If the equation x4−(k−1)x2+(2−k)=0 has three distinct real roots, then the possible value(s) of k is/are
A
{2}
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B
{√2−1,2}
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C
{√5−1}
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D
{2√2,√3−√2}
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Solution
The correct option is A{2} x4−(k−1)x2+(2−k)=0⋯(1)
Assuming x2=t, t2−(k−1)t+(2−k)=0⋯(2)
Let roots of the equation (2) be t1,t2(t1≤t2)
Equation (1) will have three distinct real roots iff for equation (2), t1=0,t2>0