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Question

# Find the values of m for which the equation x4âˆ’(1âˆ’2m)x2+(m2âˆ’1)=0 has three real distinct root

A
{2}
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B
{1}
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C
ϕ
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D
{1}
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Solution

## The correct option is B {1}Given: x4−(1−2m)x2+(m2−1)=0 has three real distinct root. To find: Value of m. Step-1: Assume x2=t Step-2: Find the roots in term of t1,t2. Step-3: Write the applicable conditions. Step-4: Solve all the conditions and combine all the results for m. Let, x2=t ⇒t2−(1−2m)t+(m2−1)=0 Roots x1,x2,x3,x4 are transformed to roots t1,t2. x2=t⇒x=±√t ∴ Four roots will be x1=+√t1,x2=+√t2,x3=−√t1,x4=−√t2. Applicable Conditions : i) D>0 ii) t2=0 iii) For t1>0,−b2a>0 Now, solve all conditions, i) D>0 ⇒(1−2m)2−4.1.(m2−1) ⇒−4m+5>0 ⇒4m<5⇒m<54 ∴m∈(−∞,54)=A ii) t2=0 For three real distinct roots, either t1 or t2 should be zero. So, here we are taking t2=0 ⇒m2−1=0⇒m=±1=B iii) For t1>0,−b2a>0 ⇒−1−2m2>0 ⇒1−2m<0⇒2m>1⇒m>12 ∴m∈(12,∞)=C ∴ Set of possible value of m, is A∩B∩C=(−∞,54)∩{−1,1}∩(12,∞)={1} i.e, m={1}

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