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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
  1. {2}
  2. {21,2}
  3. {51}
  4. {22,32}


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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