Let a be an integer such that all the real roots

Let, $f(x)=2 x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+10 x+10$

$\Rightarrow f^{\prime}(x)=10\left(x^{4}+2 x^{3}+3 x^{2}+2 x+1\right)$

$=10\left(x^{2}+\frac{1}{x^{2}}+2\left(x+\frac{1}{x}\right)+3\right)$

$=10\left(\left(x+\frac{1}{z}\right)^{2}+2\left(x+\frac{1}{x}\right)+1\right)$

$=10\left(\left(x+\frac{1}{x}\right)+1\right)^{2}>0 ; \forall x \in R$

$\therefore \mathrm{f}(\mathrm{x})$ is strictly increasing function. Since it is an odd degree polynomial it will have exactly one real root. Now, by observation

$f(-1)=3>0$

$f(-2)=-64+80-80+40-20+10$

$=-34<0$

$\Rightarrow \mathrm{f}(\mathrm{x})$ has at least one root in $(-2,-1) \equiv(\mathrm{a}, \mathrm{a}+1)$

$\Rightarrow \mathrm{f}(\mathrm{x})$ has at least one root in $(-2,-1) \equiv(\mathrm{a}, \mathrm{a}+1)$

$\Rightarrow a=-2$

$\Rightarrow|a|=2$