If ax2+(1−λ)x+(a−1−λ)=0 where a≠0, has real roots for all λ∈R, 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<a≤1
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Solution
The correct option is D0<a≤1 ax2+(1−λ)x+(a−1−λ)=0 Given equation has real roots, so D≥0⇒(1−λ)2−4a(a−1−λ)≥0⇒λ2−2λ+1−4a2+4a+4aλ≥0⇒λ2+2(2a−1)λ+(1−4a2+4a)≥0 As it is valid for all λ∈R, so D≤0⇒4(2a−1)2−4(1−4a2+4a)≤0⇒4a2+1−4a−1+4a2−4a≤0⇒8a2−8a≤0⇒a(a−1)≤0∴0<a≤1