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Question

Find the values of m for which the equation x4(12m)x2+(m21)=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(12m)x2+(m21)=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(12m)t+(m21)=0
Roots x1,x2,x3,x4 are transformed to roots t1,t2.
x2=tx=±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
(12m)24.1.(m21)
4m+5>0
4m<5m<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
m21=0m=±1=B

iii) For t1>0,b2a>0
12m2>0
12m<02m>1m>12
m(12,)=C

Set of possible value of m, is ABC=(,54){1,1}(12,)={1}

i.e, m={1}




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