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Question

If α,β are the roots of the equation x2p(x+1)c=0, then the value of α2+2α+1α2+2α+c+β2+2β+1β2+2β+c is

A
1
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B
2
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C
-1
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D
0
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Solution

The correct option is A 1
Since, α,β are the roots of x2px(p+c)=0
α+β=p and αβ=(p+c)
Now, (α+1)(β+1)=αβ+α+β+1=1c
Consider, α2+2α+1α2+2α+c+β2+2β+1β2+2β+c
=(α+1)2(α+1)2(1c)+(β+1)2(β+1)2(1c)
=(α+1)2(α+1)2(α+1)(β+1)+(β+1)2(β+1)2(α+1)(β+1).
=(α+1)2(α+1)(αβ)+(β+1)2(β+1)(βα)
=α+1αββ+1αβ
=αβαβ=1.
α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=1

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