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Question

The value of a for which the sum of the squares of the roots of 2x22(a2)xa1=0 is least is-

A
1
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B
3/2
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C
2
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D
1
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Solution

The correct option is B 3/2
2x22(a2)xa1=0

2x22(a2)x(a+1)=0

let α,β be the roots of equation. Then we can write

α+β=2(a2)2=a2

αβ=(a+1)2

α2+β2=(α+β)22αβ

α2+β2=(a2)2(2×(a+1)2)

α2+β2=(a2)2+a+1

α2+β2=a2+44a+a+1

α2+β2=a23a+5

Let α2+β2=S

Given that S is minimum

Therefore dSda=0

dda(a23a+5)=0

2a3=0

a=32

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