Question

Let $1+\sum _{r=1}^{10}(3^r\cdot {}^{10}{C}_r+r\cdot {}^{10}{C}_r)=2^{10}(\alpha \cdot 4^5+\beta )$, and $f\left(x\right)=x^2-2x-k^2+1$. If $\alpha ,\beta$ lies between the roots of $f\left(x\right)=0$, then the smallest positive integral value of $k$ is

Answer 1

We have,
$1+\sum _{r=1}^{10}(3^r\cdot {}^{10}{C}_r+r\cdot {}^{10}{C}_r)$
$=1+\sum _{r=1}^{10}{3}^r\cdot {}^{10}{C}_r+10\sum _{r=1}^{10}{}^9{C}_{r-1}$
$=1+(4^{10}-1)+(10\times 2^9)$
$=2^{10}(4^5+5)=2^{10}(\alpha \cdot 4^5+\beta )$
$\therefore \alpha =1$ and $\beta =5$

Now, $f\left(1\right)<0$ and $f\left(5\right)<0$
$\therefore f\left(1\right)<0\Rightarrow -k^2<0\Rightarrow k\in R-\left\{0\right\}$
and $f\left(5\right)<0$
$\Rightarrow 16-k^2<0$
or, $k^2-16>0$
$\Rightarrow k\in (-\infty ,-4)\cup (4,\infty )$
Hence, the smallest positive integral values of $k=5$.