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Question

If α,β are the zeros of polynomial f(x)=x2p(x+1)c, then (α+1)(β+1)=

A
c1
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B
1c
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C
c
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D
1+c
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Solution

The correct option is B 1c
f(x)=x2pxpc
α+β=+p
αβ=pc
(α+1)(β+1)=αβ+α+β+1
=1c

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