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Question

If ax2+(1λ)x+(a1λ)=0 where a0, has real roots for all λR, then

A
a=1
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B
a>1
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C
0<a<1
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D
0<a1
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Solution

The correct option is D 0<a1
ax2+(1λ)x+(a1λ)=0
Given equation has real roots, so
D0(1λ)24a(a1λ)0λ22λ+14a2+4a+4aλ0λ2+2(2a1)λ+(14a2+4a)0
As it is valid for all λR, so
D04(2a1)24(14a2+4a)04a2+14a1+4a24a08a28a0a(a1)00<a1

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