
Equation (1) will have three distinct real roots iff for equation (2),
t1=0, t2>0


(i) f(0)=02k=0k=2 (3)(ii) b2a>0k12>0k>1 (4)

From equation (3) and (4),
k=2


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