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Question

# If (1+ax+bx2)4=a0+a1x+a2x2+…+a8x8; a, b, a0, a1…a8ϵR and are such that a0+a1+a2≠0 and ∣∣ ∣∣a0a1a2a1a2a0a2a0a1∣∣ ∣∣=0, then

A

a=34, b=58

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B

a=14, b=532

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C

a=1, b=23

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D

a=1, b=532

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Solution

## The correct option is B a=14, b=532 Put x=0 in (1+ax+bx2)4=a0+a1x+a2x2+…+a8x8 …(1) ∴a0=1 Differentiate equation (1) and substitute x=0, we get a1=4a. Differentiating again and substituting x=0, we get a2=6a2+4b Now ∣∣ ∣∣a0a1a2a1a2a0a2a0a1∣∣ ∣∣=0 ⇒a30+a31+a32–3a0a1a2=0 ⇒(a0+a1+a2)((a0−a1)2+(a1−a2)2+(a2−a0)2)=0 But a0+a1+a2≠0⇒a0=a1=a2=1 ∴1=4a, 6a2+4b=1 ∴a=14,b=532

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