Question

Let $\mathrm{a}$ be an integer such that all the real roots of the polynomial $2{\mathrm{x}}^5+5{\mathrm{x}}^4+10{\mathrm{x}}^3+10{\mathrm{x}}^2$ $+10\mathrm{x}+10$ lie in the interval $(\mathrm{a},\mathrm{a}+1)$. Then, $\left|\mathrm{a}\right|$is equal to

Answer 1

Explanation for the correct answer:

Given :

$$\begin{gathered} \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^5+5{\mathrm{x}}^4+10{\mathrm{x}}^3+10{\mathrm{x}}^2+10\mathrm{x}+10. \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^5+\left({\mathrm{x}}^5+5{\mathrm{x}}^4+10{\mathrm{x}}^3+10{\mathrm{x}}^2+10\mathrm{x}+1\right)+9\dots \left(1\right) \end{gathered}$$

we have,
$$\begin{gathered} \Rightarrow (\mathrm{x}+1{)}^5={}^5\mathrm{c}_0{\mathrm{x}}^0+{}^5\mathrm{c}_1\mathrm{x}+{}^5\mathrm{c}_2{\mathrm{x}}^2+{}^5\mathrm{c}_3{\mathrm{x}}^3+{}^5\mathrm{c}_4{\mathrm{x}}^4+{}^5\mathrm{c}_5{\mathrm{x}}^5 \\ \Rightarrow (\mathrm{x}+1{)}^5=1+5\mathrm{x}+10{\mathrm{x}}^2+10{\mathrm{x}}^3+5{\mathrm{x}}^4+{\mathrm{x}}^5 \end{gathered}$$

Equation (1) becomes,

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^5+9+(\mathrm{x}+1{)}^5$

Now differentiating with respect to $\text{'}\mathrm{x}\text{'}$

$\Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=5{\mathrm{x}}^4+5(\mathrm{x}+1{)}^4·1$

$\Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)⩾0$
$\therefore \mathrm{f}\left(\mathrm{x}\right)$ is an increasing function.

Let $\mathrm{x}=-2$, the $\mathrm{f}(-2)$
$$\begin{gathered} \Rightarrow \mathrm{f}(-2)=(-2+1{)}^5+(-2{)}^5+9 \\ =-1-32+9 \\ =-24 \end{gathered}$$

Let $\mathrm{x}=-1$
$$\begin{gathered} \Rightarrow \mathrm{f}(-1)=(-1+1{)}^5+(-1{)}^5+9 \\ =0-1+9 \\ =8 \end{gathered}$$

Therefore, the roots lies between $-2$ to $-1$.

Since we have given,

$$\begin{gathered} \left(\mathrm{a},\mathrm{a}+1\right),\mathrm{a}=-2 \\ \Rightarrow \left|\mathrm{a}\right|=2 \end{gathered}$$

Therefore, the value of $\left|\mathrm{a}\right|$is $2$.