Let 1+10∑r=1(3r⋅ 10Cr+r⋅ 10Cr)=210(α⋅45+β), and f(x)=x2−2x – InterviewSolution

1.	Let 1+10∑r=1(3r⋅ 10Cr+r⋅ 10Cr)=210(α⋅45+β), and f(x)=x2−2x−k2+1. If α, β lies between the roots of f(x)=0, then the smallest positive integral value of k is
Answer» Let $1 + 10 \sum r = 1 (3^{r} \cdot^{10} C_{r} + r \cdot^{10} C_{r}) = 2^{10} (α \cdot 4^{5} + β)$ , and $f (x) = x^{2} - 2 x - k^{2} + 1$ . If $α, β$ lies between the roots of $f (x) = 0$ , then the smallest positive integral value of $k$ is

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