Let 1+ displaystyle∑ r =110(3 r ⋅ 10 C r + r ⋅ 10 C r )=210(α ⋅ 45+β) where α, β ∈ N and f ( x )= x 2-2 x - k 2+1. If α, β lies between the roots of f ( x )=0, then find the smallest positive integral value of k.

Question

Let 1+ displaystyle∑ r =110(3 r ⋅ 10 C r + r ⋅ 10 C r )=210(α ⋅ 45+β) where α, β ∈ N and f ( x )= x 2-2 x - k 2+1. If α, β lies between the roots of f ( x )=0, then find the smallest positive integral value of k. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

We have 1+∑ r =110(3 r ⋅ 10 C r + r ⋅ 10 C r )=1+∑ r =110 3 r ⋅ 10 C r +10 ∑ r =110 9 C r -1=1+410-1+10 ⋅ 29 =410+5.210=210(45+5)=210(α ⋅ 45+β) text , so α=1 text and β=5 text Now f(1)<0 text and f(5)<0 So f (1)<0 ⇒ - k 2<0 ⇒ k ≠ 0 and f (5)<0 ⇒ 16- k 2<0 ⇒ k 2-16>0 ⇒ k ∈(-∞, 4) ∪(4, ∞) Hence smallest positive integral value of k = 5

Q. Let 1+r=1∑10​(3r⋅10Cr​+r⋅10Cr​)=210(α⋅45+β) where α,β∈N and f(x)=x2−2x−k2+1. If α,β lies between the roots of f(x)=0, then find the smallest positive integral value of k.

Q. Let $1 + r = 1 \sum 10 (3^{r} \cdot^{10} C_{r} + r \cdot^{10} C_{r}) = 2^{10} (α \cdot 4^{5} + β)$ where $α, β \in N$ and $f (x) = x^{2} - 2 x - k^{2} + 1$ . If $α, β$ lies between the roots of $f (x) = 0$ , then find the smallest positive integral value of $k$ .