Question:
Let a be an integer such that all the real roots of the polynomial $2 x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+10 x+10$ lie in the interval $(a, a+1)$. Then, lal is equal to
Solution:
Let $2 x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+10 x+10=f(x)$
Now $f(-2)=-34$ and $f(-1)=3$
Hence $f(x)$ has a root in $(-2,-1)$
Further $f^{\prime}(x)=10 x^{4}+20 x^{3}+20 x^{2}+20 x+10$
$=10 x^{2}\left[\left(x^{2}+\frac{1}{x^{2}}\right)+2\left(x+\frac{1}{x}\right)+20\right]$
$=10 x^{2}\left[\left(x+\frac{1}{x}+1\right)^{2}+17\right]>0$
Hence $f(\mathrm{x})$ has only one real root, so $\mid \mathrm{al}=2$