If α, β are the zeros of the polynomial f(x) = x2 − p(x + 1) − c


If $\alpha, \beta$ are the zeros of the polynomial $f(x)=x^{2}-p(x+1)-c$ such that $(\alpha+1)(\beta+1)=0$, then $c=$

(a) 1

(b) 0

(c) $-1$

(d) 2


Since $\alpha$ and $\beta$ are the zeros of quadratic polynomial


$f(x)=x^{2}-p x-p-c$

$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$



$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$



We have


$0=\alpha \beta+(\alpha+\beta)+1$



The value of $c$ is 1 .

Hence, the correct alternative is $(a)$

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