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Question

If x21 is a factor of x4+ax3+3xb, then

A
a=3,b=1
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B
None of these
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C
a=3,b=1
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D
a=3,b=1
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Solution

The correct option is D a=3,b=1
Given: (x21) is a factor of (x4+ax3+3xb)
Let f(x)=x4+ax3+3xb
Using factor theorem, x=±1 are the zeros of f(x).
f(1)=0 and f(1)=0

Taking f(1)=0
ab+4=0 . . . . .(i)

And f(1)=0
a+b+2=0 . . . . .(ii)

On solving (i) and (ii),we get
a=3,b=1

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