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Question

If (1+ax+bx2)4=a0+a1x+a2x2++a8x8; a, b, a0, a1a8ϵR and are such that a0+a1+a20 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 C

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+a323a0a1a2=0 

(a0+a1+a2)((a0a1)2+(a1a2)2+(a2a0)2)=0

But a0+a1+a20a0=a1=a2=1

1=4a, 6a2+4b=1

a=14,b=532


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